Unformatted text preview: CHAPTER 14 CHEMICAL KINETICS
PRACTICE EXAMPLES
1A (E) The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient. A 0.3187 M 0.3629 M 1min rate of consumption of A = = = 8.93 105 M s 1 t 8.25 min 60 s rate of reaction = rate of consumption of A2 = 1B 8.93 105 M s 1 4.46 105 M s 1 2 (E) We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (2 moles A). (Recall that "M" means "moles per liter.") 0.5522 M A 0.5684 M A 3moles B rate of B formation= 1.62 104 M s 1 60s 2 moles A 2.50 min 1min (M) (a) The 2400s tangent line intersects the 1200s vertical line at 0.75 M and reaches 0 M at 3500 s. The slope of that tangent line is thus 0 M 0.75 M slope = = 3.3 104 M s 1 = instantaneous rate of reaction 3500 s 1200 s The instantaneous rate of reaction = 3.3 104 M s 1 . (b)
At 2400 s, H 2 O 2 = 0.39 M. At 2450 s, H 2 O 2 = 0.39 M + rate t
At 2450 s, H 2 O 2 = 0.39 M + 3.3 10 4 mol H 2 O 2 L1s 1 50s = 0.39 M 0.017 M = 0.37 M 2A 2B We calculate the change in H 2 O 2 and add it to H 2 O 2 0 to determine H 2 O 2 100 . H 2 O 2 = rate of reaction of H 2 O 2 t = 15.0 104 M s 1 100 s = 0.15 M (M) With only the data of Table 14.2 we can use only the reaction rate during the first 400 s, H 2 O 2 /t = 15.0 104 M s 1 , and the initial concentration, H 2 O 2 0 = 2.32 M. H 2O2 100 = H 2O 2 0 + H 2O2 = 2.32 M + 0.15 M = 2.17 M
This value differs from the value of 2.15 M determined in text Example 142b because the text used the initial rate of reaction 17.1104 M s 1 , which is a bit faster than the average
rate over the first 400 seconds.
611 Chapter 14: Chemical Kinetics 3A (M) We write the equation for each rate, divide them into each other, and solve for n. R1 = k N 2 O5 1 = 5.45 105 M s 1 = k 3.15 M n n
n n n
n R2 = k N 2 O5 2 = 1.35 105 M s 1 = k 0.78 M n k N 2 O 5 1 k 3.15 M R1 5.45 105 M s 1 n 3.15 = = 4.04 = = = = 4.04 n n 5 1 R2 1.35 10 M s 0.78 k N 2 O5 2 k 0.78 M We kept an extra significant figure (4) to emphasize that the value of n = 1. Thus, the reaction is firstorder in N 2 O5 .
3B
1 2 (E) For the reaction, we know that rate = k HgCl2 C 2O 4 . Here we will compare Expt. 4 to Expt. 1 to find the rate. 2 1 2 2 rate 4 k HgCl2 C 2 O 4 rate4 = 0.025 M 0.045 M = 0.0214 = = 2 1 2 2 rate1 k HgCl C O 1.8 105 M min 1 0.105 M 0.150 M 2 2 4 2 The desired rate is rate4 = 0.0214 1.8 105 M min 1 = 3.9 107 M min 1 .
4A (E) We place the initial concentrations and the initial rates into the rate law and solve for k . 2 2 rate = k A B = 4.78 102 M s 1 = k 1.12 M 0.87 M k= 4.78 102 M s 1 1.12 M 2 = 4.4 102 M 2 s 1
2 0.87 M 4B 1 2 (E) We know that rate = k HgCl2 C2 O 4 and k = 7.6 103 M 2 min 1 . Thus, insertion of the starting concentrations and the k value into the rate law yields: 1 2 Rate = 7.6 103 M 2 min 1 0.050 M 0.025 M = 2.4 107 M min 1 5A (E) Here we substitute directly into the integrated rate law equation. ln A t = kt + ln A 0 = 3.02 103 s1 325 s + ln 2.80 = 0.982 +1.030 = 0.048 b g At= e 0.048 = 1.0 M 5B (M) This time we substitute the provided values into text Equation 14.13. 0.443 1.49 M H 2O2 t k= ln = kt = k 600 s = ln = 0.443 = 7.38 104 s 1 600 s 2.32 M H 2O2 0 Now we choose H 2 O2 0 = 1.49 M, H 2 O 2 t = 0.62, t = 1800 s 600 s = 1200 s ln H 2O2 t H 2O 2 0 = kt = k 1200 s = ln 0.62 M = 0.88 1.49 M k= 0.88 = 7.3 104 s 1 1200 s These two values agree within the limits of the experimental error and thus, the reaction is firstorder in [H2O2]. 612 Chapter 14: Chemical Kinetics 6A (M) We can use the integrated rate equation to find the ratio of the final and initial concentrations. This ratio equals the fraction of the initial concentration that remains at time t. ln A t = kt = 2.95 103 s 1 150 s = 0.443 A 0 A t = e 0.443 = 0.642; 64.2% of A 0 remains. A 0 6B (M) After twothirds of the sample has decomposed, onethird of the sample remains. Thus H 2 O 2 t = H 2 O 2 0 3 , and we have ln H 2O2 t H2O 2 0 = kt = ln H 2O2 0 3 H 2O 2 0 = ln 1/ 3 = 1.099 = 7.30 104 s 1t t=
7A 1.099 1 min = 1.51103 s = 25.1 min 4 1 7.30 10 s 60 s (M) At the end of one halflife the pressure of DTBP will have been halved, to 400 mmHg. At the end of another halflife, at 160 min, the pressure of DTBP will have halved again, to 200 mmHg. Thus, the pressure of DTBP at 125 min will be intermediate between the pressure at 80.0 min (400 mmHg) and that at 160 min (200 mmHg). To obtain an exact answer, first we determine the value of the rate constant from the halflife.
k= ln 0.693 0.693 = = 0.00866 min 1 80.0 min t1/2 = kt = 0.00866 min 1 125 min = 1.08 PDTBP t PDTBP 0 PDTBP t = e 1.08 = 0.340 PDTBP 0 PDTBP t = 0.340 PDTBP 0 = 0.340 800 mmHg = 272 mmHg
7B (M) (a) We use partial pressures in place of concentrations in the integrated firstorder rate equation. Notice first that more than 30 halflives have elapsed, and thus the ethylene oxide 30 pressure has declined to at most 0.5 = 9 1010 of its initial value. b g ln P30 P 3600 s = kt = 2.05 104 s 1 30.0 h = 22.1 30 = e 22.1 = 2.4 1010 P0 P0 1h P30 = 2.4 1010 P0 = 2.4 1010 782 mmHg = 1.9 107 mmHg 613 Chapter 14: Chemical Kinetics (b) Pethylene oxide initially 782 mmHg 1.9 107 mmHg (~ 0). Essentially all of the ethylene oxide is converted to CH4 and CO. Since pressure is proportional to moles, the final pressure will be twice the initial pressure (1 mole gas 2 moles gas; 782 mmHg 1564 mmHg). The final pressure will be 1.56 103 mmHg. 8A (D) We first begin by looking for a constant rate, indicative of a zeroorder reaction. If the rate is constant, the concentration will decrease by the same quantity during the same time period. If we choose a 25s time period, we note that the concentration decreases 0.88 M 0.74 M = 0.14 M during the first 25 s, 0.74 M 0.62 M = 0.12 M during the second 25 s, 0.62 M 0.52 M = 0.10 M during the third 25 s, and 0.52 M 0.44 M = 0.08 M during the fourth 25s period. This is hardly a constant rate and we thus conclude that the reaction is not zeroorder. We next look for a constant halflife, indicative of a firstorder reaction. The initial concentration of 0.88 M decreases to one half of that value, 0.44 M, during the first 100 s, indicating a 100s halflife. The concentration halves again to 0.22 M in the second 100 s, another 100s halflife. Finally, we note that the concentration halves also from 0.62 M at 50 s to 0.31 M at 150 s, yet another 100s halflife. The rate is established as firstorder. The rate constant is 0.693 0.693 k= = = 6.93 103 s1 . t1/2 100 s That the reaction is firstorder is made apparent by the fact that the ln[B] vs time plot is a straight line with slope = k (k = 6.85 103 s1).
Plot of [B] versus Time
0.85 Plot of ln([B]) versus Time
0 0.2 0 200 400
6 7 Plot of 1/[B] versus Time 0.75 0.65 [B] (M) 0.55 0.45 0.35 0.25 0.15 0 100 200 300 0.4 0.6 0.8 ln([B]) 1 1.2 1.4 1.6 1.8
1 2 1/[B] (M 1) 5 4 3 2 Time(s) 0 Time(s) 100 200 Tim e (s) 300 614 Chapter 14: Chemical Kinetics 8B (D) We plot the data in three ways to determine the order. (1) A plot of [A] vs. time is linear if the reaction is zeroorder. (2) A plot of ln [A] vs. time will be linear if the reaction is firstorder. (3) A plot of 1/[A] vs. time will be linear if the reaction is secondorder. It is obvious from the plots below that the reaction is zeroorder. The negative of the slope of the line equals k = 0.083 M 0.250 M 18.00 min = 9.28 103 M/min (k = 9.30 103 M/min using a graphical approach).
0.25 0.23 0.21 1.7 0.19 Plot of [A] versus Time
1.5 0 10 20 12 11 10 9 8 7 Plot of 1/[A] versus Time Plot of ln([A]) versus Time ln([A]) 0.17 0.15 0.13 0.11 1.9 2.1 1/[A] (M1) [A] (M) 6
2.3 0.09 y = 0.00930x + 0.2494 0.07 0 10 20 2.5 5 4 0 Time (min) Time (min) 10 Time (min) 20 9A (M) First we compute the value of the rate constant at 75.0 C with the Arrhenius equation. We know that the activation energy is Ea = 1.06 105 J/mol, and that k = 3.46 105 s1 at 298 K. The temperature of 75.0 o C = 348.2 K.
ln k2 k2 E 1 1 1.06 105 J / mol 1 1 = ln = a = = 6.14 5 1 1 1 k1 3.46 10 s R T1 T2 8.3145 J mol K 298.2 K 348.2 K FG H IJ K FG H IJ K k2 = 3.46 105 s 1 e +6.14 = 3.46 105 s 1 4.6 102 = 0.016 s 1 t1/2 =
9B 0.693 0.693 = = 43 s at 75 C k 0.016 s 1 (M) We use the integrated rate equation to determine the rate constant, realizing that onethird remains when twothirds have decomposed. ln N 2 O5 t N 2O 5 0 = ln N 2 O5 0 3 N 2O 5 0 1 = ln = kt = k 1.50 h = 1.099 3 k= 1.099 1h = 2.03 104 s 1 1.50 h 3600 s 615 Chapter 14: Chemical Kinetics Now use the Arrhenius equation to determine the temperature at which the rate constant is 2.04 104 s1.
k2 2.04 104 s1 Ea 1 1 1.06 105 J / mol 1 1 ln = ln = 1.77 = = 5 1 1 1 k1 3.46 10 s R T1 T2 8.3145 J mol K 298 K T2 FG H IJ K FG H IJ K 1 1 1.77 8.3145 K 1 = = 3.22 103 K 1 5 1.06 10 T2 298 K T2 = 311 K 10A (M) The two steps of the mechanism must add, in a Hess's law fashion, to produce the overall reaction. Overall reaction: CO + NO 2 CO 2 + NO or CO + NO 2 CO 2 + NO Second step: NO3 + CO NO 2 + CO 2 or + NO 2 + CO 2 NO3 + CO First step: 2 NO 2 NO + NO 3 If the first step is the slow step, then it will be the ratedetermining step, and the rate of that step 2 will be the rate of the reaction, namely, rate of reaction = k1 NO2 .
10B (M) (1) The steps of the mechanism must add, in a Hess's law fashion, to produce the overall reaction. This is done below. The two intermediates, NO 2 F2 g and F(g), are each produced in one step and consumed in the next one. bg Fast: Slow: Fast: Net: (2) NO 2 g + F2 g NO2 F2 g NO 2 F2 g NO 2 F g + F g F g + NO 2 g NO 2 F g 2 NO 2 g + F2 g 2 NO 2 F g bg bg bg bg bg bg bg bg bg The proposed mechanism must agree with the rate law. We expect the ratedetermining step to determine the reaction rate: Rate = k3 NO 2 F2 . To eliminate NO 2 F2 , we recognize that the first elementary reaction is very fast and will have the same rate forward as reverse: R f = k1 NO 2 F2 = k2 NO 2 F2 = Rr . We solve for the concentration of intermediate: NO 2 F2 = k1 NO 2 F2 /k2 . We now substitute this expression for NO 2 F2 into the rate equation: Rate = k1k3 /k2 NO 2 F2 . Thus the predicted rate law agrees with the experimental rate law. 616 Chapter 14: Chemical Kinetics INTEGRATIVE EXAMPLE
A. (M) (a) The time required for the fixed (c) process of souring is three times as long at 3 C refrigerator temperature (276 K) as at 20 C room temperature (293 K). E 1 1 E 1 64 h 1 Ea c / t2 t 4 ln ln 1 ln 1.10 a a ( 2.10 10 ) 3 64 h c / t1 t2 R T1 T2 R 293 K 276 K R 1.10 R 1.10 8.3145 J mol1 K 1 4.4 104 J/mol 44 kJ/mol 2.10 104 K 1 2.10 104 K 1 (b) Use the Ea determined in part (a) to calculate the souring time at 40 C = 313 K. E 1 1 t 4.4 104 J/mol 1 1 t1 ln 1 a 1.15 ln 1 1 t2 R T1 T2 8.3145 J mol K 313 K 293 K 64 h Ea t1 e 1.15 0.317 t1 0.317 64 h 20. h 64 h
B. (M) The species A* is a reactive intermediate. Let's deal with this species by using a steady state approximation. d[A*]/dt = 0 = k1[A]2 k1[A*][A] k2[A*]. Solve for [A*]. k1[A*][A] + k2[A*] = k1[A]2 k1 [A]2 k 2 k1 [A]2 [A*] = The rate of reaction is: Rate = k2[A*] = k 1 [A] k 2 k 1 [A] k 2 At low pressures ([A] ~ 0 and hence k2 >> k1[A]), the denominator becomes ~ k2 and the rate law is k k [A]2 Rate = 2 1 = k1 [A]2 Secondorder with respect to [A] k2 At high pressures ([A] is large and k1[A] >> K2), the denominator becomes ~ k1[A] and the rate law is k k [A]2 k k [A] Rate = 2 1 = 2 1 Firstorder with respect to [A] k 1 [A] k 1 EXERCISES
Rates of Reactions
1. (M) 2A + B C + 3D (a) (b) (c) Rate = 1 [A] = 1/2(6.2 104 M s1) = 3.1 104 Ms1 2 t 1 [A] Rate of disappearance of B = = 1/2(6.2 104 M s1) = 3.1 104 Ms1 2 t 3 [A] Rate of appearance of D = = 3(6.2 104 M s1) = 9.3 104 Ms1 2 t [A] = 6.2 104 M s1 t 617 Chapter 14: Chemical Kinetics 2. (M) In each case, we draw the tangent line to the plotted curve. H 2 O 2 1.7 M 0.6 M (a) The slope of the line is = = 9.2 104 M s 1 t 400 s 1600 s H 2O2 reaction rate = = 9.2 104 M s 1 t (b) Read the value where the horizontal line [H2O2] = 0.50 M intersects the curve, 2150 s or 36 minutes A 0.474 M 0.485 M = 1.0 103 M s 1 = t 82.4 s 71.5 s A 0.1498 M 0.1565 M = = 0.0067 M min 1 t 1.00 min 0.00 min A 0.1433 M 0.1498 M = = 0.0065 M min 1 t 2.00 min 1.00 min 3. 4. (E) Rate = (M) (a) Rate = Rate = (b) The rates are not equal because, in all except zeroorder reactions, the rate depends on the concentration of reactant. And, of course, as the reaction proceeds, reactant is consumed and its concentration decreases, so the rate of the reaction decreases. A = A i + A = 0.588 M 0.013 M = 0.575 M A = 0.565 M 0.588 M = 0.023 M t 0.023 M t = A = = 1.0 min A 2.2 102 M/min
time = t + t = 4.40 +1.0 min = 5.4 min 5. (M) (a) (b) 6. (M) Initial concentrations are HgCl 2 = 0.105 M and C 2 O 4 2 = 0.300 M. The initial rate of the reaction is 7.1 105 M min 1 . Recall that the reaction is: 2 2 HgCl2 (aq) + C2 O 4 (aq) 2 Cl (aq) + 2 CO 2 (g) + Hg 2 Cl2 (aq) . The rate of reaction equals the rate of disappearance of C2 O 2 . Then, after 1 hour, assuming that 4 the rate is the same as the initial rate,
(a) HgCl2 = 0.105 M 7.1 105 mol C 2 O 4 2 2 mol HgCl2 60 min 1 h 0.096 M 2 L s 1h 1 mol C2 O 4 (b) C 2 O 2 = 0.300 M 7.1 105 4 FG H mol 60 min 1 h = 0.296 M L min 1h IJ K 618 Chapter 14: Chemical Kinetics 7. (M)
(a)
Rate = C A t = C 2t = 1.76 105 M s 1 (b) (c) t A C = = 1.76 105 M s1 Assume this rate is constant. t 2 t 60 s A = 0.3580 M + 1.76 105 M s1 1.00 min = 0.357 M 1min A = 1.76 105 M s 1 t A 0.3500 M 0.3580 M t = = = 4.5 102 s 5 5 1.76 10 M/s 1.76 10 M/s = 2 1.76 105 M s 1 = 3.52 105 M/s 8. (M) (a) (b) (c) n O 2 5.7 104 mol H 2 O 2 1 mol O 2 = 1.00 L soln = 2.9 104 mol O 2 /s t 1 L soln s 2 mol H 2 O 2 n O 2 mol O 2 60 s = 2.9 104 = 1.7 102 mol O 2 / min 1 min t s V O 2 mol O 2 22,414 mL O 2 at STP 3.8 102 mL O 2 at STP = 1.7 102 = 1 mol O 2 t min min 9. (M) Notice that, for every 1000 mmHg drop in the pressure of A(g), there will be a corresponding 2000 mmHg rise in the pressure of B(g) plus a 1000 mmHg rise in the pressure of C(g). (a) We set up the calculation with three lines of information below the balanced equation: (1) the initial conditions, (2) the changes that occur, which are related to each other by reaction stoichiometry, and (3) the final conditions, which simply are initial conditions + changes. A(g) 2B(g) + C(g) Initial 1000. mmHg 0. mmHg 0. mmHg Changes 1000. mmHg +2000. mmHg +1000. mmHg Final 0. mmHg 2000. MmHg 1000. mmHg Total final pressure = 0. mmHg + 2000. mmHg +1000. mmHg = 3000. mmHg A(g) 1000. mmHg 200. mmHg 800 mmHg (b) Initial Changes Final 2B(g) + 0. mmHg +400. mmHg 400. mmHg C(g) 0. mmHg +200. mmHg 200. mmHg Total pressure = 800. mmHg + 400. mmHg + 200. mmHg = 1400. mmHg 619 Chapter 14: Chemical Kinetics 10. (M) (a) We will use the ideal gas law to determine N 2 O5 pressure
1 mol N 2 O5 1.00 g 108.0 g nRT P{N 2 O5 } = = V L atm 0.08206 mol K 273 + 65 K 760 mmHg = 13 mmHg
15 L 1 atm (b) (c) After 2.38 min, one halflife passes. The initial pressure of N 2 O5 decreases by half to 6.5 mmHg. From the balanced chemical equation, the reaction of 2 mol N 2 O5 g produces 4 mol NO 2 g and 1 mol O 2 g . That is, the consumption of 2 mol of reactant gas produces 5 mol of product gas. When measured at the same temperature and confined to the same volume, pressures will behave as amounts: the reaction of 2 mmHg of reactant produces 5 mmHg of product. 5 mmHg(product) Ptotal = 13 mmHg N 2 O5 (initially) 6.5 mmHg N 2 O5 (reactant) + 6.5 mmHg(reactant) 2 mmHg(reactant) (13 6.5 16) mmHg 23 mmHg bg bg bg Method of Initial Rates
11. (M) (a) From Expt. 1 to Expt. 3, [A] is doubled, while [B] remains fixed. This causes the rate to 6.75 104 M s1 increases by a factor of = 2.01 2 . 3.35 104 M s1 Thus, the reaction is firstorder with respect to A. From Expt. 1 to Expt. 2, [B] doubles, while [A] remains fixed. This causes the rate to 1.35 103 M s1 = 4.03 4 . increases by a factor of 3.35 104 M s1 Thus, the reaction is secondorder with respect to B. (b) Overall reaction order order with respect to A order with respect to B = 1 2 = 3. The reaction is thirdorder overall. 2 (c) Rate = 3.35 104 M s1 = k 0.185 M 0.133 M b gb g k= b0.185 Mgb0.133 Mg 3.35 10 4 Ms 1 2 = 0.102 M 2 s1 12. (M) From Expt. 1 and Expt. 2 we see that [B] remains fixed while [A] triples. As a result, the initial rate increases from 4.2 103 M/min to 1.3 102 M/min, that is, the initial reaction rate triples. Therefore, the reaction is firstorder in [A]. Between Expt. 2 and Expt. 3, we see that [A] doubles, which would double the rate, and [B] doubles. As a consequence, the initial rate goes from 1.3 102 M/min to 5.2 10 2 M/min, that is, the rate quadruples. Since an additional doubling of the rate is due to the change in [B], the reaction is firstorder in [B]. Now we determine the value of the rate constant. 620 Chapter 14: Chemical Kinetics Rate = k A 1 B 1 5.2 102 M / min Rate = 5.8 103 L mol 1min 1 = k= 3.00 M 3.00 M A B The rate law is Rate = 5.8 103 L mol 1min 1 A
13. c h 1 B . 1 (M) From Experiment 1 to 2, [NO] remains constant while [Cl2] is doubled. At the same time the initial rate of reaction is found to double. Thus, the reaction is firstorder with respect to [Cl2], since dividing reaction 2 by reaction 1 gives 2 = 2x when x = 1. From Experiment 1 to 3, [Cl2] remains constant, while [NO] is doubled, resulting in a quadrupling of the initial rate of reaction. Thus, the reaction must be secondorder in [NO], since dividing reaction 3 by reaction 1 gives 4 = 2x when x = 2. Overall the reaction is thirdorder: Rate = k [NO]2[Cl2]. The rate constant may be calculated from any one of the experiments. Using data from Exp. 1, k= Rate 2.27 105 M s 1 = = 5.70 M2 s1 [NO]2 [Cl2 ] (0.0125 M) 2 (0.0255 M) 14. (M) (a) From Expt. 1 to Expt. 2, [B] remains constant at 1.40 M and [C] remains constant at 1.00 M, but [A] is halved 0.50 . At the same time the rate is halved 0.50 . Thus, the b g b g reaction is firstorder with respect to A, since 0.50 = 0.50 when x = 1. From Expt. 2 to Expt. 3, [A] remains constant at 0.70 M and [C] remains constant at 1.00 M, but [B] is halved 0.50 , from 1.40 M to 0.70 M. At the same time, the rate is quartered 0.25 . Thus, the reaction is secondorder with respect to B, since 0.50 y = 0.25 when y = 2 . From Expt. 1 to Expt. 4, [A] remains constant at 1.40 M and [B] remains constant at 1.40 M, but [C] is halved 0.50 , from 1.00 M to 0.50 M. At the same time, the rate is increased by a factor of 2.0. 1 1 Rate4 = 16 Rate3 = 16 Rate 2 = 4 Rate2 = 4 Rate1 = 2 Rate1. 4 2 Thus, the order of the reaction with respect to C is 1 , since 0.5z = 2.0 when z = 1 .
(b)
rate5 = k 0.70 M 0.70 M 0.50 M 1 2
1 x b g b g 1.40 M 1.40 M 1.00 M = k 2 2 2 1 1 2 1 k 11 2 1.40 M 1 12 2 1.40 M 2 1 1 2 1 1.00 M 2 1 = rate1 2
1 1+2 1 1 = rate1 = 1 rate1 4 2 2 This is based on rate1 = k 1.40 M 1.40 M 1.00 M 621 Chapter 14: Chemical Kinetics FirstOrder Reactions
15. (E) (a) TRUE The rate of the reaction does decrease as more and more of B and C are formed, but not because more and more of B and C are formed. Rather, the rate decreases because the concentration of A must decrease to form more and more of B and C. The time required for one half of substance A to reactthe halflifeis independent of the quantity of A present. (b) FALSE 16. (E) (a) (b) FALSE TRUE For firstorder reactions, a plot of ln [A] or log [A] vs. time yields a straight line. A graph of [A] vs. time yields a curved line. The rate of formation of C is related to the rate of disappearance of A by the stoichiometry of the reaction. 17. (M) (a) Since the halflife is 180 s, after 900 s five halflives have elapsed, and the original quantity of A has been cut in half five times. 5 final quantity of A = 0.5 initial quantity of A = 0.03125 initial quantity of A About 3.13% of the original quantity of A remains unreacted after 900 s. or More generally, we would calculate the value of the rate constant, k, using ln 2 0.693 = = 0.00385 s 1 Now ln(% unreacted) = kt = 0.00385 s1(900s) = k= 180 s t1/2 3.465 (% unreacted) = 0.0313 100% = 3.13% of the original quantity. b g (b) 18. Rate = k A = 0.00385 s1 0.50 M = 0.00193 M / s (M) (a) The reaction is firstorder, thus A t 0.100 M ln = kt = ln = 54 min(k ) 0.800 M A 0 k= 2.08 = 0.0385 min 1 54 min We may now determine the time required to achieve a concentration of 0.025 M A t 3.47 0.025 M t= ln = kt = ln = 0.0385 min 1 (t ) = 90. min 0.0385 min 1 0.800 M A 0
(b) Since we know the rate constant for this reaction (see above), 1 Rate = k A = 0.0385 min 1 0.025 M = 9.6 104 M/min 622 Chapter 14: Chemical Kinetics 19. (M) (a) The mass of A has decreased to one fourth of its original value, from 1.60 g to 1 1 1 0.40 g. Since = , we see that two halflives have elapsed. 4 2 2 Thus, 2 t1/2 = 38 min, or t1/2 = 19 min . A t 0.693 (b) k = 0.693/t1/2 = = 0.036 min 1 ln = kt = 0.036 min 1 60 min = 2.2 19 min A 0 A t = A 0 A t A 0 e 2.2 = 0.11 or A t = A 0 e kt = 1.60 g A 0.11 = 0.18 g A 20. (M) (a) (b) (c) ln = kt = ln 0.632 M = 0.256 0.816 M k= 0.256 = 0.0160 min 1 16.0 min t1/2 = 0.693 0.693 = = 43.3 min k 0.0160 min 1 We need to solve the integrated rate equation to find the elapsed time.
ln A t A 0 = kt = ln 0.235 M = 1.245 = 0.0160 min 1 t 0.816 M t= 1.245 = 77.8 min 0.0160 min 1 (d) ln A A = e kt which in turn becomes = kt becomes A0 A 0
0 A = A
21. e kt = 0.816 M exp 0.0160 min 1 2.5 h 60 min 1h = 0.816 0.0907 = 0.074 M (M) We determine the value of the firstorder rate constant and from that we can calculate the halflife. If the reactant is 99% decomposed in 137 min, then only 1% (0.010) of the initial concentration remains. A t 4.61 0.010 k= ln = kt = ln = 4.61 = k 137min = 0.0336 min 1 137 min 1.000 A 0 t1/2 = 0.0693 0.693 = = 20.6 min k 0.0336 min 1 623 Chapter 14: Chemical Kinetics 22. P is lost, 1% (0.010) of that radioactivity remains. First we 0.693 0.693 compute the value of the rate constant from the halflife. k = = = 0.0485 d 1 t1/2 14.3 d Then we use the integrated rate equation to determine the elapsed time. At At 1 1 0.010 ln = kt t = ln = = 95 days ln 1 0.0485 d 1.000 k A0 A0
(D) (a) 35 100 [A]0 3 1 ln = ln(0.35) = kt = (4.81 10 min )t t = 218 min. [A]0 Note: We did not need to know the initial concentration of acetoacetic acid to answer the question. (E) If 99% of the radioactivity of 32 23. (b) Let's assume that the reaction takes place in a 1.00L container. 10.0 g acetoacetic acid 1 mol acetoacetic acid 102.090 g acetoacetic acid = 0.09795 mol acetoacetic acid. After 575 min. (~ 4 half lives, hence, we expect ~ 6.25% remains as a rough approximation), use integrated form of the rate law to find [A]t = 575 min. [A]t 3 1 ln = kt = (4.81 10 min )(575 min) = 2.766 [A]0 [A]t [A]t = e2.766 = 0.06293 (~ 6.3% remains) = 0.063 [A]t = 6.2 103 [A]0 0.09795 moles moles. [A]reacted = [A]o [A]t = (0.098 6.2 103) moles = 0.092 moles acetoacetic acid. The stoichiometry is such that for every mole of acetoacetic acid consumed, one mole of CO2 forms. Hence, we need to determine the volume of 0.0918 moles CO2 at 24.5 C (297.65 K) and 748 torr (0.984 atm) by using the Ideal Gas law. L atm 0.0918 mol 0.08206 297.65 K K mol nRT = 2.3 L CO2 = V= P 0.984 atm
24. (M) ln
(a) A t A 0 = kt = ln 2.5 g = 3.47 = 6.2 104 s 1t 80.0 g 3.47 = 5.6 103 s 93 min 6.2 104 s 1 We substituted masses for concentrations, because the same substance (with the same molar mass) is present initially at time t , and because it is a closed system. t= 624 Chapter 14: Chemical Kinetics (b) 1 mol N 2 O5 1 mol O 2 = 0.359 mol O 2 108.0 g N 2 O5 2 mol N 2 O5 L atm 0.359 mol O 2 0.08206 (45 273) K nRT mol K V 9.56 L O 2 1 atm P 745 mmHg 760 mmHg amount O 2 = 77.5 g N 2 O5 If the reaction is firstorder, we will obtain the same value of the rate constant from several sets of data. A t 0.497 M 0.188 ln = kt = ln = k 100 s = 0.188, k = = 1.88 103 s 1 A 0 0.600 M 100 s 25. (D) (a) A t 0.344 M = kt = ln = k 300 s = 0.556, 0.600 M A 0 A t 0.285 M ln = kt = ln = k 400 s = 0.744, 0.600 M A 0 A t 0.198 M ln = kt = ln = k 600 s = 1.109, 0.600 M A 0 A t 0.094 M ln = kt = ln = k 1000 s = 1.854, 0.600 M A 0
ln k= k= k= k= 0.556 = 1.85 103 s 1 300 s 0.744 = 1.86 103 s 1 400 s 1.109 = 1.85 103 s 1 600 s 1.854 = 1.85 103 s 1 1000 s (b) (c) The virtual constancy of the rate constant throughout the time of the reaction confirms that the reaction is firstorder. For this part, we assume that the rate constant equals the average of the values obtained in part (a). 1.88 +1.85 +1.86 +1.85 k= 103 s1 = 1.86 103 s1 4 We use the integrated firstorder rate equation: A 750 = A 0 exp kt = 0.600 M exp 1.86 103 s 1 750 s A 750 = 0.600 M e 1.40 = 0.148 M 26. (D) (a) If the reaction is firstorder, we will obtain the same value of the rate constant from several sets of data. 0.167 P 264 mmHg ln t = kt = ln = k 390 s = 0.167 , k= = 4.28 104 s1 390 s P0 312 mmHg 0.331 P 224 mmHg k= = 4.26 104 s1 ln t = kt = ln = k 777 s = 0.331 , 777 s 312 mmHg P0 0.512 P 187 mmHg k= = 4.28 104 s1 ln t = kt = ln = k 1195 s = 0.512 , 1195 s 312 mmHg P0 625 Chapter 14: Chemical Kinetics Pt 78.5 mmHg 1.38 = kt = ln = k 3155 s = 1.38 , k= = 4.37 104 s 1 312 mmHg 3155 s P0 The virtual constancy of the rate constant confirms that the reaction is firstorder. ln
(b) (c) For this part we assume the rate constant is the average of the values in part (a): 4.3 104 s1 . At 390 s, the pressure of dimethyl ether has dropped to 264 mmHg. Thus, an amount of dimethyl ether equivalent to a pressure of 312 mmHg 264 mmHg = 48 mmHg has decomposed. For each 1 mmHg pressure of dimethyl ether that decomposes, 3 mmHg of pressure from the products is produced. Thus, the increase in the pressure of the products is 3 48 = 144 mmHg. The total pressure at this point is 264 mmHg +144 mmHg = 408 mmHg. Below, this calculation is done in a more systematic fashion: b g bCH g Obgg
3 2 CH 4 g bg + H2 g bg + CO g bg Initial 312 mmHg Changes 48 mmHg Final 264 mmHg 0 mmHg + 48 mmHg 48 mmHg 0 mmHg + 48 mmHg 48 mmHg 0 mmHg + 48 mmHg 48 mmHg Ptotal = PDME + Pmethane + Phydrogen + PCO 264 mmHg 48 mmHg 48 mmHg 48 mmHg 408 mmHg
(d) This question is solved in the same manner as part (c). The results are summarized below. bCH g Obgg
3 2 CH 4 g bg H2 g bg + CO g bg Initial 312 mmHg 0 mmHg Changes 312 mmHg 312 mmHg Final 0 mmHg 312 mmHg Ptotal = PDME + Pmethane + Phydrogen + PCO 0 mmHg 312 mmHg 312 mmHg 0 mmHg + 312 mmHg 312 mmHg 0 mmHg 312 mmHg 312 mmHg 312 mmHg 936 mmHg P (e) We first determine PDME at 1000 s. ln 1000 = kt = 4.3 104 s1 1000 s = 0.43 P0 P 1000 = e 0.43 = 0.65 P = 312 mmHg 0.65 = 203 mmHg 1000 P0 Then we use the same approach as was used for parts (c) and (d) CH3 2 O g Initial 312 mmHg Changes 109 mmHg CH 4 g 0 mmHg 109 mmHg + H2 g 0 mmHg 109 mmHg 109 mmHg + CO g 0 mmHg 109 mmHg 109 mmHg Final 203 mmHg 109 mmHg Ptotal = PDME + Pmethane +Phydrogen +PCO = 203 mmHg +109 mmHg +109 mmHg +109 mmHg = 530. mmHg 626 Chapter 14: Chemical Kinetics Reactions of Various Orders
27. (M) (a) Set II is data from a zeroorder reaction. We know this because the rate of set II is constant. 0.25 M/25 s = 0.010 M s1 . Zeroorder reactions have constant rates of reaction. (b) A firstorder reaction has a constant halflife. In set I, the first halflife is slightly less than 75 sec, since the concentration decreases by slightly more than half (from 1.00 M to 0.47 M) in 75 s. Again, from 75 s to 150 s the concentration decreases from 0.47 M to 0.22 M, again by slightly more than half, in a time of 75 s. Finally, two halflives should see the concentration decrease to onefourth of its initial value. This, in fact, is what we see. From 100 s to 250 s, 150 s of elapsed time, the concentration decreases from 0.37 M to 0.08 M, i.e., to slightly less than onefourth of its initial value. Notice that we cannot make the same statement of constancy of halflife for set III. The first halflife is 100 s, but it takes more than 150 s (from 100 s to 250 s) for [A] to again decrease by half. For a secondorder reaction,1/ A t 1/ A 0 = kt . For the initial 100 s in set III, we have 1 1 = 1.0 L mol1 = k 100 s, k = 0.010 L mol1 s 1 0.50 M 1.00 M For the initial 200 s, we have 1 1 = 2.0 L mol1 = k 200 s, k = 0.010 L mol1 s 1 0.33 M 1.00 M Since we obtain the same value of the rate constant using the equation for secondorder kinetics, set III must be secondorder. (c) 28. 29. (E) For a zeroorder reaction (set II), the slope equals the rate constant: k = A /t = 1.00 M/100 s = 0.0100 M/s (M) Set I is the data for a firstorder reaction; we can analyze those items of data to determine the halflife. In the first 75 s, the concentration decreases by a bit more than half. This implies a halflife slightly less than 75 s, perhaps 70 s. This is consistent with the other time periods noted in the answer to Review Question 18 (b) and also to the fact that in the 150s interval from 50 s to 200 s, the concentration decreases from 0.61 M to 0.14 M, which is a bit more than a factoroffour decrease. The factoroffour decrease, to onefourth of the initial value, is what we would expect for two successive halflives. We can determine the halflife more accurately, by obtaining a value of k from the relation ln A t / A 0 = kt followed by t1/2 = 0.693 / k For instance, ln(0.78/1.00) = k (25 s); k = 9.94 103 s1. Thus, t1/2 = 0.693/9.94 103 s1 = 70 s.
30. (E) We can determine an approximate initial rate by using data from the first 25 s. A 0.80 M 1.00 M Rate = = = 0.0080 M s 1 t 25 s 0 s 627 Chapter 14: Chemical Kinetics 31. (M) The approximate rate at 75 s can be taken as the rate over the time period from 50 s to 100 s. A 0.00 M 0.50 M (a) Rate II = = = 0.010 M s1 100 s 50 s t (b) (c) Rate I = A 0.37 M 0.61 M = = 0.0048 M s1 100 s 50 s t A 0.50 M 0.67 M = = 0.0034 M s1 100 s 50 s t Alternatively we can use [A] at 75 s (the values given in the table) in the m relationship Rate = k A , where m = 0 , 1, or 2. Rate III = (a) (b) Rate II = 0.010 M s 1 0.25 mol/L = 0.010 M s 1
0 Since t1/2 70s, k 0.693 / 70s 0.0099s 1 Rate I 0.0099 s 1 (0.47 mol/L)1 0.0047 M s 1 Rate III = 0.010 L mol1 s 1 0.57 mol/L = 0.0032 M s 1
2 (c) 32. (M) We can combine the approximate rates from Exercise 31, with the fact that 10 s have elapsed, and the concentration at 100 s. (a) A II = 0.00 M There is no reactant left after 100 s. (b) (c) AI= A A
III 100 10 s rate = 0.37 M 10 s 0.0047 M s1 = 0.32 M 10 s rate = 0.50 M 10 s 0.0032 M s1 = 0.47 M b g c h = A 100 b g c h 33. (E) Substitute the given values into the rate equation to obtain the rate of reaction. 2 0 2 0 Rate = k A B = 0.0103 M 1min 1 0.116 M 3.83 M = 1.39 10 4 M / min c hb gb g 34. (M) (a) A firstorder reaction has a constant halflife. Thus, half of the initial concentration remains after 30.0 minutes, and at the end of another halflife60.0 minutes totalhalf of the concentration present at 30.0 minutes will have reacted: the concentration has decreased to onequarter of its initial value. Or, we could say that the reaction is 75% complete after two halflives60.0 minutes. (b) A zeroorder reaction proceeds at a constant rate. Thus, if the reaction is 50% complete in 30.0 minutes, in twice the time60.0 minutesthe reaction will be 100% complete. (And in onefifth the time6.0 minutesthe reaction will be 10% complete. Alternatively, we can say that the rate of reaction is 10%/6.0 min.) Therefore, the time required for the 60.0 min reaction to be 75% complete = 75% = 45 min. 100% 628 Chapter 14: Chemical Kinetics 35. (M) For reaction: HI(g) 1/2 H2(g) 1/2 I2(g) (700 K) Time (s) 0 100 200 300 400 [HI] (M) 1.00 0.90 0.81 0.74 0.68 ln[HI] 0 0.105 0.211 0.301 0.386 1/[HI](M1) 1.00 1.11 1.235 1.35 1.47 Plot of 1/[HI] vs time
1.60 y = 0.00118x + 0.997 1/[HI] (M1) 1.40 1.20 From data above, a plot of 1/[HI] vs. t yields a straight line. The reaction is secondorder in HI at 700 K. Rate = k[HI]2. The slope of the line = k = 0.00118 M1s1 1.00 0 250 Time(s) 500 36. (D) (a) We can graph 1/[ArSO2H] vs. time and obtain a straight line. We can also graph [ArSO2H] vs. time and ln([ArSO2H]) vs. time to demonstrate that they do not yield a straight line. Only the plot of 1/[ArSO2H] versus time is shown.
50 1/[ArSO2H] (M ) 1/[ArSO2H] (M1)
1 Plot of 1/[ArSO H] versus Time Plot of 1/[ArSO2H]2versus Time 30 y = 0.137x + 9.464 10 0 50 100 150 Time (min) 200 250 300 The linearity of the line indicates that the reaction is secondorder.
(b) We solve the rearranged integrated secondorder rate law for the rate constant, using the longest time interval.
k= 1 1 = kt At A0 1 1 1 =k t At A0 F GH I JK 1 1 1 1 1 = 0.137 L mol min 300 min 0.0196 M 0.100 M (c) We use the same equation as in part (b), but solved for t , rather than k . 1 1 1 1 1 1 t= = = 73.0 min 1 1 k A t A 0 0.137 L mol min 0.0500 M 0.100 M 629 Chapter 14: Chemical Kinetics (d) We use the same equation as in part (b), but solve for t , rather than k . 1 1 1 1 1 1 t= = = 219 min 1 1 k A t A 0 0.137 L mol min 0.0250 M 0.100 M We use the same equation as in part (b), but solve for t , rather than k . 1 1 1 1 1 1 t= = = 136 min A A 0.137 L mol1min 1 0.0350 M 0.100 M k 0 t (e) 37. (M) (a) Plot [A] vs t, ln[A] vs t, and 1/[A] vs t and see which yields a straight line.
0.8 0.7 0.6 [A] (mol/L) 0.5 0.4 0.3 0.2 0.1 0 0 Time(s) 150 0
y = 0.0050x + 0.7150 R 2 = 1.0000 0.5 0 1 ln[A] 1.5 2 2.5 3 3.5 Time(s)
y = 0.0191x  0.1092 R2 = 0.9192 150 20 15 10 5 0 0 Time(s) 150
y = 0.1228x  0.9650 R2 = 0.7905 Clearly we can see that the reaction is zeroorder in reactant A with a rate constant of 5.0 103.
(b) The halflife of this reaction is the time needed for one half of the initial [A] to react. 0.358 M = 72 s. Thus, A = 0.715 M 2 = 0.358 M and t1/2 = 5.0 103 M / s 1/[A] 630 Chapter 14: Chemical Kinetics 38. (D) (a) We can either graph 1/ C 4 H 6 vs. time and obtain a straight line, or we can determine the secondorder rate constant from several data points. Then, if k indeed is a constant, the reaction is demonstrated to be secondorder. We shall use the second technique in this case. First we do a bit of algebra. IJ FG K H 1 FG 1 1 IJ = 0.875 L mol min k= 24.55 min H 0.0124 M 0.0169 M K 1 FG 1 1 IJ = 0.892 L mol min k= 42.50 min H 0.0103 M 0.0169 M K 1 FG 1 1 IJ = 0.870 L mol min k= 68.05 min H 0.00845 M 0.0169 M K
k=
1 1 1 1 1 = kt At A0 1 1 1 =k t At A0 F GH I JK 1 1 1 = 0.843 L mol 1min 1 12.18 min 0.0144 M 0.0169 M
1 1 1 (b) The fact that each calculation generates similar values for the rate constant indicates that the reaction is secondorder. The rate constant is the average of the values obtained in part (a). 0.843 + 0.875 + 0.892 + 0.870 k= L mol1min 1 = 0.87 L mol1min 1 4 We use the same equation as in part (a), but solve for t , rather than k . 1 1 1 1 1 1 t= = 2.0 102 min = 1 1 A A 0.870 L mol min 0.00423 M 0.0169 M k t 0 We use the same equation as in part (a), but solve for t , rather than k . 1 1 1 1 1 1 2 t= = = 1.6 10 min 1 1 k A t A 0 0.870 L mol min 0.0050 M 0.0169 M A 1.490 M 1.512 M = = +0.022 M / min t 1.0 min 0.0 min A 2.935 M 3.024 M initial rate = = = +0.089 M / min t 1.0 min 0.0 min initial rate = When the initial concentration is doubled 2.0 , from 1.512 M to 3.024 M, the initial rate quadruples 4.0 . Thus, the reaction is secondorder in A (since 2.0 x = 4.0 when x = 2 ). (c) (d) 39. (E) (a) (b) b g b g 631 Chapter 14: Chemical Kinetics 40. (M) (a) Let us assess the possibilities. If the reaction is zeroorder, its rate will be constant. During the first 8 min, the rate is 0.60 M 0.80 M /8 min = 0.03 M/min . Then, during the first 24 min, the rate is 0.35 M 0.80 M /24 min = 0.019 M/min . Thus, the reaction is not zeroorder. If the reaction is firstorder, it will have a constant halflife that is consistent with its rate constant. The halflife can be assessed from the fact that 40 min elapse while the concentration drops from 0.80 M to 0.20 M, that is, to onefourth of its initial value. Thus, 40 min equals two halflives and t1/2 = 20 min . This gives k = 0.693 / t1/2 = 0.693 / 20 min = 0.035 min 1 . Also At 0.35 M 0.827 kt = ln = ln = 0.827 = k 24 min k= = 0.034 min 1 0.80 M 24 min A0 The constancy of the value of k indicates that the reaction is firstorder.
(b) (c) The value of the rate constant is k = 0.034 min 1 . Reaction rate =
ln A A 0 1 2 (rate of formation of B) = k A First we need [A] at t = 30 . min 1 = kt = 0.034 min 1 30. min = 1.02 A A 0 = e 1.02 = 0.36 A = 0.36 0.80 M = 0.29 M rate of formation of B = 2 0.034 min 1 0.29 M = 2.0 102 M min 1
41. (M) The halflife of the reaction depends on the concentration of A and, thus, this reaction cannot be firstorder. For a secondorder reaction, the halflife varies inversely with the reaction rate: t1/2 = 1/ k A 0 or k = 1/ t1/2 A 0 . Let us attempt to verify the secondorder nature of this c h c h reaction by seeing if the rate constant is fixed. 1 k= = 0.020 L mol1min 1 1.00 M 50 min 1 k= = 0.020 L mol1min 1 2.00 M 25 min 1 k= = 0.020 L mol 1 min 1 0.50 M 100 min The constancy of the rate constant demonstrates that this reaction indeed is secondorder. The rate 2 equation is Rate = k A and k = 0.020 L mol 1min 1 . 632 Chapter 14: Chemical Kinetics 42. (M) (a) The halflife depends on the initial NH 3 and, thus, the reaction cannot be firstorder. Let us attempt to verify secondorder kinetics. 1 1 k= for a secondorder reaction k = = 42 M 1min 1 NH3 0 t1/2 0.0031 M 7.6 min 1 1 = 180 M 1min 1 k= = 865 M 1min 1 0.0015 M 3.7 min 0.00068 M 1.7min The reaction is not secondorder. But, if the reaction is zeroorder, its rate will be constant. A 0 / 2 0.0031 M 2 Rate = = = 2.0 104 M/min t1/2 7.6 min 0.0015 M 2 Rate = = 2.0 104 M/min 3.7 min 0.00068 M 2 Rate = 2.0 104 M / min Zeroorder reaction 1.7 min k=
(b) The constancy of the rate indicates that the decomposition of ammonia under these conditions is zeroorder, and the rate constant is k = 2.0 104 M/min. 43. [A]0 1 Secondorder: t1/2 = k[A]0 2k A zeroorder reaction has a half life that varies proportionally to [A]0, therefore, increasing [A]0 increases the halflife for the reaction. A secondorder reaction's halflife varies inversely proportional to [A]0, that is, as [A]0 increases, the halflife decreases. The reason for the difference is that a zeroorder reaction has a constant rate of reaction (independent of [A]0). The larger the value of [A]0, the longer it will take to react. In a secondorder reaction, the rate of reaction increases as the square of the [A]0, hence, for high [A]0, the rate of reaction is large and for very low [A]0, the rate of reaction is very slow. If we consider a bimolecular elementary reaction, we can easily see that a reaction will not take place unless two molecules of reactants collide. This is more likely when the [A]0 is large than when it is small.
(M) Zeroorder: t1/2 = (M) (a) (b) 44. [A]0 0.693 = k 2k [A]0 1 , 2k k[A]0 Hence, Hence, [A]0 = 0.693 or [A]0 = 1.39 M 2 [A]0 2 = 1 or [A]02 = 2.00 M 2 [A]0 = 1.414 M (c) 0.693 1 1 , Hence, 0.693 = or [A]0 = 1.44 M k k[A]o [A]0 633 Chapter 14: Chemical Kinetics Collision Theory; Activation Energy
45. (M) (a) The rate of a reaction depends on at least two factors other than the frequency of collisions. The first of these is whether each collision possesses sufficient energy to get over the energy barrier to products. This depends on the activation energy of the reaction; the higher it is, the smaller will be the fraction of successful collisions. The second factor is whether the molecules in a given collision are properly oriented for a successful reaction. The more complex the molecules are, or the more freedom of motion the molecules have, the smaller will be the fraction of collisions that are correctly oriented. (b) Although the collision frequency increases relatively slowly with temperature, the fraction of those collisions that have sufficient energy to overcome the activation energy increases much more rapidly. Therefore, the rate of reaction will increase dramatically with temperature. The addition of a catalyst has the net effect of decreasing the activation energy of the overall reaction, by enabling an alternative mechanism. The lower activation energy of the alternative mechanism, (compared to the uncatalyzed mechanism), means that a larger fraction of molecules have sufficient energy to react. Thus the rate increases, even though the temperature does not. (c) 46. (M) (a) The activation energy for the reaction of hydrogen with oxygen is quite high, too high, in fact, to be supplied by the energy ordinarily available in a mixture of the two gases at ambient temperatures. However, the spark supplies a suitably concentrated form of energy to initiate the reaction of at least a few molecules. Since the reaction is highly exothermic, the reaction of these first few molecules supplies sufficient energy for yet other molecules to react and the reaction proceeds to completion or to the elimination of the limiting reactant. (b) A larger spark simply means that a larger number of molecules react initially. But the eventual course of the reaction remains the same, with the initial reaction producing enough energy to initiate still more molecules, and so on. 47. (M) (a) The products are 21 kJ/mol closer in energy to the energy activated complex than are the reactants. Thus, the activation energy for the reverse reaction is 84 kJ / mol 21 kJ / mol = 63 kJ / mol. 634 Chapter 14: Chemical Kinetics (b) The reaction profile for the reaction in Figure 1410 is sketched below.
Potential Energy (kJ) A ....B Transiton State Ea(reverse) = 63 kJ Products C+D Ea(forward) = 84 kJ H = +21 kJ Reactants A+B Progress of Reaction 48. (M) In an endothermic reaction (right), Ea must be larger than the H for the reaction. For an exothermic reaction (left), the magnitude of Ea may be either larger or smaller than that of H . In other words, a small activation energy can be associated with a large decrease in the enthalpy, or a large Ea can be connected to a small decrease in enthalpy. reactants
H<0 products products reactants
H>0 49. (E) (a) (b) (c) (d) (e) (f) There are two intermediates (B and C). There are three transition states (peaks/maxima) in the energy diagram. The fastest step has the smallest Ea, hence, step 3 is the fastest step in the reaction with step 2 a close second. Reactant A (step 1) has the highest Ea, and therefore the slowest smallest constant Endothermic; energy is needed to go from A to B. Exothermic; energy is released moving from A to D. 635 Chapter 14: Chemical Kinetics 50. (E) (a) (b) (c) (d) (e) (f) There are two intermediates (B and C). There are three transition states (peaks/maxima) in the energy diagram. The fastest step has the smallest Ea, hence, step 2 is the fastest step in the reaction. Reactant A (step 1) has the highest Ea, and therefore the slowest smallest constant Endothermic; energy is needed to go from A to B. Endothermic, energy is needed to go from A to D. Effect of Temperature on Rates of Reaction
51. (M) k E 1 1 5.4 104 L mol 1 s1 E a 1 1 ln 1 = a = ln = 2 1 1 k2 R T2 T1 2.8 10 L mol s R 683 K 599 K FG H IJ K FG H IJ K 3.95R = Ea 2.05 104 3.95 R Ea = = 1.93 104 K 1 8.3145 J mol 1 K 1 = 1.60 105 J / mol = 160 kJ / mol 4 2.05 10
52. (M) E 1 1 k 5.0 103 L mol1 s 1 1.60 105 J/mol 1 1 ln 1 = a = ln = 2 1 1 1 1 k2 R T2 T1 2.8 10 L mol s 8.3145 J mol K 683 K T 1 1 1.72 1 1 1.72 = 1.92 104 = = 8.96 105 683 K T 683 K T 1.92 104 1 = 8.96 105 +1.46 103 = 1.55 103 T = 645 K T (D) (a) 53. First we need to compute values of ln k and 1/ T . Then we plot the graph of ln k versus 1/T. T , C 0 C 10 C 20 C 30 C T,K 273 K 283 K 293 K 303 K 1 1/ T , K 0.00366 0.00353 0.00341 0.00330 1 6 5 4 k, s 5.6 10 3.2 10 1.6 10 7.6 104 ln k 12.09 10.35 8.74 7.18 636 Chapter 14: Chemical Kinetics
Plot of ln k versus 1/T
0.00325 7.1 8.1 9.1 10.1 11.1 12.1 1/T (K1) 0.00345 0.00365 ln k y = 13520x + 37.4 (b) The slope = Ea / R . J 1 kJ kJ 1.352 104 K = 112 mol K 1000 J mol 6 1 We apply the Arrhenius equation, with k = 5.6 10 s at 0 C (273 K), k = ? at 40 C (313 K), and Ea = 113 103 J/mol. Ea = R slope = 8.3145
ln E 1 1 k = a 6 1 R T1 T2 5.6 10 s k e6.306 = 548 = 5.6 106 s 1 112 103 J/mol 1 1 = = 6.306 1 1 8.3145 J mol K 273 K 313 K k = 548 5.6 106 s 1 = 3.07 103 s 1 (c) t1/2 = 0.693 0.693 = = 2.3 10 2 s 3 1 k 3.07 10 s 54. (D) (a) Here we plot ln k vs. 1/ T . The slope of the straight line equals Ea / R . First we tabulate the data to plot. (the plot is shown below). T , C T ,K 1/ T , K1 k , M1s1 ln k
0.0027 0 2 4 15.83 288.98 0.0034604 5.03 105 9.898
0.0029 32.02 305.17 0.0032769 3.68 104 7.907
1/T (K )
1 59.75 332.90 0.0030039 6.71 103 5.004
0.0033 90.61 363.76 0.0027491 0.119 2.129
0.0035 Plot of ln k versus 1/T
0.0031 ln k 6 8 10 12 y = 10890x + 27.77 637 Chapter 14: Chemical Kinetics The slope of this graph = 1.09 104 K = Ea / R J J 1 kJ kJ Ea = 1.089 104 K 8.3145 = 9.054 104 = 90.5 mol K mol 1000 J mol
(b) We calculate the activation energy with the Arrhenius equation using the two extreme data points. E 1 1 E k 0.119 1 1 ln 2 = ln = +7.77 = a = a 5 k1 5.03 10 R T1 T2 R 288.98 K 363.76 K Ea 7.769 8.3145 J mol1 K 1 Ea = = 9.08 104 J/mol 4 1 R 7.1138 10 K Ea = 91 kJ/mol. The two Ea values are in quite good agreement, within experimental limits. = 7.1138 104 K 1 (c) We apply the Arrhenius equation, with Ea = 9.080 104 J/mol, k = 5.03 10 5 M 1 s1 at 15.83 C (288.98 K), and k = ? at 100.0 C (373.2 K). Ea 1 1 90.80 103 J/mol k 1 1 ln = = 5 1 1 1 1 5.03 10 M s R T1 T 2 8.3145 J mol K 288.98 K 373.2 K ln k
5 1 1 5.03 10 M s k = 5.05 103 5.03 105 M 1 s 1 = 0.254 M 1 s 1 = 8.528 e8.528 = 5.05 103 = k 5.03 10
5 M 1 s 1 55. (M) The halflife of a firstorder reaction is inversely proportional to its rate constant: k = 0.693 / t1/2 . Thus we can apply a modified version of the Arrhenius equation to find Ea. (a) ln (b) t1/2 1 Ea 1 1 k2 46.2 min Ea 1 1 = ln = = = ln k1 2.6 min R 298 K 102 + 273 K t1/2 2 R T1 T2 1 1 E 2.88 8.3145 J mol K 1 kJ 2.88 = a 6.89 104 Ea = = 34.8 kJ/mol 4 1 R 6.89 10 K 1000 J 10.0 min 34.8 103 J / mol 1 1 1 1 ln = = 1.53 = 4.19 103 1 1 46.2 min 8.3145 J mol K T 298 T 298 FG H IJ K FG H IJ K 1 1.53 1 = 3.65 104 = T 298 4.19 103 56. 1 = 2.99 103 T T = 334 K = 61 o C (M) The halflife of a firstorder reaction is inversely proportional to its rate constant: k = 0.693 / t1/2 . Thus, we can apply a modified version of the Arrhenius equation. (a) ln t1/2 1 Ea 1 1 k2 22.5 h Ea 1 1 = ln = = = ln k1 1.5 h R 293 K 40 + 273 K t1/2 2 R T1 T2 1 1 E 2.71 8.3145 J mol K 1 kJ 2.71 = a 2.18 104 , Ea = = 103 kJ/mol 4 1 R 2.18 10 K 1000 J 638 Chapter 14: Chemical Kinetics (b) The relationship is k = A exp Ea / RT
3 b g 103 10 J mol1 13 1 40.9 k = 2.05 1013 s 1exp 3.5 105 s 1 = 2.05 10 s e 1 1 8.3145 J mol K 273 + 30 K 57. (M) (a) It is the change in the value of the rate constant that causes the reaction to go faster. Let k1 be the rate constant at room temperature, 20 C (293 K). Then, ten degrees higher (30 C or 303 K), the rate constant k2 = 2 k1 .
E 1 1 E 1 k2 2 k1 1 4 1 Ea = ln = 0.693 = a = a = 1.13 10 K k1 k1 R T1 T2 R 293 303 K R 0.693 8.3145 J mol 1 K 1 Ea = = 5.1 104 J / mol = 51 kJ / mol 4 1 1.13 10 K Since the activation energy for the depicted reaction (i.e., N2O + NO N2 + NO2) is 209 kJ/mol, we would not expect this reaction to follow the rule of thumb. ln (b) 58. (M) Under a pressure of 2.00 atm, the boiling point of water is approximately 121 C or 394 K. Under a pressure of 1 atm, the boiling point of water is 100 C or 373 K. We assume an activation energy of 5.1 104 J / mol and compute the ratio of the two rates. Rate 2 Ea 1 1 5.1 104 J / mol 1 1 ln = = = 0.88 1 1 Rate1 R T1 T2 8.3145 J mol K 373 394 K FG H IJ K FG H IJ K Rate 2 = e0.88 Rate1 = 2.4 Rate1 . Cooking will occur 2.4 times faster in the pressure cooker. Catalysis
59. (E) (a) Although a catalyst is recovered unchanged from the reaction mixture, it does "take part in the reaction." Some catalysts actually slow down the rate of a reaction. Usually, however, these negative catalysts are called inhibitors. The function of a catalyst is to change the mechanism of a reaction. The new mechanism is one that has a different (lower) activation energy (and frequently a different A value), than the original reaction. (b) 60. (M) If the reaction is firstorder, its halflife is 100 min, for in this time period [S] decreases from 1.00 M to 0.50 M, that is, by one half. This gives a rate constant of k = 0.693 / t1/2 = 0.693 / 100 min = 0.00693 min 1 . The rate constant also can be determined from any two of the other sets of data. A 0 1.00 M 0.357 kt = ln k= = ln = 0.357 = k 60 min = 0.00595 min 1 0.70 M 60 min A t This is not a very good agreement between the two k values, so the reaction is probably not firstorder in [A]. Let's try zeroorder, where the rate should be constant. 639 Chapter 14: Chemical Kinetics 0.90 M 1.00 M 0.50 M 1.00 M = 0.0050 M/min Rate = = 0.0050 M/min 20 min 100 min 0.20 M 0.90 M 0.50 M 0.90 M Rate = = 0.0050 M/min Rate = = 0.0050 M/min 160 min 20 min 100 min 20 min Thus, this reaction is zeroorder with respect to [S]. Rate = 61. (E) Both platinum and an enzyme have a metal center that acts as the active site. Generally speaking, platinum is not dissolved in the reaction solution (heterogeneous), whereas enzymes are generally soluble in the reaction media (homogeneous). The most important difference, however, is one of specificity. Platinum is rather nonspecific, catalyzing many different reactions. An enzyme, however, is quite specific, usually catalyzing only one reaction rather than all reactions of a given class. (E) In both the enzyme and the metal surface cases, the reaction occurs in a specialized location: either within the enzyme pocket or on the surface of the catalyst. At high concentrations of reactant, the limiting factor in determining the rate is not the concentration of reactant present but how rapidly active sites become available for reaction to occur. Thus, the rate of the reaction depends on either the quantity of enzyme present or the surface area of the catalyst, rather than on how much reactant is present (i.e., the reaction is zeroorder). At low concentrations or gas pressures the reaction rate depends on how rapidly molecules can reach the available active sites. Thus, the rate depends on concentration or pressure of reactant and is firstorder. (E) For the straightline graph of Rate versus [Enzyme], an excess of substrate must be present. (E) For human enzymes, we would expect the maximum in the curve to appear around 37C, i.e., normal body temperature (or possibly at slightly elevated temperatures to aid in the control of diseases (37  41 C). At lower temperatures, the reaction rate of enzymeactivated reactions decreases with decreasing temperature, following the Arrhenius equation. However, at higher temperatures, these temperature sensitive biochemical processes become inhibited, probably by temperatureinduced structural modifications to the enzyme or the substrate, which prevent formation of the enzymesubstrate complex. 62. 63. 64. Reaction Mechanisms
65. (E) The molecularity of an elementary process is the number of reactant molecules in the process. This molecularity is equal to the order of the overall reaction only if the elementary process in question is the slowest and, thus, the ratedetermining step of the overall reaction. In addition, the elementary process in question should be the only elementary step that influences the rate of the reaction. (E) If the type of molecule that is expressed in the rate law as being firstorder collides with other molecules that are present in much larger concentrations, the reaction will seem to depend only on the amount of those types of molecules present in smaller concentration, since the larger concentration will be essentially unchanged during the course of the reaction. Such a situation is quite common, and has been given the name pseudo firstorder. It is also possible to have molecules which, do not participate directly in the reaction including product molecules 66. 640 Chapter 14: Chemical Kinetics strike the reactant molecules and impart to them sufficient energy to react. Finally, collisions of the reactant molecules with the container walls may also impart adequate energy for reaction to occur.
67. (M) The three elementary steps must sum to give the overall reaction. That is, the overall reaction is the sum of step 1 step 2 step 3. Hence, step 2 = overall reaction step 1 step 3. Note that all species in the equations below are gases. 2 NO + 2 H 2 N 2 + 2 H 2O Overall: 2 NO + 2H 2 N 2 + 2 H 2 O First: Third 2 NO N 2 O 2 N 2 O + H 2 N 2 + H 2 O or N 2 + H 2 O N 2 O + H 2 N 2 O 2 2NO The rate of this ratedetermining step is: Rate = k2 H2 N2O2 The result is the second step, which is slow: H 2 + N 2O2 H 2O + N 2O Since N 2 O 2 does not appear in the overall reaction, we need to replace its concentration with the concentrations of species that do appear in the overall reaction. To do this, recall that the first step is rapid, with the forward reaction occurring at the same rate as the reverse reaction. 2 k1 NO = forward rate = reverse rate = k 1 N 2 O 2 . This expression is solved for N 2 O 2 , which then is substituted into the rate equation for the overall reaction. 2 k1 NO k k 2 N 2O2 = Rate = 2 1 H 2 NO k 1 k 1 The reaction is firstorder in H 2 and secondorder in [NO]. This result conforms to the experimentally determined reaction order.
68. (M) Proposed mechanism:
1 I 2 (g) 2 I(g) k k Observed rate law: Rate = k[I2][H2] 1 k2 2 I(g) + H 2 (g) 2 HI(g) I2(g) + H2(g) 2 HI(g) The first step is a fast equilibrium reaction and step 2 is slow. Thus, the predicted rate law is Rate = k2[I]2[H2]. In the first step, set the rate in the forward direction for the equilibrium equal to the rate in the reverse direction. Then solve for [I]2. Rateforward = Ratereverse Use: Rateforward = k1[I2] and Ratereverse = k1[I]2 From this we see: k1[I2] = k1[I]2. Rearranging (solving for [I]2) k [I ] k [I ] [I]2 = 1 2 Substitute into Rate = k2[I]2[H2] = k2 1 2 [H2] = kobs[I2][H2] k1 k1 Since the predicted rate law is the same as the experimental rate law, this mechanism is plausible. 641 Chapter 14: Chemical Kinetics 69. (M) Proposed mechanism: 1 Cl2 (g) 2 Cl(g) k k Observed rate law: Rate = k[Cl2][NO]2 1 k2 2 Cl(g) + 2 NO(g) 2 NOCl(g) Cl2(g) + 2NO(g) 2 NOCl(g) The first step is a fast equilibrium reaction and step 2 is slow. Thus, the predicted rate law is Rate = k2[Cl]2[NO]2 In the first step, set the rate in the forward direction for the equilibrium equal to the rate in the reverse direction. Then express [Cl]2 in terms of k1, k1 and [Cl2]. This mechanism is almost certainly not correct because it involves a tetra molecular second step. Rateforward = Ratereverse Use: Rateforward = k1[Cl2] and Ratereverse = k1[Cl]2 From this we see: k1[Cl2] = k1[Cl]2. Rearranging (solving for [Cl]2) k [Cl ] k [Cl ] [Cl]2 = 1 2 Substitute into Rate = k2[Cl]2[NO]2 = k2 1 2 [NO]2 = kobs[Cl2][NO2]2 k1 k1
k1 There is another plausible mechanism. Cl2 (g) + NO(g) NOCl(g) + Cl(g) k1 k1 Cl(g) + NO(g) NOCl(g) k1 Cl2(g) + 2NO(g) 2 NOCl(g) Rateforward = Ratereverse Use: Rateforward = k1[Cl2][NO] and Ratereverse = k1[Cl][NOCl] From this we see: k1[Cl2][NO] = k1[Cl][NOCl]. Rearranging (solving for [Cl]) k [Cl ][NO] k k [Cl ][NO]2 [Cl] = 1 2 Substitute into Rate = k2[Cl][NO] = 2 1 2 k1 [NOCl] k 1 [NOCl] If [NOCl], the product is assumed to be constant (~ 0 M using method of initial rates), then k2 k1 constant = k obs Hence, the predicted rate law is kobs [Cl2 ][NO]2 which agrees with k1 [NOCl] the experimental rate law. Since the predicted rate law agrees with the experimental rate law, both this and the previous mechanism are plausible, however, the first is dismissed as it has a tetramolecular elementary reaction (extremely unlikely to have four molecules simultaneously collide).
70. (M) A possible mechanism is: Step 1: O3 O 2 + O fast k Step 2: 3 O + O3 2 O 2 slow The overall rate is that of the slow step: Rate = k 3 O O 3 . But O is a reaction intermediate, whose concentration is difficult to determine. An expression for [O] can be found by assuming that the forward and reverse "fast" steps proceed with equal speed. 2 k1 O3 k1 O3 k3 k1 O3 Rate = k3 O3 = Rate1 = Rate 2 k1 O3 = k2 O 2 O O = k2 O 2 k2 O 2 k2 O 2 Then substitute this expression into the rate law for the reaction. This rate equation has the same form as the experimentally determined rate law and thus the proposed mechanism is plausible. 642 Chapter 14: Chemical Kinetics 71. (M) k1 S1 S2 S1 : S2 k
1 S1 : S2 k2 S1 : S2 d S1 : S2 k1 S1 S2 k 1 S1 : S2 k 2 S1 : S2 dt k1 S1 S2 k 1 k 2 S1 : S2 S1 : S2 k1 S1 S2 k 1 k 2 k k S S d S1 : S2 k 2 S1 : S2 2 1 1 2 dt k 1 k 2 72. (M) k1 CH3 2 CO (aq) + OH CH3C O CH 2 (aq) + H 2O (l) k
1 k2 CH 3C O CH (aq) + CH 3 2 CO (aq) Pr od 2 We note that CH3C(O)CH2 is an intermediate species. Using the steady state approximation, while its concentration is not known during the reaction, the rate of change of its concentration is zero, except for the very beginning and towards the end of the reaction. Therefore, d CH 3C O CH 2 k CH CO OH k CH C O CH H O 1 3 2 2 2 1 3 dt k 2 CH 3C O CH 2 CH 3 2 CO 0 Rearranging the above expression to solve for CH3C(O)CH2 gives the following expression k1 CH 3 2 CO OH CH3C O CH 2 k 1 H 2 O k 2 CH 3 2 CO The rate of formation of product, therefore, is: d Pr od k 2 CH 3C O CH 2 CH 3 2 CO dt k 2 CH 3 2 CO k 1 H 2 O k 2 CH 3 2 CO 2 k1 CH 3 2 CO OH k 2 k1 CH 3 2 CO OH CH 3 2 CO k 1 H 2 O k 2 643 Chapter 14: Chemical Kinetics INTEGRATIVE AND ADVANCED EXERCISES
73. (M) The data for the reaction starting with 1.00 M being firstorder or secondorder as well as that for the firstorder reaction using 2.00 M is shown below Time (min) 0 5 10 15 25 [A]o = 1.00 M (second order) 1.00 0.63 0.46 0.36 0.25 [A]o = 1.00 M (first order) 1.00 0.55 0.30 0.165 0.05 [A]o = 2.00 M (second order) 2.00 0.91 0.59 0.435 0.286 [A]o = 2.00 M (first order) 2.00 1.10 0.60 0.33 0.10 Clearly we can see that when [A]o = 1.00 M, the firstorder reaction concentrations will always be lower than that for the secondorder case (assumes magnitude of the rate constant is the same). If, on the other hand, the concentration is above 1.00 M, the secondorder reaction decreases faster than the firstorder reaction (remember that the halflife shortens for a secondorder reaction as the concentration increases, whereas for a firstorder reaction, the halflife is constant). From the data, it appears that the crossover occurs in the case where [A]0 = 2.00 M at just over 10 minutes. Secondorder at 11 minutes: Firstorder at 11 minutes: 1 1 0.12 (11 min) [A] 2 M min [A] = 0.549 M 0.12 ln[A] ln(2) [A] = 0.534 M (11 min) min A quick check at 10.5 minutes reveals, 1 1 0.12(10.5 min) Secondorder at 10.5 minutes: [A] = 0.568 M [A] 2 M min 0.12(10.5 min) Firstorder at 10.5 minutes: ln[A] ln(2) [A] = 0.567 M M min Hence, at approximately 10.5 minutes, these two plots will share a common point (point at which the concentration versus time curves overlap).
74. (M) (a) The concentration vs. time graph is not linear. Thus, the reaction is obviously not zeroorder (the rate is not constant with time). A quick look at various half lives for this reaction shows the ~2.37 min (1.000 M to 0.5 M), ~2.32 min (0.800 M to 0.400 M), and ~2.38 min(0.400 M to 0.200 M). Since the halflife is constant, the reaction is probably firstorder. (b) average t1/ 2 (2.37 2.32 2.38) 0.693 0.693 2.36 min k 0.294 min 1 3 t1/2 2.36 min 1 or perhaps better expressed as k = 0.29 min due to imprecision. 644 Chapter 14: Chemical Kinetics (c) When t 3.5 min, [A] 0.352 M. Then, rate = k[A] = 0.294 min 1 0.352 M 0.103 M/min. (d) Slope [A] t Rate 0.1480 M 0.339 M 6.00 min 3.00 min 0.0637 M / min Rate 0.064 M/min. (e) Rate = k[A] = 0.294 min1 1.000 M = 0.294 M/min. 75. (M) The reaction being investigated is: 2 MnO 4 (aq) + 5 H 2O 2 (aq) + 6 H + (aq) 2 Mn 2+ (aq) +8 H 2O(l) + 5O 2 (g) We use the stoichiometric coefficients in this balanced reaction to determine [H2O2]. 0.1000 mmol MnO4 5 mmol H 2 O 2 37.1 mL titrant 1 mL titrant 2 mmol MnO4 [H 2 O 2 ] 1.86 M 5.00 mL 76. (D) We assume in each case that 5.00 mL of reacting solution is titrated. 2.32 mmol H 2 O 2 2 mmol MnO4 1 mL titrant volume MnO4 5.00 mL 1 mL 5 mmol H 2 O 2 0.1000 mmol MnO4 At 200 s At 400 s At 600 s At 1200 s At 1800 s At 3000 s 20.0 2.32 M H 2 O 2 46.4 mL 0.1000 M MnO4 titrant at 0 s Vtitrant = 20.0 2.01 M H2O2 = 40.2 mL 38.5 33 Vtitrant = 20.0 1.72 M H2O2 = 34.4 mL 27.5 Vtitrant = 20.0 1.49 M H2O2 = 29.8 mL 22 16.5 Vtitrant = 20.0 0.98 M H2O2 = 19.6 mL 11 Tangent line Vtitrant = 20.0 0.62 M H2O2 = 12.5.5mL 4 Vtitrant = 20.0 0.25 M H2O2 = 5.000mL
0 700 1400 2100 2800 The graph of volume of titrant vs. elapsed time is given above. This graph is of approximately the same shape as Figure 142, in which [H2O2] is plotted against time. In order to determine the rate, the tangent line at 1400 s has been drawn on the graph. The intercepts of the tangent line are at 34 mL of titrant and 2800 s. From this information we determine the rate of the reaction. 34 mL 1L 0.1000 mol MnO4 5 mol H 2 O 2 2800 s 1000 mL 1 L titrant 2 mol MnO4 6.1 104 M/s Rate 0.00500 L sample This is the same as the value of 6.1 104 obtained in Figure 142 for 1400 s. The discrepancy is due, no doubt, to the coarse nature of our plot. 645 Chapter 14: Chemical Kinetics 77. (M) First we compute the change in [H2O2]. This is then used to determine the amount, and ultimately the volume, of oxygen evolved from the given quantity of solution. Assume the O2(g) is dry. [H 2 O 2 ] 60 s [H 2 O 2 ] t 1.7 103 M/s 1.00 min 0.102 M t 1 min 0.102 mol H 2 O 2 1 mol O 2 amount O 2 0.175 L soln 0.00892 mol O 2 1L 2 mol H 2 O 2 L atm 0.00892 mol O 2 0.08206 (273 24) K nRT mol K Volume O 2 0.22 L O 2 1 atm P 757 mmHg 760 mmHg 78. (M) We know that rate has the units of M/s, and also that concentration has the units of M. The generalized rate equation is Rate k A 0 . In terms of units, this becomes M/s = {units of k} M0. Therefore {Units of k} M/s M10 s 1 M0 79. (M) (a) Comparing the third and the first lines of data, [I] and [OH]stay fixed, while [OCl] doubles. Also the rate for the third kinetics run is one half of the rate found for the first run. Thus, the reaction is firstorder in [OCl]. Comparing the fourth and fifth lines, [OCl] and [I] stay fixed, while [OH] is halved. Also, the fifth run has a reaction rate that is twice that of the fourth run. Thus, the reaction is minus firstorder in [OH]. Comparing the third and second lines of data, [OCl] and [OH] stay fixed, while the [I] doubles. Also, the second run has a reaction rate that is double that found for the third run. Thus, the reaction is firstorder in [I]. (b) The reaction is firstorder in [OCl] and [I] and minus firstorder in [OH]. Thus, the overall order = 1 + 1 1 = 1. The reaction is firstorder overall. Rate k
(c) [OCl ][I ] [OH ] Rate [OH ] 4.8 104 M/s 1.00 M 60. s 1 [OCl ][I ] 0.0040 M 0.0020 M using data from first run: 80. (M) We first determine the number of moles of N2O produced. The partial pressure of N2O(g) in the "wet" N2O is 756 mmHg 12.8 mmHg = 743 mmHg. 1 atm 743 mmHg 0.0500 L 760 mmHg PV amount N 2 O 0.00207 mol N 2 O RT 0.08206 L atm mol 1 K 1 (273 15) K Now we determine the change in [NH2NO2]. 646 Chapter 14: Chemical Kinetics 1 mol NH 2 NO 2 1 mol N 2 O [NH 2 NO 2 ] 0.0125 M 0.165 L soln 0.693 [NH 2 NO 2 ]final 0.105 M0.0125 M 0.093 M 0.00563 min 1 k 123 min [A]t 1 1 0.093 M ln 22 min elapsed time t ln 1 0.00563 min 0.105 M k [A]0 0.00207 mol N 2 O 81. (D) We need to determine the partial pressure of ethylene oxide at each time in order to determine the order of the reaction. First, we need the initial pressure of ethylene oxide. The pressure at infinite time is the pressure that results when all of the ethylene oxide has decomposed. Because two moles of product gas are produced for every mole of reactant gas, this infinite pressure is twice the initial pressure of ethylene oxide. Pinitial = 249.88 mmHg 2 = 124.94 mmHg. Now, at each (CH 2 ) 2 O(g) CH 4 (g) CO(g) time we have the following. Initial: 124.94 mmHg Changes: x mmHg +x mmHg +x mmHg Final: 124.94 + x mmHg Thus, x = Ptot 124.94 and PEtO = 124.94 x = 124.94 (Ptot 124.94) = 249.88 Ptot Hence, we have, the following values for the partial pressure of ethylene oxide. t, min 0 10 20 40 60 100 200 PEtO, mmHg 124.94 110.74 98.21 77.23 60.73 37.54 11.22 For the reaction to be zeroorder, its rate will be constant. (110.74 124.94) mmHg The rate in the first 10 min is: Rate 1.42 mmHg/min 10 min (77.23 124.94) mmHg The rate in the first 40 min is: Rate 1.19 mmHg/min 40 min We conclude from the nonconstant rate that the reaction is not zeroorder. For the reaction to be firstorder, its halflife must be constant. From 40 min to 100 mina period of 60 minthe partial pressure of ethylene oxide is approximately halved, giving an approximate halflife of 60 min. And, in the first 60 min, the partial pressure of ethylene oxide is approximately halved. Thus, the reaction appears to be firstorder. To verify this tentative conclusion, we use the integrated firstorder rate equation to calculate some values of the rate constant.
1 P 1 110.74 mmHg k ln ln 0.0121 min 1 t P0 10 min 124.94 mmHg k 1 100 min ln 37.54 mmHg 124.94 mmHg 0.0120 min 1 k 1 60 min ln 60.73 mmHg 124.94 mmHg 0.0120 min 1 The constancy of the firstorder rate constant suggests that the reaction is firstorder.
82. (M) For this firstorder reaction
Pt 1 P kt Elapsed time is computed as: t ln t P0 k P0 We first determine the pressure of DTBP when the total pressure equals 2100 mmHg. ln 647 Chapter 14: Chemical Kinetics Reaction: Initial: Changes: C8 H18O 2 (g) 2C3 H 6 O(g) C2 H 6 (g) [Equation 15.16] 800.0 mmHg x mmHg 2x mmHg x mmHg x mmHg Final: (800.0 x) mmHg 2x mmHg Total pressure (800.0x) 2x x 800.0 2 x 2100. x 650. mmHg P{C8 H18O 2 (g)} 800. mmHg 650. mmHg 150. mmHg 1 P 1 150. mmHg 192 min 1.9 102 min t ln t ln 3 1 k P0 8.7 10 min 800. mmHg
83. (D) If we compare Experiment 1 with Experiment 2, we notice that [B] has been halved, and also that the rate, expressed as [A]/ t , has been halved. This is most evident for the times 5 min, 10 min, and 20 min. In Experiment 1, [A] decreases from 1.000 103 M to 0.779 103 M in 5 min, while in Experiment 2 this same decrease in [A] requires 10 min. Likewise in Experiment 1, [A] decreases from 1.000 103 M to 0.607 M 103 in 10 min, while in Experiment 2 the same decrease in [A] requires 20 min. This dependence of rate on the first power of concentration is characteristic of a firstorder reaction. This reaction is firstorder in [B]. We now turn to the order of the reaction with respect to [A]. A zeroorder reaction will have a constant rate. Determine the rate (0.951 1.000) 10 3 M Rate 4.9 10 5 M/min After over the first minute: 1 min (0.779 1.000) 10 3 M Rate 4.4 10 5 M/min After over the first five minutes: 5 min (0.368 1.000) 10 3 M 3.2 10 5 M/min After over the first twenty minutes: Rate 20 min This is not a very constant rate; we conclude that the reaction is not zeroorder. There are no clear halflives in the data with which we could judge the reaction to be firstorder. But we can determine the value of the firstorder rate constant for a few data. 1 [A] 1 0.951 mM 0.0502 min 1 k ln ln t [A]0 1 min 1.000 mM 1 0.607 mM 1 0.368 mM 0.0499 min 1 0.0500 min 1 ln k ln 10 min 1.000 mM 20 min 1.000 mM The constancy of the firstorder rate constant indicates that the reaction indeed is firstorder in [A]. (As a point of interest, notice that the concentrations chosen in this experiment are such that the reaction is pseudozeroorder in [B]. Here, then it is not necessary to consider the variation of [B] with time as the reaction proceeds when determining the kinetic dependence on [A].) k 648 Chapter 14: Chemical Kinetics 84. (M) In Exercise 79 we established the rate law for the iodinehypochlorite ion reaction: Rate k[OCl ][I ][OH ]1 . In the mechanism, the slow step gives the rate law; Rate k3 [I ][HOCl] . We use the initial fast equilibrium step to substitute for [HOCl] in this rate equation. We assume in this fast step that the forward rate equals the reverse rate. k1[OCl ][H 2 O] k1[HOCl][OH ] Rate k2 [I ] k1[OCl ][H 2 O] [HOCl] k1[OH ] k1[OCl ][H 2 O] k2 k1[H 2 O] [OCl ][I ] [OCl ][I ] k k1[OH ] k1 [OH ] [OH ] This is the same rate law that we established in Exercise 79. We have incorporated [H2O] in the rate constant for the reaction because, in an aqueous solution, [H2O] remains effectively constant during the course of the reaction. (The final fast step simply involves the neutralization of the acid HOI by the base hydroxide ion, OH.)
85. (M) It is more likely that the cisisomer, compound (I), would be formed than the transisomer, compound (II). The reason for this is that the reaction will involve the adsorption of both CH3CCCH3 and H2 onto the surface of the catalyst. These two molecules will eventually be adjacent to each other. At some point, one of the bonds in the CC bond will break, the HH bond will break, and two CH bonds will form. Since these two CH bonds form on the same side of the carbon chain, compound (I) will be produced. In the sketches below, dotted lines (...) indicate bonds forming or breaking.
H H CH3 C C CH3 CH3 H H C C CH3 H H CH3 C C CH3 86. (D) Hg2 2+ + Tl 2 Hg + Tl Experimental rate law = k
3+ 2+ +
k [Hg 2 2+ ][Tl3+ ] [Hg 2+ ] Possible mechanism: 1 Hg22+ + Tl3+ Hg2+ + HgTl3+ (fast) k1 k2 (slow) HgTl3+ Hg2+ + Tl3+ 2+ 3+ 2+ + Rate = k2[HgTl3+] Hg2 + Tl 2 Hg + Tl k1[Hg22+][Tl3+] = k1 [Hg2+][HgTl3+] Rate = k2[HgTl3+] = rearrange [HgTl3+ ] = k1 [Hg 2 2+ ][Tl3+ ] k1 [Hg 2+ ] k2 k1 [Hg 2 2+ ][Tl3+ ] [Hg 2 2+ ][Tl3+ ] = kobs k1 [Hg 2+ ] [Hg 2+ ] 649 Chapter 14: Chemical Kinetics 87. (M) CCl 3 rateformation ratedisappearance 0 t so rateformation ratedecomposition k2 [CHCl 3 ] k3 k 2 [Cl(g)][CHCl 3 ] k 3[CCl 3 ][[Cl(g)] and, simplifying, [CCl 3 ] k since rate k 3[CCl 3 ][Cl(g)] k3 2 [CHCl 3 ] [Cl(g)] k 2 [CHCl 3 ][Cl(g)] k3 k We know: [Cl(g)] 1 [Cl 2 (g)] k 1 1/ 2 k then rateoverall = k 2 [CHCl 3 ] 1 [Cl 2 (g)] k 1 1/2 1/2 k and the rate constant k will be: k k2 1 k 1 88. (D)
Rate = k[A]3  4.8 103 (1.3 10 ) 3 3.6 10 2 1/2 0.015 d[A] d[A] Rearrange: kt = dt [A]3 Integrate using the limits time (0 to t) and concentration ([A]0 to [A]t )
t 1 1 1 1 2 2 [A]t 2 [A]o 2 0 1 1 1 1 Simplify : kt Rearrange : kt 2 2 2 2[A]t 2[A]o 2[A]t 2[A]0 2 1 1 Multiply through by 2 to give the integrated rate law: 2kt 2 [A]t [A]o 2 k t = d[A] [A]3 [A]o
[A]t kt (k (0)) To derive the half life(t1 2 ) substitute t =t1 2 and [A]t = 1 [A]0 2 2kt1 2 2 [A]o 2 2kt1 2 1 1 4 2 2 [A]o [A]o 2 [A]o 4 Collect terms 1 3 4 2 2 [A]o [A]o [A]o 2 Solve for t1 2 t1 2 3 2k[A]o 2 89. (D) Consider the reaction: A + B products (firstorder in A, firstorder in B). The initial concentration of each reactant can be defined as [A]0 and [B]0 Since the stoichiometry is 1:1, we can define x as the concentration of reactant A and reactant B that is removed (a variable that changes with time). The [A]t = ([A]o x) and [B]t = ([B]o x). Algebra note: [A]o x = (x [A]o) and [B]o x = (x [B]o). 650 Chapter 14: Chemical Kinetics As well, the calculus requires that we use the absolute value of x [A]o and x [B]o when taking the integral of the reciprocal of x [A]o and x [B]o
Rate = d[A] dt = d[B] dx = k [A]t [B]t =k ([A]o x)([B]o x) dt dt or dx ([A]o x)([B]o x) = kdt In order to solve this, partial fraction decomposition is required to further ease integration:
dx 1 1 kdt ([B]o [A]o ) ([A]o x) ([B]o x) Note: this is a constant 1 From the point of view of integration, a further rearrangement is desirable (See algebra note above). dx 1 1 kdt ([B]o [A]o ) x[A]o x[B]o t
1 Integrate both sides ln x[A]o  (ln x[B]o ) = kt + C ([B]o [A]o ) Substitute and simplify x[A]o = [A]o x and x[B]o = [B]o x
1 ([B]o x) 1 ln = kt + C Determine C by setting x = 0 at t = 0 ([B]o [A]o ) [A]o x [B]o ([B]o x) [B]o 1 1 Hence: ln ln = kt + ln ([B]o [A]o ) [A]o ([B]o [A]o ) ([A]o x) ([B]o [A]o ) [A]o ([B]o x) [B]o Multiply both sides by ([B]o [A]o ), hence, ln = ([B]o [A]o )kt + ln [A]o ([A]o x) ([B]o x) ([A] x) ([B]o x) [B]o = ln [A]o ([B]o x) o ln ln =([B]o [A]o )kt = ln [B]o [B]o ([A]o x) ([A]o x) [A]o [A]o
C= 1 Set ([A]o x) =[A]t and ([B]o x) =[B]t to give ln [A]o [B]t = ([B]o [A]o )kt [B]o [A]t 651 Chapter 14: Chemical Kinetics 90. (D) Let 2502x equal the partial pressure of CO(g) and x be the partial pressure of CO2(g). 2CO CO 2 C(s) Ptot =PCO + PCO2 = 250 2x +x = 250 x x 2502x Ptot [torr] 250 238 224 210 Time [sec] 0 398 1002 1801 PCO2 0 12 26 40 PCO [torr] 250 226 198 170 The plots that follow show that the reaction appears to obey a secondorder rate law. Rate = k[CO]2
250 240 230 220 210 200 190 180 R = 0.9867 170 0 1000 time (S) 2000
2 5.52 0.006 PCO (mmHg) 5.32 ln PCO 0.005 5.12 0 R2 = 0.9965 500 1000 time (S) 1500 2000 1/PCO (mmHg ) 1 R2 = 1 0.004 0 1000 time (S) 2000 ZERO ORDER PLOT
T 0 398 1002 1801 PCO 250 226 198 170 1st ORDER PLOT
T 0 398 1002 1801 lnPCO 5.521461 5.420535 5.288267 5.135798 2nd ORDER PLOT
T 0 398 1002 1801 1/CO 0.004 0.004425 0.005051 0.005882 (Best correlation coefficient) 652 Chapter 14: Chemical Kinetics 91. (D) Let 1004x equal the partial pressure of PH3(g), x be the partial pressure of P4(g)and 6x be the partial pressure of H2 (g) 4 PH 3 (g) P 4 ( g ) 6 H 2 (g)
100 4x x 6x Ptot =PPH3 + PP4 + PH2 = 100 4x + x + 6x = 100+ 3x Ptot [torr] 100 150 167 172 Time [sec] 0 40 80 120 PP4 [torr] 0 50/3 67/3 72/3 PPH3 [torr] 100 100(4)(50/3) 100(4)(67/3) 100(4)(72/3) The plots to follow show that the reaction appears to obey a firstorder rate law. Rate = k[PH3]
100 90 80 70 PPH3 (mmHg) 60 50 40 30 20 1.85 10
0.010 1/PPH3 (mmHg 1 ) 4.35 R = 0.8369 3.85
0.160
2 R2 = 0.8652 R = 0.9991 2 0.210 3.35 ln PPH3 2.85 0.110 2.35
0.060 0 50 50 time (S) 150 1.35 0 100 time (S) 200 0 50 100 150 time (S) ZERO ORDER PLOT
T 0 40 80 120 PPH3 100 33.3 10.7 4 1st ORDER PLOT
T 0 40 80 120 lnPPH3 4.61 3.51 2.37 1.39 2nd ORDER PLOT
T 0 40 80 120 1/PPH3 0.010 0.030 0.093 0.250 653 Chapter 14: Chemical Kinetics 92. (D) Consider the following equilibria. k1 KI k2 E + S ES E + P E + I EI k 1 Product production d[P] = k 2 [ES] dt Use the steady state approximation for [ES] d[ES] = k1 [E][S] k 1 [ES] k 2 [ES] = k1 [E][S] [ES] k 1 k 2 = 0 dt k k2 k [E][S] [E][S] [ES] = 1 = solve for [ES] Keep in mind K M = 1 k1 k 1 k 2 KM [E][I] [E][I] Formation of EI: K I = [EI] = [EI] KI [E o ] = [E] + [ES] + [EI] = [E] + [S] [I] [E o ] = [E] 1 + + KM KI [E][S] [E][I] + KM KI
[E] = [E o ] [S] [I] + 1 + KM KI Solve for [E] From above: [ES] = [E][S] KM [S] [E o ] KM [ES] = [S] [I] + 1 + KM KI = [E o ][S] [I] K M 1 + + [S] KI k 2 [E o ][S] multiplication by KM affords KM [ES] = [E o ][S] [I] K M 1 + + [S] KI Vmax [S] d[P] If we substitute k 2 [E o ] = Vmax then = dt [I] K M 1 + + [S] KI Vmax [S] Thus, as [I] increases, the ratio decreases; [I] K M 1 + + [S] KI i.e., the rate of product formation decreases as [I] increases. [I]K M K M + [S] + KI d[P] = k 2 [ES] = Remember dt 654 Chapter 14: Chemical Kinetics 93. (D) In order to determine a value for KM, we need to rearrange the equation so that we may obtain a linear plot and extract parameters from the slope and intercepts. k [E ][S] K M +[S] KM 1 [S] V= 2 o = = + K M +[S] V k 2 [E o ][S] k 2 [E o ][S] k 2 [E o ][S] KM 1 1 = + V k 2 [E o ][S] k 2 [E o ] Plot We need to have this in the form of y = mx+b 1 1 on the yaxis and on the xaxis. See result below: V [S]
1 [V]
K M o] [E = k2 The plot of 1/V vs 1/[S] should yield a slope of yintercept of
0= KM and a k 2 [E o ] pe lo s xintercept = 1 KM 1 . The xintercept = 1/KM k 2 [E o ]
1 [S] = k 2 [E o ] k 2 [E o ] K M 1 KM yintercept = 1 k2[Eo] 1 [S] KM 1 1 [S] k [E ] k 2 [E o ] 2 o To find the value of KM take the negative inverse of xintercept. To find k2, invert the yintercept and divide by the [Eo]. 94. (M) a) The first elementary step HBr O 2 k1 HOOBr is ratedetermining if the reaction obeys reaction rate = k [HBr][O2] since the rate of this step is identical to that of the experimental rate law. b) No, mechanisms cannot be shown to be absolutely correct, only consistent with experimental observations. c) Yes; the sum of the elementary steps (3 HBr + O2 HOBr + Br2 + H2O) is not consistent with the overall stoichiometry (since HOBr is not detected as a product) of the reaction and therefore cannot be considered a valid mechanism. 95. (M) (a) Both reactions are firstorder, because they involve the decomposition of one molecule. (b) k2 is the slow reaction. (c) To determine the concentration of the product, N2, we must first determine how much reactant remains at the end of the given time period, from which we can calculate the amount of reactant consumed and therefore the amount of product produced. Since this is a firstorder reaction, the concentration of the reactant, N2O after time t is determined as follows: A t A 0 e kt N 2O0.1 2.0 M exp 25.7 s1 0.1 s 0.153 M N 2O remaining
655 Chapter 14: Chemical Kinetics The amount of N2O consumed = 2.0 M 0.153 M =1.847 M 1 M N2 0.9235 M N 2 [N 2 ] 1.847 M NO 2 M NO (d) The process is identical to step (c). N 2O0.1 4.0 M exp 18.2 s1 0.025 s 2.538 M N 2O remaining The amount of N2O consumed = 4.0 M 2.538 M =1.462 M 1 M N2O 0.731 M N 2 O [N 2 O] 1.462 M NO 2 M NO FEATURE PROBLEMS
96. (D) (a) To determine the order of the reaction, we need C 6 H 5 N 2 Cl at each time. To determine to C 6 H 5 N 2 Cl = 0.000 M . this value, note that 58.3 mL N 2 g evolved corresponds to total depletion of C6 H 5 N 2 Cl , Thus, at any point in time, C 6 H 5 N 2 Cl = 0.071 M volume N 2 g F GH bg 0.071 M C 6 H 5 N 2 Cl 58.3 mL N 2 g I b g JK 0.071 M C6 H 5 N 2 Cl Consider 21 min: C6 H 5 N 2 Cl = 0.071 M 44.3 mL N 2 = 0.017 M 58.3 mL N 2 g The numbers in the following table are determined with this method.
Time, min VN2 , mL 0 0 71 3 6 9 12 15 18 21 24 27 30 10.8 19.3 26.3 32.4 37.3 41.3 44.3 46.5 48.4 50.4 58.3 58 47 39 32 26 21 17 14 12 10
27 30 C H
6 5 N 2 Cl , mM 0 (b) [The concentration is given in thousandths of a mole per liter (mM).] time(min) 0 3 6 9 12 15 18 21 24 T(min) 3 3 3 3 3 3 3 3 3 3 [C6H5N2Cl](mM) 71 58 47 39 32 26 21 17 14 12 10 [C6H5N2Cl](mM) Reaction Rate (mM min )
1 13 11 8 7 6 2.0 5 1.7 4 1.3 3 1.0 2 0.7 2 0.7 4.3 3.7 2.7 2.3 656
Chapter 14: Chemical Kinetics (c) The two graphs are drawn on the same axes.
[C6H5N2Cl (mM) or VN2 (mL) Plot of [C 6H 5N 2Cl] and VN versus time
2 mL of N2 (m L)
N 60 2 Cl] (m M) or V 40 N5 H6 [C 2 20 [C6H5N2Cl] 0 0 5 10 15 20 25 30 time (min) (d) The rate of the reaction at t = 21 min is the slope of the tangent line to the C 6 H 5 N 2 Cl curve. The tangent line intercepts the vertical axis at about C 6 H 5 N 2 Cl = 39 mM and the horizontal axis at about 37 min 39 103 M = 1.05 103 M min 1 = 1.1 103 M min 1 37 min The agreement with the reported value is very good. Reaction rate = (e) The initial rate is the slope of the tangent line to the C 6 H 5 N 2 Cl curve at t = 0 . The intercept with the vertical axis is 71 mM, of course. That with the horizontal axis is about 13 min. 71103 M Rate = = 5.5 103 M min 1 13 min The firstorder rate law is Rate = k C6 H 5 N 2 Cl , which we solve for k:
k= Rate C6 H5 N 2Cl (f) k0 = 5.5 103 M min 1 = 0.077 min 1 3 71 10 M 1.1103 M min 1 k21 = = 0.065 min 1 3 17 10 M An average value would be a reasonable estimate: k avg = 0.071 min 1
(g) The estimated rate constant gives one value of the halflife: 0.693 0.693 t1/2 = = = 9.8 min 0.071 min 1 k The first halflife occurs when C6 H 5 N 2 Cl drops from 0.071 M to 0.0355 M. This occurs at about 10.5 min. (h) The reaction should be threefourths complete in two halflives, or about 20 minutes.
657 Chapter 14: Chemical Kinetics (i) The graph plots ln C6 H 5 N 2 Cl (in millimoles/L) vs. time in minutes.
Time (min) Plot of ln[C6H5N2Cl] versus time
10 20 30 0 2.75 ln[C6H5N2Cl] 3.25 3.75 y = 0.0661x  2.66
4.25 4.75 The linearity of the graph demonstrates that the reaction is firstorder.
(j) k = slope = 6.61102 min 1 = 0.0661 min 1 t1/2 = 0.693 = 10.5 min , in good agreement with our previously determined values. 0.0661 min 1 97. (D) (a) In Experiments 1 & 2, [KI] is the same (0.20 M), while bNH g S O
4 2 2 8 is halved, from 0.20 M to 0.10 M. As a consequence, the time to produce a color change doubles (i.e., the rate is 2 halved). This indicates that reaction (a) is firstorder in S2 O8 . Experiments 2 and 3 produce a similar conclusion. In Experiments 4 and 5, bNH g S O
4 2 2 8 is the same (0.20 M) while [KI] is halved, from 0.10 to 0.050 M. As a consequence, the time to produce a color change nearly doubles, that is, the rate is halved. This indicates that reaction (a) is also firstorder in I . Reaction (a) is (1 + 1) secondorder overall.
(b) The blue color appears when all the S2 O 2 has been consumed, for only then does reaction 3 2 (b) cease. The same amount of S2 O 3 is placed in each reaction mixture. 0.010 mol Na 2S2 O3 1 mol S2 O32 1L amount S2 O32 = 10.0 mL = 1.0 104 mol 1000 mL 1L 1 mol Na 2 S2 O3 Through stoichiometry, we determine the amount of each reactant that reacts before this 2 amount of S2 O3 will be consumed. 658 Chapter 14: Chemical Kinetics amount S2 O8 2 1.0 10 4 mol S2 O3 2 1 mol I3 1 mol S2 O8 2 2 mol S2 O3 1 mol I3 2 = 5.0 105 mol S2 O8 amount I = 5.0 105 mol S2 O8 2 2 2 mol I = 1.0 104 mol I 2 1 mol S2 O8 Note that we do not use "3 mol I " from equation (a) since one mole has not been oxidized; it simply complexes with the product I 2 . The total volume of each solution is 25.0 mL + 25.0 mL +10.0 mL + 5.0 mL = 65.0 mL , or 0.0650 L. The amount of S2 O8 S2 O8 2 =
2 that reacts in each case is 5.0 105 mol and thus 5.0 105 mol = 7.7 104 M 0.0650 L 2 S2 O8 +7.7 104 M Thus, Rate1 = = = 3.7 105 M s1 t 21 s +7.7 104 M = 1.8 105 M s1 t 42 s To determine the value of k , we need initial concentrations, as altered by dilution. 25.0 mL 25.0 mL 2 S2 O8 = 0.20 M I = 0.20 M = 0.077 M = 0.077 M 1 1 65.0 mL total 65.0 mL =
2 Rate1 = 3.7 105 M s1 = k S2 O8 1 (c) For Experiment 2, Rate 2 = S2 O8 2 I = k 0.077 M 1 b g b0.077 Mg
1 1 3.7 105 M s 1 k= = 6.2 103 M 1 s 1 0.077 M 0.077 M 25.0 mL 2 S2 O8 = 0.10 M = 0.038 M 2 65.0 mL total
1 1 25.0 mL I = 0.20 M = 0.077 M 2 65.0 mL
1 1 2 Rate 2 = 1.8 105 M s 1 = k S2 O8 I = k 0.038 M 0.077 M 5 1 1.8 10 M s k= = 6.2 103 M 1 s1 0.038 M 0.077 M (d) First we determine concentrations for Experiment 4. 25.0 mL 25.0 mL S2 O8 2 = 0.20 M I = 0.10 = 0.077 M = 0.038 M 4 4 65.0 mL total 65.0 mL We have two expressions for Rate; let us equate them and solve for the rate constant. 2 S2 O8 +7.7 104 M 1 2 1 = Rate 4 = = k S2O8 I = k 0.077 M 0.038 M 4 4 t t k= 7.7 104 M 0.26 M 1 = t 0.077 M 0.038 M t k3 = 0.26 M 1 = 0.0014 M 1 s 1 189 s 659 Chapter 14: Chemical Kinetics k13 = 0.26 M 1 = 0.0030 M 1 s 1 88 s 0.26 M 1 = 0.012 M 1 s 1 21 s k24 = 0.26 M 1 = 0.0062 M 1 s 1 42 s k33 =
(e) We plot ln k vs. 1/ T The slope of the line = Ea / R .
Plot of ln k versus 1/T
0.00325 3.5
1 1/T (K ) 0.0034 0.00355 4 4.5 ln k
5 y = 6135x + 16 5.5 6 Ea = +6135 K 8.3145 J mol1 K 1 = 51.0 103 J/mol = 51.0 kJ/mol The scatter of the data permits only a two significant figure result: 51 kJ/mol
(f) For the mechanism to agree with the reaction stoichiometry, the steps of the mechanism must sum to the overall reaction, in the manner of Hess's law. 2 3 (slow) I + S2 O8 IS2 O8 3 (fast) IS2 O8 2 SO 2 + I + 4 (fast) I + + I I 2 (fast) I 2 + I I 3 2 (net) 3 I + S2 O8 2 SO 2 + I 3 4
3 Each of the intermediates cancels: IS2 O8 is produced in the first step and consumed in the second, I + is produced in the second step and consumed in the third, I 2 is produced in the third step and consumed in the fourth. The mechanism is consistent with the stoichiometry. The rate of the slow step of the mechanism is 2 Rate1 = k1 S2 O8 1 I 1 This is exactly the same as the experimental rate law. It is reasonable that the first step be slow since it involves two negatively charged species coming together. We know that like charges repel, and thus this should not be an easy or rapid process. 660 Chapter 14: Chemical Kinetics SELFASSESSMENT EXERCISES
98. (E) (a) [A]0: Initial concentration of reactant A (b) k: Reaction rate constant, which is the proportionality constant between reaction rate and reactant concentration (c) t1/2: Halflife of the reaction, the amount of time that the concentration of a certain reactant is reduced by half (d) Zeroorder reaction: A reaction in which the rate is not dependent on the concentration of the reactant (e) Catalyst: A substance which speeds up the reaction by lowering the activation energy, but it does not itself get consumed (E) (a) Method of initial rates: A study of the kinetics of the reaction by measuring the initial reaction rates, used to determine the reaction order (b) Activated complex: Species that exist in a transitory state between the reactants and the products (c) Reaction mechanism: Sequential elementary steps that show the conversion of reactant(s) to final product(s) (d) Heterogeneous Catalyst: A catalyst which is in a different physical phase than the reaction medium (e) Ratedetermining step: A reaction which occurs more slowly than other reactions in a mechanism and therefore usually controls the overall rate of the reaction 99. 100. (E) (a) Firstorder and secondorder reactions: In a firstorder reaction, the rate of the reaction depends on the concentration of only one substrate and in a 1to1 manner (doubling the concentration of the reactant doubles the rate of the reaction). In a secondorder reaction, the rate depends on two molecules reacting with each other at the elementary level. (b) Rate law and integrated rate law: Rate law describes how the rate relates to the concentration of the reactants and the overall rate of a reaction, whereas the integrated rate law expresses the concentration of a reactant as a function of time (c) Activation energy and enthalpy of reaction: Activation energy is the minimum energy required for a particular reaction to take place, whereas enthalpy of reaction is the amount of heat generated (or consumed) by a reaction when it happens (d) Elementary process and overall reaction: Individual steps of a reaction mechanism, which describes any molecular event that significantly alters a molecule's energy or geometry or produces a new molecule (e) Enzyme and substrate: An enzyme is a protein that acts as a catalyst for a biological reaction. A substrate is the reactant that is transformed in the reaction (in this context, by the enzyme). 101. (E) The answer is (c). The rate constant k is only dependent on temperature, not on the concentration of the reactants 661 Chapter 14: Chemical Kinetics 102. (E) The answers are (b) and (e). Because halflife is 75 seconds, the quantity of reactant left at two halflives (75 + 75 = 150) equals onehalf of the level at 75 seconds. Also, if the initial concentration is doubled, after one halflife the remaining concentration would have to be twice as much as the original concentration. 103. (E) The answer is (a). Halflife t = 13.9 min, k = ln 2/t = 0.050 min1. Rate of a firstorder reaction is as follows: d A k A 0.050 min 1 0.40 M 0.020 M min 1 dt 104. (E) The answer is (d). A secondorder reaction is expressed as follows: d A 2 k A dt If the rate of the reaction when [A] =0.50 is k(0.50)2 = k(0.25). If [A] = 0.25 M, then the rate is k(0.0625), which is of the rate at [A] =0.50. 105. (M) The answer is (b). Going to slightly higher temperatures broadens the molecular speed distribution, which in turn increases the fraction of molecules at the high kinetic energy range (which are those sufficiently energetic to make a reaction happen). 106. (E) The answer is (c). Since the reaction at hand is described as an elementary one, the rate of the reaction is k[A][B]. 107. (E) We note that from the given data, the halflife of the reaction is 100 seconds (at t = 0, [A] = 0.88 M/s, whereas at t = 100, [A] = 0.44 M/s). Therefore, the rate constant k is: k = ln 2/100 s = 0.00693 s1. We can now calculate instantaneous rate of the reaction: d[A]/dt = (0.00693 s1)(0.44 M) = 3.0103 Ms1 108. (M) (a) For a firstorder reaction, ln 2 0.693 0.0231 min 1 k t1/2 30 ln 0.25 0 ln[A]t ln[A]0 60.0 min k 0.0231 min 1 (b) For a zeroorder reaction, 0.5 0.5 0.0167 min 1 k t1/2 30 t
[A]t [A]0 kt t [A]t [A]0 0.25 1.00 45.0 min k 0.0167 min 1 ln[A]t ln[A]0 kt 662 Chapter 14: Chemical Kinetics 109. (M) The reaction is secondorder, because the halflife doubles with each successive halflife period. 110. (M) (a) The initial rate = M/t = (1.204 M 1.180 M)/(1.0 min) = 0.024 M/min (b) In experiment 2, the initial concentration is twice that of experiment A. For a secondorder reaction: Rate = k [Aexp 2]2 = k [2Aexp 1]2 = 4 k [Aexp 1]2 This means that if the reaction is second order, its initial rate of experiment 2 will be 4 times that of experiment 1 (that is, 4 times as many moles of A will be consumed in a given amount of time). The initial rate is 40.024 M/min = 0.096 M/s. Therefore, at 1 minute, [A] = 2.408 0.0960 = 2.312 M.
(c) The halflife of the reaction, obtained from experiment 1, is 35 minutes. If the reaction is firstorder, then k = ln 2/35 min = 0.0198 min1. For a firstorder reaction, A t A 0 e kt A 35min 2.408 exp 0.0198 min 1 30 min 1.33 M
111. (D) The overall stoichiometry of the reaction is determined by adding the two reactions with each other: A + 2B C + D (a) Since I is made slowly but is used very quickly, its rate of formation is essentially zero. The amount of I at any given time during the reaction can be expressed as follows:
d I dt 0 k1 A B k 2 B I k1 A k2 I Using the above expression for [I], we can now determine the overall reaction rate law: d C k k 2 I B k 2 1 A B k1 A B dt k2
(b) Adding the two reactions given, we still get the same overall stoichiometry as part (a). However, with the given proposed reaction mechanisms, the rate law for the product(s) is given as follows: 663 Chapter 14: Chemical Kinetics d B2 dt k1 B k 1 B2 k 2 A B 0
2 B2 d C dt k 1 k 2 A k 2 k1 A B
2 k1 B 2 Therefore, k 2 A B2 k 1 k 2 A which does not agree with the observed reaction rate law.
112. (M) The answer is (b), firstorder, because only in a firstorder reaction is the halflife independent of the concentration of the reacting species. 113. (E) The answer is (a), zeroorder, because in a zeroorder reaction the relationship between concentration and time is: [A]t = kt + [A]0 114. (M) The answer is (d). The relationship between rate constant (and thus rate) between two reactions can be expressed as follows: Ea 1 1 k2 exp R T T k1 2 1 If T2 is twice T1, the above expression gets modified as follows: k Ea 1 1 ln 2 R T1 2T1 k1 R k 2 T1 1 , ln E a k1 2T1 k Ea T 1 ln 2 1 2T1 k1 R For reasonably high temperatures, k Ea T 1 ln 2 1 2T1 2 k1 Therefore, k 2 Ea 1/2 e 1.64 k1 R R 664 Chapter 14: Chemical Kinetics 115. (E) The answer is (c), remain the same. This is because for a zeroorder reaction, d[A]/dt = k[A]0 = k. Therefore, the reaction rate is independent of the concentration of the reactant.
116. (M) The overarching concept for this concept map is kinetics as a result of successful collision. The subtopics are collision theory, molecular transition theory. Molecular speed and orientation derive from collision theory. Transition complexes and partial bonds fall under the molecular transition theory heading. Deriving from the collision theory is another major topic, the Arrhenius relationship. The Arrhenius relationship encompasses the ideas of activation energy, Arrhenius collision factor, and exponential relationship between temperature and rate constant. 665 CHAPTER 15 PRINCIPLES OF CHEMICAL EQUILIBRIUM
PRACTICE EXAMPLES
1A (E) The reaction is as follows: 2Cu 2 (aq) Sn 2 (aq) 2Cu (aq) Sn 4 (aq) Therefore, the equilibrium expression is as follows: Cu Sn 4 K 2 2 2 Cu Sn Rearranging and solving for Cu2+, the following expression is obtained: Cu + 2 Sn 4+ 2+ Cu = K Sn 2+ 1B
1/ 2 2 x2 x 1.48 x 1/2 x 1.22 (E) The reaction is as follows: 2 2Fe3 (aq) Hg 2 (aq) 2Fe2 (aq) 2Hg 2 (aq) Therefore, the equilibrium expression is as follows:
2 2 Fe2 Hg 2 0.0025 0.0018 9.14 106 K 2 2 2 0.015 x Fe3 Hg 2 2 2 Rearranging and solving for Hg22+, the following expression is obtained:
2 2 Fe2 Hg 2 0.0025 0.0018 0.009847 0.0098 M 2 Hg 2 2 2 0.015 9.14 106 Fe3 K 2 2 2A (E) The example gives Kc = 5.8105 for the reaction N 2 g + 3 H 2 g 2 NH 3 g . The reaction we are considering is onethird of this reaction. If we divide the reaction by 3, we should take the cube root of the equilibrium constant to obtain the value of the equilibrium constant for the "divided" reaction: K c3 3 K c 3 5.8 105 8.3 102 2B (E) First we reverse the given reaction to put NO 2 g on the reactant side. The new equilibrium constant is the inverse of the given one. bg NO 2 g NO g + 1 O 2 g 2 K c ' = 1/ (1.2 102 ) = 0.0083 Then we double the reaction to obtain 2 moles of NO 2 g as reactant. The equilibrium constant is then raised to the second power. 2 NO 2 g 2 NO g + O 2 g K c = 0.00833 = 6.9 105
2 bg 666 Chapter15: Principles of Chemical Equilibrium 3A (E) We use the expression K p = K c RT ngas . In this case, ngas = 3 +1 2 = 2 and thus we
2 have K p = K c RT = 2.8 109 0.08314 298 = 1.7 106
2 3B (M) We begin by writing the Kp expression. We then substitute
P n / V RT = [ concentration ]RT for each pressure. We collect terms to obtain an expression relating Kc and Kp , into which we substitute to find the value of Kc . Kp { P( H 2 )}2 { P(S2 )} ([ H 2 ]RT ) 2 ([S2 ]RT ) [H 2 ]2 [S2 ] RT Kc RT { P( H 2S)}2 ([ H 2S]RT ) 2 [H 2S]2 The same result can be obtained by using Kp = Kc RT
Kc = Kp RT = 1.2 102 = 1.1 104 0.08314 1065 + 273 b g ngas , since ngas = 2 +1 2 = +1. But the reaction has been reversed and halved. Thus Kfinal= 1 1 9091 95 1.1104 Kc 4A (E) We remember that neither solids, such as Ca 5 ( PO 4 ) 3 OH(s) , nor liquids, such as H 2 O(l) , appear in the equilibrium constant expression. Concentrations of products appear in the numerator, those of reactants in the denominator. Kc = Ca 2+ 5 HPO 4
4 2 3 H+ 4B (E) First we write the balanced chemical equation for the reaction. Then we write the equilibrium constant expressions, remembering that gases and solutes in aqueous solution appear in the Kc expression, but pure liquids and pure solids do not. 3 Fe s + 4 H 2 O g Fe3O 4 s + 4 H 2 g { P H 2 }4 Kp = { P H 2 O }4 b g b g Kc = H2 H 2O 4 4 Because ngas = 4 4 = 0, Kp = Kc 5A (M) We compute the value of Qc. Each concentration equals the mass m of the substance divided by its molar mass (this quotient is the amount of the substance in moles) and further divided by the volume of the container. bg 667 Chapter15: Principles of Chemical Equilibrium 1 44.0 g CO 2 2.0 g H 2 28.0 18.0 CO 2 H 2 V V Qc = 44.0 2.0 = 5.7 1.00 = K c 1 44.0 2.0 CO H 2 O m 1 mol CO m 1 mol H 2O 18.0 g H 2 O 28.0 18.0 28.0 g CO V V m 1 mol CO 2 m 1 mol H 2 (In evaluating the expression above, we cancelled the equal values of V , and we also cancelled the equal values of m .) Because the value of Qc is larger than the value of Kc , the reaction will proceed to the left to reach a state of equilibrium. Thus, at equilibrium there will be greater quantities of reactants, and smaller quantities of products than there were initially.
5B (M) We compare the value of the reaction quotient, Qp , to that of K p .
Qp = {P (PCl3 )}{P(Cl2 )} 2.19 0.88 = = 0.098 {P(PCl5 } 19.7
2 1 K p = K c RT = K c RT = 0.0454 0.08206 (261 273) = 1.99
1 1 Because Qc Kc , the net reaction will proceed to the right, forming products and consuming reactants.
6A (E) O 2 g is a reactant. The equilibrium system will shift right, forming product in an bg attempt to consume some of the added O2(g) reactant. Looked at in another way, O 2 is increased above its equilibrium value by the addition of oxygen. This makes Qc smaller than Kc . (The O 2 is in the denominator of the expression.) And the system shifts right to drive Qc back up to Kc, at which point equilibrium will have been achieved.
6B (M) (a) The position of an equilibrium mixture is affected only by changing the concentration of substances that appear in the equilibrium constant expression, Kc = CO 2 . Since CaO(s) is a pure solid, its concentration does not appear in the equilibrium constant expression and thus adding extra CaO(s) will have no direct effect on the position of equilibrium. (b) The addition of CO 2 g will increase CO 2 above its equilibrium value. The reaction will shift left to alleviate this increase, causing some CaCO 3 s to form. bg bg (c) Since CaCO3(s) is a pure solid like CaO(s), its concentration does not appear in the equilibrium constant expression and thus the addition of any solid CaCO3 to an equilibrium mixture will not have an effect upon the position of equilibrium. 668 Chapter15: Principles of Chemical Equilibrium 7A (E) We know that a decrease in volume or an increase in pressure of an equilibrium mixture of gases causes a net reaction in the direction producing the smaller number of moles of gas. In the reaction in question, that direction is to the left: one mole of N 2 O 4 g is formed when two moles of NO 2 g combine. Thus, decreasing the cylinder volume would have the initial bg bg effect of doubling both N 2 O 4 and NO 2 . In order to reestablish equilibrium, some NO2 will then be converted into N2O4. Note, however, that the NO2 concentration will still ultimately end up being higher than it was prior to pressurization.
7B (E) In the balanced chemical equation for the chemical reaction, ngas = 1+1 1+1 = 0 . As a consequence, a change in overall volume or total gas pressure will have no effect on the position of equilibrium. In the equilibrium constant expression, the two partial pressures in the numerator will be affected to exactly the same degree, as will the two partial pressures in the denominator, and, as a result, Qp will continue to equal Kp . (E) The cited reaction is endothermic. Raising the temperature on an equilibrium mixture favors the endothermic reaction. Thus, N 2 O 4 g should decompose more completely at higher temperatures and the amount of NO 2 g formed from a given amount of N 2 O 4 g will be greater at high temperatures than at low ones. b g b g 8A bg bg bg 8B (E) The NH3(g) formation reaction is 1 2 N 2 g + 3 H 2 g NH 3 g , H o = 46.11 kJ/mol. 2 This reaction is an exothermic reaction. Lowering temperature causes a shift in the direction of this exothermic reaction to the right toward products. Thus, the equilibrium NH 3 g will bg be greater at 100 C .
9A (E) We write the expression for Kc and then substitute expressions for molar concentrations. 0.22 0.11 [H 2 ]2 [S2 ] 3.00 3.00 Kc 2.3 10 4 2 2 [H 2 S] 2.78 3.00 2 9B (M) We write the equilibrium constant expression and solve for N 2 O 4 . = = 0.121 M N 2 O4 = N 2 O4 4.61 103 4.61 103 Then we determine the mass of N 2 O 4 present in 2.26 L. 0.121 mol N 2 O 4 92.01 g N 2 O 4 N 2 O 4 mass = 2.26 L = 25.2 g N 2 O 4 1L 1 mol N 2 O 4 K c = 4.61 10
3 NO2 = 2 NO2 2 0.0236 2 669 Chapter15: Principles of Chemical Equilibrium 10A (M) We use the initialchangeequilibrium setup to establish the amount of each substance at equilibrium. We then label each entry in the table in the order of its determination (1st, 2nd, 3rd, 4th, 5th), to better illustrate the technique. We know the initial amounts of all substances (1st). There are no products at the start. Because ``initial''+ ``change'' = ``equilibrium'' , the equilibrium amount (2nd) of Br2 g enables us to determine "change" (3rd) for Br2 g . We then use stoichiometry to write other entries (4th) on the "change" line. And finally, we determine the remaining equilibrium amounts (5th). bg 2 NO g st Initial: 1.86 mol (1 ) 0.00 mol (1st) Change: 0.164 mol (4th) +0.164 mol (4th) th Equil.: 1.70 mol (5 ) 0.164 mol (5th) 2 0.164 0.082 mol 5.00 [NO]2 [Br2 ] 5.00 1.5 104 Kc 2 2 [NOBr] 1.70 5.00 Reaction:
Here, ngas = 2 +1 2 = +1.
+1 2 NOBr g + Br2 g 0.00 mol (1st) +0.082 mol (3rd) 0.082 mol (2nd) K p = K c RT = 1.5 104 0.08314 298 = 3.7 103 10B (M) Use the amounts stated in the problem to determine the equilibrium concentration for each substance. 2 SO3 g 2 SO 2 g + O2 g Reaction: Initial: 0 mol 0.100 mol 0.100 mol 0.0916 mol 0.0916 / 2 mol Changes: +0.0916 mol Equil.: 0.0916 mol 0.0084 mol 0.0542 mol 0.0084 mol 0.0542 mol 0.0916 mol Concentrations: 1.52 L 1.52 L 1.52 L Concentrations: 0.0603 M 0.0055 M 0.0357 M We use these values to compute Kc for the reaction and then the relationship
Kp = Kc RT b g
2 ngas (with ngas = 2 +1 2 = +1) to determine the value of Kp . SO2 O2 = 0.00552 0.0357 = 3.0 104 Kc = 2 2 0.0603 SO3 K p = 3.0 104 (0.08314 900) 0.022 670 Chapter15: Principles of Chemical Equilibrium 11A (M) The equilibrium constant expression is Kp = P{H 2 O} P{CO 2 } = 0.231 at 100 C . From the balanced chemical equation, we see that one mole of H 2 O g is formed for each mole of CO 2 g produced. Consequently, P{H 2 O} = P{CO 2 } and Kp = P{CO 2 } . We solve this expression for P{CO 2 } : P{CO 2 } ( P{CO 2 }) 2 K p 0.231 0.481 atm. bg bg b g 2 11B (M) The equation for the reaction is NH 4 HS s NH3 g + H 2S g , K p = 0.108 at 25 C.
The two partial pressures do not have to be equal at equilibrium. The only instance in which they must be equal is when the two gases come solely from the decomposition of NH 4 HS s . In this case, some of the NH 3 g has come from another source. We can obtain the pressure of H 2S g by substitution into the equilibrium constant expression, since we are given the equilibrium pressure of NH 3 g . 0.108 K p = P{H 2S} P{NH 3 } = 0.108 = P{H 2S} 0.500 atm NH 3 P{H 2S} = = 0.216 atm 0.500 So, Ptotal = PH S PNH3 = 0.216 atm + 0.500 atm = 0.716 atm bg bg bg bg 2 12A (M) We set up this problem in the same manner that we have previously employed, namely designating the equilibrium amount of HI as 2x . (Note that we have used the same multipliers for x as the stoichiometric coefficients.) Equation: Initial: Changes: Equil: H2 g + I2 g 2 HI g 0 mol +2x mol 2 x mol 0.150 mol x mol 0.150 x mol b g 0.200 mol x mol 0.200 x mol b g 2x 2 2x 15.0 Kc = = 50.2 0.150 x 0.200 x 0.150 x 0.200 x 15.0 15.0 2 We substitute these terms into the equilibrium constant expression and solve for x. 4 x 2 = 0.150 x 0.200 x 50.2 = 50.2 0.0300 0.350 x + x 2 = 1.51 17.6 x + 50.2 x 2 0 = 46.2 x 2 17.6 x +1.51
x= b gb g c h Now we use the quadratic equation to determine the value of x. b b 2 4ac 17.6 (17.6) 2 4 46.2 1.51 17.6 5.54 = = = 0.250 or 0.131 2a 2 46.2 92.4 671 Chapter15: Principles of Chemical Equilibrium The first root cannot be used because it would afford a negative amount of H 2 (namely, 0.1500.250 = 0.100). Thus, we have 2 0.131 = 0.262 mol HI at equilibrium. We check by substituting the amounts into the Kc expression. (Notice that the volumes cancel.) The slight disagreement in the two values (52 compared to 50.2) is the result of rounding error. 0.0686 b0.262g K = = b0.150 0.131gb0.200 0.131g 0.019 0.069 = 52
2 c 12B (D) (a) The equation for the reaction is N 2 O 4 g 2 NO 2 g and K c = 4.61103 at 25 C .
In the example, this reaction is conducted in a 0.372 L flask. The effect of moving the mixture to the larger, 10.0 L container is that the reaction will be shifted to produce a greater number of moles of gas. Thus, NO 2 g will be produced and N 2 O 4 g will dissociate. Consequently, the amount of N 2 O 4 will decrease. bg bg (b) The equilibrium constant expression, substituting 10.0 L for 0.372 L, follows. 2 2x [ NO 2 ]2 4x2 10.0 4.61 103 Kc 0.0240 x 10.0(0.0240 x ) [N 2 O 4 ] 10.0 This can be solved with the quadratic equation, and the sensible result is x = 0.0118 moles. We can attempt the method of successive approximations. First, assume that x 0.0240 . We obtain: FG IJ H K x 4.61 103 10.0 (0.0240 0) 4.61 103 2.50 (0.0240 0) 0.0166 4 Clearly x is not much smaller than 0.0240. So, second, assume x 0.0166 . We obtain: x 4.61 103 2.50 (0.0240 0.0166) 0.00925 This assumption is not valid either. So, third, assume x 0.00925 . We obtain: x 4.61 103 2.50 (0.0240 0.00925) 0.0130 Notice that after each cycle the value we obtain for x gets closer to the value obtained from the roots of the equation. The values from the next several cycles follow. Cycle
x value 5th 6th 7th 8th 9th 10th 11th 4th 0.0112 0.0121 0.0117 0.0119 0.01181 0.01186 0.01183 0.01184 The amount of N 2 O 4 at equilibrium is 0.0118 mol, less than the 0.0210 mol N 2 O 4 at equilibrium in the 0.372 L flask, as predicted. 672 Chapter15: Principles of Chemical Equilibrium 13A (M) Again we base our solution on the balanced chemical equation. Fe3+ aq + Ag s K c = 2.98 Equation: Ag + aq + Fe 2+ aq Initial: 0M Changes: +x M x M Equil: Kc = Fe 3+ Ag
+ 0M +x M x M = 2.98 = 1.20 x x2 b 1.20 M x M 1.20 x M g Fe 2+ 2.98 x 2 = 1.20 x 0 = 2.98 x 2 + x 1.20 We use the quadratic formula to obtain a solution.
2 b b 2 4ac 1.00 (1.00) + 4 2.98 1.20 1.00 3.91 x= = = = 0.488 M or 0.824 M 2a 2 2.98 5.96 A negative root makes no physical sense. We obtain the equilibrium concentrations from x. Ag + = Fe 2+ = 0.488 M Fe3+ = 1.20 0.488 = 0.71 M 13B (M) We first calculate the value of Qc to determine the direction of the reaction. Qc = V 2+ Cr 3+ V
3+ Cr 2+ = 0.150 0.150 = 225 7.2 102 = Kc 0.0100 0.0100 Because the reaction quotient has a smaller value than the equilibrium constant, a net reaction to the right will occur. We now set up this solution as we have others, heretofore, based on the balanced chemical equation. + Cr 2+ aq V 2+ aq + Cr 3+ aq initial 0.0100 M 0.0100 M 0.150 M 0.150 M changes x M x M +x M +x M equil (0.0100 x)M (0.0100 x)M (0.150 + x)M (0.150 + x)M
Kc = V 2+ Cr 3+ V 3+ Cr 2+ V 3+ aq b0.150 + xg b0.150 + xg = 7.2 10 = FG 0.150 + x IJ = H 0.0100 x K b0.0100 xg b0.0100 xg
2 2 If we take the square root of both sides of this expression, we obtain 0.150 x 27 7.2 102 0.0100 x 0.150 + x = 0.27 27x which becomes 28 x = 0.12 and yields 0.0043 M. Then the equilibrium concentrations are: V 3+ = Cr 2+ = 0.0100 M 0.0043 M = 0.0057 M V 2+ = Cr 3+ = 0.150 M + 0.0043 M = 0.154 M 673 Chapter15: Principles of Chemical Equilibrium INTEGRATIVE EXAMPLE
A. (E) We will determine the concentration of F6P and the final enthalpy by adding the two reactions: C6 H12 O6 ATP G6P ADP
G6P F6P C6 H12 O6 ATP ADP F6P HTOT = 19.74 kJmol1 + 2.84 kJmol1 = 16.9 kJmol1 Since the overall reaction is obtained by adding the two individual reactions, then the overall reaction equilibrium constant is the product of the two individual K values. That is, K K1 K 2 1278 The equilibrium concentrations of the reactants and products is determined as follows: C6H12O6 1.20106 x 1.20106x + ATP 1104 x 1104x Initial Change Equil
K ADP 1102 +x 1102+x + F6P 0 +x x ADP F6P C6 H12O6 ATP 1278 1.20 10 110
6 2 x 1 104 x x x 1.0 102 x x 2 1.2 1010 1.012 104 x x 2 Expanding and rearranging the above equation yields the following secondorder polynomial: 1277 x2 0.1393 x + 1.534107 = 0 Using the quadratic equation to solve for x, we obtain two roots: x = 1.113106 and 1.080104. Only the first one makes physical sense, because it is less than the initial value of C6H12O6. Therefore, [F6P]eq = 1.113106. During a fever, the body generates heat. Since the net reaction above is exothermic, Le Chtelier's principle would force the equilibrium to the left, reducing the amount of F6P generated. 674 Chapter15: Principles of Chemical Equilibrium B. (E) (a) The ideal gas law can be used for this reaction, since we are relating vapor pressure and concentration. Since K = 3.31029 for decomposition of Br2 to Br (very small), then it can be ignored. 1 nRT 0.100 mol 0.08206 L atm K 298.15 K V 8.47 L P 0.289 atm (b) At 1000 K, there is much more Br being generated from the decomposition of Br2. However, K is still rather small, and this decomposition does not notably affect the volume needed. EXERCISES
Writing Equilibrium Constants Expressions
1. (E) (a) 2 COF2 g CO 2 g + CF4 g Cu s + 2 Ag
+ Kc = CO2 CF4 2 COF2 (b) aq Cu aq + 2 Ag s 2+ Cu 2+ Kc = + 2 Ag 2 3+ SO 4 Fe Kc = 2 S2 O8 2 Fe 2+ 2 2 (c) S2 O8 2 aq + 2 Fe2+ aq 2 SO4 2 aq + 2 Fe3+ aq 2. (E) (a) 4 NH 3 g + 3 O 2 g 2 N 2 g + 6 H 2 O g 7 H 2 g + 2 NO 2 g 2 NH 3 g + 4 H 2 O g N 2 g + Na 2 CO3 s + 4 C s 2 NaCN s + 3 CO g KP = KP = Kc P N 2 P H 2 O
2 4 2 6 3 4 P NH 3 P O 2 7 (b) P NH 3 P H 2 O P H 2 P NO 2 3 2 (c) CO = N2 3. (E) (a) NO2 Kc = 2 NO O2 2 (b) Kc = Zn 2+ Ag +
2 (c) Kc = OH 2 CO 32 675 Chapter15: Principles of Chemical Equilibrium 4. (E) (a) Kp = P{CH 4 }P{H 2S}2 P{CS2 }P{H 2 }4 (b) Kp = P O 2 l q 1/2 (c) Kp = P{CO 2 } P{H 2 O} 5. (E) In each case we write the equation for the formation reaction and then the equilibrium constant expression, Kc, for that reaction. HF 1 1 (a) Kc = 2 H 2 g + 2 F2 g HF g 1/2 1/2 H 2 F2 (b) (c) (d) N 2 g + 3 H 2 g 2 NH 3 g 2 N 2 g + O2 g 2 N 2 O g 1 2 3 Cl2 (g) + 2 F2 (g) ClF3 (l) NH3 Kc = 3 N 2 H 2 2 [N 2 O]2 [N 2 ]2 [O 2 ] 1 Kc 1/ 2 [Cl2 ] [F2 ]3 / 2 Kc = 6. (E) In each case we write the equation for the formation reaction and then the equilibrium constant expression, Kp, for that reaction. PNOCl 1 1 1 (a) KP = 2 N 2 g + 2 O 2 g + 2 Cl 2 g NOCl g 1/2 1/2 1/2 PN2 PO2 PCl2 (b) N 2 g + 2 O 2 g + Cl2 g 2 ClNO 2 g N2 g + 2 H2 g N2 H4 g 1 2 KP = KP = KP = PClNO2 2 2 (c) (d) PN 2 PO2 PCl2 PN2 H4 PN2 PH2 1
2 N2 g + 2 H2 g + 1 2 Cl2 g NH 4 Cl s PN2 1/2 PH 2 PCl2 2 1/2 7. (E) Since K p = K c RT (a) (b)
Kc = ng , it is also true that K c = K p RT p ng . SO2 [Cl2 ] = K
SO 2 Cl 2 2 RT ( 1) = 2.9 102 (0.08206 303) 1 = 0.0012 (c) NO2 = K RT ( 1) = 1.48 104 Kc = p 2 NO [O2 ] 3 H 2S = K RT 0 = K = 0.429 Kc = p p 3 (0.08206 303) = 5.55 105 H2 676 Chapter15: Principles of Chemical Equilibrium 8. (E) K p = K c RT (a) (b) (c) ng , with R = 0.08206 L atm mol1 K 1 Kp = Kp Kp P{NO 2 }2 K c (RT) +1 = 4.61 103 (0.08206298)1 0.113 P{N 2 O 4 } P{C2 H 2 }P{H 2 }3 K c (RT)( 2) (0.154) (0.08206 2000) 2 4.15 103 2 P{CH 4 } P{H 2 }4 P{CS2 } K c (RT)( 2) (5.27 108 ) (0.08206 973) 2 3.36 104 2 P{H 2 S} P{CH 4 }
ng 9. (E) The equilibrium reaction is H 2 O l H 2 O g with ngas = +1. K p = K c RT K c = K p RT ng gives . = Kp RT Kp = P{H 2 O} = 23.8 mmHg = 1 atm = 0.0313 760 mmHg K c = K p RT 10. 1 0.0313 = 1.28 103 0.08206 298
ng (E) The equilibrium rxn is C6 H 6 l C6 H 6 g with ngas = +1. Using K p = K c RT K p =K c RT = 5.12 103 0.08206 298 = 0.125 = P{C6 H 6 } , P{C6 H 6 } = 0.125 atm 11. 760 mmHg = 95.0 mmHg 1 atm (E) Add onehalf of the reversed 1st reaction with the 2nd reaction to obtain the desired reaction. 1 1 1 Kc = 2 N 2 g + 2 O 2 g NO g 2.11030 NO g + 1 Br2 g NOBr g K c = 1.4 2 net : 1 N 2 g + 1 O 2 g + 1 Br2 g NOBr g 2 2 2
12. Kc = 1.4 2.1 10
30 = 9.7 1016 (M) We combine the several given reactions to obtain the net reaction. 1 Kc = 2 N 2 O g 2 N 2 g O 2 (g) 2 2.7 1018 4 NO 2 g 2 N 2 O 4 g 2 N 2 (g) + 4 O 2 (g) 4NO 2 (g) Kc = 4.6 10 1 3 2 K c (4.1 109 ) 4 net : 2 N 2 O g + 3 O 2 g 2 N 2 O 4 g K c ( Net ) 4.1 10 = 2.7 10 4.6 10 9 4 18 2 3 2 = 1.8 106 677 Chapter15: Principles of Chemical Equilibrium 13. (M) We combine the Kc values to obtain the value of Kc for the overall reaction, and then convert this to a value for Kp. 2 2 CO 2 g + 2H 2 g 2 CO g + 2 H 2O g K c = 1.4 2 C graphite + O2 g 2 CO g K c = 1 108 2 4 CO g 2 C graphite + 2 CO 2 g net: 2H 2 (g) + O 2 (g) 2H 2 O(g)
n Kc = 1 0.64 2 K c(Net) (1.4) 2 (1108 ) 2 5 106 2 (0.64) 14. Kc 5 1016 = = 5 1014 RT 0.08206 1200 (M) We combine the Kp values to obtain the value of Kp for the overall reaction, and then convert this to a value for Kc. K p = K c RT = 2 NO 2 Cl g 2 NOCl g + O 2 g 2 NO 2 g + Cl2 g 2 NO 2 Cl g N 2 g + 2 O 2 g 2 NO 2 g 1 Kp = 2 1.1 10 K p = 0.3
2 2 K p = (1.0 109 ) 2 K p(Net) 2 3 net: N 2 (g) + O 2 (g) + Cl2 (g) 2 NOCl(g) K p = K c RT n (0.3) 2 (1.0 109 )2 7.4 1024 (1.1 102 ) 2 = 2 1022 Kc = Kp RT n = 7.4 1024 0.08206 298 15. (E) CO 2 (g) H 2 O(l) H 2CO3 (aq) In terms of concentration, K = a(H2CO3)/a(CO2) H CO In terms of concentration and partial pressure, K 2 3 c PCO2 P 16. (E) 2 Fe(s) 3 O2 (g) Fe 2 O3 (s) a Fe2O3 . Since activity of solids and liquids is defined as 1, then the expression K a Fe a O2 simplifies to K 1 a O2 Similarly, in terms of pressure and concentration, K 1 PO2 / P 678 Chapter15: Principles of Chemical Equilibrium Experimental Determination of Equilibrium Constants
17. (M) First, we determine the concentration of PCl5 and of Cl 2 present initially and at equilibrium, respectively. Then we use the balanced equation to help us determine the concentration of each species present at equilibrium. 1.00 103 mol PCl5 9.65 104 mol Cl2 = 0.00400 M = 0.00386 M PCl5 initial = Cl2 equil = 0.250 L 0.250 L Equation: PCl5 g PCl3 (g) + Cl2 g Initial: 0.00400M Changes: xM Equil: 0.00400MxM 0M +xM xM 0M +xM xM 0.00386 M (from above) At equilibrium, [Cl2] = [PCl3] = 0.00386 M and [PCl5] = 0.00400M xM = 0.00014 M Kc 18. PCl3 Cl2 (0.00386 M) (0.00386 M) 0.00014 M PCl5 = 0.106 (M) First we determine the partial pressure of each gas. Pinitial {H 2 g }= nRT = V 1.00 g H 2 1mol H 2 0.08206 L atm 1670 K 2.016 g H 2 mol K 136 atm 0.500 L 1mol H 2S 0.08206 L atm 1670 K 34.08g H 2S mol K 8.52 atm 0.500 L 0.08206 L atm 1670 K mol K 2.19 103 atm 0.500 L 2 H 2 S g Pinitial {H 2S g }= nRT = V 1.06 g H 2S Pequil {S2 g }=
Equation: Initial : Changes : Equil : nRT = V 8.00 106 mol S2 S2 g 2 H2 g + 136 atm 0 atm 8.52 atm +0.00438atm 0.00219 atm 0.00438 atm 136 atm 0.00219 atm 8.52 atm
2 8.52 Kp = = =1.79 2 P{H 2 g } P{S2 (g)} 136 2 0.00219 P{H 2S g }2 679 Chapter15: Principles of Chemical Equilibrium 19. (M) (a) . 0105 g PCl 5 1 mol PCl5 PCl5 2.50 L 208.2 g Kc = L 26.3 OP LMCl 2 OP = 0.220 g PCl 1 mol PCl 2.12 g Cl 2 1 mol Cl 2 NM PCl 3 Q N Q 3 3 2.50 L 137.3 g 2.50 L 70.9 g Kp = Kc RT (b) 20. (M) b g FG H n = 26.3 0.08206 523 b g IJ FG K H IJ K 1 = 0.613 0.682 g ICl [ICl]initial = Reaction: Initial: Change Equilibrium 1 mol ICl 162.36 g ICl = 6.72103 M 0.625 L I2(g) 0M +x x + Cl2(g) 0M +x x 2 ICl(g) 6.72 103 M 2x 6.72 103 M 2x 0.0383 g I 2 [I 2 ]equil = 1 mol I 2 253.808 g I 2 = 2.41104 M = x 0.625 L xx (2.41104 ) 2 Kc 9.31106 3 3 4 (6.72 10 2 x) (6.72 10 2(2.4110 )) 21. (E)
3 Fe3 9.1103 Fe K 3 7 3 H 1.0 10 Fe3 9.11018 M 22. (E) K NH3 (aq) PNH3 (g) 57.5 5 109 PNH3 (g) PNH3 (g) 5 109 57.5 8.7 1011 680 Chapter15: Principles of Chemical Equilibrium Equilibrium Relationships
23. (M) K c =281= SO3 2 SO 2 O 2 2 SO3 = 2 SO 2 2 0.185 L 0.00247 mol
2 SO2 = SO 3 0.185 = 0.516 0.00247 281 24. 0.37 mol I I V 1 0.14 (M) K c 0.011 1.00 mol I 2 V I2 V
2 V 0.14 13L 0.011 25. (M) (a) A possible equation for the oxidation of NH 3 g to NO 2 g follows. NH 3 g + 7 O 2 g NO 2 g + 3 H 2 O g 4 2 b g b g (b) We obtain K p for the reaction in part (a) by appropriately combining the values of K p given in the problem. NH 3 g + 5 O 2 g NO g + 3 H 2 O g 4 2 K p = 2.111019
1 0.524 2.111019 Kp = = 4.03 1019 0.524 NO g + 1 O 2 g NO 2 g 2
3 net: NH3 g + 7 O 2 g NO 2 g + 2 H 2 O g 4 Kp = 26. (D) (a) We first determine [H2] and [CH4] and then [C2H2]. [CH4] = [H2] = 0.10 mol = 0.10 M Kc = C2 H 2 H 2 3 CH 4 2 C2 H 2 = K c CH 4 1.0 L 2 H 2 3 = 0.154 0.10 = 1.54 M 0.103
2 In a 1.00 L container, each concentration numerically equals the molar quantities of the substance. 1.54 mol C2 H 2 {C2 H 2 } = = 0.89 1.54 mol C2 H 2 + 0.10 mol CH 4 + 0.10 mol H 2
(b) The conversion of CH4(g) to C2H2(g) is favored at low pressures, since the conversion reaction has a larger sum of the stoichiometric coefficients of gaseous products (4) than of reactants (2). Initially, all concentrations are halved when the mixture is transferred to a flask that is twice as large. To reestablish equilibrium, the system reacts to the right, forming more moles of gas (to compensate for the drop in pressure). We base our solution on the balanced chemical equation, in the manner we have used before. (c) 681 Chapter15: Principles of Chemical Equilibrium Equation: Initial: 2 CH 4 (g) 0.10 mol 2.00 L = 0.050 M 2 x M Changes: Equil: C 2 H 2 (g) 1.5 mol 2.00 L = 0.75 M +x M + 3 H2 0.10 mol 2.00 L = 0.050 M +3x M Kc 0.050 2 x M 0.0750 + x M 3 3 C H H 0.050 + 3x 0.750 + x = 0.154 = 2 2 22 = 2 0.050 2 x CH 4 0.050 + 3x M We can solve this 4thorder equation by successive approximations. First guess: x = 0.010 M . x = 0.010 Qc 0.050 + 3 0.010 0.750 + 0.010 0.080 0.760 = = = 0.433 0.154 0.030 0.050 2 0.010 3 3 2 2 x = 0.020 Qc 0.050 + 3 0.020 0.750 + 0.020 0.110 0.770 = = = 10.2 0.154 0.010 0.050 2 0.020 3 3 2 2 x = 0.005 Qc 0.050 + 3 0.005 0.750 + 0.005 0.065 0.755 = = = 0.129 0.154 0.040 0.050 2 0.005 3 3 2 2 x = 0.006 Qc 0.050 + 3 0.006 0.750 + 0.006 0.068 0.756 = = = 0.165 0.154 0.038 0.050 2 0.006 3 3 2 2 This is the maximum number of significant figures our system permits. We have x = 0.006 M . CH 4 = 0.038 M ; C2 H 2 = 0.756 M; H 2 = 0.068 M Because the container volume is 2.00 L, the molar amounts are double the values of molarities.
2.00 L 0.756 mol C 2 H 2 1L = 1.51 mol C 2 H 2 2.00 L 0.038 mol CH 4 1L = 0.076 mol CH 4 2.00 L 0.068 mol H 2 = 0.14 mol H 2 1L Thus, the increase in volume results in the production of some additional C2H2. 682 Chapter15: Principles of Chemical Equilibrium 27. (M) (a) n CO n H 2 O CO H 2O V V Kc CO2 H 2 n{CO2 } n H 2 V V Since V is present in both the denominator and the numerator, it can be stricken from the expression. This happens here because ng = 0. Therefore, Kc is independent of V.
Note that Kp = Kc for this reaction, since ngas = 0 . (b) Kc = Kp =
28. 0.224 mol CO 0.224 mol H 2 O = 0.659 0.276 mol CO 2 0.276 mol H 2 (M) For the reaction CO(g) + H2O(g) CO2(g) + H2(g) the value of Kp = 23.2 [CO 2 ][H 2 ] The expression for Qp is . Consider each of the provided situations [CO][H 2 O] (a) PCO = PH2O = PH2 = PCO2 ; Qp = 1 Not an equilibrium position
(b) (c) (d)
PH2 PH2O
CO = PCO2 PCO =x; Qp = x2 If x = 23.2 , this is an equilibrium position. P PH2O = PCO2 PH2 ; Qp = 1
= PCO2 PH2O =x; Not an equilibrium position 23.2 , this is an equilibrium position. PH2 PCO Qp = x2 If x = Direction and Extent of Chemical Change
29. (M) We compute the value of Qc for the given amounts of product and reactants. 0.82 K c 100 2 SO 2 2.2 mol O 2 7.2 L 7.2 L The mixture described cannot be maintained indefinitely. In fact, because Qc K c , the reaction will proceed to the right, that is, toward products, until equilibrium is established. We do not know how long it will take to reach equilibrium. SO3 Qc = 2 SO2 O2 3.6 mol
2 1.8 mol SO3 7.2 L 2 683 Chapter15: Principles of Chemical Equilibrium 30. (M) We compute the value of Qc for the given amounts of product and reactants. 0.0205 mol NO 2 2 5.25 L NO2 1.07 104 K 4.61 103 Qc c 0.750 mol N 2 O 4 N 2 O4 5.25 L The mixture described cannot be maintained indefinitely. In fact, because Qc < Kc, the reaction will proceed to the right, that is, toward products, until equilibrium is established. If Ea is large, however, it may take some time to reach equilibrium.
31. (M) (a) We determine the concentration of each species in the gaseous mixture, use these concentrations to determine the value of the reaction quotient, and compare this value of Qc with the value of Kc . 2 SO =
2 0.455 mol SO 2 1.90 L = 0.239 M O =
2 0.183 mol O 2 1.90 L = 0.0963 M
2 SO =
3 0.568 mol SO 3 1.90 L = 0.299 M Qc = SO 3 2 SO 2 2 O2 0.299 = = 16.3 2 0.239 0.0963 Since Qc = 16.3 2.8 102 = Kc , this mixture is not at equilibrium.
(b) Since the value of Qc is smaller than that of Kc , the reaction will proceed to the right, forming product and consuming reactants to reach equilibrium. 32. (M) We compute the value of Qc . Each concentration equals the mass m of the substance divided by its molar mass and further divided by the volume of the container.
m Qc 1 mol CO 2 44.0 g CO 2 V 1 mol CO 28.0 g CO V m 1 mol H 2 2.0 g H 2 1 bg CO H = CO H O
2 2 2 m 28.0 18.0 V 44.0 2.0 = 5.7 31.4(value of K c ) 1 mol H 2 O 1 44.0 2.0 m 28.0 18.0 18.0 g H 2 O V (In evaluating the expression above, we cancelled the equal values of V , along with, the equal values of m .) Because the value of Qc is smaller than the value of Kc , (a) the reaction is not at equilibrium and (b) the reaction will proceed to the right (formation of products) to reach a state of equilibrium. 684 Chapter15: Principles of Chemical Equilibrium 33. (M) The information for the calculation is organized around the chemical equation. Let x = mol H 2 (or I 2 ) that reacts. Then use stoichiometry to determine the amount of HI formed, in terms of x , and finally solve for x . Equation: Initial: Changes: Equil: H2 g 0.150 mol x mol 0.150 x bg I2 g 0.150 mol x mol 0.150 x bg 2 HI g 0.000 mol +2 x mol 2x bg 2x 2 HI 3.25 L Kc H 2 I2 0.150 x 0.150 x 3.25 L 3.25 L 2 Then take the square root of both sides: 2 x = 1.06 7.09 x Kc 50.2 2x 7.09 0150 x . 1.06 = 0.117 mol, amount HI = 2 x = 2 0.117 mol = 0.234 mol HI 9.09 amount H 2 = amount I 2 = 0.150 x mol = 0.150 0.117 mol = 0.033 mol H 2 (or I 2 ) x= 34. (M) We use the balanced chemical equation as a basis to organize the information Equation: SbCl5 g SbCl3 g + Cl2 g Initial: Initial: Changes: Equil: 0.00 mol 2.50 L 0.000 M +x M xM K c = 0.025= SbCl3 Cl2 = 0.112 x 0.0640 x = 0.00717 0.176x + x 2
SbCl5 b 0.280 mol 2.50 L 0.112 M xM 0.112 x M g 0.160 mol 2.50 L 0.0640 M xM 0.0640 x M b g x x 0.025 x = 0.00717 0.176 x + x 2 x 2 0.201x + 0.00717 = 0 b b 2 4ac 0.201 0.0404 0.0287 x= = = 0.0464 or 0.155 2a 2 The second of the two values for x gives a negative value of Cl2 = 0.091 M , and thus is physically meaningless in our universe. Thus, concentrations and amounts follow. SbCl5 = x = 0.0464 M SbCl3 = 0.112 x = 0.066 M
Cl2 = 0.0640 x = 0.0176 M amount SbCl5 = 2.50 L 0.0464 M = 0.116 mol SbCl5 amount SbCl3 = 2.50 L 0.066 M = 0.17 mol SbCl3 amount Cl2 = 2.50 L 0.0176 M = 0.0440 mol Cl2 685 Chapter15: Principles of Chemical Equilibrium 35. (M) We use the chemical equation as a basis to organize the information provided about the reaction, and then determine the final number of moles of Cl2 g present. Equation: CO g + Cl2 g COCl2 g Initial: 0.3500 mol 0.0000 mol Changes: + x mol + x mol Equil.: 0.3500 + x mol x mol 0.05500 mol x mol 0.05500 x mol COCl2 3.050 L K c 1.2 103 CO Cl2 (0.3500 + x)mol 3.050 L (0.0550 x) mol x mol 3.050 L 1.2 103 0.05500 x 3.050 (0.3500 x) x 1.2 103 0.05500 = 3.050 0.3500 x Assume x 0.0550 This produces the following expression. 3.050 0.05500 = 4.0 104 mol Cl2 3 0.3500 1.2 10 x= We use the first value we obtained, 4.0 104 (= 0.00040), to arrive at a second value. x= 3.050 0.0550 0.00040 = 4.0 104 mol Cl2 3 0.3500 + 0.00040 1.2 10 Because the value did not change on the second iteration, we have arrived at a solution.
36. (M) Compute the initial concentration of each species present. Then determine the equilibrium concentrations of all species. Finally, compute the mass of CO 2 present at equilibrium. COint = 1.00 g 1 mol CO = 0.0253 M 1.41 L 28.01 g CO 1.00 g 1 mol H 2 = 0.352 M 1.41 L 2.016 g H 2 + H2 O g 0.0394 M x M H 2Oint = 1.00 g 1 mol H 2 O = 0.0394 M 1.41 L 18.02 g H 2 O H 2 int = Equation : CO g Initial : 0.0253 M Changes : x M Equil : CO 2 g 0.0000 M +x M + H2 g 0.352 M +x M 0.0253 x M 0.0394 x M xM 0.352 + x M Kc = x 0.352 + x CO2 H 2 = 23.2 = 0.352 x + x 2 = 0.0253 x 0.0394 x 0.000997 0.0647 x + x 2 CO H 2O
22.2 x 2 1.852 x + 0.0231 = 0 0.0231 1.50 x + 23.2 x 2 = 0.352 x + x 2 686 Chapter15: Principles of Chemical Equilibrium x= b b 2 4ac 1.852 3.430 2.051 = = 0.0682 M, 0.0153 M 2a 44.4 The first value of x gives negative concentrations for reactants ([CO] = 0.0429 M and [H2O] = 0.0288 M). Thus, x = 0.0153 M = CO 2 . Now we can find the mass of CO 2 . 1.41 L 37. 0.0153 mol CO 2 44.01 g CO 2 = 0.949 g CO 2 1 L mixture 1 mol CO 2 (D) We base each of our solutions on the balanced chemical equation. (a)
Equation : Initial : Changes : Equil : PCl5 g PCl3 g + Cl2 g 0 mol 2.50 L + x mol 2.50 L x mol 2.50 L
x(0.550 x) 3.8 102 2.50(0.550 x) 0.550 mol 2.50 L x mol 2.50 L 0.550 x mol 2.50 L 0.550 mol 2.50 L + x mol 2.50 L 0.550 + x mol 2.50 L 0.550 x mol x mol [PCl3 ][Cl2 ] 2.50L 2.50L Kc 3.8 102 (0.550 x)mol [PCl5 ] 2.50L x 2 + 0.550 x = 0.052 0.095x
x= x 2 + 0.645x 0.052 = 0 b b 2 4ac 0.645 0.416 + 0.208 = = 0.0725 mol, 0.717 mol 2a 2 The second answer gives a negative quantity. of Cl 2 , which makes no physical sense. n PCl5 = 0.550 0.0725 = 0.478 mol PCl5
n Cl2 = x = 0.0725 mol Cl2 n PCl3 = 0.550 + 0.0725 = 0.623 mol PCl3 (b) Equation : Initial : Changes : Equil : PCl5 g PCl3 g 0M + x mol 2.50 L x mol 2.50 L + Cl2 g 0M + x mol 2.50 L x mol 2.50 L 0.610 mol 2.50 L x mol 2.50 L 0.610 x mol 2.50 L 687 Chapter15: Principles of Chemical Equilibrium ( x mol) ( x mol) PCl3 Cl2 = 3.8 102 2.50 L 2.50 L Kc = 0.610 x mol PCl5 2.50 L x2 0.095 2.50 3.8 10 0.610 x
2 0.058 0.095 x = x 2 x 2 + 0.095 x 0.058 = 0 x= b b 2 4ac 0.095 0.0090 + 0.23 = = 0.20 mol, 0.29 mol 2a 2 amount PCl 3 = 0.20 mol = amount Cl 2 ; amount PCl5 = 0.610 0.20 = 0.41 mol
38. (D) (a) We use the balanced chemical equation as a basis to organize the information we have about the reactants and products.
Equation: Initial: Initial: 2 COF2 g 0.145 mol 5.00 L 0.0290 M CO 2 g 0.262 mol 5.00 L 0.0524 M + CF4 g 0.074 mol 5.00 L 0.0148 M And we now compute a value of Qc and compare it to the given value of K c . Qc = CO2 CF4 = 0.0524 0.0148 = 0.922 2.00 = K c 2 2 0.0290 COF2 Because Qc is not equal to K c , the mixture is not at equilibrium.
(b) Because Qc is smaller than K c , the reaction will shift right, that is, products will be formed at the expense of COF2, to reach a state of equilibrium. We continue the organization of information about reactants and products. Equation: Initial: Changes: Equil: 2 COF2 g 0.0290 M 2x M (c) CO 2 g 0.0524 M +x M + CF4 g 0.0148 M +x M 0.0290 2x M 0.0524 + x M 0.0148 + x M 688 Chapter15: Principles of Chemical Equilibrium Kc = CO2 CF4 = 0.0524 + x 0.0148 + x = 2.00 = 0.000776 + 0.0672 x + x 2 2 2 0.000841 0.1160 x + 4 x 2 0.0290 2 x COF2 0.00168 0.232 x + 8 x 2 = 0.000776 + 0.0672 x + x 2 7 x 2 0.299 x + 0.000904 = 0
x= b b 2 4ac 0.299 0.0894 0.0253 = = 0.0033 M ,0.0394 M 2a 14 The second of these values for x (0.0394) gives a negative COF2 = 0.0498 M , clearly a nonsensical result. We now compute the concentration of each species at equilibrium, and check to ensure that the equilibrium constant is satisfied. COF2 = 0.0290 2 x = 0.0290 2 0.0033 = 0.0224 M CO2 = 0.0524 + x = 0.0524 + 0.0033 = 0.0557 M CF4 = 0.0148 + x = 0.0148 + 0.0033 = 0.0181 M
Kc = CO2 CF4 = 0.0557 M 0.0181 M = 2.01 2 2 0.0224 M COF2 The agreement of this value of Kc with the cited value (2.00) indicates that this solution is correct. Now we determine the number of moles of each species at equilibrium. mol COF2 = 5.00 L 0.0224 M = 0.112 mol COF2 mol CO 2 = 5.00 L 0.0557 M = 0.279 mol CO 2 mol CF4 = 5.00 L 0.0181 M = 0.0905 mol CF4 But suppose we had incorrectly concluded, in part (b), that reactants would be formed in reaching equilibrium. What result would we obtain? The setup follows. Equation : Initial : Changes : Equil : 2 COF2 g 0.0290 M +2 y M CO2 g 0.0524 M y M + CF4 g 0.0148 M yM 0.0290 + 2 y M 0.0524 y M 0.0148 y M Kc = CO2 CF4 = 0.0524 y 0.0148 y = 2.00 = 0.000776 0.0672 y + y 2 2 2 0.000841 + 0.1160 y + 4 y 2 0.0290 + 2 y COF2 0.00168 + 0.232 y + 8 y 2 = 0.000776 0.0672 y + y 2 7 y 2 + 0.299 y + 0.000904 = 0
y= b b 2 4ac 0.299 0.0894 0.0253 = = 0.0033 M , 0.0394 M 2a 14 689 Chapter15: Principles of Chemical Equilibrium The second of these values for x 0.0394 gives a negative COF2 = 0.0498 M , clearly a nonsensical result. We now compute the concentration of each species at equilibrium, and check to ensure that the equilibrium constant is satisfied. b g COF2 = 0.0290 + 2 y = 0.0290 + 2 0.0033 = 0.0224 M CO2 = 0.0524 y = 0.0524 + 0.0033 = 0.0557 M CF4 = 0.0148 y = 0.0148 + 0.0033 = 0.0181 M
These are the same equilibrium concentrations that we obtained by making the correct decision regarding the direction that the reaction would take. Thus, you can be assured that, if you perform the algebra correctly, it will guide you even if you make the incorrect decision about the direction of the reaction.
39. (D) (a) We calculate the initial amount of each substance. n {C 2 H 5OH n{CH 3CO 2 H = 17.2 g C2 H5OH 1 mol C2 H 5OH = 0.373 mol C 2 H 5OH 46.07 g C 2 H 5OH 1 mol CH 3CO 2 H = 0.396 mol CH 3CO 2 H 60.05 g CH 3CO 2 H = 23.8 g CH3CO2 H n{CH 3 CO 2 C2 H 5 } = 48.6 g CH3 CO 2 C2 H 5 1 mol CH 3 CO 2 C2 H 5 88.11 g CH 3 CO 2 C 2 H 5 n{CH 3 CO 2 C2 H 5 } = 0.552 mol CH 3 CO 2 C2 H 5 n H 2O = 71.2 g H 2O 1 mol H 2 O = 3.95 mol H 2 O 18.02 g H 2 O Since we would divide each amount by the total volume, and since there are the same numbers of product and reactant stoichiometric coefficients, we can use moles rather than concentrations in the Qc expression. Qc = n{CH 3CO 2 C2 H 5 n H 2 O 0.552 mol 3.95 mol = = 14.8 K c = 4.0 n{C2 H 5OH n CH 3CO 2 H 0.373 mol 0.396 mol Since Qc K c the reaction will shift to the left, forming reactants, as it attains equilibrium.
(b)
Equation: Initial Changes Equil C2 H5 OH + 0.373 mol +x mol (0.373+x) mol CH3 CO2 H CH 3 CO 2 C2 H 5 + 0.396 mol 0.552 mol +x mol x mol (0.396+x) mol (0.552x) mol H2O 3.95 mol x mol (3.95x) mol 690 Chapter15: Principles of Chemical Equilibrium Kc (0.552 x)(3.95 x) 2.18 4.50 x x 2 4.0 (0.373 x)(0.396 x) 0.148 0.769 x x 2 3x 2 + 7.58 x 1.59 = 0 x 2 4.50 x + 2.18 = 4 x 2 + 3.08 x + 0.59 b b 2 4ac 7.58 57 +19 x= = = 0.19 moles, 2.72 moles 2a 6 Negative amounts do not make physical sense. We compute the equilibrium amount of each substance with x = 0.19 moles.
n{C 2 H 5OH = 0.373 mol + 0.19 mol = 0.56 mol C2 H5OH
46.07 g C 2 H 5OH 26 g C 2 H 5OH 1 mol C2 H 5OH mass C 2 H 5OH = 0.56 mol C 2 H 5OH n{CH 3CO 2 H = 0.396 mol + 0.19 mol = 0.59 mol CH3CO2 H
60.05g CH3CO 2 H 35g CH3CO2 H 1 mol CH 3CO 2 H mass CH 3CO 2 H = 0.59 mol CH3CO2 H n{CH 3CO 2C 2 H 5 = 0.552 mol 0.19 mol = 0.36 CH3CO2C2 H5
88.10 g CH3CO 2 C2 H 5 32 g CH 3CO 2 C 2 H 5 1 mol CH 3CO 2 C2 H 5 mass CH 3CO 2 C 2 H 5 = 0.36 mol CH 3CO 2 C2 H 5 n{H 2 O = 3.95 mol 0.19 mol = 3.76 mol H 2O
18.02 g H 2 O 68g H 2 O 1 mol H 2 O mass H 2O = 3.76 mol H 2O To check K c = 0.36 mol 3.76 mol = = 4.1 n C 2 H 5 OH n CH 3 CO 2 H 0.56 mol 0.59 mol
H2O n CH 3 CO 2 C 2 H 5 n 40. (M) The final volume of the mixture is 0.750 L + 2.25 L = 3.00 L . Then use the balanced chemical equation to organize the data we have concerning the reaction. The reaction should shift to the right, that is, form products in reaching a new equilibrium, since the volume is greater. Equation: Initial: Initial: Changes: Equil : N 2 O4 g 2 NO 2 g 0.0580 mol 3.00 L 0.0193M 2x M (0.0193 2x)M 0.971 mol 3.00 L 0.324 M x M (0.324 x)M 691 Chapter15: Principles of Chemical Equilibrium Kc NO2 = 2 N2 O4 = 0.0193 + 2 x 0.324 x 2 0.000372 + 0.0772 x + 4 x 2 = = 4.61 103 0.324 x 0.000372 + 0.0772 x + 4 x 2 = 0.00149 0.00461x 4 x 2 + 0.0818 x 0.00112 = 0 b b 2 4ac 0.0818 0.00669 + 0.0179 x= = = 0.00938 M, 0.0298 M 2a 8 NO 2 = 0.0193 + 2 0.00938 = 0.0381 M
amount NO 2 = 0.0381 M 3.00 L = 0.114 mol NO 2 N 2O4 = 0.324 0.00938 = 0.3146 M
amount N 2 O 4 = 0.3146 M 3.00 L = 0.944 mol N 2 O 4 41. (M) HCONH 2 init =
Equation: Initial : Changes: Equil : 0.186 mol = 0.0861 M 2.16 L
CO g 0M +x M HCONH 2 g NH 3 g +
0.0861 M x M (0.0861x) M 2 0M +x M xM xM Kc = NH3 CO = HCONH xx = 4.84 0.0861 x x 2 = 0.417 4.84 x 0 = x 2 + 4.84 x 0.417 x= b b 2 4ac 4.84 23.4 +1.67 = = 0.084 M, 4.92 M 2a 2 The negative concentration obviously is physically meaningless. We determine the total concentration of all species, and then the total pressure with x = 0.084. total = NH 3 + CO + HCONH 2 = x + x + 0.0861 x = 0.0861+ 0.084 = 0.170 M
Ptot = 0.170 mol L1 0.08206 L atm mol1 K 1 400. K = 5.58 atm
42. (E) Compare Qp to K p . We assume that the added solids are of negligible volume so that the initial partial pressures of CO2(g) and H2O(g) do not significantly change. 1 atm P{H 2 O = 715 mmHg = 0.941 atm H 2 O 760 mmHg Qp = P CO 2 P H 2 O = 2.10 atm CO 2 0.941 atm H 2 O = 1.98 0.23 = K p Because Qp is larger than K p , the reaction will proceed left toward reactants to reach equilibrium. Thus, the partial pressures of the two gases will decrease. 692 Chapter15: Principles of Chemical Equilibrium 43. (M) We organize the solution around the balanced chemical equation. Equation: 2 Cr 3+ aq + Cd(s) 2 Cr 2+ aq Cd 2 aq Initial : Changes: Equil: 1.00 M 2 x M (1.00 2 x) M
2 Cr 2+ Cd 2+ = (2 x) ( x) = 0.288 Kc = 2 2 1.00 2 x Cr 3 2 0M 2 x 2x 0M x x
Via successive approximations , one obtains x = 0.257 M Therefore, at equilibrium, [Cd2+] = 0.257 M, [Cr2+] = 0.514 M and [Cr3+] = 0.486 M Minimum mass of Cd(s) = 0.350L 0.257 M 112.41 g/mol = 10.1 g of Cd metal
44. (M) Again we base the setup of the problem around the balanced chemical equation.
Kc =3.210 10 Equation: Initial: Changes: Equil : Kc =
3 Pb s + 2 Cr 3+ aq 0.100 M 2x M (0.100 2x)M Pb 2+ aq + 2 Cr 2+ aq 0M xM xM 0M 2x M 2x M x (2 x) 2 = 3.2 1010 2 (0.100) 4 x 3 (0.100) 2 3.2 1010 3.2 1012 3.2 1012 x 9.3 105 M Assumption that 2 x 0.100 , is valid and thus 4 Pb 2+ = x = 9.3 10 5 M , [Cr2+] = 1.9 104 M and [Cr3+] = 0.100 M
45. (M) We are told in this question that the reaction SO2(g) + Cl2(g) SO2Cl2(g) has Kc = 4.0 at a certain temperature T. This means that at the temperature T, [SO2Cl2] = 4.0 [Cl2] [SO2]. Careful scrutiny of the three diagrams reveals that sketch (b) is the best representation because it contains numbers of SO2Cl2, SO2, and Cl2 molecules that are consistent with the Kc for the reaction. In other words, sketch (b) is the best choice because it contains 12 SO2Cl2 molecules (per unit volume), 1 Cl2 molecule (per unit volume) and 3 SO2 molecules (per unit volume), which is the requisite number of each type of molecule needed to generate the expected Kc value for the reaction at temperature T. 46. (M) In this question we are told that the reaction 2 NO(g) + Br2(g) 2 NOBr(g) has Kc = 3.0 at a certain temperature T. This means that at the temperature T, [NOBr]2 = 3.0 [Br2][NO]2. Sketch (c) is the most accurate representation because it contains 18 NOBr molecules (per unit volume), 6 NO molecules (per unit volume), and 3 Br2 molecules (per unit volume), which is the requisite number of each type of molecule needed to generate the expected Kc value for the reaction at temperature T. 693 Chapter15: Principles of Chemical Equilibrium 47. (E) K Q aconitate citrate
4.0 105 0.031 0.00128 Since Q = K, the reaction is at equilibrium,
48. (E) CO2 NADred oxoglut. citrate NADox 0.00868 0.00132 0.00868 0.00895 Q 0.00128 0.00868
K Since Q < K, the reaction needs to proceed to the right (products). Partial Pressure Equilibrium Constant, Kp
49. (M) The I 2 s maintains the presence of I 2 in the flask until it has all vaporized. Thus, if enough HI(g) is produced to completely consume the I 2 s , equilibrium will not be achieved. 1 atm P{H 2S} = 747.6 mmHg = 0.9837 atm 760 mmHg H 2S g + I 2 s 2 HI g + S s Equation: bg bg Initial: Changes: Equil:
Kp = 0.9837 atm x atm 0.9837 x atm b g 0 atm +2x atm 2x atm 2x P{HI}2 4x2 = = 1.34 105 = P{H 2 S} 0.9837 x 0.9837
2 1.34 105 0.9837 x= = 1.82 103 atm 4 The assumption that 0.9837 x is valid. Now we verify that sufficient I2(s) is present by computing the mass of I 2 needed to produce the predicted pressure of HI(g). Initially, 1.85 g I2 is present (given). mass I 2 = 1.82 103 atm 0.725 L 1 mol I 2 253.8 g I 2 = 0.00613 g I 2 1 1 0.08206 L atm mol K 333 K 2 mol HI 1 mol I 2 Ptot = P{H 2S} + P{HI} = 0.9837 x + 2 x = 0.9837 + x = 0.9837 + 0.00182 = 0.9855 atm Ptot = 749.0 mmHg 694 Chapter15: Principles of Chemical Equilibrium 50. (M) We first determine the initial pressure of NH 3 . P{NH 3 g }= Equation: Initial: Changes: Equil: nRT 0.100 mol NH 3 0.08206 L atm mol1 K 1 298 K = =0.948 atm V 2.58 L + H2S(g) NH4HS(s) NH3(g) 0.948 atm 0 atm +x atm +x atm x atm 0.948 + x atm Kp = P{NH 3} P{H 2S} = 0.108 = 0.948 + x x = 0.948 x + x 2 0 = x 2 + 0.948 x 0.108 x=
b b 2 4ac 0.948 0.899 + 0.432 = = 0.103 atm, 1.05 atm 2a 2 b g The negative root makes no physical sense. The total gas pressure is obtained as follows. Ptot = P{NH 3} + P{H 2 S} = 0.948 + x + x = 0.948 + 2 x = 0.948 + 2 0.103 = 1.154 atm b g 51. (M) We substitute the given equilibrium pressure into the equilibrium constant expression P{O 2 }3 P{O 2 }3 = 28.5 = and solve for the other equilibrium pressure. K p = 2 P{CO 2 }2 0.0721 atm CO2 P{O 2 } 3 P{O 2 }3 3 28.5(0.0712 atm) 2 0.529 atm O 2 Ptotal = P{CO 2 } + P{O 2 } = 0.0721 atm CO 2 + 0.529 atm O 2 = 0.601 atm total
52. (M) The composition of dry air is given in volume percent. Division of these percentages by 100 gives the volume fraction, which equals the mole fraction and also the partial pressure in atmospheres, if the total pressure is 1.00 atm. Thus, we have P{O 2 } = 0.20946 atm and P{CO 2 } = 0.00036 atm. We substitute these two values into the expression for Qp . 0.20946 atm O2 = 6.4 104 28.5 = K P{O 2 }3 Qp = = p 2 2 P{CO 2 } 0.00036 atm CO2 3 The value of Qp is much larger than the value of K p . Thus this reaction should be spontaneous in the reverse direction until equilibrium is achieved. It will only be spontaneous in the forward direction when the pressure of O2 drops or that of CO 2 rises (as would be the case in selfcontained breathing devices). 695 Chapter15: Principles of Chemical Equilibrium 53. (M) (a) We first determine the initial pressure of each gas. nRT 1.00 mol 0.08206 L atm mol 1 K 1 668 K = = 31.3 atm V 1.75 L Then we calculate equilibrium partial pressures, organizing our calculation around the balanced chemical equation. We see that the equilibrium constant is not very large, meaning that we must solve the polynomial exactly (or by successive approximations). + Cl2 g COCl2 g Kp = 22.5 Equation CO(g) P{CO} = P{Cl 2 } = Initial: Changes: Equil:
Kp 31.3 atm x atm 31.3 x atm 31.3 atm x atm 31.3 x atm 0 atm +x atm x atm P{COCl2 } x x 22.5 2 P{CO} P{Cl2 } (31.3 x) (979.7 62.6 x x 2 ) 22.5(979.7 62.6 x x 2 ) x 22043 1408.5x 22.5 x 2 x 22043 1409.5x 22.5 x 2 0 (Solve by using the quadratic equation) b b 2 4ac (1409.5) (1409.5) 2 4(22.5)(22043) = x= 2a 2(22.5)
x=
1409.5 2818 30.14, 32.5(too large) 45 P{CO} = P{Cl2 } = 31.3 atm 30.14 atm = 1.16 atm P{COCl2 } = 30.14 atm (b) Ptotal = P{CO}+ P{Cl2 }+ P{COCl2 } = 1.16 atm +1.16 atm + 30.14 atm = 32.46 atm 54. (M) We first find the value of Kp for the reaction. 2 NO 2 g 2 NO g + O 2 g , K c = 1.8 106 at 184 C = 457 K . For this reaction ngas = 2 +1 2 = +1. K p K c (RT)
n g = 1.8 106 (0.08206 457) 1 = 6.8 105 To obtain the required reaction NO g + 1 O 2 g NO 2 g from the initial reaction, 2 that initial reaction must be reversed and then divided by two. Thus, in order to determine the value of the equilibrium constant for the final reaction, the value of Kp for the initial reaction must be inverted, and the square root taken of the result.
K p, final 1 6.8 105 1.2 102 696 Chapter15: Principles of Chemical Equilibrium Le Chtelier's Principle
55. (E) Continuous removal of the product, of course, has the effect of decreasing the concentration of the products below their equilibrium values. Thus, the equilibrium system is disturbed by removing the products and the system will attempt (in vain, as it turns out) to reestablish the equilibrium by shifting toward the right, that is, to generate more products. (E) We notice that the density of the solid ice is smaller than is that of liquid water. This means that the same mass of liquid water is present in a smaller volume than an equal mass of ice. Thus, if pressure is placed on ice, attempting to force it into a smaller volume, the ice will be transformed into the lessspaceoccupying water at 0 C . Thus, at 0 C under pressure, H2O(s) will melt to form H2O(l). This behavior is not expected in most cases because generally a solid is more dense than its liquid phase. (M) (a) This reaction is exothermic with H o = 150 . kJ. Thus, high temperatures favor the reverse reaction (endothermic reaction). The amount of H 2 g present at high 56. 57. temperatures will be less than that present at low temperatures.
(b) position to shift to the right, favoring products. The amount of H 2 g will increase.
(c) H 2 O g is one of the reactants involved. Introducing more will cause the equilibrium bg Doubling the volume of the container will favor the side of the reaction with the largest sum of gaseous stoichiometric coefficients. The sum of the stoichiometric coefficients of gaseous species is the same (4) on both sides of this reaction. Therefore, increasing the volume of the container will have no effect on the amount of H 2 g present at equilibrium. A catalyst merely speeds up the rate at which a reaction reaches the equilibrium position. The addition of a catalyst has no effect on the amount of H 2 g present at equilibrium. (d) 58. (M) (a) This reaction is endothermic, with H o = +92.5 kJ. Thus, a higher temperature will favor the forward reaction and increase the amount of HI(g) present at equilibrium. (b) (c) The introduction of more product will favor the reverse reaction and decrease the amount of HI(g) present at equilibrium. The sum of the stoichiometric coefficients of gaseous products is larger than that for gaseous reactants. Increasing the volume of the container will favor the forward reaction and increase the amount of HI(g) present at equilibrium. A catalyst merely speeds up the rate at which a reaction reaches the equilibrium position. The addition of a catalyst has no effect on the amount of HI(g) present at equilibrium. (d) 697 Chapter15: Principles of Chemical Equilibrium (e) The addition of an inert gas to the constantvolume reaction mixture will not change any partial pressures. It will have no effect on the amount of HI(g) present at equilibrium. 59. (M) (a) The formation of NO(g) from its elements is an endothermic reaction ( H o = +181 kJ/mol). Since the equilibrium position of endothermic reactions is shifted toward products at higher temperatures, we expect more NO(g) to be formed from the elements at higher temperatures. (b) Reaction rates always are enhanced by higher temperatures, since a larger fraction of the collisions will have an energy that surmounts the activation energy. This enhancement of rates affects both the forward and the reverse reactions. Thus, the position of equilibrium is reached more rapidly at higher temperatures than at lower temperatures. 60. (M) If the reaction is endothermic (H > 0), the forward reaction is favored at high temperatures. If the reaction is exothermic (H < 0), the forward reaction is favored at low temperatures. (a) H o = H fo PCl5 g H fo PCl3 g H fo Cl2 g H o = 374.9 kJ 287.0 kJ 0.00 kJ = 87.9 kJ/mol (favored at low temperatures) (b) H o = 2 241.8 kJ + 3 0.00 kJ 296.8 kJ 2 20.63 kJ H o = 145.5 kJ/mol (favored at low temperatures)
(c) H o = 2 Hfo [ H 2 O(g)] 3Hfo [S(rhombic)] Hfo [SO 2 (g)] 2 Hfo [ H 2S(g)] H o = 4H fo NOCl g + 2H fo H 2 O g 2H fo N 2 g 3H fo O 2 g 4H fo HCl g H o = 4 51.71 kJ + 2 241.8 kJ 2 0.00 kJ 3 0.00 kJ 4 92.31 kJ H o = +92.5 kJ/mol (favored at higher temperatures)
61. (E) If the total pressure of a mixture of gases at equilibrium is doubled by compression, the equilibrium will shift to the side with fewer moles of gas to counteract the increase in pressure. Thus, if the pressure of an equilibrium mixture of N2(g), H2(g), and NH3(g) is doubled, the reaction involving these three gases, i.e., N2(g) + 3 H2(g) 2 NH3(g), will proceed in the forward direction to produce a new equilibrium mixture that contains additional ammonia and less molecular nitrogen and molecular hydrogen. In other words, P{N2(g)} will have decreased when equilibrium is reestablished. It is important to note, however, that the final equilibrium partial pressure for the N2 will, nevertheless, be higher than its original partial pressure prior to the doubling of the total pressure. 698 Chapter15: Principles of Chemical Equilibrium 62. (M) (a) Because H o = 0 , the position of the equilibrium for this reaction will not be affected by temperature. Since the equilibrium position is expressed by the value of the equilibrium constant, we expect Kp to be unaffected by, or to remain constant with, temperature.
(b) From part (a), we know that the value of Kp will not change when the temperature is changed. The pressures of the gases, however, will change with temperature. (Recall the ideal gas law: P = nRT / V .) In fact, all pressures will increase. The stoichiometric coefficients in the reaction are such that at higher pressures the formation of more reactant will be favored (the reactant side has fewer moles of gas). Thus, the amount of D(g) will be smaller when equilibrium is reestablished at the higher temperature for the cited reaction. 1 A s B s + 2 C g + D g 2 63. (M) Increasing the volume of an equilibrium mixture causes that mixture to shift toward the side (reactants or products) where the sum of the stoichiometric coefficients of the gaseous species is the larger. That is: shifts to the right if ngas 0 , shifts to the left if ngas 0 , and does not shift if ngas = 0 .
(a) (b) (c) 64. C s + H 2 O g CO g + H 2 g , ngas 0, shift right, toward products Ca OH 2 s + CO 2 g CaCO3 s + H 2 O g , ngas = 0, no shift, no change in equilibrium position. 4 NH 3 g + 5 O 2 g 4 NO g + 6 H 2 O g , ngas 0, shifts right, towards products (M) The equilibrium position for a reaction that is exothermic shifts to the left (reactants are favored) when the temperature is raised. For one that is endothermic, it shifts right (products are favored) when the temperature is raised. 2 (a) NO g 1 N 2 g + 1 O 2 g H o = 90.2 kJ shifts left, % dissociation 2 (b) (c) (d) SO3 g SO 2 g + 1 O 2 g 2 N2 H4 g N2 g + 2 H2 g COCl2 g CO g + Cl2 g H o = +98.9 kJ shifts right, % dissociation H o = 95.4 kJ shifts left, % dissociation H o = +108.3 kJ shifts right, % dissociation 65. (E) (a) Hb:O2 is reduced, because the reaction is exothermic and heat is like a product. (b) No effect, because the equilibrium involves O2 (aq). Eventually it will reduce the Hb:O2 level because removing O2(g) from the atmosphere also reduces O2 (aq) in the blood. (c) Hb:O2 level increases to use up the extra Hb. 699 Chapter15: Principles of Chemical Equilibrium 66. (E) (a) CO2 (g) increases as the equilibrium is pushed toward the reactant side (b) Increase CO2 (aq) levels, which then pushes the equilibrium to the product side (c) It has no effect, but it helps establish the final equilibrium more quickly (d) CO2 increases, as the equilibrium shifts to the reactants (E) The pressure on N2O4 will initially increase as the crystal melts and then vaporizes, but over time the new concentration decreases as the equilibrium is shifted toward NO2. (E) If the equilibrium is shifted to the product side by increasing temperature, that means that heat is a "reactant" (or being consumed). Therefore, HI decomposition is endothermic. (E) Since H is >0, the reaction is endothermic. If we increase the temperature of the reaction, we are adding heat to the reaction, which shifts the reaction toward the decomposition of calcium carbonate. While the amount of calcium carbonate will decrease, its concentration will remain the same because it is a solid. (E) The amount of N2 increases in the body. As the pressure on the body increases, the equilibrium shifts from N2 gas to N2 (aq). 67. 68. 69. 70. Integrative and Advanced Exercises
71. (E) In a reaction of the type I2(g) 2 I(g) the bond between two iodine atoms, the II bond, must be broken. Since I2(g) is a stable molecule, this bond breaking process must be endothermic. Hence, the reaction cited is an endothermic reaction. The equilibrium position of endothermic reactions will be shifted toward products when the temperature is raised. 72. (M) (a) In order to determine a value of Kc, we first must find the CO2 concentration in the gas phase. Note, the total volume for the gas is 1.00 L (moles and molarity are numerically equal)
[CO2 ] n P V RT 1.00 atm 0.0409 M L atm 298 K 0.08206 mol K Kc [CO2 (aq)] 3.29 102 M 0.804 [CO2 (g)] 0.0409 M (b) It does not matter to which phase the radioactive 14CO2 is added. This is an example of a Le Chtelier's principle problem in which the stress is a change in concentration of the reactant CO2(g). To find the new equilibrium concentrations, we must solve and I.C.E. table. Since Qc < Kc, the reaction shifts to the product, CO2 (aq) side.
Reaction: Initial: Stress Changes: Equilibrium: CO 2 (g) CO 2 (aq) 3.29 103 mol x mol 0.0409 mol +0.01000 mol x mol (0.05090 x ) mol 3.29 103 x mol 700 Chapter15: Principles of Chemical Equilibrium 3.29 103 x mol [CO 2 (aq)] 3.29 102 10 x 0.1000 L KC 0.804 [CO2 (g)] (0.05090 x) mol 0.05090 x 1.000 L x 7.43 104 mol CO 2 Total moles of CO2 in the aqueous phase (0.1000 L)(3.29102 +7.43103) = 4.03 103 moles Total moles of CO2 in the gaseous phase (1.000 L)(5.090102 7.43104) = 5.02 102 moles Total moles of CO2 = 5.02 102 moles + 4.03 103 moles = 5.42102 moles There is continuous mixing of the 12C and 14C such that the isotopic ratios in the two phases is the same. This ratio is given by the mole fraction of the two isotopes. For
14 CO 2 in either phase its mole fraction is 0.010002mol 100 18.45 %
5.419 10 mol Moles of Moles of
73. CO 2 in the gaseous phase = 5.02102 moles0.1845 = 0.00926 moles 14 CO 2 in the aqueous phase = 4.03 103 moles0.1845 = 0.000744 moles
14 (M) Dilution makes Qc larger than Kc. Thus, the reaction mixture will shift left in order to regain equilibrium. We organize our calculation around the balanced chemical equation.
Equation: Equil: Ag (aq) 0.31 M Fe 2 (aq) 0.21 M Fe3 (aq) Ag(s) 0.19 M 0.076 M x M (0.076 x) M K c 2.98 Dilution: 0.12 M 0.084 M Changes: xM xM New equil: (0.12 x) M (0.084 x) M Kc 3 [Fe ] 0.076 x 2.98 2 [Ag ][Fe ] (0.12 x) (0.084 x) 2.98 (0.12 x) (0.084 x) 0.076 x x 0.046 0 0.076 x 0.030 0.61x 2.98 x 2 2.98 x 2 1.61 1.61 2.59 0.55 0.027, 0.57 Note that the negative root makes no physical 5.96 sense; it gives [Fe 2 ] 0.084 0.57 0.49 M . Thus, the new equilibrium concentrations are x
[Fe 2 ] 0.084 0.027 0.111 M [Ag ] 0.12 0.027 0.15 M [Fe3 ] 0.076 0.027 0.049 M We can check our answer by substitution. Kc =
74. 0.049 M = 2.94 2.98 (within precision limits) 0.111 M 0.15 M (M) The percent dissociation should increase as the pressure is lowered, according to Le Chtelier's principle. Thus the total pressure in this instance should be more than in Example 1512, where the percent dissociation is 12.5%. The total pressure in Example 1512 was computed starting from the total number of moles at equilibrium. The total amount = (0.0240 0.00300) moles N2O4 + 2 0.00300 mol NO2 = 0.027 mol gas. 701 Chapter15: Principles of Chemical Equilibrium Ptotal nRT 0.0270mol 0.08206 L atm K 1 mol1 298 K = = = 1.77 atm (Example 1512) V 0.372 L We base our solution on the balanced chemical equation. We designate the initial pressure of N2O4 as P. The change in P{N2O4}is given as 0.10 P atm. to represent the 10.0 % dissociation. 2 NO 2 (g) Equation: N 2 O 4 (g) Initial: Changes: Equil: P atm 0.10 P atm 0.90 P atm 0 atm +2(0.10 P atm) 0.20 P atm P{NO 2 }2 (0.20 P) 2 0.040 P 0.113 0.90 = = = 0.113 P= = 2.54 atm. P{N 2 O 4 } 0.90 P 0.90 0.040 Thus, the total pressure at equilibrium is 0.90 P + 0.20 P and 1.10 P (where P = 2.54 atm) Therefore, total pressure at equilibrium = 2.79 atm. Kp =
75. (M) Equation: Initial:
2 SO3 (g) 1.00 atm 2 SO 2 (g) 0 atm O 2 (g) 0 atm Changes: Equil: 2 x atm
(1.00 2 x)atm 2x atm
2x atm x atm
x atm Because of the small value of the equilibrium constant, the reaction does not proceed very far toward products in reaching equilibrium. Hence, we assume that x << 1.00 atm and calculate an approximate value of x (small K problem). KP P{SO 2 }2 P{O 2 } (2 x) 2 x 4 x3 1.6 105 P{SO3 }2 (1.00 2 x) 2 (1.00) 2 x 0.016 atm A second cycle may get closer to the true value of x. 4 x3 5 1.6 10 x 0.016 atm (1.00 0.032) 2 Our initial value was sufficiently close. We now compute the total pressure at equilibrium. Ptotal P{SO 3 } P{SO 2 } P{O 2 } (1.00 2 x) 2 x x 1.00 x 1.00 0.016 1.02 atm
76. (M) Let us start with one mole of air, and let 2x be the amount in moles of NO formed.
Equation: Initial: Changes: Equil: N 2 (g) 0.79 mol x mol (0.79 x)mol O 2 (g) 0.21 mol x mol (0.21 x)mol 2 NO(g) 0 mol 2x mol 2x mol 702 Chapter15: Principles of Chemical Equilibrium NO 2x 2x n{NO} 0.018 n{N 2 } n{O 2 } n{NO} (0.79 x) (0.21 x) 2 x 1.00 0.79 x 0.78 mol N 2
2 x 0.0090 mol 2x 0.018 mol NO 0.21 x 0.20 mol O 2 n{NO} RT 2 (0.018) 2 P{NO} n{NO}2 Vtotal 2.1 103 Kp n{N 2 ) RT n{O 2 } RT n{N 2 } n{O 2 } 0.78 0.20 P (N 2 } P{O 2 } Vtotal Vtotal 77. (D) We organize the data around the balanced chemical equation. Note that the reaction is stoichimoetrically balanced. (a)
Equation: 2 SO 2 (g) O 2 (g) 2 SO3 (g) Equil : Add SO 3 Initial : 0.32 mol 0.32 mol 0.32 mol 10.0 L 0.16 mol 0.16 mol 0.16 mol 10.0 L 0.68 mol 1.68 mol 1.68 mol 10.0 L Initial : To right : Changes : Equil : 0.032 M 0.000 M 2x M 2x M 0.016 M 0.000 M xM xM 0.168 M 0.200 M 2x M (0.200 2 x)M In setting up this problem, we note that solving this question exactly involves finding the roots for a cubic equation. Thus, we assumed that all of the reactants have been converted to products. This gives the results in the line labeled "To right." We then reach equilibrium from this position by converting some of the product back into reactants. Now, we substitute these expressions into the equilibrium constant expression, and we solve this expression approximately by assuming that 2 x 0.200 . Kc [SO 3 ] 2 (0.200 2 x) 2 (0.200) 2 2.8 10 2 or x 0.033 [SO 2 ] 2 [O 2 ] ( 2 x) 2 x 4x3 (0.200 0.066) 2 2.8 10 2 or x 0.025 3 4x We then substitute this approximate value into the expression for Kc. Kc (0.200 0.050) 2 Let us try one more cycle. K c 2.8 10 2 or x 0.027 3 4x 703 Chapter15: Principles of Chemical Equilibrium This gives the following concentrations and amounts of each species. [SO 3 ] 0.200 (2 0.027) 0.146 M [SO 2 ] 2 0.027 0.054 M [O 2 ] 0.027 M
(b)
Equation: Equil : 2 SO 2 (g) 0.32 mol amount SO 3 10.0 L 0.146 M 1.46 mol SO 3 amount SO 2 10.0 L 0.054 M 0.54 mol SO 2 amount O 2 10.0 L 0.027 M 0.27 mol O 2 O 2 (g) 2 SO3 (g) 0.16 mol 0.68 mol Equil : Equil : 0.10 V : To right : Changes : Equil : 0.32 mol 10.0 L 0.032 M 0.32 M 0.00 M 2x M 2x M 0.16 mol 10.0 L 0.016 M 0.16 M 0.00 M x M xM 0.68 mol 10.0 L 0.068 M 0.68 M 1.00 M 2 x M (1.00 2 x)M Again, notice that an exact solution involves finding the roots of a cubic. So we have taken the reaction 100% in the forward direction and then sent it back in the reverse direction to a small extent to reach equilibrium. We now solve the Kc expression for x, obtaining first an approximate value by assuming 2x << 1.00. Kc [SO 3 ] 2 [SO 2 ] 2 [O 2 ] (1.00 2 x) 2 (1.00) 2 2.8 10 2 or x 0.096 (2 x) 2 x 4x 3 We then use this approximate value of x to find a second approximation for x. (1.00 0.19) 2 Kc 2.8 10 2 3 4x Another cycle gives K c or x 0.084 (1.00 0.17) 2 2.8 10 2 or x 0.085 3 4x Then we compute the equilibrium concentrations and amounts. amount SO3 1.00 L 0.83 M 0.83 mol SO3 amount SO 2 1.00 L 0.17 M 0.17 mol SO 2 amount O 2 1.00 L 0.085 M 0.085 mol O 2 [SO3 ] 1.00 (2 0.085) 0.83 M [SO 2 ] 2 0.085 0.17 M [O 2 ] 0.085 M
78. (M) Equation: HOC 6 H 4 COOH(g) C 6 H 5 OH(g) CO 2 (g) n co 2 730mmHg 1L 48.2 48.5 PV 760mmHg/atm 2 1000 mL 1.93 103 mol CO 2 RT 0.0821L atm/mol K 293K 704 Chapter15: Principles of Chemical Equilibrium Note that moles of CO 2 moles phenol n salicylic acid 0.300 g 2.17 103 mol salicylic acid 138 g / mol
2 1.93 mmol [C 6 H 5 OH ] [CO 2 (g)] 50.0 mL 0.310 Kc (2.17 1.93) mmol [HOC 6 H 4 COOH] 50.0 mL L atm K p K c (RT)(21) (0.310) 0.08206 (473 K) 12.0 mol K 79. (D) (a) This reaction is exothermic and thus, conversion of synthesis gas to methane is favored at lower temperatures. Since n gas (1 1) (1 3) 2 , high pressure favors the products. (b) The value of Kc is a large number, meaning that almost all of the reactants are converted to products (note that the reaction is stoichiometrically balanced). Thus, after we set up the initial conditions we force the reaction to products and then allow the system to reach equilibrium. Equation: 3 H 2 (g) Initial: Initial: 3.00 mol 15.0 L 0.200 M CO(g) 1.00 mol 15.0 L 0.0667 M 0.000 M xM xM CH 4 (g) H 2 O(g) 0M 0M 0M 0M To right: 0.000 M Changes: 3x M Equil: 3x M 0.0667 M 0.0667 M x M xM (0.0667 x)M (0.0667 x)M
190. 0.0667 x 27 x 2 [CH 4 ][H 2 O] (0.0667 x) 2 190. Kc [H 2 ]3 [CO] (3 x)3 x 190 27 x 2 0.0667 x 71.6 x 2 x 71.6 x 2 x 0.0667 0 b b 2 4ac 1.00 1.00 19.1 0.0244 M 2a 143 [CH 4 ] [H 2 O] 0.0667 0.0244 0.0423 M [H 2 ] 3 0.0244 0.0732 M [CO] 0.0244 M We check our calculation by computing the value of the equilibrium constant. (0.0423) 2 Kc 187 [H 2 ] 3 [CO] (0.0732) 3 0.0244 [CH 4 ] [H 2 O] 705 Chapter15: Principles of Chemical Equilibrium Now we compute the amount in moles of each component present at equilibrium, and finally the mole fraction of CH4. amount CH 4 amount H 2O 0.0423 M 15.0 L 0.635 mol amount H 2 0.0732 M 15.0 L 1.10 mol
amount CO 0.0244 M 15.0 L 0.366 mol CH 4 0.635 mol 0.232 0.635 mol 0.635 mol 1.10 mol 0.366 mol 80. (M) We base our calculation on 1.00 mole of PCl5 being present initially. Equation: PCl5 (g) PCl3 (g) Initial: 1.00 mol 0M Changes: mol mol Equil: (1.00 ) mol mol ntotal 1.00 1.00 Cl2 (g) 0M mol mol Equation: PCl5 (g) Mol fract. 1.00 1.00 PCl3 (g) Cl2 (g) 1.00 1.00 2 P P{Cl2 } P{PCl3 } [ {Cl2 }Ptotal ] [{PCl3 }Ptotal ] 1.00 total 2 Ptotal 2 Ptotal Kp 1.00 P{PCl5 } [ {PCl5 }Ptotal ] (1.00 )(1.00 ) 1 2 Ptotal 1.00 81. (M) We assume that the entire 5.00 g is N2O4 and reach equilibrium from this starting point. 1 mol N 2 O 4 5.00 g [ N 2 O 4 ]i 0.109 M 0.500 L 92.01 g N 2 O 4 2 NO2 (g) Equation: N2O4 (g) Initial: 0.109 0M + 2x M Changes: x M 2x M Equil: (0.0109 x) M KC [NO 2 ]2 (2 x) 2 4.61103 [N 2 O 4 ] 0.109 x 4x 2 5.02 104 4.61103 x 4 x 2 0.00461 x 0.000502 0
x b b 2 4ac 0.00461 2.13 105 8.03 103 0.0106 M, 0.0118 M 2a 8 706 Chapter15: Principles of Chemical Equilibrium (The method of successive approximations yields 0.0106 after two iterations) amount N 2 O 4 0.500 L (0.109 0.0106) M 0.0492 mol N 2O 4 amount NO 2 0.500 L 2 0.0106 M 0.0106 mol NO 2 mol fraction NO 2 82. 0.0106 mol NO 2 0.177 0.0106 mol NO 2 0.0492 mol N 2 O 4 (M) We let P be the initial pressure in atmospheres of COCl2(g). Equation: COCl 2 (g) CO(g) Cl2 (g) Initial: Changes: Equil: P x Px 0M x x 0M x x P 3.00 x Total pressure 3.00 atm P x x x P x P{COCl 2 } P x 3.00 x x 3.00 2x Kp P{CO}P{Cl2 } x x 0.0444 P(COCl2 } 3.00 2x x 2 0.133 0.0888 x x 2 0.0888 x 0.133 0 x
b b 2 4ac 0.0888 0.00789 0.532 0.323, 0.421 2a 2 Since a negative pressure is physically meaningless, x = 0.323 atm. (The method of successive approximations yields x = 0.323 after four iterations.) P{CO} P{Cl2 } 0.323 atm P{COCl2 } 3.00 2 0.323 2.35 atm The mole fraction of each gas is its partial pressure divided by the total pressure. And the contribution of each gas to the apparent molar mass of the mixture is the mole fraction of that gas multiplied by the molar mass of that gas.
M avg P{CO} P{Cl2 } P{COCl2 } M {CO} M {Cl 2 } M {COCl2 } Ptot Ptot Ptot 2.32 atm 0.323 atm 0.323 atm 28.01 g/mol 70.91 g/mol 98.92 g/mol 3.00 atm 3.00 atm 3.00 atm 87.1 g/mol 707 Chapter15: Principles of Chemical Equilibrium 83. (M) Each mole fraction equals the partial pressure of the substance divided by the total pressure. Thus {NH 3 } P{NH 3 } / Ptot or P{NH 3 } {NH 3 } Ptot
Kp P{NH 3 } 2 P{N 2 } P{H 2 }3 ( {NH 3 }Ptot ) 2 ( {N 2 }Ptot ) ( {H 2 }Ptot ) 3 {NH 3 }2 ( Ptot ) 2 {N 2 } {H 2 }3 ( Ptot ) 4 {NH 3 } 2 1 3 {N 2 } {H 2 } ( Ptot ) 2 This is the expression we were asked to derive.
84. (D) Since the mole ratio of N2 to H2 is 1:3, {H 2 } 3 {N 2 } . Since Ptot = 1.00 atm, it follows. {NH 3 } 2 1 Kp 9.06 10 2 0.0906 3 2 {N 2 } (3 {N 2 }) (1.00)
3 3 0.0906 {NH 3 } 2 {NH 3 } 2 {N 2 } {N 2 }3 {N 2 } 4 {NH 3 } 3 3 0.0906 1.56 2 {N 2 } We realize that {NH 3 } {N 2 } {H 2 } 1.00 {NH 3 } {N 2 } 3 {N 2 } And we have This gives {NH 3 } 1.00 4 {N 2 }
1.56 1.00 4 {N 2 } For ease of solving, we let x {N 2 } {N 2 }2 1.00 4 x 1.56 1.56 x 2 1.00 4 x 1.56 x 2 4 x 1.00 0 2 x b b 2 4 ac 4.00 16.00 6.24 0.229, 2.794 x 2a 3.12 Thus {N 2 } 0.229 Mole % NH3 = (1.000 mol (4 0.229 mol)) 100% = 8.4% 85. (M) Since the initial mole ratio is 2 H2S(g) to 1 CH4(g), the reactants remain in their stoichiometric ratio when equilibrium is reached. Also, the products are formed in their stoichiometric ratio. 1 mol CH 4 4.77 103 mol CH 4 amount CH 4 9.54 103 mol H 2 S 2 mol H 2S 1 mol CS2 1 mol S amount CS2 1.42 103 mol BaSO4 7.10 104 mol CS2 1 mol BaSO 4 2 mol S 4 mol H 2 2.84 103 mol H 2 amount H 2 7.10 104 mol CS2 1 mol CS2
total amount 9.54 103 mol H 2 S 4.77 103 mol CH 4 7.10 104 mol CS2 2.84 103 mol H 2 17.86 103 mol The partial pressure of each gas equals its mole fraction times the total pressure. 708 Chapter15: Principles of Chemical Equilibrium 9.54 103 mol H 2S P{H 2S} 1.00 atm 0.534 atm 17.86 103 mol total 4.77 103 mol CH 4 P{CH 4 } 1.00 atm 0.267 atm 17.86 103 mol total 7.10 104 mol CS2 P{CS2 } 1.00 atm 0.0398 atm 17.86 103 mol total 2.84 103 mol H 2 P{H 2 } 1.00 atm 0.159 atm 17.86 103 mol total P{H 2 }4 P{CS2 } 0.1594 0.0398 Kp 3.34 104 2 2 P{H 2S} P{CH 4 } 0.534 0.267
86. (D) 81. We base our calculation on an I.C.E. the direction of the reaction by computing : table, after we first determine (0.03000) 2 (0.03000) 2 Qc 6.48 10 6 2 2 3 2 (0.5000) (0.5000) [Fe ] [Hg 2 ] Because this value is smaller than Kc, the reaction will shift to the right to reach equilibrium. Since the value of the equilibrium constant for the forward reaction is quite small, let us assume that the reaction initially shifts all the way to the left (line labeled "to left:"), and then reacts back in the forward direction to reach a position of equilibrium.
Equation: 2 Fe3 (aq) Initial: To left: 0.5000 M 0.03000 M 0.5300 M Hg 2 2 (aq) 0.5000 M 0.0150 M 0.5150 M
2 2 Fe (aq) 0.03000 M [Fe 2 ] 2 [Hg 2 ] 2 2 Hg 2 (aq) 0.03000 M 0.03000 M 0 M 2x M 2x M Changes: 2 x M x M Equil: (0.5300 2 x)M (0.5150 x)M Kc 0.03000 M 0 M 2x M 2x M [Fe 2 ]2 [Hg 2 ]2 (2 x) 2 (2 x) 2 4 x 2 4x 2 9.14 106 [Fe3 ]2 [Hg 2 2 ] (0.5300 2 x) 2 (0.5150 x) (0.5300) 2 0.5150 Note that we have assumed that 2 x 0.5300 and x 0.5150 9.14 10 6 (0.5300) 2 (0.5150) 8.26 10 8 x 0.0170 4 4 Our assumption, that 2 x ( 0.0340) 0.5300 , is reasonably good. x4 [Fe3 ] 0.5300 2 0.0170 0.4960 M [Fe 2 ] [Hg 2 ] 2 0.0170 0.0340 M [Hg 2 2 ] 0.5150 0.0170 0.4980 709 Chapter15: Principles of Chemical Equilibrium We check by substituting into the K c expression. 9.14 106 K c [Fe 2 ]2 [Hg 2 ]2 (0.0340) 2 (0.0340) 2 11106 Not a substantial difference. 3 2 2 2 [Fe ] [Hg 2 ] (0.4960) 0.4980 Mathematica (version 4.0, Wolfram Research, Champaign, IL) gives a root of 0.0163.
87. (D) Again we base our calculation on an I.C.E. table. In the course of solving the Integrative Example, we found that we could obtain the desired equation by reversing equation (2) and adding the result to equation (1)
(2) (1)
(2) (1) H 2 O(g) CO(g) CH 4 (g) H 2 O(g) 3 H 2 (g) CO(g) CO 2 (g) H 2 (g) K 1/190 K 1.4 Equation:CH 4 (g)
H 2 O(g) CO(g) 2 H 2 O(g) CO 2 (g) 4 H 2 (g) K 1.4 /190 0.0074 CH 4 (g) CO(g) 3 H 2 (g) K 1/190 CO 2 (g) CO 2 (g) 0.025 mol 0.075 mol 0.015 M x mol (0.015 x) M H 2 (g) K 1.4 4 H 2 (g) K 1.4 /190 0.0074 0.100 mol 0.100 mol 0.000 mol 0.000 M 4 x mol 4x mol H 2 O(g) Equation: CH 4 (g) 2 H 2 O(g) Initial: 0.100 mol 0.100 mol To left: Concns: 0.025 mol 0.125 mol 0.0250 M 0.050 mol 0.150 mol 0.0300 M 0.100 mol Changes: x M 2 x mol Equil: (0.0250 x) M (0.0300 2 x) Notice that we have a fifth order polynomial to solve. Hence, we need to try to approximate its final solution as closely as possible. The reaction favors the reactants because of the small size of the equilibrium constant. Thus, we approach equilibrium from as far to the left as possible.
K c 0.0074 x4 [CO 2 ][H 2 ] 4 [CH 4 ][H 2 O] 2 (0.0150 x)(4 x) 4 0.0150 (4 x) 4 (0.0250 x) (0.0300 2 x) 2 0.0250 (0.0300) 2 0.0250 (0.0300) 2 0.0074 0.014 M 0.0150 256 Our assumption is terrible. We substitute to continue successive approximations. (0.0150 0.014) (4x) 4 (0.029)(4 x) 4 (0.0250 0.014) (0.0300 2 0.014) 2 (0.011)(0.002) 2 Next, try x2 = 0.0026 0.0074 0.074 (0.0150 0.0026)(4x)4 (0.0250 0.0026) (0.0300 2 0.0026)2 710 Chapter15: Principles of Chemical Equilibrium then, try x3 = 0.0123. After 18 iterations, the x value converges to 0.0080. Considering that the equilibrium constant is known to only two significant figures, this is a pretty good result. Recall that the total volume is 5.00 L. We calculate amounts in moles. CH 4 (g) H 2 O(g) CO 2 (g) H 2 (g)
88. (0.0250 0.0080) 5.00 L 0.017 M 5.00 L 0.085 moles CH 4 (g) (0.0300 2 0.0080) M 5.00 L 0.014 M 5.00 L 0.070 moles H 2 O(g) (0.015 0.0080) M 5.00 L 0.023 M 5.00 L 0.12 mol CO 2 (4 0.0080) M 5.00 L 0.032 M 5.00 L 0.16 mol H 2 (M) The initial mole fraction of C2H2 is i 0.88 . We use molar amounts at equilibrium to compute the equilibrium mole fraction of C2H2, eq . Because we have a 2.00L container, molar amounts are double the molar concentrations.
(2 0.756) mol C2 H 2 0.877 (2 0.756) mol C2 H 2 (2 0.038) mol CH 4 (2 0.068) mol H 2 Thus, there has been only a slight decrease in mole fraction. eq 89. (M) (a) Keq = 4.6104
P{NO} =
P{NOCl}2 (4.125) 2 = P{NO}2 P{Cl2 } P{NO}2 (0.1125) (4.125) 2 = 0.0573 atm 4.6 104 (0.1125) (b) Ptotal = PNO+ PCl2 + PNOCl = 0.0573 atm + 0.1125 atm + 4.125 atm = 4.295 atm 90. (M) We base our calculation on an I.C.E. table. Reaction: Initial: Change Equilibrium N 2 (g) 0.424 mol 10.0 L  x mol 10.0 L (0.424x) mol 10.0 L + 3H 2 (g) 1.272 mol 10.0 L 3x mol 10.0 L (1.2723x) mol 10.0 L 2NH 3 (g) 0 mol 10.0 L +2x mol 10.0 L 2x mol 10.0 L 711 Chapter15: Principles of Chemical Equilibrium 2x mol 2 [NH 3 ] 10.0 L Kc = =152 = 3 [N 2 ][H 2 ]3 (0.424x) mol (1.2723x) mol 10.0 L 10.0 L 2 (0.424x) mol 3 0.424x) mol 2 2 100 2x mol 2x mol Kc = 3 = 152 4 4 3 0.424x) mol 0.424x) mol 3 Kc = 100 2x mol 2 = 41.04 Take root of both sides 2x mol 0.424x) mol 2 = 6.41 6.41(0.424x) 2 = 2x 3.20(0.180 0.848 x x 2 ) x 3.20 x 2 2.71x 0.576 3.20 x 2 3.71x 0.576 0 Now solve using the quadratic equation: x = 0.1846 mol or 0.9756 mol (too large) amount of NH 3 2 x 2(0.1846 mol) = 0.369 mol in 10.0 L or 0.0369 M ([H 2 ] = 0.0718 M and [ N 2 ] 0.0239 M ) 91. (D) Equation: 2 H 2 (g) K p K c (RT)
n CO(g) CH3 OH(g) K c 14.5 at 483 K
2 Latm 14.5 0.08206 483K molK 9.23 103 We know that mole percents equal pressure percents for ideal gases. P
CO 0.350 100 atm 35.0 atm CO(g) P H 2 0.650 100 atm 65.0 atm Equation: 2 H 2 (g) CH 3 OH(g) Initial: 65 atm 35 atm Changes: 2P atm P atm P atm 652P 35P P Equil: P CH 3 OH P 9.23 103 Kp 2 2 (35.0 P)(65.0 2P) P CO P H 2 By successive approximations, P 24.6 atm = PCH3OH at equilibrium.
Mathematica (version 4.0, Wolfram Research, Champaign, IL) gives a root of 24.5. 712 Chapter15: Principles of Chemical Equilibrium FEATURE PROBLEMS
92. (M) We first determine the amount in moles of acetic acid in the equilibrium mixture. 0.1000 mol Ba OH 2 2 mol CH 3CO 2 H 1L amount CH 3CO 2 H = 28.85 mL 1000 mL 1L 1 mol Ba OH 2 complete equilibrium mixture = 0.5770 mol CH 3CO 2 H 0.01 of equilibrium mixture 0.423mol 0.423mol CH3 CO2 C2 H5 H 2 O = V V 0.423 0.423 4.0 Kc C2 H5 OH CH3 CO2 H 0.077 mol 0.577 mol 0.077 0.577 V V 93. (D) In order to determine whether or not equilibrium has been established in each bulb, we need to calculate the concentrations for all three species at the time of opening. The results from these calculations are tabulated below and a typical calculation is given beneath this table. Bulb No. 1 2 3 4 5 Time Initial Amount of I2(g) Amount HI(g) [HI] [I2] & and H2(g) at at Time of (mM) [H2] Bulb Amount Time of Opening (mM) Opening Opened HI(g) (in mmol) (in mmol) (hours) (in mmol) 2 2.345 0.1572 2.03 5.08 0.393 4 2.518 0.2093 2.10 5.25 0.523 12 2.463 0.2423 1.98 4.95 0.606 20 3.174 0.3113 2.55 6.38 0.778 40 2.189 0.2151 1.76 4.40 0.538 [H 2 ][I 2 ] [HI]2 0.00599 0.00992 0.0150 0.0149 0.0150 Consider, for instance, bulb #4 (opened after 20 hours). 1 mole HI = 0.003174 mol HI(g) or 3.174 mmol Initial moles of HI(g) = 0.406 g HI(g) 127.9 g HI moles of I2(g) present in bulb when opened. 1 mol I 2 0.0150 mol Na 2S2O3 = 0.04150 L Na2S2O3 = 3.113 104 mol I 2 2 mol Na 2 S 2 O 3 1 L Na 2S2 O3 millimoles of I2(g) present in bulb when opened = 3.113 104 mol I 2 moles of H2 present in bulb when opened = moles of I2(g) present in bulb when opened. 2 mole HI = 6.226 104 mol HI (0.6226 mmol HI) 1 mol I 2 moles of HI(g) in bulb when opened= 3.174 mmol HI 0.6226 mmol HI = 2.55 mmol HI HI reacted = 3.113 104 mol I2 Concentrations of HI, I2, and H2 713 Chapter15: Principles of Chemical Equilibrium [HI] = 2.55 mmol HI 0.400 L = 6.38 mM [I2] = [H2] = 0.3113 mmol 0.400 L = 0.778 mM [H 2 ][I 2 ] (0.778 mM)(0.778 mM) = 0.0149 Ratio: [HI]2 (6.38 mM) 2 As the time increases, the ratio [H 2 ][I 2 ] initially climbs sharply, but then plateaus at [HI]2 0.0150 somewhere between 4 and 12 hours. Consequently, it seems quite reasonable to conclude that the reaction 2HI(g) H2(g) + I2(g) has a Kc ~ 0.015 at 623 K.
94. (D) We first need to determine the number of moles of ammonia that were present in the sample of gas that left the reactor. This will be accomplished by using the data from the titrations involving HCl(aq). Original number of moles of HCl(aq) in the 20.00 mL sample = 0.01872 L of KOH 0.0523 mol KOH 1 mol HCl 1 L KOH 1 mol KOH = 9.7906 104 moles of HCl(initially) Moles of unreacted HCl(aq) 0.0523 mol KOH 1 mol HCl = 1 L KOH 1 mol KOH 8.0647 104 moles of HCl(unreacted) = 0.01542 L of KOH Moles of HCl that reacted and /or moles of NH3 present in the sample of reactor gas = 9.7906 104 moles 8.0647 104 moles = 1.73 104 mole of NH3 (or HCl). The remaining gas, which is a mixture of N2(g) and H2(g) gases, was found to occupy 1.82 L at 273.2 K and 1.00 atm. Thus, the total number of moles of N2 and H2 can be found via the ideal gas law: nH 2 N2 =
PV (1.00 atm)(1.82 L) = = 0.08118 moles of (N2 + H2) RT (0.08206 L atm )(273.2 K) K mol According to the stoichiometry for the reaction, 2 parts NH3 decompose to give 3 parts H2 and 1 part N2. Thus the nonreacting mixture must be 75% H2 and 25% N2. So, the number of moles of N2 = 0.25 0.08118 moles = 0.0203 moles N2 and the number of moles of H2 = 0.75 0.08118 moles = 0.0609 moles H2. 714 Chapter15: Principles of Chemical Equilibrium Before we can calculate Kc, we need to determine the volume that the NH3, N2, and H2 molecules occupied in the reactor. Once again, the ideal gas law (PV = nRT) will be employed. ngas = 0.08118 moles (N2 + H2 ) + 1.73 104 moles NH3 = 0.08135 moles
nRT Vgases = = P (0.08135 mol)( 0.08206 L atm )(1174.2 K) K mol = 0.2613 L 30.0 atm
2 So, Kc = 1.73 104 moles 0.2613 L 3 0.0609 moles 0.0203 moles 0.2613 L 0.2613 L 1 = 4.46 104 To calculate Kp at 901 C, we need to employ the equation K p K c RT n gas ; ngas 2 Kp = 4.46 104 [(0.08206 L atm K1mol1)] (1174.2 K)]2 = 4.80 108 at 901C for the reaction N2(g) + 3 H2(g) 2 NH3(g)
95. (M) For step 1, rate of the forward reaction = rate of the reverse reaction, so, k1 [I]2 = = Kc (step 1) k1 [I 2 ] Like the first step, the rates for the forward and reverse reactions are equal in the second step and thus, k [HI]2 = Kc (step 2) k2[I]2[H2] = k2[HI]2 or 2 = 2 k2 [I] [H 2 ] k1[I2] = k1[I]2 or Now we combine the two elementary steps to obtain the overall equation and its associated equilibrium constant. k [I]2 Kc = 1 = (STEP 1) I2(g) 2 I(g) k1 [I 2 ] and k2 [HI]2 2 HI(g) (STEP 2) H2(g) + 2 I(g) Kc = = k2 [I]2 [H 2 ] H2(g) + I2(g) 2 HI(g) Kc(overall) = Kc(step 1) Kc(step 2) Kc(overall) = k k1 [HI]2 [I]2 2 = 2 [I 2 ] [I] [H 2 ] k1 k 2 k1k2 [I]2 [HI]2 [HI]2 Kc(overall) = = 2 = [I] [I 2 ][H 2 ] [I 2 ][H 2 ] k1k2 715 Chapter15: Principles of Chemical Equilibrium 96. (M) The equilibrium expressions for the two reactions are: 2 H HCO3 ; K H CO3 K1 2 H 2 CO3 HCO3 First, start with [H+] = 0.1 and [HCO3] = 1. This means that [H+]/[HCO3] = 0.1, which means that [CO32] = 10K2. By adding a small amount of H2CO3 we shift [H+] by 0.1 and [HCO3] by 0.1. This leads to [H+]/[HCO3] 0.2, which means that [CO32] = 5K2. Note that [CO32] has decreased as a result of adding H2CO3 to the solution.
97. (D) First, it is most important to get a general feel for the direction of the reaction by determining the reaction quotient: Q C(aq) 0.1 10 A(aq) B(aq) 0.1 0.1 Since Q>>K, the reaction proceeds toward the reactants. Looking at the reaction in the aqueous phase only, the equilibrium can be expressed as follows: Initial Change Equil. A(aq) 0.1 x 0.1  x + B(aq) 0.1 x 0.1  x C(aq) 0.1 +x 0.1 + x We will do part (b) first, which assumes the absence of an organic layer for extraction: K 0.1 x 0.01 0.1 x 0.1 x Expanding the above equation and using the quadratic formula, x = 0.0996. Therefore, the concentration of C(aq) and equilibrium is 0.1 + (0.0996) = 4104 M. If the organic layer is present for extraction, we can add the two equations together, as shown below: A(aq) + B(aq) C(aq) A(aq) + B(aq) K = K1 K2 = 0.1 15 = 0.15. Since the organic layer is present with the aqueous layer, and K2 is large, we can expect that the vast portion of C initially placed in the aqueous phase will go into the organic phase. C(aq) C(or) C(or) 716 Chapter15: Principles of Chemical Equilibrium Therefore, the initial [C] = 0.1 can be assumed to be for C(or). The equilibrium can be expressed as follows A(aq) 0.1 x 0.1  x + B(aq) 0.1 x 0.1  x Initial Change Equil. C(or) 0.1 +x 0.1 + x We will do part (b) first, which assumes the absence of an organic layer for extraction: K 0.1 x 0.15 0.1 x 0.1 x Expanding the above equation and using the quadratic formula, x = 0.0943. Therefore, the concentration of C(or) and equilibrium is 0.1 + (0.0943) = 6104 M. This makes sense because the K for the overall reaction is < 1, which means that the reaction favors the reactants. SELFASSESSMENT EXERCISES
98. (E) (a) Kp: The equilibrium constant of a reaction where the pressures of gaseous reactants and products are used instead of their concentrations (b) Qc: The reaction quotient using the molarities of the reactants and products (c) ngas: The difference between the number of moles (as determined from a balanced reaction) of product and reactant gases (E) (a) Dynamic equilibrium: In a dynamic equilibrium (which is to say, real equilibrium), the forward and reverse reactions happen, but at a constant rate (b) Direction of net chemical change: In a reversible reaction, if the reaction quotient Qc > Kc, then the net reaction will go toward the reactants, and vice versa (c) Le Chtelier's principle: When a system at equilibrium is subjected to external change (change in partial pressure of reactants/products, temperature or concentration), the equilibrium shifts to a side to diminish the effects of that external change (d) Effect of catalyst on equilibrium: A catalyst does not affect the final concentrations of the reactants and products. However, since it speeds up the reaction, it allows for the equilibrium concentrations to be established more quickly 99. 100. (E) (a) Reaction that goes to completion and reversible reaction: In a reversible reaction, the products can revert back to the reactants in a dynamic equilibrium. In a reaction that goes to completion, the formation of products is so highly favored that there is practically no reverse reaction (or the reverse is practically impossible, such as a combustion reaction). 717
Chapter15: Principles of Chemical Equilibrium (b) Kp and Kc: Kp is the equilibrium constant using pressures of products and reactants, while Kc is the constant for reaction using concentrations. (c) Reaction quotient (Q) and equilibrium constant expression (K): The reaction quotient Q is the ratio of the concentrations of the reactants and products expressed in the same format as the equilibrium expression. The equilibrium constant expression is the ratio of concentrations at equilibrium. (d) Homogeneous and heterogeneous reaction: In a homogeneous reaction, the reaction happens within a single phase (either aqueous or gas). In a heterogeneous reaction, there is more than one phase present in the reaction. 101. (E) The answer is (c). Because the limiting reagent is I2 at one mole, the theoretical yield of HI is 2 moles. However, because there is an established equilibrium, there is a small amount of HI which will decompose to yield H2 and I2. Therefore the total moles of HI created is close, but less than 2. 102. (E) The answer is (d). The equilibrium expression is: K P SO3 2 2 P SO 2 P O 2 100 If equilibrium is established, moles of SO3 and SO2 cancel out of the equilibrium expression. Therefore, if K = 100, the moles of O2 have to be 0.01 to make K = 100.
103. (E) The answer is (a). As the volume of the vessel is expanded (i.e., pressure is reduced), the equilibrium shifts toward the side with more moles of gas. 104. (E) The answer is (b). At half the stoichiometric values, the equilibrium constant is K1/2. If the equation is reversed, it is K1. Therefore, the K' = K1/2 = (1.8106)1/2 = 7.5102. 105. (E) The answer is (a). We know that Kp = Kc (RT)n. Since n = (32) = 1, Kp = Kc (RT). Therefore, Kp > Kc. 106. (E) The answer is (c). Since the number of moles of gas of products is more than the reactants, increasing the vessel volume will drive the equilibrium more toward the product side. The other options: (a) has no effect, and (b) drives the equilibrium to the reactant side. 107. (E) The equilibrium expression is: C 0.432 1.9 K 2 2 B A 0.55 0.33
2 108. (E) (a) As more O2 (a reactant) is added, more Cl2 is produced. (b) As HCl (a reactant) is removed, equilibrium shifts to the left and less Cl2 is made. (c) Since there are more moles of reactants, equilibrium shifts to the left and less Cl2 is made. (d) No change. However, the equilibrium is reached faster. (e) Since the reaction is exothermic, increasing the temperature causes less Cl2 to be made. 718 Chapter15: Principles of Chemical Equilibrium 109. (E) SO2 (g) will be less than SO2 (aq), because K > 1, so the equilibrium lies to the product side, SO2 (aq). 110. (E) Since K >>1, there will be much more product than reactant 111. (M) The equilibrium expression for this reaction is: SO3 35.5 K 2 SO2 O2 2 (a) If [SO3]eq = [SO2]eq, then [O2] = 1/35.5 = 0.0282 M. moles of O2 = 0.0282 2.05 L = 0.0578 moles (b) Plugging in the new concentration values into the equilibrium expression: SO3 2 SO2 4 35.5 K 2 2 SO2 O2 SO2 O2 O2 2 2 [O2] = 0.113 M moles of O2 = 0.113 2.05 L = 0.232 moles
112. (M) This concept map involves the various conditions that affect equilibrium of a reaction, and those that don't. Under the category of conditions that do cause a change, there is changing the partial pressure of gaseous products and reactants, which includes pressure and vessel volume. The changes that do not affect partial pressure are changing the concentration of reactants or products in an aqueous solution, through dilution or concentration. Changing the temperature can affect both aqueous and gaseous reactions. Under the category of major changes that don't affect anything is the addition of a nonreactive gas. 719 CHAPTER 16 ACIDS AND BASES
PRACTICE EXAMPLES
1A (E) (a) In the forward direction, HF is the acid (proton donor; forms F), and H2O is the base (proton acceptor; forms H3O). In the reverse direction, F is the base (forms HF), accepting a proton from H3O, which is the acid (forms H2O). (b) In the forward direction, HSO4 is the acid (proton donor; forms SO42), and NH3 is the base (proton acceptor; forms NH4). In the reverse direction, SO42 is the base (forms HSO4), accepting a proton from NH4, which is the acid (forms NH3). (c) In the forward direction, HCl is the acid (proton donor; forms Cl), and C2H3O2 is the base (proton acceptor; forms HC2H3O2). In the reverse direction, Cl is the base (forms HCl), accepting a proton from HC2H3O2, which is the acid (forms C2H3O2). 1B (E) We know that the formulas of most acids begin with H. Thus, we identify HNO 2 and HCO 3 as acids. HNO 2 aq + H 2O(l) NO 2 aq + H 3O + aq ; 2 HCO3 aq + H 2O(l) CO3 aq + H 3O + aq A negatively charged species will attract a positively charged proton and act as a base. 3 3 Thus PO 4 and HCO 3 can act as bases. We also know that PO 4 must be a base because it cannot act as an acidit has no protons to donateand we know that all three species have acidbase properties. aq + H 2O(l) HPO4 2 aq + OH aq ; HCO3 aq + H 2O(l) H 2 CO3 (aq) CO 2 H 2 O aq + OH aq PO 4 3 Notice that HCO 3 is the amphiprotic species, acting as both an acid and a base.
2A (M) H 3O + is readily computed from pH: H 3O + = 10 pH H 3O + = 102.85 = 1.4 103 M. OH can be found in two ways: (1) from Kw = H 3O + OH , giving
Kw 1.0 1014 12 OH = = H O + 1.4 103 = 7.110 M, or (2) from pH + pOH = 14.00 , giving 3 pOH = 14.00 pH = 14.00 2.85 = 11.15 , and then OH = 10 pOH = 1011.15 = 7.1 1012 M. 720 Chapter 16: Acids and Bases 2B (M) H 3O + is computed from pH in each case: H 3O + = 10 pH
H 3O +
conc . = 102.50 = 3.2 103 M H 3O + dil . = 103.10 = 7.9 10 4 M All of the H 3O + in the dilute solution comes from the concentrated solution. 3.2 103 mol H 3O + = 3.2 103 mol H 3O + 1 L conc. soln Next we calculate the volume of the dilute solution. 1 L dilute soln volume of dilute solution = 3.2 103 mol H 3O + = 4.1 L dilute soln 7.9 104 mol H 3O + Thus, the volume of water to be added is = 3.1 L. Infinite dilution does not lead to infinitely small hydrogen ion concentrations. Since dilution is done with water, the pH of an infinitely dilute solution will approach that of pure water, namely pH = 7. amount H 3O + = 1.00 L conc. soln 3A (E) pH is computed directly from the equation below. H 3O + , pH = log H 3O + = log 0.0025 = 2.60 . We know that HI is a strong acid and, thus, is completely dissociated into H 3O + and I . The consequence is that I = H 3O + = 0.0025 M. OH is most readily computed from pH: pOH = 14.00 pH = 14.00 2.60 = 11.40;
3B OH = 10 pOH = 1011.40 = 4.0 1012 M (M) The number of moles of HCl(g) is calculated from the ideal gas law. Then H 3O + is calculated, based on the fact that HCl(aq) is a strong acid (1 mol H 3O + is produced from each mole of HCl). 1atm 747 mmHg 0.535 L 760 mmHg moles HCl(g) = 0.08206 L atm (26.5 273.2) K mol K moles HCl(g) 0.0214 mol HCl(g) = 0.0214 mol H 3O + when dissolved in water [H 3O + ] 0.0214 mol
4A pH = log(0.0214) = 1.670 (E) pH is most readily determined from pOH = log OH . Assume Mg OH b g 2 is a strong base. OH = 9.63 mg Mg OH 2 1000 mL 1 mol Mg OH 2 1g 2 mol OH 100.0 mL soln 1 L 1000 mg 58.32 g Mg OH 2 1 mol Mg OH 2 OH = 0.00330 M; pOH = log 0.00330 = 2.481 pH = 14.000 pOH = 14.000 2.481 = 11.519 721 Chapter 16: Acids and Bases 4B (E) KOH is a strong base, which means that each mole of KOH that dissolves produces one mole of dissolved OH aq . First we calculate OH and the pOH. We then use b g pH + pOH = 14.00 to determine pH. OH = 3.00 g KOH 1 mol KOH 1 mol OH 1.0242 g soln 1000 mL = 0.548 M 100.00 g soln 56.11 g KOH 1 mol KOH 1 mL soln 1L
pH = 14.000 pOH = 14.000 0.261 = 13.739 pOH = log 0.548 = 0.261 5A (M) H 3O + = 10 pH = 104.18 = 6.6 105 M. Organize the solution using the balanced chemical equation. Equation: Initial: Changes: Equil:
HOCl aq + H 2 O(l)
   0.150 M 6.6 105 M 0.150 M + H 3O + aq 0 M + 6.6 105 M 6.6 105 M b g OCl aq 0M + 6.6 105 M 6.6 105 M b g H 3O + OCl 6.6 105 6.6 105 = Ka 2.9 108 0.150 HOCl
5B (M) First, we use pH to determine OH . pOH = 14.00 pH = 14.00 10.08 = 3.92 . OH = 10 pOH = 10 3.92 = 1.2 104 M. We determine the initial concentration of cocaine and then organize the solution around the balanced equation in the manner we have used before. 0.17 g C17 H 21O 4 N 1000 mL 1mol C17 H 21O 4 N [C17 H 21O 4 N] = = 0.0056 M 100 mL soln 1L 303.36 g C17 H 21O 4 N Equation: Initial: Changes: Equil: Kb C17 H 21O 4 N(aq) + H 2 O(l) 0.0056 M 1.2 104 M 0.0055 M
  
4 C17 H 21O 4 NH + aq 0M +1.2 104 M 1.2 104 M
4 6 b g + OH aq 0 M +1.2 104 M 1.2 104 M b g C17 H 21O 4 NH + OH C17 H 21O 4 N c1.2 10 hc1.2 10 h 2.6 10 0.0055 722 Chapter 16: Acids and Bases 6A (M) Again we organize our solution around the balanced chemical equation. Equation: Initial: Changes: Equil: b HC2 H 2 FO 2 (aq) + H 2 O(l)  0.100 M  x M  0.100 x M g H 3O + (aq) + 0M +x M x M C 2 H 2 FO 2 (aq) 0M +x M x M H 3O + C2 H 2 FO 2 xx ; therefore, 2.6 103 Ka = HC2 H 2 FO 2 0.100 x We can use the 5% rule to ignore x in the denominator. Therefore, x = [H3O+] = 0.016 M, and pH = log (0.016) = 1.8. Thus, the calculated pH is considerably lower than 2.89 (Example 166).
6B (M) We first determine the concentration of undissociated acid. We then use this value in a setup that is based on the balanced chemical equation. 0.500 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 1 2 aspirin tablets = 0.0171M tablet 180.155g HC9 H 7 O 4 0.325 L Equation: HC9 H 7 O 4 (aq) + H2O(l) H 3O + (aq) + C9 H 7 O 4 (aq)  Initial: 0M 0.0171 M 0M  Changes: x M +x M +x M  Equil: x M x M 0.0171 x M b g H 3 O + C9 H 7 O 4 xx = 3.3 104 = Ka = 0.0171 x HC9 H 7 O 4 x 2 + 3.3 104 5.64 106 = 0 (find the physically reasonable roots of the quadratic equation)
3.3 104 1.1 107 2.3 105 x 0.0022 M; 2 pH log (0.0022) 2.66 7A (M) Again we organize our solution around the balanced chemical equation. Equation: HC 2 H 2 FO 2 (aq) + Initial: 0.015 M Changes x M Equil: 0.015 x M H 2 O(l)
   b g H3O+ (aq) + 0M +x M x M C 2 H 2 FO 2 (aq) 0M +x M x M 2 H 3 O + C2 H 2 FO 2 = 2.6 103 = x x x Ka = 0.015 x 0.015 HC2 H 2 FO2 x = x 2 0.015 2.6 103 0.0062 M = [H 3O + ] Our assumption is invalid: 723 Chapter 16: Acids and Bases 0.0062 is not quite small enough compared to 0.015 for the 5% rule to hold. Thus we use another cycle of successive approximations.
Ka = x x 0.015 0.0062 x x = 2.6 103 x = (0.015 0.0048) 2.6 103 0.0051 M = [H O+ ] Ka = 3 0.015 0.0048 = 2.6 103 x = (0.015 0.0062) 2.6 103 0.0048 M = [H 3O + ] Ka = x x 0.015 0.0051 = 2.6 103 x = (0.015 0.0051) 2.6 103 0.0051 M = [H 3O + ] Two successive identical results indicate that we have the solution. pH = log H 3O + = log 0.0051 = 2.29 . The quadratic equation gives the same result (0.0051 M) as this method of successive approximations.
7B (M) First we find [C5H11N]. We then use this value as the starting base concentration in a setup based on the balanced chemical equation. 114 mg C5 H 11N 1 mmol C5 H 11N = 0.00425 M C5 H 11N = 315 mL soln 85.15 mg C5 H 11N C H NH + (aq) + OH (aq) Equation: C5 H11 N(aq) + H 2 O(l) 5 11  0 M 0.00425 M 0M Initial:  +x M +x M x M Changes: x M  x M 0.00425 x M Equil:
C5 H11 NH + OH xx xx = 1.6 103 = Kb = 0.00425 x 0.00425 C5 H11 N b g We assumed that x 0.00425 x 0.0016 0.00425 0.0026 M The assumption is not valid. Let's assume x 0.0026 Let's try again, with x 0.0016 Yet another try, with x 0.0021 The last time, with x 0.0019 x 0.0016 (0.00425 0.0026) 0.0016 x 0.0016 (0.00425 0.0016) 0.0021 x 0.0016 (0.00425 0.0021) 0.0019 x 0.0016 (0.00425 0.0019) 0.0019 M = [OH ] pOH= log[H3 O ] log(0.0019) 2.72 pH = 14.00 pOH = 14.00 2.72 = 11.28 We could have solved the problem with the quadratic formula roots equation rather than by successive approximations. The same answer is obtained. In fact, if we substitute 2 x = 0.0019 into the Kb expression, we obtain 0.0019 / 0.00425 0.0019 = 1.5 103 b g b g compared to Kb = 1.6 10 . The error is due to rounding, not to an incorrect value. Using x = 0.0020 gives a value of 1.8 103 , while using x = 0.0018 gives 1.3 103 . 3 724 Chapter 16: Acids and Bases 8A (M) We organize the solution around the balanced chemical equation; a M is [HF]initial. Equation: HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq)  a M 0M 0 M Initial:  +x M +x M x M Changes:  xM x M a x M Equil: H 3O + F x x x 2 = x = a 6.6 104 = 6.6 104 Ka = ax a HF For 0.20 M HF, a = 0.20 M % dissoc= 0.011 M 100%=5.5% 0.20 M x = 0.20 6.6 104 0.011 M For 0.020 M HF, a = 0.020 M x = 0.020 6.6 104 0.0036 M We need another cycle of approximation: x = (0.020 0.0036) 6.6 104 0.0033 M Yet another cycle with x 0.0033 M : x = (0.020 0.0033) 6.6 104 0.0033 M 0.0033M % dissoc = 100% = 17% 0.020M As expected, the weak acid is more dissociated.
8B (E) Since both H 3O + and C 3 H 5O 3 come from the same source in equimolar amounts, their concentrations are equal. H 3O + = C 3 H 5O 3 = 0.067 0.0284 M = 0.0019 M
Ka = H 3 O + C 3 H 5O 3 LM HC3 H 5O 3 OP Q N = b0.0019gb0.0019g = 1.4 10
0.0284 0.0019 4 9A (M) For an aqueous solution of a diprotic acid, the concentration of the divalent anion is very close to the second ionization constant: OOCCH 2 COO K a 2 = 2.0 106 M . We organize around the chemical equation. Equation: CH 2 COOH (aq) + H 2 O(l) 2 Initial:  1.0 M Change:  x M Equil:  (1.0 x) M
Ka [H 3O ][HCH 2 (COO) 2 ] [CH 2 (COOH) 2 ] H 3 O + (aq) 0M +x M xM + HCH 2 COO 2 (aq) 0M +x M xM ( x)( x) x 2 1.4 103 1.0 x 1.0 x 1.4 103 3.7 102 M [H3 O+ ] [HOOCCH 2 COO ] x 1.0M is a valid assumption. 725 Chapter 16: Acids and Bases 9B (M) We know K a 2 doubly charged anion for a polyprotic acid. Thus, K a 2 = 5.3 105 C2 O 4 2 . From the pH, H 3O + = 10 pH = 100.67 = 0.21 M . We also recognize that HC 2 O 4 = H 3O + , since the second ionization occurs to only a very small extent. We note as well that HC 2 O 4 is produced by the ionization of H 2 C 2 O 4 . Each mole of HC 2 O 4 present results from the ionization of 1 mole of H 2 C2 O 4 . Now we have sufficient information to determine the K a1 . H 3O + HC2O 4 0.21 0.21 = K a1 = = 5.3 102 1.05 0.21 H 2 C2 O4 10A (M) H 2SO 4 is a strong acid in its first ionization, and somewhat weak in its second, with K a 2 = 1.1102 = 0.011 . Because of the strong first ionization step, this problem involves determining concentrations in a solution that initially is 0.20 M H 3O + and 0.20 M HSO 4 . We base the setup on the balanced chemical equation. Equation: Initial: Changes: Equil:
HSO 4 aq + H 2 O (l)  0.20 M  x M b g H 3O + aq 0.20 M +x M b g
M + 0.20 x M  0.20 + x SO 4 (aq) 0M +x M x M 2 H 3O + SO 4 2 0.20 + x x 0.20 x = = 0.011 , assuming that x 0.20M. Ka2 = 0.20 x 0.20 HSO 4 Try one cycle of approximation: x = 0.011M
0.011 b0.20 + 0.011gx = 0.21x b0.20 0.011g 0.19 x= 0.19 0.011 = 0.010 M 0.21 2 The next cycle of approximation produces the same answer 0.010M = SO 4 , HSO 4 = 0.20 0.010 M = 0.19 M H 3O + = 0.010 + 0.20 M = 0.21 M, 10B (M) We know that H 2SO 4 is a strong acid in its first ionization, and a somewhat weak acid in its second, with K a 2 = 1.1 102 = 0.011 . Because of the strong first ionization step, the problem essentially reduces to determining concentrations in a solution that initially is 0.020 M H 3O + and 0.020 M HSO 4 . We base the setup on the balanced chemical equation. The result is solved using the quadratic equation. 726 Chapter 16: Acids and Bases Equation: Initial: Changes: Equil: HSO 4 aq + H 2 O (l)  0.020 M  x M 0.020 x M  b g H 3O + aq 0.020 M +x M 0.020 + x M b g + SO 4 aq 0M +x M x M 2 b g H 3O + SO 4 2 0.020 + x x = = 0.011 Ka2 = 0.020 x HSO 4 x 2 + 0.031x 0.00022 = 0 x= 0.020 x + x 2 = 2.2 104 0.011x 0.031 0.00096 + 0.00088 2 = 0.0060 M = SO 4 2 HSO 4 = 0.020 0.0060 = 0.014 M H 3O + = 0.020 + 0.0060 = 0.026 M (The method of successive approximations converges to x = 0.006 M in 8 cycles.)
11A (E) (a) CH 3 NH 3 NO 3 is the salt of the cation of a weak base. The cation, CH 3 NH 3+ , will
+ hydrolyze to form an acidic solution CH 3 NH 3 + H 2 O CH 3 NH 2 + H 3O + , while + NO3 , by virtue of being the conjugate base of a strong acid will not hydrolyze to a detectable extent. The aqueous solutions of this compound will thus be acidic.
(b) NaI is the salt composed of the cation of a strong base and the anion of a strong acid, neither of which hydrolyzes in water. Solutions of this compound will be pH neutral. (c) NaNO 2 is the salt composed of the cation of a strong base that will not hydrolyze in water and the anion of a weak acid that will hydrolyze to form an alkaline solution NO 2 + H 2 O HNO 2 + OH . Thus aqueous solutions of this compound will be basic (alkaline).
11B (E) Even without referring to the K values for acids and bases, we can predict that the reaction that produces H 3O + occurs to the greater extent. This, of course, is because the pH is less than 7, thus acid hydrolysis must predominate. We write the two reactions of H 2 PO 4 with water, along with the values of their equilibrium constants. H 2 PO 4 aq + H 2 O(l) H 3O + aq + HPO 4 2 aq K a 2 = 6.3 108 H 2 PO 4 aq + H 2 O(l) OH aq + H 3 PO4 aq Kb = K w 1.0 1014 = = 1.4 1012 3 K a1 7.1 10 As predicted, the acid ionization occurs to the greater extent. 727 Chapter 16: Acids and Bases 12A (M) From the value of pKb we determine the value of Kb and then Ka for the cation. cocaine: Kb = 10 pK = 10 8.41 = 3.9 10 9 Kw 1.0 1014 = 2.6 106 Ka = = 9 Kb 3.9 10 Ka = Kw 1.0 1014 = 9.1 107 = 8 Kb 1.1 10 codeine: Kb = 10 pK = 107.95 = 1.1 108 (This method may be a bit easier: pKa = 14.00 pKb = 14.00 8.41 = 5.59, Ka = 105.59 = 2.6 106 ) The acid with the larger Ka will produce the higher H + , and that solution will have the lower pH. Thus, the solution of codeine hydrochloride will have the higher pH (i.e., codeine hydrochloride is the weaker acid).
12B (E) Both of the ions of NH 4 CN aq react with water in hydrolysis reactions.
+ NH 4 aq + H 2 O(l) NH3 aq + H 3O + aq b g Ka = K w 1.0 1014 = = 5.6 1010 K b 1.8 105 CN aq + H 2 O(l) HCN aq + OH aq K w 1.0 1014 = = 1.6 105 Kb = 10 K a 6.2 10 Since the value of the equilibrium constant for the hydrolysis reaction of cyanide ion is larger than that for the hydrolysis of ammonium ion, the cyanide ion hydrolysis reaction will proceed to a greater extent and thus the solution of NH 4 CN aq will be basic (alkaline). b g 13A (M) NaF dissociates completely into sodium ions and fluoride ions. The released fluoride ion hydrolyzes in aqueous solution to form hydroxide ion. The determination of the equilibrium pH is organized around the balanced equation. Equation: F (aq) + H 2 O(l) HF(aq) + OH (aq)  Initial: 0M 0M 0.10 M  Changes: +x M +x M x M Equil: xM xM 0.10 x M 
K w 1.0 1014 [HF][OH ] ( x)( x) x2 11 Kb 1.5 10 K a 6.6 104 [F ] (0.10 x) 0.10
x = 0.10 1.5 1011 1.2 106 M=[OH ]; pOH= log(1.2 106 ) 5.92 pH = 14.00 pOH = 14.00 5.92 = 8.08 (As expected, pH > 7) 728 Chapter 16: Acids and Bases 13B (M) The cyanide ion hydrolyzes in solution, as indicated in Practice Example 1612B. As a consequence of the hydrolysis, OH = HCN . OH can be found from the pH of the solution, and then values are substituted into the Kb expression for CN , which is then solved for CN . pOH = 14.00 pH = 14.00 10.38 = 3.62 OH = 10pOH = 103.62 = 2.4 104 M = HCN Kb =
HCN OH CN = 1.6 10 5 c2.4 10 h =
CN 4 2 CN c2.4 10 h =
1.6 10
5 4 2 = 3.6 103 M 14A (M) First we draw the Lewis structures of the four acids. Lone pairs have been omitted since we are interested only in the arrangements of atoms.
O H O N O O H O Cl O O F H O C H C O H Br H O C C O H H HClO 4 should be stronger than HNO 3 . Although Cl and N have similar electronegativities, there are more terminal oxygen atoms attached to the chlorine in perchloric acid than to the nitrogen in nitric acid. By virtue of having more terminal oxygens, perchloric acid, when ionized, affords a more stable conjugate base. The more stable the anion, the more easily it is formed and hence the stronger is the conjugate acid from which it is derived. CH 2 FCOOH will be a stronger acid than CH 2 BrCOOH because F is a more electronegative atom than Br. The F atom withdraws additional electron density from the O  H bond, making the bond easier to break, which leads to increased acidity.
14B (M) First we draw the Lewis structures of the first two acids. Lone pairs are not depicted since we are interested in the arrangements of atoms. O H O P O H H O O S O H O H than P. This suggests that H 2SO3 K a1 = 1.3 102 should be a stronger acid than H 3 PO 4 and H 2SO 3 both have one terminal oxygen atom, but S is more electronegative H 3PO 4 K a1 = 7.1103 , and it is. The only difference between CCl 3CH 2 COOH and CCl 2 FCH 2 COOH is the replacement of Cl by F. Since F is more electronegative than Cl, CCl 3CH 2 COOH should be a weaker acid than CCl 2 FCH 2 COOH . 729 Chapter 16: Acids and Bases 15A (M) We draw Lewis structures to help us decide. (a) Clearly, BF3 is an electron pair acceptor, a Lewis acid, and NH 3 is an electron pair donor, a Lewis base.
F F H F F H B N H F H O H H B N H F H (b) H 2 O certainly has electron pairs (lone pairs) to donate and thus it can be a Lewis base. It is unlikely that the cation Cr 3+ has any accessible valence electrons that can be donated to another atom, thus, it is the Lewis acid. The product of the reaction, [Cr(H2O)6]3+, is described as a water adduct of Cr3+.
OH2 Cr3+(aq) + 6 O H H H2 O H2 O Cr OH2 OH2 OH2 3+ 15B (M) The Lewis structures of the six species follow.
O H O + Al O H H O H H O H O Al O H O H 2 Cl Cl Cl Sn Cl Cl Cl Cl Cl Cl Sn Cl
2 2 Cl Both the hydroxide ion and the chloride ion have lone pairs of electrons that can be donated to electronpoor centers. These two are the electron pair donors, or the Lewis bases. Al OH 3 and SnCl 4 have additional spaces in their structures to accept pairs of b g electrons, which is what occurs when they form the complex anions [Al(OH)4] and 2[SnCl6] . Thus, Al OH 3 and SnCl 4 are the Lewis acids in these reactions.  b g 730 Chapter 16: Acids and Bases INTEGRATIVE EXAMPLE
A. (D) To confirm the pH of rainwater, we have to calculate the concentration of CO2(aq) in water and then use simple acidbase equilibrium to calculate pH. Concentration of CO2 in water at 1 atm pressure and 298 K is 1.45 g/L, or
1.45 g CO 2 1 mol CO 2 0.0329 M CO 2 L 44.0 g CO 2 Furthermore, if the atmosphere is 0.037% by volume CO2, then the mole fraction of CO2 is 0.00037, and the partial pressure of CO2 also becomes 0.00037 atm (because PCO2 CO2 Patm ) From Henry's law, we know that concentration of a gas in a liquid is proportional to its partial pressure. C(mol / L) H PCO2 , which can rearrange to solve for H: H 0.0329 M 0.0329 mol L1 atm 1 1 atm Therefore, concentration of CO2(aq) in water under atmospheric pressures at 298 K is:
CCO2 (mol / L) 0.0329 mol L1 atm 1 0.00037 atm = 1.217 105 M Using the equation for reaction of CO2 with water: CO2 (aq) + 5 1.21710 x 1.217105x 2 H2O (l) H3O+ (aq) + HCO3 (aq) 0 0 +x +x x X x2 K a1 4.4 107 5 1.217 10 x Using the quadratic formula, x = 2.104 106 M pH log H 3O log 2.104 106 5.68 5.7 We can, of course, continue to refine the value of [H3O+] further by considering the dissociation of HCO3, but the change is too small to matter. 731 Chapter 16: Acids and Bases B. (D) (a) For the acids given, we determine values of m and n in the formula EOm(OH)n. HOCl or Cl(OH) has m = 0 and n = 1. We expect Ka 107 or pKa 7, which is in good agreement with the accepted pKa = 7.52. HOClO or ClO(OH) has m = 1 and n = 1. We expect Ka 102 or pKa 2, in good agreement with the pKa = 1.92. HOClO2 or ClO2(OH) has m = 2 and n = 1. We expect Ka to be large and in good agreement with the accepted value of pKa = 3, Ka = 10pKa = 103. HOClO3 or ClO3(OH) has m = 3 and n = 1. We expect Ka to be very large and in good agreement with the accepted Ka = 8, pKa = 8, Ka = 10pKa = 108 which turns out to be the case. (b) The formula H3AsO4 can be rewritten as AsO(OH)3, which has m = 1 and n = 3. The expected value is Ka = 102. (c) The value of pKa = 1.1 corresponds to Ka = 10pKa = 101.1 = 0.08, which indicates that m = 1. The following Lewis structure is consistent with this value of m.
OH O P OH H EXERCISES
BrnstedLowry Theory of Acids and Bases
1. (E) (a) (b) (c) (d) (e) 2. HNO 2 is an acid, a proton donor. Its conjugate base is NO 2 . OCl is a base, a proton acceptor. Its conjugate acid is HOCl. NH 2 is a base, a proton acceptor. Its conjugate acid is NH 3 . NH 4 is an acid, a proton donor. It's conjugate base is NH 3 . CH 3 NH 3 is an acid, a proton donor. It's conjugate base is CH 3 NH 2 .
+ + (E) We write the conjugate base as the first product of the equilibrium for which the acid is the first reactant. (a) (b) (c) (d) HIO3 aq + H 2 O(l) IO3 aq + H 3O + aq C6 H 5COOH aq + H 2 O(l) C6 H 5COO aq + H 3O + aq HPO 4
2 aq + H 2O(l) PO43 aq + H3O+ aq C2 H 5 NH 3 aq + H 2 O(l) C 2 H 5 NH 2 aq + H 3O + aq 732 Chapter 16: Acids and Bases 3. (E) The acids (proton donors) and bases (proton acceptors) are labeled below their formulas. Remember that a proton, in BrnstedLowry acidbase theory, is H + . (a) HOBr(aq) + H 2O(l) H 3O + (aq) + OBr (aq) acid base acid base 2 HSO 4 (aq) + H 2 O(l) H 3O + (aq) + SO 4 (aq) acid base acid base (b) (c) HS (aq) + H 2 O(l) H 2S(aq) base acid acid + OH (aq) base (d) C6 H 5 NH 3 (aq) + OH (aq) C6 H 5 NH 2 (aq) + H 2 O(l) acid base base acid 4. (E) For each amphiprotic substance, we write both its acid and base hydrolysis reaction. Even for the substances that are not usually considered amphiprotic, both reactions are written, but one of them is labeled as unlikely. In some instances we have written an oxygen as to keep track of it through the reaction. H + H 2 O 2 + H 3O + unlikely H + H 2 O H 2 + OH NH 4 + H 2 O NH 3 + H 3O + H 2 + H 2 O H + H 3O + HS + H 2 O S2 + H 3O + NH 4 + has no e pairs that can be donated H 2 + H 2 O H 3 + + OH HS + H 2 O H 2S + OH NO 2 + H 2 O HNO 2 + OH HCO3 + H 2 O H 2 CO3 + OH HBr + H 2 O H 2 Br + + OH unlikely NO 2 cannot act as an acid, (no protons) HCO3 + H 2 O CO3 + H 3O + HBr + H 2 O H 3O + + Br 2 5. (E) Answer (b), NH 3 , is correct. HC 2 H 3O 2 will react most completely with the strongest base. NO3 and Cl are very weak bases. H 2 O is a weak base, but it is amphiprotic, acting as an acid (donating protons), as in the presence of NH 3 . Thus, NH 3 must be the strongest base and the most effective in deprotonating HC 2 H 3O 2 .
6. (M) Lewis structures are given below each equation. + (a) 2NH 3 l NH 4 + NH 2 H 2H N H H N H + H N H
H H 733 Chapter 16: Acids and Bases (b) 2HF l H 2 F+ + F
H H F + H F F + F H (c) + 2CH 3OH l CH 3OH 2 + CH 3O H 2H C H
(d) H H O H H C O H H H H C O H + 2HC2 H 3O 2 l H 2 C2 H 3O 2 + C2 H 3O 2 H 2H C H
(e) O C O H H O H H C C O H H H O H C C O H 2 H 2 SO 4 (l ) 2 H3SO 4 HSO 4
O H O S O H + H O O O S O H O O O O S O H O H + S O H + H O O 7. (E) The principle we will follow here is that, in terms of their concentrations, the weaker acid and the weaker base will predominate at equilibrium. The reason for this is that a strong acid will do a good job of donating its protons and, having done so, its conjugate base will be left behind. The preferred direction is: strong acid + strong base weak (conjugate) base + weak (conjugate) acid
(a) The reaction will favor the forward direction because OH (a strong base) NH 3 (a weak base) and NH 4 (relatively strong weak acid) H 2 O (very weak acid) . The reaction will favor the reverse direction because HNO 3 HSO 4 (a weak acid in the second ionization) (acting as acids), and SO 4
2 (b) NO 3 (acting as bases). (c) The reaction will favor the reverse direction because HC2 H 3O 2 CH 3OH (not usually thought of as an acid) (acting as acids), and CH 3O C 2 H 3O 2 (acting as bases). 734 Chapter 16: Acids and Bases 8. (M) The principle we follow here is that, in terms of their concentrations, the weaker acid and the weaker base predominate at equilibrium. This is because a strong acid will do a good job of donating its protons and, having done so, its conjugate base will remain in solution. The preferred direction is: strong acid + strong base weak (conjugate) base + weak (conjugate) acid
(a) The reaction will favor the forward direction because HC 2 H 3O 2 (a moderate acid) HCO 3 (a rather weak acid) (acting as acids) and CO 32 C 2 H 3O 2 (acting as bases). (b) The reaction will favor the reverse direction because HClO 4 (a strong acid) HNO 2 (acting as acids), and NO 2 ClO 4 (acting as bases). (c) The reaction will favor the forward direction, because H 2 CO3 HCO3 (acting as acids) (because K1 K2 ) and CO 3
2 HCO 3 (acting as bases). Strong Acids, Strong Bases, and pH
9. (M) All of the solutes are strong acids or strong bases. (a) 1 mol H 3O + H 3O + = 0.00165 M HNO3 0.00165 M 1 mol HNO3 Kw 1.0 1014 12 OH = = H O + 0.00165M = 6.110 M 3 (b) 1 mol OH OH = 0.0087 M KOH 0.0087 M 1 mol KOH 1.0 1014 12 H3 O = OH = 0.0087M = 1.1 10 M + Kw (c) 2 mol OH OH = 0.00213 M Sr OH 2 0.00426 M 1mol Sr OH 2 Kw 1.0 10 12 H3 O+ = OH = 0.00426 M = 2.3 10 M 14 (d) 1mol H 3O + H 3O + = 5.8 104 M HI 5.8 104 M 1mol HI Kw 1.0 1014 11 OH = = H O + 5.8 104 M 1.7 10 M 3 735 Chapter 16: Acids and Bases 10. (M) Again, all of the solutes are strong acids or strong bases. + (a) H O + = 0.0045 M HCl 1 mol H 3O = 0.0045 M 3 1 mol HCl
pH = log 0.0045 = 2.35 (b) 1 mol H 3O + H 3O + = 6.14 104 M HNO3 = 6.14 104 M 1 mol HNO pH = log 6.14 10
4 = 3.21 3 (c) 1 mol OH OH = 0.00683 M NaOH = 0.00683M 1 mol NaOH pOH = log 0.00683 = 2.166 and pH = 14.000 pOH = 14.000 2.166 = 11.83 (d) 2 mol OH OH = 4.8 103 M Ba OH 2 = 9.6 103 M 1 mol Ba OH pOH = log 9.6 10
3 = 2.02 2 pH = 14.00 2.02 = 11.98 11. (E) OH = 3.9 g Ba OH 2 8H 2 O 1000 mL 1 mol Ba OH 2 8H 2 O 2 mol OH 100 mL soln 1L 315.5 g Ba OH 2 8H 2 O 1 mol Ba OH 2 8H 2 O
K 1.0 1014 = 0.25 M
14 w H 3O + = OH = 0.25 M OH = 4.0 10 M pH = log 4.0 1014 = 13.40 12. (M) The dissolved Ca OH 2 is completely dissociated into ions. pOH = 14.00 pH = 14.00 12.35 = 1.65 b g OH = 10 pOH = 101.65 = 2.2 102 M OH = 0.022 M OH 0.022 mol OH 1 mol Ca OH 2 74.09 g Ca OH 2 1000 mg Ca OH 2 solubility = 1 L soln 2 mol OH 1 mol Ca OH 2 1 g Ca OH 2 = 8.1 102 mg Ca OH 2 1 L soln 8.1102 mg Ca OH 2 1L In 100 mL the solubility is 100 mL = 81 mg Ca OH 2 1000 mL 1 L soln 736 Chapter 16: Acids and Bases 13. (M) First we determine the moles of HCl, and then its concentration. 1 atm 751 mmHg 0.205 L 760 mmHg PV moles HCl = 8.34 103 mol HCl 1 1 RT 0.08206 L atm mol K 296 K H 3O + = 14. 8.34 103 mol HCl 1 mol H 3O + = 1.96 103 M 4.25 L soln 1 mol HCl (E) First determine the concentration of NaOH, then of OH . 0.125 L OH = 0.606 mol NaOH 1 mol OH 1L 1 mol NaOH = 0.00505 M 15.0 L final solution
pH = 14.00 2.297 = 11.703 pOH = log 0.00505 M = 2.297 15. (E) First determine the amount of HCl, and then the volume of the concentrated solution required. 102.10 mol H 3O + 1 mol HCl amount HCl = 12.5 L = 0.099 mol HCl 1 L soln 1 mol H 3O + Vsolution = 0.099 mol HCl 16. 36.46 g HCl 100.0 g soln 1 mL soln = 8.5 mL soln 1 mol HCl 36.0 g HCl 1.18 g soln (E) First we determine the amount of KOH, and then the volume of the concentrated solution required. pOH = 14.00 pH = 14.00 11.55 = 2.45 amount KOH = 25.0 L OH = 10 pOH = 102.45 = 0.0035 M 0.0035 M mol OH 1 mol KOH = 0.088 mol KOH 1 L soln 1 mol OH 56.11 g KOH 100.0 g soln 1 mL soln = 29 mL soln 1 mol KOH 15.0 g KOH 1.14 g soln Vsolution = 0.088 mol KOH 17. (E) The volume of HCl(aq) needed is determined by first finding the amount of NH 3 aq present, and then realizing that acid and base react in a 1:1 molar ratio. b g VHCl = 1.25 L base 0.265 mol NH 3 1 mol H 3 O + 1 mol HCl 1 L acid + 1 L base 1 mol NH 3 1 mol H 3 O 6.15 mol HCl = 0.0539 L acid or 53.9 mL acid. 737 Chapter 16: Acids and Bases 18. (M) NH 3 and HCl react in a 1:1 molar ratio. According to Avogadro's hypothesis, equal volumes of gases at the same temperature and pressure contain equal numbers of moles. Thus, the volume of H2(g) at 762 mmHg and 21.0 C that is equivalent to 28.2 L of H2(g) at 742 mmHg and 25.0 C will be equal to the volume of NH3(g) needed for stoichiometric neutralization. VNH3 g = 28.2 L HCl g 742 mmHg 273.2 + 21.0 K 1 L NH 3 g = 27.1 L NH 3 g 762 mmHg 273.2 + 25.0 K 1 L HCl g Alternatively, we can solve the ideal gas equation for the amount of each gas, and then, by equating these two expressions, we can find the volume of NH 3 needed.
n{HCl} = 742 mmHg 28.2 L R 298.2 K n{NH 3} 762 mmHg V {NH 3} R 294.2 K 742 mmHg 28.2 L 762 mmHg V {NH 3} This yields V {NH 3} 27.1 L NH 3 (g) R 298.2 K R 294.2 K
19. (M) Here we determine the amounts of H 3O + and OH and then the amount of the one that is in excess. We express molar concentration in millimoles/milliliter, equivalent to mol/L. 0.0155 mmol HI 1 mmol H 3O + 50.00 mL = 0.775 mmol H 3O + 1 mL soln 1 mmol HI 75.00 mL 0.0106 mmol KOH 1 mmol OH = 0.795 mmol OH 1 mL soln 1 mmol KOH The net reaction is H 3O + aq + OH aq 2H 2 O(l) . There is an excess of OH of ( 0.795 0.775 = ) 0.020 mmol OH . Thus, this is a basic solution. The total solution volume is ( 50.00 + 75.00 = ) 125.00 mL. 0.020 mmol OH OH = = 1.6 104 M, pOH = log 1.6 104 = 3.80, 125.00 mL pH = 10.20 20. (M) In each case, we need to determine the H 3O + or OH of each solution being mixed, and then the amount of H 3O + or OH , so that we can determine the amount in excess. We express molar concentration in millimoles/milliliter, equivalent to mol/L. H 3O + = 102.12 = 7.6 103 M pOH = 14.00 12.65 = 1.35 moles H 3O + = 25.00 mL 7.6 103 M = 0.19 mmol H 3O + OH = 101.35 = 4.5 10 2 M amount OH = 25.00 mL 4.5 10 2 M = 1.13 mmol OH 738 Chapter 16: Acids and Bases There is excess OH in the amount of 0.94 mmol (=1.13 mmol OH 0.19 mmol H3O+).
2 OH = 25.00 mL + 25.00 mL = 1.88 10 M 0.94 mmol OH pOH = log 1.88 102 = 1.73 pH=12.27 Weak Acids, Weak Bases, and pH
21. (M) We organize the solution around the balanced chemical equation. Equation: HNO2 (aq) + H 2 O(l) H 3 O + (aq) + NO 2 (aq)  Initial: 0.143 M 0 M 0M  Changes: x M +x M +x M Equil:  xM xM 0.143 x M H 3 O + NO 2 x2 x2 = 7.2104 (if x 0.143) Ka 0.143 x 0.143 HNO 2 x = 0.143 7.2 104 0.010 M We have assumed that x 0.143 M , an almost acceptable assumption. Another cycle of approximations yields:
x = (0.143 0.010) 7.2 104 0.0098 M [H 3 O + ] pH log(0.0098) 2.01 This is the same result as is determined with the quadratic formula roots equation.
22. (M) We organize the solution around the balanced chemical equation, and solve first for OH Equation : Initial : Changes : Equil : C 2 H 5 NH 2 (aq) 0.085 M xM + H 2 O(l) OH (aq) + 0M +xM xM C2 H 5 NH 3+ (aq) 0M +xM xM 0.085 x M OH C2 H 5 NH 3+ x2 x2 = Kb = = 4.3 104 assuming x 0.085 0.085 x 0.085 C 2 H 5 NH 2 x = 0.085 4.3 104 0.0060 M We have assumed that x 0.085 M, an almost acceptable assumption. Another cycle of approximations yields:
x = (0.085 0.0060) 4.3 104 0.0058 M = [OH ] Yet another cycle produces: 739 Chapter 16: Acids and Bases x (0.085 0.0058) 4.3 10 4 0.0058 M [OH ] pOH log(0.0058) 2.24 pH = 14.00 2.24 = 11.76 H 3O + = 10 pH = 1011.76 = 1.7 1012 M This is the same result as determined with the quadratic formula roots equation.
23. (M) (a) The setup is based on the balanced chemical equation. H O + (aq) + C H O (aq) Equation: HC8 H 7 O 2 (aq) + H 2 O(l) 3 8 7 2  Initial: 0.186 M 0 M 0M  Changes: x M +x M +x M Equil:  xM xM 0.186 x M K a 4.9 10 5 [H 3 O ][C8 H 7 O 2 ] xx x2 [HC8 H 7 O 2 ] 0.186 x 0.186 x = 0186 4.9 105 0.0030 M = [H 3O + ] [C8 H 7 O 2 ] . 0.0030 M is less than 5 % OF 0.186 M, thus, the approximation is valid.
(b) The setup is based on the balanced chemical equation. Equation: Initial: Changes: Equil: HC8 H 7 O 2 (aq)
0.121 M x M 0.121 x M + H 2 O(l)
   H 3 O + (aq) 0 M +x M xM C8 H 7 O 2 (aq) 0M +x M xM H 3 O + C8 H 7 O 2 x2 x2 = K a = 4.9 10 = 0.121 x 0.121 HC8 H 7 O 2 5 x = 0.0024 M = H 3O + Assumption x 0.121, is correct. pH = log 0.0024 = 2.62
24. (E) b g HC3H5O2 initial = HC3H5O2 equilibrium = 0.440 M 0.00239 M = 0.438 M
Ka = H 3O + C 3 H 5O 2 0.275 mol 1000 mL = 0.440 M 625 mL 1 L LM HC3 H 5O 2 OP N Q b0.00239g =
0.438 2 = 1.30 105 740 Chapter 16: Acids and Bases 25. (M) We base our solution on the balanced chemical equation. H 3O + = 101.56 = 2.8 102 M
Equation : Initial : Changes : Equil : CH 2 FCOOH aq + H 2 O(l) CH 2 FCOO aq + H 3 O + aq 0.318 M 0.028 M 0.290 M 0M + 0.028 M 0.028 M
3 0M + 0.028 M 0.028 M Ka = H 3O + CH 2 FCOO CH 2 FCOOH = b0.028gb0.028g = 2.7 10
0.290 26. (M) HC 6 H 11O 2 = 11 g 1 mol HC 6 H 11O 2 = 9.5 102 M = 0.095 M 1 L 116.2 g HC 6 H 11O 2 2.94 H3 O+ = 1.1 103 M = 0.0011 M equil = 10 The stoichiometry of the reaction, written below, indicates that H 3O + = C 6 H 11O 2 Equation : Initial : Changes : Equil :
Ka = HC 6 H11O 2 (aq) + H 2 O(l) C6 H11O 2 (aq) 0.095 M 0.0011 M 0.094 M + H 3O + (aq) 0 M + 0.0011 M 0.0011 M 2 0M +0.0011 M 0.0011 M C 6 H 11O 2 LM HC6 H11O 2 OP Q N H 3O + b0.0011g =
0.094 = 1.3 10 5 27. (M) Here we need to find the molarity S of the acid needed that yields H 3 O + = 102.85 = 1.4 103 M Equation: Initial: Changes: Equil:
Ka = HC7 H 5O 2 (aq) + H 2 O(l)
S 0.0014 M S 0.0014M   
2 H 3O + (aq) 0M +0.0014M 0.0014 M
5 + C7 H 5O 2 (aq) 0M +0.0014M 0.0014 M
2 H 3O + C 7 H 5O 2 LM HC7 H 5O 2 OP Q N b0.0014g = S 0.0014 = 6.3 10 b0.0014g S 0.0014 = 6.3 105 = 0.031 S = 0.031+ 0.0014 = 0.032 M = HC7 H 5O 2 350.0 mL 0.032 mol HC7 H 5O 2 122.1 g HC7 H 5O 2 1L = 1.4 g HC7 H 5O 2 1000 mL 1 L soln 1 mol HC7 H 5O 2 741 Chapter 16: Acids and Bases 28. (M) First we find OH and then use the Kb expression to find (CH 3 )3 N pOH = 14.00 pH = 14.00 11.12 = 2.88
K b = 6.3 10
5 OH = 10 pOH = 102.88 = 0.0013 M 2 (CH 3 )3 NH + OH = 0.0013 = (CH 3 )3 N equil (CH 3 )3 N equil (CH3 )3 N equil = 0.027 M (CH 3 )3 N CH 3 3 N initial = 0.027 M(equil. concentration) + 0.0013 M(concentration ionized) CH 3 3 N initial = 0.028 M
29. (M) We use the balanced chemical equation, then solve using the quadratic formula.
Equation : Initial : Changes : Equil : HClO 2 (aq) 0.55 M x M + H 2 O(l) H3O+ (aq) 0M +x M + ClO 2 (aq) 0M +x M 0.55 x M xM xM 2 H 3O + ClO 2 = x Ka = = 1.1102 = 0.011 HClO 2 0.55 x x 2 = 0.0061 0.011x x 2 + 0.011x 0.0061 = 0
b b 2 4ac 0.011 0.000121+ 0.0244 0.073 M H 3O + 2a 2 The method of successive approximations converges to the same answer in four cycles. pH = log H 3O + = log 0.073 = 1.14 pOH = 14.00 pH = 14.00 1.14 = 12.86
x b g 30. OH = 10 pOH = 1012.86 = 1.4 1013 M (M) Organize the solution around the balanced chemical equation, and solve first for OH . Equation : Initial : Changes : Equil : CH 3 NH 2 (aq) 0.386 M x M + H 2O(l) OH (aq) 0M + CH 3 NH 3+ (aq) 0M +x M
xM M +x M
xM 0.386 x OH CH 3 NH 3+ x2 x2 = Kb = = 4.2 104 0.386 x 0.386 CH 3 NH 2 x = 0.386 4.2 104 0.013 M = [OH  ] assuming x 0.386 pOH = log(0.013) = 1.89 742 Chapter 16: Acids and Bases pH = 14.00 1.89 = 12.11 H 3O + = 10 pH = 1012.11 = 7.8 1013 M This is the same result as is determined with the quadratic equation roots formula.
31. (M) C10 H 7 NH 2 = 1 g C10 H 7 NH 2 1.00 g H 2 O 1000 mL 1 mol C10 H 7 NH 2 590 g H 2O 1 mL 1L 143.2 g C10 H 7 NH 2 = 0.012 M C10 H 7 NH 2
K b = 10 pKb = 103.92 = 1.2 104 Equation: Initial: Changes: Equil: C10 H 7 NH 2 (aq) + H 2 O(l) 0.012 M x M 0.012 x M
   OH (aq) + 0 M +x M xM C10 H 7 NH 3 (aq) 0M +x M xM OH C10 H 7 NH 3 x2 x2 = Kb = = 1.2 104 0.012 x 0.012 C10 H 7 NH 2 assuming x 0.012 x = 0.012 12 104 0.0012 This is an almost acceptable assumption. Another . approximation cycle gives: x = (0.012 0.0012) 12 104 0.0011 . Yet another cycle seems necessary. x = (0.012 0.0011) 1.2 104 0.0011M [OH ] The quadratic equation roots formula provides the same answer. pOH = log OH = log 0.0011 = 2.96 H 3 O 10 pH 1011.04 9.1 1012 M
32. b g pH = 14.00 pOH = 11.04 (M) The stoichiometry of the ionization reaction indicates that H 3O + = OC6 H 4 NO 2 , K a = 10 pKb = 107.23 = 5.9 108 and H 3O + = 104.53 = 3.0 105 M . We let S = the molar solubility of o nitrophenol. Equation: Initial: Changes: Equil: HOC6 H 4 NO 2 (aq) + H 2 O(l) S 3.0 105 M ( S 3.0 105 ) M
   OC6 H 4 NO 2 (aq) + 0M +3.0 105 M 3.0 105 M H 3O + (aq) 0 M +3.0 105 M 3.0 105 M 743 Chapter 16: Acids and Bases Ka = 5.9 108 = OC 6 H 4 NO 2 H 3O + LM HOC 6 H 5 NO 2 OP Q N
5 2 8 c3.0 10 h = 5 2 S 3.0 105
S = 1.5 102 M + 3.0 105 M = 1.5 102 M S 3.0 10 5 c3.0 10 h =
5.9 10 = 1.5 102 M Hence,
solubility = 1.5 102 mol HOC 6 H 4 NO 2 139.1 g HOC 6 H 4 NO 2 = 2.1 g HOC 6 H 4 NO 2 / L soln 1 L soln 1 mol HOC 6 H 4 NO 2 33. (M) Here we determine H 3O + which, because of the stoichiometry of the reaction, equals C2 H 3O 2 . H 3O + = 10 pH = 104.52 = 3.0 105 M = C2 H 3O 2 We solve for S , the concentration of HC 2 H 3O 2 in the 0.750 L solution before it dissociates. Equation: Initial: Changes: Equil: HC2 H 3 O 2 (aq) + SM 3.0 105 M S 3.0 105 M H 2 O(l)
   C2 H 3 O 2 (aq) 0M +3.0 10 5 M 3.0 105 M + H 3 O + (aq) 0M +3.0 105 M 3.0 105 M [H 3O ][C2 H 3O 2 ] (3.0 105 ) 2 5 1.8 10 Ka [HC2 H 3O 2 ] ( S 3.0 105 ) c3.0 10 h
S= 5 2 = 1.8 105 S 3.0 105 = 9.0 1010 = 1.8 105 S 5.4 1010 c h 9.0 1010 + 5.4 1010 = 8.0 105 M Now we determine the mass of vinegar needed. 1.8 105 8.0 105 mol HC2 H 3O 2 60.05 g HC 2 H 3O 2 100.0 g vinegar 1 L soln 1 mol HC2 H 3O 2 5.7 g HC2 H 3O 2 mass vinegar = 0.750 L = 0.063 g vinegar
34. (M) First we determine OH which, because of the stoichiometry of the reaction, equals NH 4 .
pOH = 14.00 pH = 14.00 11.55 = 2.45 OH = 10 pOH = 102.45 = 3.5 103 M = NH 4 We solve for S , the concentration of NH 3 in the 0.625 L solution prior to dissociation. 744 Chapter 16: Acids and Bases Equation: Initial: Changes: Equil:
Kb = NH 3 (aq ) + H 2 O(l) SM 0.0035 M S 0.0035 M    NH 4 (aq) 0M +0.0035 M 0.0035 M
2 + OH (aq) 0 M +0.0035 M 0.0035 M NH 4 OH LM NH 3 OP N Q = 1.8 10 = 5 bS 0.0035g b0.0035g 0.0035
S= 2 = 1.8 105 S 0.0035 = 1.225 105 = 1.8 105 S 6.3 108 1.225 105 + 6.3 108 = 0.68 M 1.8 105 Now we determine the volume of household ammonia needed Vammonia = 0.625 L soln 0.68 mol NH 3 17.03 g NH 3 100.0 g soln 1 mL soln 1 L soln 1 mol NH 3 6.8 g NH 3 0.97 g soln = 1.1 10 2 mL household ammonia solution 35. (D) 1 atm 316 Torr 0.275 L PV 760 Torr = 4.67 103 mol propylamine = (a) nproplyamine = 1 1 RT 0.08206 L atm K mol 298.15 K [propylamine] =
n 4.67 103 moles = 9.35 103 M = V 0.500 L Kb = 10pKb = 103.43 = 3.7 104 CH3CH2CH2NH2(aq) + H2O(l)  Initial 9.35 103 M  Change x M  Equil. (9.35 103 x) M Kb = 3.7 104 = CH3CH2CH2NH3+(aq) + OH (aq) 0M 0M +x M +x M xM xM x2 or 3.5 106 3. 7 104(x) = x2 9.35 103 x x2 + 3.7 104x 3.5 106 = 0 745 Chapter 16: Acids and Bases Find x with the roots formula:
x 3.7 104 (3.7 104 ) 2 4(1)( 3.5 106 ) 2(1) Therefore x = 1.695 103 M = [OH ] pOH = 2.77 and pH = 11.23
(b) [OH] = 1.7 103 M = [NaOH] (MMNaOH = 39.997 g mol1) nNaOH = (C)(V) = 1.7 103 M 0.500 L = 8.5 104 moles NaOH mass of NaOH = (n)(MMNaOH) = 8.5 104 mol NaOH 39.997 g NaOH/mol NaOH mass of NaOH = 0.034 g NaOH(34 mg of NaOH)
36. (M) Kb = 10pKb = 109.5 = 3.2 1010 MMquinoline = 129.16 g mol1 Solubility(25 C) = 0.6 g 100 mL 0.6 g 1 mol 1000 mL = 0.046 M (note: only 1 sig fig) 100 mL 129.16 g 1L Molar solubility(25 C) = C9H7N(aq) + H2O(l) C9H7NH+(aq) + OH(aq)  0M Initial 0.046 M 0M  Change +x M +x M x M  xM xM Equil. (0.046 x) M 3.2 1010 =
x2 x2 , 0.046 x 0.046 x = 3.8 106 M (x << 0.046, valid assumption) Therefore, [OH ] = 4 106 M
37. pOH = 5.4 and pH = 8.6 (M) If the molarity of acetic acid is doubled, we expect a lower initial pH (more H3O+(aq) in solution) and a lower percent ionization as a result of the increase in concentration. The ratio between [H3O+] of concentration "M" and concentration 2 M is 2 1.4 . Therefore, (b), containing 14 H3O+ symbols best represents the conditions (~(2)1/2 times greater). (M) If NH3 is diluted to half its original molarity, we expect a lower pH (a lower [OH]) and a higher percent ionization in the diluted sample. The ratio between [OH] of concentration "M" and concentration 0.5 M is 0.5 0.71 . Since the diagram represent M has 24 OH symbols, then diagram (c), containing 17 OH symbols would be the correct choice. 38. 746 Chapter 16: Acids and Bases Percent Ionization
39. (M) Let us first compute the H 3O + in this solution. Equation: HC3 H 5 O 2 (aq) + H 2 O(l) H 3 O + (aq)  Initial: 0.45 M 0 M  Changes: x M +x M Equil:  xM 0.45 x M + C3 H 5 O 2 (aq) 0 M +x M xM H 3 O + C3 H 5 O 2 x2 x2 = 4.89 5 Ka = = 10 = 1.3 10 0.45 x 0.45 HC3 H 5 O 2 x = 2.4 103 M ; We have assumed that x 0.45 M , an assumption that clearly is correct.
(a) (b) H 3O + 2.4 103 M equil = = = 0.0053 = degree of ionization 0.45 M HC3 H 5 O 2 initial % ionization = 100% = 0.0053 100% 0.53% 40. (M) For C2H5NH2 (ethylamine), Kb = 4.3 104 C2H5NH2(aq) + H2O(l) Initial Change Equil. 0.85 M x M (0.85 x) M
   C2H5NH3+(aq) + OH(aq) 0M +x M x M 0M +x M x M 4.3 104 = x2 x2 (0.85 x) M 0.85 M x = 0.019 M (x << 0.85, thus a valid assumption)
Percent ionization = 100% = 2.2 % Degree of ionization =
41. 0.019 M = 0.022 0.85 M (M) Let x be the initial concentration of NH3, hence, the amount dissociated is 0.042 x NH3(aq) Initial xM Change 0.042x M Equil. (1x0.042x) M = 0.958 x + H2O(l)
   Ka = 1.8 10 5 NH4+(aq) 0M +0.042x M 0.042x M + OH(aq) 0M +0.042x M 0.042x M 747 Chapter 16: Acids and Bases NH 4 OH 0.042 x 2 = K b = 1.8 105 = = 0.00184x [0.958 x] NH 3 equil NH3 initial = x 42. (M) 1.8 105 0.00978 M 0.0098 M 0.00184
Ka = 1.3 10 5 HC3H5O2(aq) + H2O(l) Initial 0.100 M Change x M Equil. (0.100 x) M
5 + C3H5O2(aq) + H3O (aq)    0M +x M xM 0M +x M xM x2 x2 1.3 10 = x = 1.1 103 (x << 0.100, thus a valid assumption) 0.100 M x 0.100 0.0011 100% = 1.14 % Percent ionization = 0.100 Next we need to find molarity of acetic acid that is 1.1% ionized
HC2H3O2(aq) Initial Change Equil. + H2O(l)

Ka =1.8 10 5 C2H3O2(aq) + 0M
1.14 XM + 100 1.14 XM 100 H3O+(aq) 0M
1.14 + XM 100 1.14 XM 100 XM
1.14 XM 100 1.14 X X 100   = 0.9886 X M = 0.0114 M = 0.0114 M 2 1.14 X = 1.3 104(X); 5 100 1.8 10 = X = 0.138 M 0.9886 X Consequently, approximately 0.14 moles of acetic acid must be dissolved in one liter of water in order to have the same percent ionization as 0.100 M propionic acid.
43. (E) We would not expect these ionizations to be correct because the calculated degree of ionization is based on the assumption that the [HC2H3O2]initial [HC2H3O2]initial [HC2H3O2]equil., which is invalid at the 13 and 42 percent levels of ionization seen here. (M) HC2Cl3O2 (trichloroacetic acid) pKa = 0.52 Ka = 100.52 = 0.30 For a 0.035 M solution, the assumption will not work (the concentration is too small and Ka is too large) Thus, we must solve the problem using the quadratic equation. 44. HC2Cl3O2(aq) + H2O(l)  Initial 0.035 M  Change x M  Equil. (0.035 x) M + C2Cl3O2(aq) + H3O (aq) 0M 0M +x M +x M xM xM 748 Chapter 16: Acids and Bases 0.30 = x2 or 0.0105 0.30(x) = x2 or x2 + 0.30x 0.0105 = 0 ; solve quadratic: 0.035 x
(0.30) 2 4(1)(0.0105) 2(1) x 0.30 Therefore x = 0.032 M = [H3O+] Degree of ionization = 0.032 M = 0.91 0.035 M Percent ionization = 100% = 91. % Polyprotic Acids
45. (E) Because H3PO4 is a weak acid, there is little HPO42 (produced in the 2nd ionization) compared to the H3O+ (produced in the 1st ionization). In turn, there is little PO43 (produced in the 3rd ionization) compared to the HPO42, and very little compared to the H3O+. (E) The main estimate involves assuming that the mass percents can be expressed as 0.057 g of 75% H 3 PO 4 per 100. mL of solution and 0.084 g of 75% H 3 PO 4 per 100. mL of solution. That is, that the density of the aqueous solution is essentially 1.00 g/mL. Based on this assumption, the initial minimum and maximum concentrations of H 3 PO 4 is: 46. 0.057 g impure H 3 PO 4 [H 3 PO 4 ] 75g H 3 PO 4 1mol H 3 PO 4 100 g impure H3 PO4 98.00 g H3 PO 4 0.0044 M 1L 100 mL soln 1000 mL 0.084 g impure H 3 PO 4 [H 3 PO 4 ] 75g H 3 PO 4 1mol H 3 PO 4 100 g impure H 3 PO 4 98.00 g H 3 PO 4 0.0064 M 1L 100 mL soln 1000 mL + H 2 O(l)
   Equation: Initial: Changes: Equil: H 3 PO 4 (aq) H 2 PO 4 (aq) 0M +x M x M + H 3 O + (aq) 0M +x M x M 0.0044 M x M ( 0.0044 x ) M H 2 PO 4 H 3O + x2 = = 7.1103 K a1 = H 3PO 4 0.0044 x x 2 + 0.0071x 3.1 105 = 0 b b 2 4ac 0.0071 5.0 105 +1.2 104 x= = = 3.0 103 M = H 3O + 2a 2 749 Chapter 16: Acids and Bases The setup for the second concentration is the same as for the first, with the exception that 0.0044 M is replaced by 0.0064 M. H 2 PO 4 H 3O + x2 K a1 = x 2 + 0.0071x 4.5 105 = 0 = = 7.1103 H 3PO 4 0.0064 x x=
b b 2 4ac 0.0071 5.0 105 +1.8 104 = = 4.0 103 M = H 3O + 2a 2 The two values of pH now are determined, representing the pH range in a cola drink. pH = log 3.0 103 = 2.52
47. (D) (a) pH = log 4. 103 = 2.40 Equation: H 2S(aq) + H 2O(l)
   HS (aq) 0M +x M x M + H3O + (aq) 0M +x M +x M Initial: 0.075 M Changes: x M Equil: ( 0.075 x ) M HS H 3O + x2 x2 = 1.0 107 = K a1 = 0.075 x 0.075 H 2S
5 19 2 HS = 8.7 10 M and S = K a 2 = 110 M x = 8.7 105 M = H 3O + (b) The setup for this problem is the same as for part (a), with 0.0050 M replacing 0.075 M as the initial value of H 2S . HS H 3O + x2 x2 = 1.0 107 = K a1 = x = 2.2 105 M = H 3O + H 2S 0.0050 x 0.0050 5 19 2 HS = 2.2 10 M and S = K a 2 = 1 10 M (c) The setup for this part is the same as for part (a), with 1.0 105 M replacing 0.075 M as the initial value of H 2S . The solution differs in that we cannot assume x 1.0 105 . Solve the quadratic equation to find the desired equilibrium concentrations. HS H 3 O + = 1.0 107 = K a1 = x2 1.0 10 5 x H S
2 x 2 1.0 107 x 1.0 1012 0 750 Chapter 16: Acids and Bases 1.0 107 1.0 1014 2 1 x= 4 1.0 10 12 x = 9.5 107 M = H 3O + 48. HS = 9.5 107 M S2 = K a 2 = 1 1019 M (M) For H 2 CO 3 , K1 = 4.4 107 and K2 = 4.7 1011 (a) The first acid ionization proceeds to a far greater extent than does the second and determines the value of H 3O + . Equation: Initial: Changes: Equil:
K1 = H 2 CO3 (aq) + 0.045 M x M 0.045 x M H 2 O(l)
   b g H 3O + (aq) 0 M +x M x M + HCO3 (aq) 0M +x M x M H 3O + HCO 3 LM H 2 CO 3 OP Q N x2 x2 7 = = 4.4 10 0.045 x 0.045 x = 1.4 104 M = H 3O + (b) Since the second ionization occurs to only a limited extent, HCO3 = 1.4 104 M = 0.00014 M We use the second ionization to determine CO 32 .
2 Equation: HCO3 (aq) + H 2 O(l) H 3 O + (aq) CO3 (aq)  0.00014 M 0.00014 M 0M  +x M x M +x M  0.00014 + x M x M 0.00014 x M Initial: Changes: Equil: H 3O + CO32 0.00014 + x x 0.00014 x = K2 = = 4.7 1011 0.00014 x 0.00014 HCO3 (c) x = 4.7 1011 M = CO 32 Again we assumed that x 0.00014 M, which clearly is the case. We also note that our original assumption, that the second ionization is much less significant than the first, also is a valid one. As an alternative to all of this, we could have recognized that the concentration of the divalent anion equals the second ionization constant: CO 32 = K2 . 751 Chapter 16: Acids and Bases 49. (D) In all cases, of course, the first ionization of H 2SO 4 is complete, and establishes the initial values of H 3O + and HSO 4 . Thus, we need only deal with the second ionization in each case. (a) Equation: HSO 4 (aq) + H 2 O(l) SO 4 2 (aq) + H 3O + (aq) 0.75 M +x M ( 0.75 + x ) M  Initial: 0M 0.75 M  Changes: x M +x M  Equil: x M ( 0.75 x ) M SO 4 2 H 3O + x 0.75 + x 0.75 x Ka2 = = 0.011 = 0.75 x 0.75 HSO 4 2 x = 0.011 M = SO 4 We have assumed that x 0.75 M, an assumption that clearly is correct. HSO 4 = 0.75 0.011 = 0.74 M H 3O + = 0.75 + 0.011 = 0.76 M (b) The setup for this part is similar to part (a), with the exception that 0.75 M is replaced by 0.075 M.
SO 4 2 H 3O + x 0.075 + x Ka2 = = 0.011 = 0.075 x HSO 4 0.011 0.075 x = 0.075 x + x 2 x 2 + 0.086 x 8.3 104 = 0 b b 2 4ac 0.086 0.0074 + 0.0033 = = 0.0087 M 2a 2 2 x = 0.0087 M = SO 4 x= HSO 4 = 0.075 0.0087 = 0.066M (c) H 3O + = 0.075 + 0.0088M = 0.084 M Again, the setup is the same as for part (a), with the exception that 0.75 M is replaced by 0.00075 M
Ka2 x= x(0.00075 x) [SO 4 ][H 3O ] 0.011 0.00075 x [HSO 4 ]
2 0.011(0.00075 x) 0.00075 x x 2 x 2 0.0118 x 8.3 106 0 b b 2 4ac 0.0118 1.39 104 + 3.3 105 = 6.6 104 = 2 2a 2 x = 6.6 104 M = SO 4 HSO 4 = 0.00075 0.00066 = 9 105 M H 3O + = 0.00075 + 0.00066M = 1.41 103 M H 3O + is almost twice the initial value of H 2SO 4 . Thus, the second ionization of H 2SO 4 is nearly complete in this dilute solution. 752 Chapter 16: Acids and Bases 50. (D) First we determine H 3O + , with the calculation based on the balanced chemical equation. Equation: HOOC CH 2 4 COOH aq + H 2 O(l) HOOC CH 2 4 COO aq + H 3 O + aq Initial: Changes: Equil: H 3O + Ka1 = HOOCF CH 2 I 4 COOH H K g HOOCbCH g COO
2 4 0.10 M x M 0.10 x M    b 0M +x M x M
xx x2 0.10 x 0.10 0M +x M x M = 3.9 105 = x 010 3.9 105 2.0 103 M = [H 3O + ] . We see that our simplifying assumption, that x 0.10 M, is indeed valid. Now we consider the second ionization. We shall see that very little H 3O + is produced in this ionization because of the small size of two numbers: Ka 2 and HOOC CH 2 4 COO . Again we base our calculation on the balanced chemical equation. b g Equation: HOOC CH 2 4 COO aq + H 2 O(l) OOC CH 2 4 COO aq + H 3O + aq Initial: Changes: Equil:
Ka 2 = b H 3O + HOOCF CH 2 I 4 COO H K g OOCbCH g COO
2 4 2.0 103 M y M 0.0020 y M    0M +y M y M b g 2.0 10 3 M +y M 0.0020 + y M g = 3.9 106 = y 0.0020 + y 0.0020 y 0.0020 y 0.0020 b y 3.9 106 M = [ OOC(CH 2 ) 4 COO ] Again, we see that our assumption, that y 0.0020 M, is valid. In addition, we also note that virtually no H 3O + is created in this second ionization. The concentrations of all species have been calculated above, with the exception of OH .
Kw 1.0 1014 12 OH = = H O + 2.0 103 = 5.0 10 M; 3 HOOC CH 2 4 COOH = 0.10 M OOC CH 2 4 COO = 3.9 106 M H 3O + = HOOC CH 2 4 COO = 2.0 103 M; 753 Chapter 16: Acids and Bases 51. (M) (a) Recall that a base is a proton acceptor, in this case, accepting H+ from H 2 O . First ionization : C 20 H 24O 2 N 2 + H 2O C 20 H 24 O 2 N 2 H + + OH pK b1 = 6.0 Second ionization : C 20 H 24O 2 N 2 H + + H 2O C 20 H 24O 2 N 2 H 2 + OH pK b2 = 9.8
2+ (b) 1.00 g quinine [C20 H 24 O 2 N 2 ] K b1 106.0 1 106 1mol quinine 324.4 g quinine 1.62 103 M 1L 1900 mL 1000 mL Because the OH produced in (a) suppresses the reaction in (b) and since Kb1 >> Kb2, the solution's pH is determined almost entirely by the first base hydrolysis of the first reaction. Once again, we setup the I.C.E. table and solve for the [OH ] in this case: Equation: C20 H 24 O 2 N 2 (aq) + H 2 O(l) C20 H 24 O 2 N 2 H + (aq) + OH (aq) Initial: Changes: Equil: 0.00162 M x M 0.00162 x M
   b g 0M +x M x M 0 M +x M x M
x 4 105 M C20 H 24 O 2 N 2 H + OH x2 x2 = 1 106 = K b1 = 0.00162 x 0.00162 C 20 H 24 O 2 N 2 The assumption x 0.00162 , is valid. pOH = log 4 105 = 4.4 pH = 9.6 52. (M) Hydrazine is made up of two NH2 units held together by a nitrogennitrogen single bond:
H N H
+ H N H (a) (b) N2H4(aq) + H2O(l) N2H5 (aq) + OH(aq) Kb1 = 106.07 = 8.5 107 N2H5+(aq) + H2O(l) N2H62+(aq) + OH(aq) Kb2 = 1015.05 = 8.9 1016 Because the OH produced in (a) suppresses the reaction in (b) and since Kb1 >> Kb2, the solution's pH is determined almost entirely by the first base hydrolysis reaction (a). Thus we need only solve the I.C.E. table for the hydrolysis of N2H4 in order to find the [OH]equil, which will ultimately provide us with the pH via the relationship [OH][H3O+] = 1.00 1014. 754 Chapter 16: Acids and Bases Reaction: Initial: Changes: Equil: N2H4(aq) + 0.245 M x M (0.245 x ) M H2O(l)
   N2H5+(aq) + 0M +x M x M OH(aq) 0 M +x M x M N 2 H 5 + OH x2 x2 = K b1 = = 8.5 107 = 0.245 x 0.245 N2H4 x = 4.56 104 M = [OH  ]equil 1.0 1014 Thus, [H 3O ] = = 2.19 1011 M 4 4.56 10
+ pH = log(2.19 1011 ) = 10.66 53. (E) Protonated codeine hydrolyzes water according to the following reaction: C18 H 21O3 NH H 2 O C18 H 21O3 N H 3O pK a 6.05 pK b 14 pK a 7.95 54. (E) Protonated quinoline is, of course, an acid, which hydrolyzes water to give a hydronium ion. The reaction is as follows:
C9 H 7 NH H 2 O C9 H 7 N H 3O Kb KW 1 1014 1.6 105 K a 6.3 1010 pK b log 1.6 105 4.8 55. (E) The species that hydrolyze are the cations of weak bases, namely, NH 4 and + C 6 H 5 NH 3 , and the anions of weak acids, namely, NO 2 and C 7 H 5O 2 .
(a) (b) (c) (d) (e) + NH 4 aq + NO3 aq + H 2O(l) NH 3 aq + NO3 aq + H 3O + aq + Na + aq + NO 2 aq + H 2O(l) Na + aq + HNO 2 aq + OH aq K + aq + C7 H 5O 2 aq + H 2 O(l) K + aq + HC7 H 5O 2 aq + OH aq K + aq + Cl aq + Na + aq + I aq + H 2 O(l) no reaction
+ C6 H 5 NH 3 aq + Cl aq + H 2 O(l) C6 H 5 NH 2 aq + Cl aq + H 3O + aq 755 Chapter 16: Acids and Bases 56. (E) Recall that, for a conjugate weak acidweak base pair, Ka Kb = Kw (a) Ka = K w 1.0 1014 = = 6.7 106 for C5 H 5 NH + 9 K b 1.5 10 (b) K w 1.0 1014 Kb = = 5.6 1011 for CHO 2 4 K a 1.8 10 Kb = K w 1.0 1014 = 1.0 104 for C6 H 5O 10 K a 1.0 10 (c) 57. (E) (a) Because it is composed of the cation of a strong base, and the anion of a strong acid, KCl forms a neutral solution (i.e., no hydrolysis occurs). (b) KF forms an alkaline (basic) solution. The fluoride ion, being the conjugate base of a relatively strong weak acid, undergoes hydrolysis, while potassium ion, the weakly polarizing cation of a strong base, does not react with water. F (aq) + H 2 O(l) HF(aq) + OH (aq)
(c) (d) NaNO 3 forms a neutral solution, being composed of a weakly polarizing cation and anionic conjugate base of a strong acid. No hydrolysis occurs. Ca OCl b g 2 forms an alkaline (basic) solution, being composed of a weakly polarizing cation and the anion of a weak acid Thus, only the hypochlorite ion hydrolyzes. OCl (aq) + H 2 O(l) HOCl(aq) + OH (aq)
(e) NH 4 NO 2 forms an acidic solution. The salt is composed of the cation of a weak base and the anion of a weak acid. The ammonium ion hydrolyzes: NH 4 (aq) + H 2O(l) NH 3 aq + H 3O + (aq)
+ K a = 5.6 1010 as does the nitrite ion: NO 2 (aq) + H 2 O(l) HNO 2 (aq) + OH (aq) K b = 1.4 1011 Since NH 4 is a stronger acid than NO 2 is a base (as shown by the relative sizes of the K values), the solution will be acidic. The ionization constants were computed from data in Table 163 and with the relationship Kw = Ka Kb . For NH 4 + , Ka = 1.0 1014 1.0 1014 = 1.4 1011 = 5.6 1010 For NO 2 , Kb = 5 4 7.2 10 1.8 10 + Where 1.8 105 = K b of NH3 and 7.2 104 = K a of HNO2 756 Chapter 16: Acids and Bases 58. (E) Our list in order of increasing pH is also in order of decreasing acidity. First we look for the strong acids; there is just HNO 3 . Next we look for the weak acids; there is only one, HC 2 H 3O 2 . Next in order of decreasing acidity come salts with cations from weak bases and anions from strong acids; NH 4 ClO 4 is in this category. Then come salts in which both ions hydrolyze to the same degree; NH 4 C 2 H 3O 2 is an example, forming a pHneutral solution. Next are salts that have the cation of a strong base and the anion of a weak acid; NaNO 2 is in this category. Then come weak bases, of which NH 3 aq is an example. And finally, we terminate the list with strong bases: NaOH is the only one. Thus, in order of increasing pH of their 0.010 M aqueous solutions, the solutes are: HNO 3 HC 2 H 3O 2 NH 4 ClO 4 NH 4 C 2 H 3O 2 NaNO 2 NH 3 NaOH b g 59. (M) NaOCl dissociates completely in aqueous solution into Na + aq , which does not hydrolyze, and OCl(aq), which undergoes base hydrolysis. We determine [OH] in a 0.089 M solution of OCl , finding the value of the hydrolysis constant from the ionization constant of HOCl, Ka = 2.9 108 .
Equation : Initial : Changes : Equil : OCl (aq) 0.089 M
x M b g + H 2O(l) HOCl(aq) 0M +x M
xM + OH (aq) 0M +x M
xM 0.089 x M Kb = HOCl OH Kw 1.0 1014 x2 x2 = 3.4 107 = = = 0.089 x 0.089 Ka 2.9 108 OCl 1.7 104 M 0.089, the assumption is valid.
x = 1.7 104 M = OH ; pOH = log 1.7 104 = 3.77 , pH = 14.00 3.77 = 10.23 60. (M) dissociates completely in aqueous solution into NH 4 + aq , which hydrolyzes, and Cl aq , which does not. We determine H 3O H 2 O(l) b g b g + in a 0.123 M solution of NH 4 , finding
NH 3 (aq) 0M +x M
xM + the value of the hydrolysis constant from the ionization constant of NH 3 , K b = 1.8 105 .
Equation : Initial : Changes : Equil : NH 4 + (aq) 0.123M
x M + H 3O + (aq) 0M +x M
xM 0.123 x M Kb = Kw 1.0 1014 x2 x2 [ NH 3 ][ H 3O ] = 5.6 1010 = = 0123 x 0123 . . Ka 1.8 105 [ NH 4 ] 757 Chapter 16: Acids and Bases 8.3 106 M << 0.123, the assumption is valid
x = 8.3 106 M = H 3O + ; 61. pH = log 8.3 106 = 5.08 (M) KC 6 H 7 O 2 dissociates completely in aqueous solution into K + (aq), which does not hydrolyze, and the ion C 6 H 7 O 2 (aq) , which undergoes base hydrolysis. We determine OH in 0.37 M KC 6 H 7 O 2 solution using an I.C.E. table. Note: K a = 10 pK = 104.77 = 1.7 105
Equation : Initial : Changes : Equil : C6 H 7 O 2 (aq) 0.37 M xM + H 2O(l) HC6 H 7 O 2 (aq) 0M + xM
xM + OH (aq) 0M +x M
xM 0.37 x M Kb [ HC 6 H 7 O 2 ][OH ] Kw 10 1014 x2 x2 . 5.9 1010 . 0.37 x 0.37 Ka 17 105 [C6 H 7 O 2 ] pH = 14.00 4.82 = 9.18 x = 1.5 105 M = OH , pOH = log 1.5 105 = 4.82, 62. (M) Equation :
Initial : Changes : Equil :
Ka = C5 H 5 NH + (aq) + H 2O(l) C5 H 5 N(aq)
0.0482 M xM + H 3O + (aq) 0M 0M + xM
xM + xM
xM 0.0482 x M C5 H 5 N H 3 O + K w 1.0 1014 x2 x2 = = = 6.7 106 = K b 1.5 109 0.0482 x 0.0482 C5 H 5 NH + pH = log 5.7 104 = 3.24 5.7 104 M << 0.0482 M, the assumption is valid
x = 5.7 104 M = H 3O + , 63. (M) 2 (a) HSO3 + H 2 O H 3O + + SO3 K a = K a 2 ,Sulfurous = 6.2 108
Kw K a1 ,Sulfurous HSO3 + H 2 O OH + H 2SO3 K b = = 1.0 1014 = 7.7 1013 2 1.3 10 Since Ka Kb , solutions of HSO 3 are acidic. 758 Chapter 16: Acids and Bases (b) HS + H 2 O H 3O + + S2 HS + H 2 O OH + H 2S K a = K a 2 ,Hydrosulfuric = 11019
Kb = Kw K a1 ,Hydrosulfuric = 1.0 1014 = 1.0 107 1.0 107 Since Ka Kb , solutions of HS are alkaline, or basic.
(c) HPO 4 + H 2 O H 3O + + PO 4
2 2 3 K a = K a 3 ,Phosphoric = 4.2 1013 HPO 4 + H 2 O OH + H 2 PO 4 Kb = Kw K a 2 ,Phosphoric = 1.0 1014 = 1.6 107 8 6.3 10 Since Ka Kb , solutions of HPO 4
64. 2 are alkaline, or basic. (M) pH = 8.65 (basic). We need a salt made up of a weakly polarizing cation and an anion of a weak acid, which will hydrolyze to produce a basic solution. The salt (c) KNO2 satisfies these requirements. (a) NH 4 Cl is the salt of the cation of a weak base and the anion of a strong acid, and should form an acidic solution. (b) KHSO 4 and (d) NaNO 3 have weakly polarizing cations and anions of strong acids; they form pHneutral solutions. pOH = 14.00 8.65 = 5.35 OH = 105.35 = 4.5 106 M Equation: Initial: Changes: Equil: NO 2 (aq) S 4.5 106 M + H 2O(l) HNO 2 (aq) 0M +4.5 106 M 4.5 106 M + OH (aq) 0 M +4.5 106 M 4.5 106 M S 4.5 10 6 M HNO 2 OH 4.5 106 K 1.0 1014 Kb = w = = 1.4 1011 = = Ka 7.2 104 S 4.5 106 NO 2 c h 2 S 4.5 10 6 4.5 10 =
1.4 1011 6 2 = 1.4 M S = 1.4 M + 4.5 10 6 = 1.4 M = KNO 2 Molecular Structure and AcidBase Behavior
65. (M) (a) HClO 3 should be a stronger acid than is HClO 2 . In each acid there is an H O Cl grouping. The remaining oxygen atoms are bonded directly to Cl as terminal O atoms. Thus, there are two terminal O atoms in HClO 3 and only one in HClO 2 . For oxoacids of the same element, the one with the higher number of terminal oxygen atoms is the stronger. With more oxygen atoms, the negative charge on the conjugate base is more effectively spread out, which affords greater stability. HClO 3 : K a1 = 5 102 and HClO 2 : Ka = 1.1 102 . 759 Chapter 16: Acids and Bases (b) electronegative than C, which makes HNO2 K a = 7.2 104 , a stronger acid than H2CO3 K a1 = 4.4 107 . HNO 2 and H 2 CO 3 each have one terminal oxygen atom. The difference? N is more (c) H 3 PO 4 and H 2SiO 3 have the same number (one) of terminal oxygen atoms. They differ in P being more electronegative than Si, which makes H 3 PO 4 K a1 = 7.1 103 a stronger acid than H 2SiO3 K a1 = 1.7 1010 . 66. (E) CCl 3COOH is a stronger acid than CH 3COOH because in CCl 3COOH there are electronegative (electrowithdrawing) Cl atoms bonded to the carbon atom adjacent to the COOH group. The electronwithdrawing Cl atoms further polarize the OH bond and enhance the stability of the conjugate base, resulting in a stronger acid. (E) (a) (b) 67. HI is the stronger acid because the H  I bond length is longer than the H  Br bond length and, as a result, H  I is easier to cleave. HOClO is a stronger acid than HOBr because (i) there is a terminal O in HOClO but not in HOBr (ii) Cl is more electronegative than Br. H 3CCH 2 CCl 2 COOH is a stronger acid than I 3CCH 2 CH 2 COOH both because Cl is more electronegative than is I and because the Cl atoms are closer to the acidic hydrogen in the COOH group and thus can exert a stronger e withdrawing effect on the OH bond than can the more distant I atoms. (c) 68. (M) The weakest of the five acids is CH 3CH 2 COOH . The reasoning is as follows. HBr is a strong acid, stronger than the carboxylic acids. A carboxylic acidsuch as CH 2ClCOOH and CH 2 FCH 2 COOH with a strongly electronegative atom attached to the hydrocarbon chain will be stronger than one in which no such group is present. But the I atom is so weakly electronegative that it barely influences the acid strength. Acid strengths of some of these acids (as values of pKa ) follow. (Larger values of pKa indicate weaker acids.) Strength: CH 3CH 2 COOH 4.89 CI3COOH CH 2 FCH 2 COOH CH 2 ClCOOH HBr 8.72 (E) The largest Kb (most basic) belongs to (c) CH3CH2CH2NH2 (hydrocarbon chains have the lowest electronegativity). The smallest Kb (least basic) is that of (a) ochloroaniline (the nitrogen lone pair is delocalized (spread out over the ring), hence, less available to accept a proton (i.e., it is a poorer Brnsted base)). (E) The most basic species is (c) CH3O (methoxide ion). Methanol is the weakest acid, thus its anion is the strongest base. Most acidic: (b) orthochlorophenol. In fact, it is the only acidic compound shown. 69. 70. 760 Chapter 16: Acids and Bases Lewis Theory of Acids and Bases
71. (E) (a) acid is CO2 and base is H2O
O O C O O + H H O C O H + H O H H (b) acid is BF3 and base is H2O F F B F F
2 + O H H F B F O H H (c) acid is H2O and base is O
2O O + O 2 H H 2 (d) acid is SO3 and base is S O H 2S S + O O S O O O S O S 2 72. (E) (a) SOI2 is the acid, and BaSO3 the base. (b) HgCl3 is the acid, Cl the base. 73. (E) A Lewis base is an electron pair donor, while a Lewis acid is an electron pair acceptor. We draw Lewis structures to assist our interpretation. (a) O H
H H C H H C H B H C H C H H The lone pairs on oxygen can readily be donated; this is a Lewis base. (b) The incomplete octet of B provides a site for acceptance of an electron pair, this is a Lewis acid. C H2 H C H3 761 Chapter 16: Acids and Bases (c) H H H N C H H The incomplete octet of B provides a site for acceptance of an electron pair, this is a Lewis acid. 74. (M) A Lewis base is an electron pair donor, while a Lewis acid is an electron pair acceptor. We draw Lewis structures to assist our interpretation. (a) According to the following Lewis structures, SO 3 appears to be the electron pair acceptor (the Lewis acid), and H 2 O is the electron pair donor (the Lewis base). (Note that an additional sulfurtooxygen bond is formed upon successful attachment of water to SO3.)
O H O H O S O H O O S O O H (b) The Zn metal center in Zn OH 2 accepts a pair of electrons. Thus, Zn OH b g b g
2 2 is a Lewis acid. OH donates the pair of electrons that form the covalent bond. Therefore, OH is a Lewis base. We have assumed here that Zn has sufficient covalent character to form coordinate covalent bonds with hydroxide ions. H O 2 O H + H O Zn O H
Acid: e pair acceptor H O Zn O H O H Base: e pair donor
75. (M) (a) H O O H + H O B O H O H H O B O H O H  Base: e pair donor Acid: e pair acceptor 762 Chapter 16: Acids and Bases (b)
H H N N H H H H H O H H N N H H H H H O The actual Lewis acid is H+ , which is supplied by H3O+ (c)
F H H H H H H F F B F H H F H H H H B F H H H H H C C O C C H H C C O C C H Acid: electron pair acceptor Base: electron pair donor 76. (E) The C in a CO 2 molecule can accept a pair of electrons. Thus, CO2 is the Lewis acid. The hydroxide anion is the Lewis base as it has pairs of electrons it can donate.  O O C O acid + O H base O C H O 77. (E) I 2 (aq) I (aq) I3 (aq) I I I Lewis acid Lewis base (e pair acceptor) (e pair donor) I I I 78. (E) (a)
H F F F F Sb F H F F F Sb F F Lewis base (e pair donor) F Lewis acid (e pair acceptor) F (b) F H F F Lewis acid Lewis base (e pair donor) (e pair acceptor) B F H F F F B F 763 Chapter 16: Acids and Bases 79. (E)
H O H O H O S O H O S O H O O H S O Lewis base Lewis acid (e pair donor) (e pair acceptor) 80. (E)
H H Ag+ electron pair acceptor H H H + 2 H N
H electron pair donor H N Ag N H ammonia aduct ammonia adduct INTEGRATIVE AND ADVANCED EXERCISES
81. (E) We use (ac) to represent the fact that a species is dissolved in acetic acid. (a) C2 H 3 O 2 (ac) HC 2 H 3 O 2 (l) HC 2 H 3 O 2 (l) C 2 H 3 O 2 (ac) C2H3O2 is a base in acetic acid. (b) H 2 O(ac) HC 2 H 3 O 2 (l ) H 3 O (ac) C 2 H 3 O 2 (ac) Since H2O is a weaker acid than HC2H3O2 in aqueous solution, it will also be weaker in acetic acid. Thus H2O will accept a proton from the solvent acetic acid, making H2O a base in acetic acid. (c) HC2 H 3 O 2 (l) HC2 H 3 O 2 (l ) H 2 C 2 H 3 O 2 (ac) C 2 H 3 O 2 (aq) Acetic acid can act as an acid or a base in acetic acid. (d) HClO 4 (ac) HC2 H 3 O 2 (l) ClO 4 (ac) H 2 C2 H 3 O 2 (ac) Since HClO4 is a stronger acid than HC2H3O2 in aqueous solution, it will also be stronger in acetic acid. Thus, HClO4 will donate a proton to the solvent acetic acid, making HClO4 an acid in acetic acid. 82. (E) Sr(OH)2 (sat'd) Sr2+(aq) + 2 OH(aq) pOH 14 pH=1413.12 0.88 [OH ] 100.88 0.132 M 764 Chapter 16: Acids and Bases [OH ]dilute 0.132M ( [HCl] 83. (M) 10.0 ml concentrate 250.0 mL total ) 5.28 10 3 M (10.0 ml ) (5.28 10 3 M) 25.1 mL 2.10 103 M (a) H2SO4 is a diprotic acid. A pH of 2 assumes no second ionization step and the second ionization step alone would produce a pH of less than 2, so it is not matched. (b) pH 4.6 matched; ammonium salts hydrolyze to form solutions with pH < 7. (c) KI should be nearly neutral (pH 7.0) in solution since it is the salt of a strong acid (HI) and a strong base (KOH). Thus it is not matched. (d) pH of this solution is not matched as it's a weak base. A 0.002 M solution of KOH has a pH of about 11.3. A 0.0020 M methylamine solution would be lower (~10.9). (e) pH of this solution is matched, since a 1.0 M hypochlorite salt hydrolyzes to form a solution of ~ pH 10.8. (f) pH of this solution is not matched. Phenol is a very weak acid (pH >5). (g) pH of this solution is 4.3. It is close, but not matched. (h) pH of this solution is 2.1, matched as it's a strong organic acid. (i) pH of this solution is 2.5, it is close, but not matched. 84. (M) We let [H3O+]a = 0.500 [H3O+]i pH = log[H3O+]i pHa = log[H3O+]a = log(0.500 [H3O+]i) = log [H3O+]i log (0.500) = pHi + 0.301 The truth of the statement has been demonstrated. Remember, however, that the extent of ionization of weak acids and weak bases in water increases as these solutions become more and more dilute. Finally, there are some solutions whose pH doesn't change with dilution, such as NaCl(aq), which has pH = 7.0 at all concentrations.
85. (M) Since a strong acid is completely ionized, increasing its concentration has the effect of increasing [H3O+] by an equal degree. In the case of a weak acid, however, the solution of the Ka expression gives the following equation (if we can assume that [H3O+] is negligible compared to the original concentration of the undissociated acid, [HA]). x2 = Ka[HA] (where x = [H3O+]), or x K a [HA] . Thus, if [HA] is doubled, [H3O+] increases by 2 . 86. (E) H2O(l) + H2O(l) H3O+(aq) + OH(aq) Ho = (230.0 kJ mol) (285.8 kJ mol1) = 55.8 kJ Since Ho is positive, the value of Keq is larger at higher temperatures. Therefore, the ionic products increase with increasing temperature. Le Ch telier's principle or the Van't Hoff equation would show this to be true. 765 Chapter 16: Acids and Bases 87. (M) First we assume that molarity and molality are numerically equal in this dilute aqueous solution. Then compute the molality that creates the observed freezing point, with Kf = 1.86 C/m for water.
m
Tf 0.096 C 0.052 m K f 1.86 C/m concentration 0.052 M This is the total concentration of all species in solution, ions as well as molecules. We base our remaining calculation on the balanced chemical equation. Equation: HC4 H 5 O 2 (aq) H 2 O C 4 H 5 O 2 (aq) H 3 O (aq) Initial: 0.0500 M 0M 0M
Changes: Equil:
xM xM xM (0.0500 x) M xM xM Total conc. 0.0516 M (0.0500 x) M x M x M 0.0500 M xM
Ka x 0.0016 M [C4 H 5 O 2 ][H3 O ] (0.0016 ) xx 5 105 [HC4 H5 O2 ] 0.0500 x 0.0500 0.0016
2 88. (D) [H3O+] = 10pH = 105.50 = 3.2 106 M (a) To produce this acidic solution we could not use 15 M NH3(aq), because aqueous solutions of NH3 have pH values greater than 7 owing to the fact that NH3 is a weak base. (b)
VHCl 100.0 mL 3.2 10 6 mol H 3 O 1L 1 mol HCl 1 mol H 3 O 1 L soln 12 mol HCl 2.7 10 5 mL 12 M HCl soln (very unlikely, since this small a volume is hard to measure)
(c) Ka = 5.6 1010 for NH4+(aq)
Equation: Initial: Changes: Equil: NH 4 (aq) c M 3.2 10
6 H 2 O(l) NH 3 (aq) 0M 3.2 10
6 H3 O (aq) 0M M 3.2 106 M 3.2 106 M M (c 3.2 106 ) M 3.2 106 M K a 5.6 1010 [NH 3 ][H 3 O ] (3.2 106 )(3.2 106 ) [NH 4 ] c 3.2 106 766 Chapter 16: Acids and Bases (3.2 10 6 ) 2 1.8 10 2 M [NH 4 ] 0.018 M c 3.2 10 10 5.6 10 0.018 mol NH 4 1 mol NH 4 Cl 53.49 g NH 4 Cl NH 4 Cl mass 100.0 mL 0.096 g NH 4 Cl 1000 mL 1 mol NH 4 Cl 1 mol NH 4 This would be an easy mass to measure on a laboratory three decimal place balance.
6 (d) The setup is similar to that for NH4Cl. Equation: Initial: Changes: Equil: HC2 H 3 O 2 (aq) c M 3.2 10
6 H 2 O (l) C2 H3 O2 (aq) 0 M 3.2 10
6 H3 O (aq) 0 M 3.2 106 M 3.2 106 M M M (c 3.2 106 ) M 3.2 106 M K a 1.8 105 c 3.2 106 [C2 H3O 2 ][H 3O ] (3.2 106 )(3.2 106 ) [HC2 H 3O 2 ] c 3.2 106 (3.2 106 ) 2 5.7 107 c [HC2 H 3O 2 ] 3.8 106 5 1.8 10 3.8 106 mol HC2 H 3O 2 60.05 g HC2 H 3O 2 HC2 H 3O 2 mass 100.0 mL 1 mol HC2 H 3O 2 1000 mL 2.3 105 g HC2 H3O2 (an almost impossibly small mass to measure using conventional laboratory scales) 89. (M) (a) 1.0 105 M HCN: HCN(aq) +
K a 6.210 H2O(l) H3O+(aq) 10 + CN(aq) 0M +x x Initial: Change: 1.0 105 M x 1.0 107 M +x (1.0 107+x) M
6.2 1015 6.2 1010 x 1.0 107 x x 2 Equilibrium: (1.0 105x) M
6.2 1010 x(1.0 107 x) (1.0 105 x) x 2 1.006 107 6.2 1015 0 1.006 107 (1.006 107 ) 2 4(1)( 6.2 1015 ) 1.44 107 2(1) + Final pH = log[H3O ] = log(1.0 107+ 1.44 107) = 6.61 x 767 Chapter 16: Acids and Bases (b) 1.0 105 M C6H5NH2: K a 7.410 C6H5NH2(aq) + H2O(l) 5 Initial: 1.0 10 M Change: x 5 Equilibrium: (1.0 10 x) M 10 OH(aq) + C6H5NH3+(aq) 1.0 107 M 0M +x +x 7 (1.0 10 +x) M x 7.4 1010 x(1.0 107 x) (1.0 105 x) 7.4 1015 7.4 1010 x 1.0 107 x x 2 x 2 1.0074 107 7.4 1015 0 (1.0074 107 ) 2 4(1)( 7.4 1015 ) 4.93 108 M 2(1) Final pOH = log[OH ] = log(1.0 108+ 4.93 107) = 6.83 Final pH = 7.17 x 1.0074 107 90. (D) (a) For a weak acid, we begin with the ionization constant expression for the weak acid, HA. [H O ][A ] [H ]2 [H ]2 [H ]2 Ka 3 [HA] [HA] M [H ] M The first modification results from realizing that [H+] = [A]. The second modification is based on the fact that the equilibrium [HA] equals the initial [HA] minus the amount that dissociates. And the third modification is an approximation, assuming that the amount of dissociated weak acid is small compared to the original amount of HA in solution. Now we take the logarithm of both sides of the equation. [H ] 2 log K a log log [H ] 2 log M 2 log [H ] log M c 2 log [H ] log K a log M 2 pH pK a log M pH 1 pK a 1 log M 2 2 For a weak base, we begin with the ionization constant expression for a weak base, B. [BH ][OH ] [OH ] 2 [OH ] 2 [OH ] 2 Kb [B] [B] M M [OH ] The first modification results from realizing that [OH] = [BH+]. The second modification is based on the fact that the equilibrium [B] equals the initial [B] minus the amount that dissociates. And the third modification is an approximation, assuming that the amount of dissociated weak base is small compared to the original amount of B in solution. Now we take the logarithm of both sides of the equation. [OH ]2 log K b log log [OH ]2 log M 2 log [OH ] log M M 2 pOH pK b log M pOH 1 pK b 1 logM 2 log [OH ] log K b log M 2 2 1 1 pH 14.00 pOH 14.00 2 pK b 2 log M The anion of a relatively strong weak acid is a weak base. Thus, the derivation is the same as that for a weak base, immediately above, with the exception of the value for Kb. We now obtain an expression for pKb. 768 Chapter 16: Acids and Bases Kb Kw Ka log K b log Kw log K w log K a Ka log K b log K w log K a pK b pK w pK a Substitution of this expression for pKb into the expression above gives the following result. pH 14.00 1 pK w 1 pK a 1 log M 2 2 2 (b) 0.10 M HC2H3O2 pH = 0.500 4.74 0.500 log 0.10 = 2.87 H 2 O(l) C2 H 3 O 2 (aq) 0M Equation: HC2 H 3 O 2 (aq) Initial: Changes: Equil: 0.10 M xM (0.10 x) M H 3 O (aq) 0M xM xM x M xM [H 3 O ][C2 H 3 O 2 ] x x x2 5 Ka 1.8 10 [HC2 H 3 O 2 ] 0.10 x 0.10 x 0.10 1.8 105 1.3 103 M, pH log(1.3 103 ) 2.89
0.10 M NH 3 pH 14.00 0.500 4.74 0.500 log 0.10 11.13
Equation: Initial: Changes: Equil:
Kb NH 3 (aq) 0.10 M xM (0.10 x) M H 2 O(l) NH 4 (aq) 0M x M OH (aq) 0M xM x M 1.3 103 M xM [NH 4 ][OH ] x x x2 1.8 105 x 0.10 1.8 105 [NH 3 ] 0.10 x 0.10 pH 14.00 2.89 11.11 pOH log(1.3 103 ) 2.89
0.10 M NaC2 H 3O 2 pH 14.00 0.500 14.00 0.500 4.74 0.500 log 0.10 8.87 Equation: C 2 H 3 O 2 (aq) H 2 O(l) HC2 H 3 O 2 (aq) Initial: 0.10 M 0M Changes: x M xM xM Equil: (0.10 x) M Ka x [OH ][HC2 H 3 O 2 ] 1.0 1014 x x x2 [C2 H 3 O 2 ] 1.8 105 0.10 x 0.10 OH (aq) 0M xM xM 0.10 1.0 1014 7.5 106 M 5 1.8 10 pH 14.00 5.12 8.88 pOH log(7.5 106 ) 5.12 769 Chapter 16: Acids and Bases 91. (D) (a) We know K a , and finally % ionized = 100%. [HA]i [HA]eq Let us see how far we can get with no assumptions. Recall first what we mean by [HA]eq. [HA] eq [HA]i [A ] eq or [HA]i [HA]eq [A ] eq We substitute this expression into the expression for . [ A ]eq [HA]eq [ A ]eq [H 3 O ] eq [A ] eq [A ]eq We solve both the and the Keq expressions for [HA]eq, equate the results, and solve for . [H 3O ][A _ ] [A ] [A ] [HA]eq [A ] [HA]eq [A ] [HA]eq Ka [H 3O ][A _ ] [A ](1 ) R Ka R 1 R 1 (1 R) 1 1 R [H 3 O] 10 pH pK 10 ( pK pH) Ka 10 Momentarily, for ease in writing, we have let R Once we realize that 100 = % ionized, we note that we have proven the cited formula. Notice that no approximations were made during the course of this derivation.
(b) Formic acid has pKa = 3.74. Thus 10(pKpH) = 10(3.742.50) = 101.24 = 17.4 % ionization 100 5.43% 1 17.4 % ionization 1.4 10 3 M 100% 0.93% 0.150 M (c) [H3O+] = 102.85 = 1.4 103 M 0.93 100 100 1 10 ( pK pH) 108 10 ( pK pH) 107 ( pK pH) 0.93 1 10 2.03 pK pH pK pH 2.03 2.85 2.03 4.88 K a 10 4.88 1.3 10 5 92. (M) We note that, as the hydrocarbon chain between the two COOH groups increases in length, the two values of pKa get closer together. The reason for this is that the ionized COOH group, COO, is negatively charged. When the two COOH groups are close together, the negative charge on the molecule, arising from the first ionization, inhibits the second ionization, producing a large difference between the two ionization constants. But as the hydrocarbon chain lengthens, the effect of this negative COO group on the second ionization becomes less pronounced, due to the increasing separation between the two carbonyl groups, and the values of the two ionization constants are more alike. 770 Chapter 16: Acids and Bases 93. (M) Consider only the 1st dissociation. Equation: Initial: Changes: Equil: 6.210 H 2 C4 H 4 O 4 (aq) H 2 O(l) H 3 O (aq) HC4 H 4 O 4 (aq) 0.100 M 0M 0M xM xM xM (0.100 x) M xM xM
5 x2 6.2 105 x 0.00246 M Ka 0.100 x Solve using the quadratic equation. Hence pH = log 0.00246 = 2.61 Consider only the 2nd dissociation. Equation: Initial: Changes: Equil:
Ka HC4 H 4 O 4 (aq) 0.0975 M
x M (0.0975 x) M 2.310 H 2 O(l) 5 H3O (aq) 0.00246 M
x M 0.00246 + x M C4 H 4 O4 2 (aq) 0M
x M xM x(0.00246 x) x(0.00246) 2.3 106 0.0975 x 0.0975 x 0.000091 M or x 0.000088 M (solve quadratic) Hence pH 2.59 Thus, the pH is virually identical, hence, we need only consider the 1st ionization.
94. (M) To have the same freezing point, the two solutions must have the same total concentration of all particles of soluteions and molecules. We first determine the concentrations of all solute species in 0.150 M HC2H2ClO2, Ka = 1.4 103
Equation: Initial: Changes: Equil: HC2 H 2 ClO 2 (aq) H 2 O(l) 0.150 M xM (0.150 x) M H 3 O (aq) C2 H 2 ClO 2 (aq) 0M xM xM 0M xM xM [H 3 O ][C2 H 2 ClO 2 ] x x 1.4 103 Ka [HC2 H 2 ClO 2 ] 0.150 x x 2 2.1 104 1.4 103 x x 2 1.4 103 x 2.1 104 0 x b b 2 4ac 1.4 103 2.0 106 8.4 104 0.014 M 2a 2 total concentration = (0.150 x) + x + x = 0.150 + x = 0.150 + 0.014 = 0.164 M Now we determine the [HC2H3O2] that has this total concentration.
771 Chapter 16: Acids and Bases Equation: Initial: Changes: Equil: HC 2 H 3 O 2 (aq) H 2 O(l) zM yM (z y ) M H3 O (aq) C2 H3 O 2 (aq ) 0M yM yM 0M yM yM Ka [H 3 O ][C2 H 3 O 2 ] y y 1.8 105 [HC2 H 3 O 2 ] z y We also know that the total concentration of all solute species is 0.164 M. y2 z y y y z y 0.164 z 0.164 y 1.8 10 5 0.164 2 y We assume that 2 y 0.164 y 2 0.164 1.8 10 5 3.0 10 6 y 1.7 10 3 The assumption is valid: 2 y 0.0034 0.164 Thus, [HC2H3O2] = z = 0.164 y = 0.164 0.0017 = 0.162 M
mass HC 2 H 3 O 2 1.000 L 0.162 mol HC 2 H 3 O 2 60.05 g HC 2 H 3 O 2 9.73 g HC 2 H 3 O 2 1 L soln 1 mol HC 2 H 3 O 2 95. (M) The first ionization of H2SO4 makes the major contribution to the acidity of the solution. Then the following two ionizations, both of which are repressed because of the presence of H3O+ in the solution, must be solved simultaneously.
Equation: HSO 4  (aq) Initial: Changes: Equil: K2 0.68 M x M (0.68 x) M
2 H 2 O(l) x(0.68 x) 0.68 x SO 4 2 (aq) H 3O (aq) 0M x M xM 0.68 M x M (0.68 x)M [H 3 O ][SO 4 ] [HSO 4 ] 0.011 Let us solve this expression for x. 0.011 (0.68 x) = 0.68x + x2 = 0.0075 0.011x x 2 0.69 x 0.0075 0 x b b2 4ac 0.69 0.48 0.030 0.01 M 2a 2 772 Chapter 16: Acids and Bases This gives [H3O+] = 0.68 + 0.01 = 0.69 M. Now we solve the second equilibrium. Equation: HCHO 2 (aq) H 2 O(l ) Initial: Changes: Equil: 1.5 M xM (1.5 x) M CHO 2 (aq) H 3 O (aq) 0M xM xM 0.69 M xM (0.69 x) M [H 3 O ][CHO 2 ] x(0.69 x) 0.69 x 1.8 104 Ka [HCHO 2 ] 1.5 x 1.5 x 3.9 104 M We see that the second acid does not significantly affect the [H3O+], for which a final value is now obtained. [H3O+] = 0.69 M, pH = log(0.69) = 0.16
96. (D)
Let [ HA1 ] M Let [ HA2 ] M K HA1 2 K HA1 [ H ]2 M [ H ]2 M K HA1 [ H ][ A1 ] [ H ]2 [ H ]2 [ HA1 ] [ H ] [ HA1 ] M [ H ][ A2 ] [ H ]2 [ H ]2 [ HA2 ] [ H ] [ HA2 ] M [ H ] K HA1 M K HA2 2 K HA1 [ H ]2 K HA1 M [ H ]2 2 K HA1 M K HA1 M 1/2 [ H ] 2 K HA1 M 2 K HA1 M
1/2 1/2 [ H ]overall K HA1 M log([ H ]overall ) log K HA1 M pH log K HA1 pH log 2 K M 1 1 M log K M log K M log 2 K 2 2
1/2 1/2
HA1 1/2 2 K HA1 M (take the negative log10 of both sides) 1/ 2 1/2 HA1 HA1 HA1 M 1 log K HA1 M log 2 K HA1 M 2 1 1 pH 3K HA1 M 3MK HA1 2 2 1 log K HA1 M 2 K HA1 M 2 773 Chapter 16: Acids and Bases 97. (M) HSbF6 H+ + SbF6H+ F F Sb F F F FFhas 2p orbital 2p orbitals uses one of its Sb uses all six of its sp3d2 hybrid orbitals Sb has 6 sp3d2 hybrid orbitals The six bonds are all sigma bonds (Sbsp3d2  F2p) F HBF4 H+ + BF4H+ F
FF hasone orbital orbitals uses 2p of its 2p B uses allsp3 hybrid sp3 hybrid orbitals B has 4 four of its orbitals B F F F The four bonds are all sigma bonds (Bsp3  F2p) H 98. (M) The structure of H3PO3 is shown on the right. H O P O H O The two ionizable protons are bound to oxygen.
99. (M) (a) H2SO3 > HF >N2H5+ > CH3NH3+ > H2O (b) OH > CH3NH2 > N2H4 > F> HSO3 (c) (i) to the right and (ii) to the left. 774 Chapter 16: Acids and Bases FEATURE PROBLEMS
100. (D) (a) From the combustion analysis we can determine the empirical formula. Note that the mass of oxygen is determined by difference. 1 mol CO 2 1 mol C amount C = 1.599 g CO 2 = 0.03633 mol C 44.01 g CO 2 1 mol CO 2 12.011 g C = 0.4364 g C 1 mol C 1 mol H 2 O 2 mol H amount H = 0.327 g H 2 O = 0.03629 mol H 18.02 g H 2 O 1 mol H 2 O mass of C = 0.03633 mol C mass of H = 0.03629 mol H 1.008 g H = 0.03658 g H 1 mol H 1 mol O = 0.0364 mol O 16.00 g O There are equal moles of the three elements. The empirical formula is CHO. The freezingpoint depression data are used to determine the molar mass. Tf 0.82 C Tf = Kf m m = = = 0.21 m Kf 3.90 C / m 1 kg 0.21 mol solute amount of solute = 25.10 g solvent = 0.0053 mol 1000 g 1 kg solvent 0.615 g solute M= = 1.2 102 g/mol 0.0053 mol solute The formula mass of the empirical formula is: 12.0g C +1.0g H +16.0g O = 29.0g / mol Thus, there are four empirical units in a molecule, the molecular formula is C4 H 4 O 4 , and the molar mass is 116.1 g/mol. amount O = 1.054 g sample 0.4364 g C 0.03568 g H (b) Here we determine the mass of maleic acid that reacts with one mol OH , mass 0.4250 g maleic acid 58.03g/mol OH 1L 0.2152 mol KOH mol OH 34.03mL base 1000 mL 1L base This means that one mole of maleic acid (116.1 g/mol) reacts with two moles of hydroxide ion. Maleic acid is a diprotic acid: H 2 C 4 H 2 O 4 . (c) b g Maleic acid has twoCOOH groups joined together by a bridging C2H2 group. A plausible Lewis structure is shown below: O H H O
H O C C C C O H 775 Chapter 16: Acids and Bases (d) We first determine H 3O + = 10 pH = 101.80 = 0.016 M and then the initial concentration of acid. 0.215g 1000 mL 1 mol CHCOOH 2 initial = 50.00 mL 1L 116.1 g = 0.0370 M We use the first ionization to determine the value of K a1 Equation: CHCOOH 2 (aq) + H 2 O(l) H CHCOO 2 (aq) + H 3 O + (aq) 0 M + 0.016 M 0.016 M Initial: 0.0370 M Changes: 0.016 M Equil: 0.021M 0M + 0.016 M 0.016 M H CHCOO H 3O + 0.016 0.016 2 Ka = = = 1.2 102 0.021 CHCOOH 2 2 ions in solution, or if we could determine CHCOO 2 K a 2 Ka2 could be determined if we had some way to measure the total concentration of all (e) Practically all the H 3O + arises from the first ionization. Equation: CHCOOH 2 (aq) + H 2O(l) H CHCOO 2 (aq) + H 3O + (aq) Initial: 0.0500 M Changes: x M Equil:
Ka 0M +x M
xM 0 M +x M
xM 0.0500 x M [H(CHCOO) 2 ][H 3O ] x2 1.2 102 [(CHCOOH)2 ] 0.0500 x
x 2 0.012 x 0.00060 0 x 2 0.00060 0.012 x x= b b 2 4ac 0.012 0.00014 + 0.0024 = = 0.019 M = H 3O + 2a 2 pH = log 0.019 = 1.72 b g 776 Chapter 16: Acids and Bases 101. (D) (a)
x2 x2 4.2 104 x1 0.00102 0.00250 x 0.00250 x2 x2 4.2 104 x2 0.000788 0.00250 0.00102 0.00148 x2 x2 4.2 104 x3 0.000848 0.00250 0.000788 0.00171 x2 x2 4.2 104 x4 0.000833 0.00250 0.000848 0.00165 x2 x2 4.2 104 x5 0.000837 0.00250 0.000833 0.00167 x2 x2 4.2 104 x6 0.000836 0.00250 0.000837 0.00166 x6 0.000836 or 8.4 104 , which is the same as the value obtained using the quadratic equation. (b) We organize the solution around the balanced chemical equation, as we have done before. Equation: HClO 2 (aq) + H 2 O(l) ClO 2 (aq) + H 3 O + (aq)
Initial: Changes: Equil: 0.500 M xM xM 0M +x M xM 0 M +x M xM ClO 2 H 3O + x2 x2 Ka = = = 1.1 102 0.500 x 0.500 HClO 2 x= x 2 0.500 1.1 102 0.074 M Assuming x 0.500 M , Not significantly smaller than 0.500 M. Assume x 0.074 x (0.500 0.0774) 1.1102 0.068 M Try once more. Assume x 0.068 x (0.500 0.068) 1.110 2 0.069 M Assume x 0.069 x (0.500 0.069) 1.1102 0.069 M One more time. Final result! H 3O + = 0.069 M, pH = log H 3O + = log 0.069 = 1.16 777 Chapter 16: Acids and Bases 102. (D) (a) Here, two equilibria must be satisfied simultaneously. The common variable, H 3O + = z . Equation: HC2 H3O2 (aq) + Initial: Changes: Equil: 0.315 M +x M H 2 O(l) C2 H 3O 2 (aq) 0M +x M xM + H 3O + (aq) 0M +z M zM 0.315 x M H 3O + C 2 H 3 O 2 xz = Ka = = 1.8 105 0.315 x HC 2 H 3O 2 Equation: Initial: Changes: Equil:
+ HCHO 2 (aq) 0.250 M y M (0.250 y ) M + H 2 O(l) CHO 2 (aq) + 0M +y M yM H 3 O + (aq) 0M +z M zM H 3 O CHO 2 yz = Ka = = 1.8 104 0.250 y HCHO 2 In this system, there are three variables, x = C 2 H 3O 2 , y = CHO 2 , and z = H 3O + . These three variables also represent the concentrations of the only charged species in the reaction, and thus x + y = z . We solve the two Ka expressions for x and y. (Before we do so, however, we take advantage of the fact that x and y are quite small: x 0.315 and y 0.250 .) Then we substitute these results into the expression for z , and solve to obtain a value of that variable. xz = 1.8 105 0.315 = 5.7 106 yz = 1.8 104 0.250 = 4.5 105 5.7 106 4.5 105 y= z z 6 5 5 5.7 10 4.5 10 5.07 10 z = x+ y = + = z = 5.07 105 = 7.1103 M = [H3O + ] z z z pH = log 7.1 103 = 2.15 We see that our assumptions about the sizes of x and y x= c h must be valid, since each of them is smaller than z . (Remember that x + y = z .)
(b) We first determine the initial concentration of each solute. 12.5g NH 2 1 mol NH 3 = 1.96M NH3 = 0.375L soln 17.03g NH 3 1.55 g CH 3 NH 2 1 mol CH 3 NH 2 = 0.133M CH3 NH 2 = 0.375L soln 31.06 g CH 3 NH 2 K b = 1.8 105 K b = 4.2 104 Now we solve simultaneously two equilibria, which have a common variable, OH = z . 778 Chapter 16: Acids and Bases Equation: NH 3 aq 1.96 M Changes: x M Equil: Initial: + H2O NH 4 + aq 0M +x M xM + OH aq 0M +z M zM 1.96 x M NH 4 + OH xz xz K b = = 1.8 105 = 1.96 x 1.96 NH 3 Eqn: CH 3 NH 2 aq + H 2 O(l) CH 3 NH3+ aq 0M yM yM + OH aq 0M z M zM Initial: 0.133M Changes: y M Equil: (0.133 y ) M CH 3 NH 3+ OH yz yz = 4.2 104 = Kb = = 0.133 y 0.133 CH 3 NH 2 In this system, there are three variables, x = NH 4 + , y = CH 3 NH 3 + , and z = OH . These three variables also represent the concentrations of the only charged species in solution in substantial concentration, and thus x + y = z . We solve the two Kb expressions for x and y . Then we substitute these results into the expression for z , and solve to obtain the value of that variable. xz = 1.96 1.8 105 = 3.53 105 x= 3.53 105 z yz = 0.133 4.2 105 = 5.59 105 y= 5.59 105 z 3.53 105 5.59 105 9.12 105 + = z 2 = 9.12 105 z z z 3 z = 9.5 10 M = OH pOH = log 9.5 103 = 2.02 pH = 14.00 2.02 = 11.98 z = x+ y = We see that our assumptions about x and y (that x 1.96 M and y 0.133 M) must be valid, since each of them is smaller than z . (Remember that x + y = z .)
(c) In 1.0 M NH 4 CN there are the following species: NH 4 + aq , NH 3 aq , CN aq , HCN(aq), H 3O aq , and OH aq . These species are related by the following six equations.
+ b g b g b g b g b g 1 K w = 1.0 1014 = H3O+ OH 1.0 1014 OH = H O+ 3 b2g K = 6.2 10
a 10 = H 3O + CN HCN b3g
779 Kb = 1.8 10 = 5 NH 4 + OH LM NH 3 OP Q N Chapter 16: Acids and Bases 4 NH3 + NH 4 + = 1.0 M 5 HCN + CN = 1.0 M + 6 NH 4 + H3O+ = CN + OH NH3 = 1.0 NH 4 + HCN = 1.0 CN or NH 4 + CN Equation (6) is the result of charge balance, that there must be the same quantity of positive and negative charge in the solution. The approximation is the result of remembering that not much H 3O + or OH will be formed as the result of hydrolysis of ions. Substitute equation (4) into equation (3), and equation (5) into equation (2). b2'gK = 4.0 10 = 1.0 CN Now substitute equation (6) into equation b2'g , and equation (1) into equation b3'g . NH 1.8 10 b2''gK = 6.2 10 = H O NH b3''g 1.0 10 = H O NH NH 1.0 e1.0 j
b b3'gK = 1.8 10 = 5 NH 4 + OH + 10 H 3O + CN 1.0 NH 4
+ a + 10 3 4 5 + 4 a + 14 + + 4 3 4 Now we solve both of these equations for NH 4 + . b`2g:6.2 10 b`3g:1.8 10 10 6.2 10
+ 10 NH 4
9 + = H 3O
+ + NH 4
+ + NH 4
+ + 6.2 1010 = 6.2 1010 + H 3O + = 1.8 109 H 3O + 1.00 +1.8 109 H 3O + 9 H 3O 1.8 10 H 3O NH 4 = NH 4 NH 4 + We equate the two results and solve for 1.8 109 H 3O + 6.2 1010 + H 3O . = 6.2 10 10 + H 3O + 1.00 +1.8 109 H 3O + 6.2 1010 +1.1 H 3O + = 1.1 H 3O + +1.8 109 H 3O +
+ 2 6.2 1010 = 1.8 109 H 3O + 2 6.2 10 10 H 3O = 5.9 1010 M pH = log 5.9 1010 = 9.23 9 18 10 . Ka K w Note that H 3 O + = or pH=0.500 (pK a pK w pK b ) Kb c h 780 Chapter 16: Acids and Bases SELFASSESSMENT EXERCISES
103. (E) (a) KW: Dissociation constant of water, which is defined as [H3O+][OH] (b) pH: A logarithmic scale of expressing acidity of a solution (H3O+); it is defined as log [H3O+] (c) pKa: A logarithmic scale expressing the strength of an acid (which is how much an acid dissociates in water at equilibrium); it is defined as log Ka (d) Hydrolysis: A reaction involving ionization of water (e) Lewis acid: A compound or species that accepts electrons 104. (E) (a) Conjugate base: A base that is derived by removing an abstractable proton from the acid (b) Percent ionization of acid/base: The ratio between the conjugate of the acid or base under study and the acid/base itself times 100 (c) Selfionization: A reaction involving ionization of two molecules of water to give [H3O+] and [OH] (d) Amphiprotic behavior: A substance acting as either an acid or a base 105. (E) (a) BrnsteadLowry acid and base: A BrnstedLowry acid is a substance that when placed in water will cause the formation of hydronium ion by donating a proton to water. The base is one that abstracts a proton from water and results in a hydroxyl ion. (b) [H3O+] and pH: [H3O+] is the molar concentration of hydronium ion in water, whereas pH is the log [H3O+] (c) Ka for NH4+ and Kb for NH3: Kb of NH3 is the equilibrium constant for hydrolysis of water with NH3 to generate NH4+ and OH. Ka of NH4+ is for deprotonation of NH4+ with water to give NH3 and OH. (d) Leveling effect and electronwithdrawing effect: The leveling effect is the result of reaction of water with any acid or base such that no acid or base can exist in water that are more acidic than H3O+ or more basic than OH. The electronwithdrawing effect is the tendency for a central atom with high electronegativity in an oxoacid to withdraw electrons from the OH bond. 106. (E) The answer is (a), HCO3, because it can donate a proton to water to become CO32, or it can be protonated to give H2CO3 (i.e., CO2(aq)). 107. (E) The answer is (c). CH3CH2COOH is a weak acid, so the concentration of H3O+ ions it will generate upon dissociation in water will be significantly less than 0.1 M. Therefore, its pH will be higher the 0.10 M HBr, which dissociates completely in water to give 0.10 M H3O+. 108. (E) The answer is (d). CH3NH2 is a weak base. 781 Chapter 16: Acids and Bases 109. (M) The answer is (e) because CO32 is the strongest base and it drives the dissociation of acetic acid furthest. 110. (M) The answer is (c). H2SO4 dissociation is nearly complete for the first proton, giving a [H3O+] = 0.10 M (pH = 1). The second dissociation is not complete. Therefore, (c) is the only choice that makes sense. 111. (M) The answer is (b). You can write down the stepwise equations for dissociation of H2SO3 and then the dissociation of HSO3 and calculate the equilibrium concentration of each species. However, you can determine the answer without detailed calculations. The two dissociation equations are given below: Eq.1 Eq.2 H 2SO3 H 2 O HSO3 H 3O Ka1 = 1.3102 Ka2 = 6.3108 2 HSO3 H 2 O SO3 H 3O For the first equation, the equilibrium concentrations, of HSO3 and H3O+ are the same. They become the initial concentration values for the second dissociation reaction. Since Ka2 is 6.3108 (which is small), the equilibrium concentrations of HSO3 and H3O+ won't change significantly. So, the equation simplifies as follows:
22SO3 H 3O + +x SO3 H 3O + K a2 6.8 10 = HSO3 x HSO3 8 2K a2 =6.8 108 SO3 112. (E) Since both the acid and the base in this reaction are strong, they dissociate completely. The pH is determined as follows: mol H3O+: 0.248 M HNO3 0.02480 L = 0.00615 mol mol OH: 0.394 M KOH 0.01540 L = 0.00607 mol Final: 0.00615 0.00607 = 8.00105 mol H3O+ 8.00 105 mol H 3O 0.00200 M 0.02480 L + 0.01540 L pH log 0.00200 2.70
113. (M) Since pH = 3.25, the [H3O+] = 103.25 = 5.62104 M. Using the reaction below, we can determine the equilibrium concentrations of other species Initial Change Equil. HC2H3O2 C x Cx + H2O C2H3O2 0 +x x + H3O+ 0 +x x 782 Chapter 16: Acids and Bases Since x = 5.62104 M, the equilibrium expression becomes H 3O C2 H 3O 2 5.62 104 M 5.62 104 M Ka 1.8 105 4 HC2 H3O2 C 5.62 10 M Solving C gives a concentration of 0.0181 M. Since concentrated acetic acid is 35% by weight and the density is 1.044 g/mL, the concentration of acetic acid is: 35 g HAc 1.044 g Conc. HAc 1 mol HAc 6.084 M HAc 100 g Conc. HAc sol'n 0.001 L sol'n 60.06 g HAc Therefore, volume of concentrated solution required to make 12.5 L of 0.0181 M HAc solution is: M V dilute M V conc. 0.0181 M 12500 mL 37.2 mL V conc 6.084 M 114. (M) Table 16.3 has the Ka value for chloroacetic acid, HC2H2ClO2. Ka = 1.4103. The solution is made of the chloroacetate salt, which hydrolyzes water according to the following reaction: Initial Change Equil. C2H2ClO2 + H2O 2.05 x 2.05 x HC2H2ClO2 + OH 0 0 +x +x x x K W 1.0 1014 7.14 1012 Kb 3 K a 1.4 10 C2 H 2 ClO 2 xx 7.14 1012 2.05 x Solving the above simplified expression for x, we get x = OH  =3.826106 pH=14pOH=14 log 3.826106 =8.58 Kb HC2 H 2ClO2 OH  783 Chapter 16: Acids and Bases 115. (M) 0.10 M HI < 0.05 M H2SO4 < 0.05 M HC2H2ClO2 < 0.50 M HC2H3O2 < 0.50 M NH4Cl < 1.0 M NaBr < 0.05 M KC2H3O2 < 0.05 M NH3 < 0.06 M NaOH < 0.05 M Ba(OH)2 116. (M) Since pH = 5 pOH, pH must be significantly larger than pOH, which means that the solution is basic. The pH can be determined by solving two simultaneous equations: pH 5 pOH 0 pH + pOH = 14 pOH = 2.333 pH = 14 2.33 = 11.67. Therefore, [H3O+] = 2.141012 M (and [OH] = 4.67103 M) The solute must be NH3, because it is the only basic species (the other two are acidic and neutral, respectively). Since the Kb of NH3 is 1.8105, the concentration of NH3 can be determined as follows:
+ OH  NH 4 4.67 103 4.67 103 1.8 105 Kb = x NH3 x 1.2 M 117. (M) The answer is (a). If HC3H5O2 is 0.42% ionized for a 0.80 M solution, then the concentration of the acid at equilibrium is (0.800.0042) = 0.00336. The equilibrium expression for the dissociation of HC3H5O2 is: H 3O C3 H 5O 2 0.00336 0.00336 Ka 1.42 105 HC3 H 5O 2 0.800 0.00336 118. (E) The answer is (b), because H2PO4 is a result of addition of one proton to the base HPO42. 119. (M) The answer is (d). This is because, in the second equation, HNO2 will give its proton to ClO (rather than HClO giving its proton to NO2), which means that HNO2 is a stronger acid than HClO. Also, in the first equation, ClO is not a strong enough base to abstract a proton from water, which means that HClO is in turn a stronger acid than H2O. 120. (M) The dominant species in the solution is ClO2. Therefore, determine its concentration and ionization constant first: 3.00 mol CaClO 2 2 mol ClO 2 ClO = 2.40 M 2.50 L 1 mol CaClO 2 2 K W 1.0 1014 9.1 1013 Kb 2 Ka 1.1 10 784 Chapter 16: Acids and Bases The reaction and the dissociation of ClO2 is as follows: ClO2 2.40 x 2.40 x + H2O Initial Change Equil. HClO2 0 +x x + OH 0 +x x HClO 2 OH  Kb ClO 2 9.1 1013 xx 2.40 x Solving the above simplified expression for x, we get x = OH  =1.478106 . pH=14pOH=14 log 1.478106 =8.2 121. (M) Section 168 discusses the effects of structure on acid/base behavior. The major subtopics are effects of structure on binary acids, oxo acids, organic acids, and amine bases. For binary acids, acidity can be discussed in terms of electronegativity of the main atom, bond length, and heterolytic bond dissociation energy. For oxo acids, the main things affecting acidity are the electronegativity of the central atom and number of oxygen atoms surrounding the central atom. For organic acids, the main topic is the electronwithdrawing capability of constituent groups attached to the carboxyl unit. For amines, the basicity is discussed in terms of electron withdrawing or donating groups attached to the nitrogen in the amine, and the number of resonance structures possible for the base molecule. 785 CHAPTER 17 ADDITIONAL ASPECTS OF ACIDBASE EQUILIBRIA
PRACTICE EXAMPLES
1A (D) Organize the solution around the balanced chemical equation, as we have done before. Equation: HF(aq) H 2 O(l) Initial: 0.500 M Changes: xM Equil: (0.500 x) M Ka H 3 O + (aq) F (aq) 0M +x M xM 0M +x M xM [ H 3O + ][ F ] ( x ) ( x ) x2 6.6 104 [ HF] 0.500 x 0.500 assuming x 0.500 x 0.500 6.6 104 0.018 M One further cycle of approximations gives: x (0.500 0.018) 6.6 104 0.018 M [H 3 O + ] Thus, [HF] = 0.500 M 0.018 M 0.482 M Recognize that 0.100 M HCl means H 3O + initial 0.100 M , since HCl is a strong acid. Equation: HF(aq) + H 2 O(l) Initial: 0.500 M Changes: x M Equil: (0.500 x ) M H 3 O + (aq) + F (aq) 0.100 M +x M (0.100+x) M 0M +x M xM 0100 x . [ H 3O + ][ F ] ( x ) (0100 + x ) . Ka assuming x 0.100 6.6 104 0.500 0.500 x [HF] 6.6 104 0.500 = 3.3 103 M = F x= 0.100 The assumption is valid. HF = 0.500 M 0.003M = 0.497 M
[H3O+] = 0.100 M + x = 0.100 M + 0.003 M = 0.103 M 786 Chapter 17: Additional Aspects of AcidBase Equilibria 1B (M) From Example 176 in the text, we know that [ H 3O + ] [C 2 H 3O 2 ] 13 103 M in . 0.100 M HC 2 H 3O 2 . We base our calculation, as usual, on the balanced chemical equation. The concentration of H 3O + from the added HCl is represented by x. H 3 O + (aq) + C 2 H 3 O 2 (aq) Equation: HC 2 H 3 O 2 (aq) + H 2 O(l) Initial: 0.100 M Changes: 0.00010 M From HCl: Equil: 0.100 M 0M +0.00010 M +x M 0M +0.00010 M (0.00010 x) M 0.00010 M Ka = [ H 3O + ][C 2 H 3O 2 ] (0.00010 + x ) 0.00010 . 18 105 [ HC 2 H 3O 2 ] 0100 . 1.8 105 0.100 0.018 M 0.00010
0.018 mol H 3O + 1L 1mol HCl 1mol H 3 O
+ 0.00010 x x 0.018 M 0.00010 M 0.018 M 1 L soln 12 mol HCl 1000 mL 1L 1 drop 0.050 mL 30.drops V12 M HCl =1.00 L Since 30. drops corresponds to 1.5 mL of 12 M solution, we see that the volume of solution does indeed remain approximately 1.00 L after addition of the 12 M HCl.
2A (M) We again organize the solution around the balanced chemical equation. Equation: HCHO 2 aq + H 2 O(l) CHO 2 aq + H 3 O + aq Initial: 0.100 M 0.150 M 0M Changes: xM Equil: (0.100 x) M +x M (0.150 + x) M +x M xM
assuming x 0.100 [CHO 2 ][H 3O ] (0.150 + x)( x) 0.150 x 1.8 104 Ka [HCHO 2 ] 0.100 x 0.100
x= 0.100 1.8 104 = 1.2 104 M = H 3 O + , x <<0.100, thus our assumption is valid 0.150 CHO 2 = 0.150 M + 0.00012 M = 0.150 M 787 Chapter 17: Additional Aspects of AcidBase Equilibria 2B (M) This time, a solid sample of a weak base is being added to a solution of its conjugate acid. We let x represent the concentration of acetate ion from the added sodium acetate. Notice that sodium acetate is a strong electrolyte, thus, it completely dissociates in aqueous solution. H 3O + = 10 pH = 10 5.00 = 1.0 10 5 M = 0.000010 M Equation: HC 2 H 3 O 2 aq + H 2 O(l) C 2 H 3 O 2 aq + H 3 O + aq Initial: 0.100 M 0M 0M Changes: 0.000010 M +0.000010 M +0.000010 M From NaAc: +x M Equil: 0.100 M (0.000010 + x) M 0.000010 M H 3O + C2 H 3O 2 0.000010 0.000010 + x = Ka = = 1.8 105 0.100 HC 2 H 3O 2 0.000010 + x 1.8 105 0.100 0.18 M 0.000010 x 0.18 M 0.000010 M = 0.18 M mass of NaC2 H 3O 2 = 1.00 L 0.18 mol C2 H 3O 2 1mol NaC2 H3O 2 82.03g NaC2 H3O 2 1L 1mol NaC2 H 3O 2 1mol C2 H 3O 2 =15g NaC2 H 3O 2 3A (M) A strong acid dissociates essentially completely, and effectively is a source of H 3O + . NaC2 H 3O 2 also dissociates completely in solution. The hydronium ion and the acetate ion react to form acetic acid: H 3O (aq) C2 H 3O 2 (aq) HC 2 H 3O 2 (aq) H 2 O(l) All that is necessary to form a buffer is to have approximately equal amounts of a weak acid and its conjugate base together in solution. This will be achieved if we add an amount of HCl equal to approximately half the original amount of acetate ion. 3B (M) HCl dissociates essentially completely in water and serves as a source of hydronium ion. This reacts with ammonia to form ammonium ion: NH3 aq + H3O+ aq NH 4+ aq + H 2 O(l) . Because a buffer contains approximately equal amounts of a weak base (NH3) and its conjugate acid (NH4+), to prepare a buffer we simply add an amount of HCl equal to approximately half the amount of NH3(aq) initially present. 788 Chapter 17: Additional Aspects of AcidBase Equilibria 4A (M) We first find the formate ion concentration, remembering that NaCHO 2 is a strong electrolyte, existing in solution as Na + aq and CHO 2 aq . b g [CHO 2 ] 23.1g NaCHO 2 1000 mL 1 mol NaCHO 2 1 mol CHO 2 0.679 M 500.0 mL soln 1L 68.01g NaCHO 2 1 mol NaCHO 2 As usual, the solution to the problem is organized around the balanced chemical equation. Equation: HCHO 2 aq + H 2 O(l) CHO 2 aq + H 3 O + aq 0M Initial: 0.432 M 0.679 M Changes: x M Equil: (0.432 x) M +x M (0.679 + x) M +x M xM 0.432 1.8 104 0.679 H 3O + CHO 2 x 0.679 + x 0.679 x = Ka = = 1.8 104 0.432 x 0.432 HCHO 2 x= This gives H 3O + = 1.14 104 M . The assumption that x 0.432 is clearly correct. pH log H 3O log 1.14 104 3.94 3.9 4B (M) The concentrations of the components in the 100.0 mL of buffer solution are found via the dilution factor. Remember that NaC2 H 3O 2 is a strong electrolyte, existing in solution as Na + aq and C 2 H 3O 2 aq . b g b g [HC 2 H 3O 2 ] 0.200 M 63.0 mL 100.0 mL 0.126 M [C 2 H 3O 2 ] 0.200 M 37.0 mL 100.0 mL 0.0740 M As usual, the solution to the problem is organized around the balanced chemical equation. Equation: Initial: Changes: Equil: HC2 H 3O 2 aq + H 2O(l) C2 H3O 2 aq + H 3O + aq 0.0740 M 0 M 0.126 M x M x M x M (0126 x ) M . 0.0740 + x M xM b g [H O + ][C2 H 3O 2 ] x (0.0740 x) 0.0740 x Ka 3 1.8 105 [HC2 H3O 2 ] 0.126 x 0.126
x 18 105 0126 . . 31 105 M = [H 3O + ] ; pH = log [ H 3O + ] log 31 105 4.51 . . 0.0740 Note that the assumption is valid: x 0.0740 0.126. Thus, x is neglected when added or subtracted 789 Chapter 17: Additional Aspects of AcidBase Equilibria 5A (M) We know the initial concentration of NH 3 in the buffer solution and can use the pH to find the equilibrium [OH]. The rest of the solution is organized around the balanced chemical equation. Our first goal is to determine the initial concentration of NH 4 . pOH = 14.00 pH = 14.00 9.00 5.00 Equation: Initial: Changes: Equil: [OH ] 10 pOH 105.00 10 105 M . NH3 aq + H 2 O(l) NH 4 + aq + OH aq 0 M 1.0 105 M 0.35 M 1.0 105 M (0.35 1.0 105 ) M xM 1.0 105 M ( x 10 105 ) M . 10 105 M . Kb [NH 4 ][OH ] ( x 1.0 105 )(1.0 105 ) 1.0 105 x 1.8 105 [NH 3 ] 0.35 1.0 105 0.35 0.35 1.8 105 = 0.63 M = initial NH 4 + concentration 5 1.0 10
0.63 mol NH 4 + 1 mol NH 4 2 SO 4 132.1 g NH 4 2 SO 4 1 L soln 1 mol NH 4 2 SO 4 2 mol NH 4+ Assume x 1.0 105 x = mass NH 4 2 SO 4 = 0.500 L Mass of (NH4)2SO4 = 21 g
5B (M) The solution is composed of 33.05 g NaC2H3O2 3 H2O dissolved in 300.0 mL of 0.250 M HCl. NaC2H3O2 3 H2O , a strong electrolyte, exists in solution as Na+(aq) and C2H3O2(aq) ions. First we calculate the number of moles of NaC2H3O2 3 H2O, which, based on the 1:1 stoichiometry, is also equal to the number of moles of C2HO that are released into solution. From this we can calculate the initial [C2H3O2] assuming the solution's volume remains at 300. mL. moles of NaC2 H 3O 2 3 H2O (and moles of C2H3O2) 33.05 g NaC2 H 3 O 2 3H 2 O = 0.243moles NaC 2 H 3 O 2 3H 2 O moles C 2 H 3 O 2 1 mole NaC2 H3 O2 3H 2 O 136.08 g NaC2 H3 O2 3H 2 O 0.243 mol C2 H 3O 2 0.810 M [C2 H 3O 2 ] 0.300 L soln (Note: [HCl] is assumed to remain unchanged at 0.250 M) We organize this information around the balanced chemical equation, as before. We recognize that virtually all of the HCl has been hydrolyzed and that hydronium ion will react to produce the much weaker acetic acid. 790 Chapter 17: Additional Aspects of AcidBase Equilibria HC2 H 3O 2 (aq) + H 2 O (l) C2 H 3O 2 (aq) + H 3O + (aq) 0.810 M 0.250 M 0M +0.250 M 0.250 M 0.250 M 0.250 M 0.560 M 0 M +x M +x M Changes: x M Equil: (0.250 x) M (0.560 + x) M +x M + [H O ][C2 H 3O 2 ] x (0.560 x) 0.560 x Ka 3 1.8 105 [HC2 H 3O 2 ] 0.250 x 0.250 Equation: Initial: Form HAc: 1.8 105 0.250 8.0 106 M = [H 3O + ] 0.560 (The approximation was valid since x << both 0.250 and 0.560 ) pH = log [H3 O + ] log 8.0 106 5.09 5.1
x
6A (D) (a) (b) For formic acid, pKa log (18 104 ) 3.74. The HendersonHasselbalch equation . provides the pH of the original buffer solution: [CHO 2 ] 0.350 pH = pKa log 3.74 log 354 . [ HCHO 2 ] 0.550 The added acid can be considered completely reacted with the formate ion to produce formic acid. Each mole/L of added acid consumes 1 M of formate ion and forms 1 M of + formic acid: CHO2 (aq) + H3 O (aq) HCHO2 (aq) + H 2 O(l). Kneut = Kb/Kw 5600. Thus, CHO 2 0.350 M 0.0050 M = 0.345 M and HCHO2 = 0.550 M + 0.0050 M = 0.555 M . By using the HendersonHasselbalch equation [CHO 2 ] 0.345 pH = pK a + log = 3.74 + log = 3.53 [HCHO2 ] 0.555 (c) Added base reacts completely with formic acid producing, an equivalent amount of formate ion. This continues until all of the formic acid is consumed. Each 1 mole of added base consumes 1 mol of formic acid and forms 1 mol of formate ion: HCHO 2 + OH CHO 2 + H 2O . Kneut = Ka/Kw 1.8 1010. Thus, CHO 2 = 0.350 M + 0.0050 M = 0.355 M HCHO2 = 0.550 0.0050 M = 0.545 M . With the HendersonHasselbalch equation HCHO 2 CHO 2 = 3.74 + log 0.355 = 3.55 we find pH = pK a + log 0.545 791 Chapter 17: Additional Aspects of AcidBase Equilibria 6B (D) The buffer cited has the same concentration as weak acid and its anion, as does the buffer of Example 176. Our goal is to reach pH = 5.03 or H 3O + = 10 pH = 105.03 = 9.3 106 M . Adding strong acid H 3O + , of course, produces HC2 H 3O 2 at the expense of C 2 H 3O 2 . Thus, adding H+ drives the reaction to the left. Again, we use the data around the balanced chemical equation. Equation: HC2 H 3O 2 aq + H 2O(l) C 2 H 3O 2 aq + H 3O + aq c h b g b g Initial: Add acid: Form HAc: Changes: Equil:
Ka 0.250 M 0.560 M b0.250 + yg M +y M b0.560 yg M y M b
b x M 0.250 + y x M g b x M 0.560 y + x M g 8.0 106 M +y M y M 0 M x M 9.3 106 M [H 3 O + ][C 2 H 3 O 2 ] 9.3 106 (0.560 y x) 9.3 10 6 (0.560 y ) 1.8 105 [HC 2 H 3 O 2 ] 0.250 y x 0.250 y (Assume that x is negligible compared to y) 1.8 105 0.250 + y 0.560 0.484 = 0.484 +1.94 y = 0.560 y y= = 0.026 M 6 1.94 +1.00 9.3 10 Notice that our assumption is valid: x 0.250 + y = 0.276 0.560 y = 0.534 .
VHNO3 300.0 mL buffer 0.026 mmol H 3O + 1 mL HNO3 aq = 1.3 mL of 6.0 M HNO3 1 mL buffer 6.0 mmol H 3O + g Instead of the algebraic solution, we could have used the HendersonHasselbalch equation, since the final pH falls within one pH unit of the pKa of acetic acid. We let z indicate the increase in HC 2 H 3O 2 , and also the decrease in C 2 H 3O 2 pH = pKa + log L = 4.74 + log = 5.03 0.250 + z MN HC 2 H 3O 2 OPQ C 2 H 3O 2 0.560 z 0.560 z = 105.03 4.74 = 1.95 0.250 + z 0.560 0.488 = 0.024 M 1.95 +1.00 This is, and should be, almost exactly the same as the value of y we obtained by the I.C.E. table method. The slight difference is due to imprecision arising from rounding errors. 0.560 z = 1.95 0.250 + z = 0.488 +1.95 z b g z= 7A (D) (a) The initial pH is the pH of 0.150 M HCl, which we obtain from H 3O + of that strong acid solution. [H 3O + ] . 0150 mol HCl 1 mol H 3O + 0150 M, . 1 L soln 1 mol HCl pH = log [H 3O + ] log (0150) 0.824 . 792 Chapter 17: Additional Aspects of AcidBase Equilibria (b) To determine H 3O + and then pH at the 50.0% point, we need the volume of the solution and the amount of H 3O + left unreacted. First we calculate the amount of hydronium ion present and then the volume of base solution needed for its complete neutralization. amount H 3O + = 25.00 mL Vacid = 3.75 mmol H 3O + 15.0 mL titrant 0.150 mmol HCl 1 mmol H 3O + = 3.75 mmol H 3O + 1 mL soln 1 mmol HCl 1 mmol OH 1 mmol NaOH 1 mL titrant + 1 mmol H3O 1 mmol OH 0.250 mmol NaOH At the 50.0% point, half of the H3O+ (1.88 mmol H 3O + ) will remain unreacted and only half (7.50 mL titrant) of the titrant solution will be added. From this information, and the original 25.00mL volume of the solution, we calculate H 3O + and then pH.
1.88 mmol H 3O left H 3O + = 25.00 mL original + 7.50 mL titrant = 0.0578 M pH = log 0.0578 = 1.238 (c) Since this is the titration of a strong acid by a strong base, at the equivalence point, the pH = 7.00 . This is because the only ions of appreciable concentration in the equivalence point solution are Na+(aq) and Cl(aq), and neither of these species undergoes detectable hydrolysis reactions. Beyond the equivalence point, the solution pH is determined almost entirely by the concentration of excess OH(aq) ions. The volume of the solution is 40.00 mL +1.00 mL = 41.00 mL. The amount of hydroxide ion in the excess titrant is calculated and used to determine OH , from which pH is computed.
amount of OH = 1.00 mL OH = 0.250 mmol NaOH = 0.250 mmol OH 1 mL (d) 0.250 mmol OH = 0.006098 M 41.00 mL pOH = log 0.006098 = 2.215; pH = 14.00 2.215 = 11.785 7B (D) (a) The initial pH is simply the pH of 0.00812 M Ba OH 2 , which we obtain from OH b g for the solution. 1 L soln OH = 0.00812 mol Ba OH b g 2 2 mol OH 1 mol Ba OH b g = 0.01624 M
2 pOH = log OH = log 0.0162 = 1.790; pH = 14.00 pOH = 14.00 1.790 = 12.21 793 Chapter 17: Additional Aspects of AcidBase Equilibria (b) To determine OH and then pH at the 50.0% point, we need the volume of the solution and the amount of OH unreacted. First we calculate the amount of hydroxide ion present and then the volume of acid solution needed for its complete neutralization.
amount OH = 50.00 mL 0.00812 mmol Ba OH 2 1mL soln
+ 2 mmol OH 1mmol Ba OH 2 = 0.812 mmol OH Vacid = 0.812 mmol OH 1mmol H 3 O 1mmol HCl 1mL titrant = 32.48 mL titrant + 1mmol OH 1mmol H 3 O 0.0250 mmol HCl At the 50.0 % point, half (0.406 mmol OH ) will remain unreacted and only half (16.24 mL titrant) of the titrant solution will be added. From this information, and the original 50.00mL volume of the solution, we calculate OH and then pH. OH = 0.406 mmol OH left = 0.00613 M 50.00 mL original +16.24 mL titrant pOH = log 0.00613 = 2.213; pH = 14.00 pOH = 11.79 (c) Since this is the titration of a strong base by a strong acid, at the equivalence point,pH = 7.00. The solution at this point is neutral because the dominant ionic species in solution, namely Ba2+(aq) and Cl(aq), do not react with water to a detectable extent. Initial pH is just that of 0.150 M HF ( pKa = log 6.6 104 = 3.18 ). [Initial solution contains 20.00 mL Equation : HF aq Initial : 0.150 M Changes : x M Equil: (0.150 x) M
+  8A (D) (a) c h 0.150 mmol HF 1 mL =3.00 mmol HF]
+ F aq 0M +x M xM + H 2O(l) H 3O + aq 0M +x M xM 2 H3 O F = x x x = 6.6 104 Ka = 0.150 x 0.150 HF approximation, H 3O + = 9.6 103 M ; pH = log 9.6 103 = 2.02 x 0.150 6.6 104 9.9 103 M x 0.05 0.150 . The assumption is invalid. After a second cycle of 794 Chapter 17: Additional Aspects of AcidBase Equilibria (b) When the titration is 25.0% complete, there are (0.253.00=) 0.75 mmol F for every 3.00 mmol HF that were present initially. Therefore, (3.000.75=) 2.25 mmol HF remain untitrated. We designate the solution volume (the volume holding these 3.00 mmol total) as V and use the HendersonHasselbalch equation to find the pH. F 0.75 mmol / V = 2.70 pH = pKa + log = 3.18 + log HF 2.25 mmol / V
(c) (d) At the midpoint of the titration of a weak base, pH = pKa = 3.18. At the endpoint of the titration, the pH of the solution is determined by the conjugate base hydrolysis reaction. We calculate the amount of anion and the volume of solution in order to calculate its initial concentration. 0.150 mmol HF 1 mmol F amount F = 20.00 mL = 3.00 mmol F 1 mL soln 1 mmol HF 1mmol OH 1 mL titrant volume titrant = 3.00 mmol HF = 12.0 mL titrant 1mmol HF 0.250 mmol OH 3.00 mmol F F = 20.00 mL original volume +12.0 mL titrant = 0.0938 M We organize the solution of the hydrolysis problem around its balanced equation.
Equation : Initial: Changes : Equil : + H 2O(l) 0.0938M xM F aq HF aq 0M +x M xM + OH aq 0M +x M xM 0.0938 x M F Kb = HF OH K w 1.0 1014 x x x2 11 = = = 1.5 10 = Ka 6.6 104 0.0938 x 0.0938 x 0.0934 1.5 10 11 1.2 10 6 M = [OH ] The assumption is valid (x 0.0934). pOH = log 1.2 106 = 5.92; pH = 14.00 pOH = 14.00 5.92 = 8.08
8B (D) (a) The initial pH is simply that of 0.106 M NH 3 . + Equation: NH 3 aq + H 2 O(l) NH 4 aq + OH aq Initial: 0.106 M Changes: x M Equil: 0.106 x M b g 0M +x M xM 0 M +x M x M 795 Chapter 17: Additional Aspects of AcidBase Equilibria NH 4 + OH = Kb = NH 3 xx 0.106 x x2 0.106 = 1.8 105
 x 0.106 1.8 10 4 1.4 103 M = [OH ] The assumption is valid (x << 0.106). pOH = log 0.0014 = 2.85 pH = 14.00 pOH = 14.00 2.85 = 11.15
(b) When the titration is 25.0% complete, there are 25.0 mmol NH 4 for every 100.0 mmol of NH 3 that were present initially (i.e., there are 1.33 mmol of NH4+ in solution), 3.98 mmol NH 3 remain untitrated. We designate the solution volume (the volume holding these 5.30 mmol total) as V and use the basic version of the HendersonHasselbalch equation to find the pH.
1.33 mmol NH 4 + = 4.74 + log V pOH = pK b + log = 4.26 3.98 mmol NH 3 V + pH = 14.00 4.26 = 9.74
(c) (d) At the midpoint of the titration of a weak base, pOH = pK b = 4.74 and pH = 9.26 At the endpoint of the titration, the pH is determined by the conjugate acid hydrolysis reaction. We calculate the amount of that cation and the volume of the solution in order to determine its initial concentration. amount NH 4 + = 50.00 mL 0.106 mmol NH 3 1 mmol NH 4 1 mL soln 1 mmol NH 3
+ + amount NH 4 + = 5.30 mmol NH 4
Vtitrant = 5.30 mmol NH 3 1 mmol H 3O + 1 mL titrant = 23.6 mL titrant 1 mmol NH 3 0.225 mmol H 3O +
+ 5.30 mmol NH 4 NH 4 + = 50.00 mL original volume + 23.6 mL titrant = 0.0720 M We organize the solution of the hydrolysis problem around its balanced chemical equation.
+ Equation: NH 4 aq + H 2 O(l) NH3 aq + H 3O + aq 0M 0 M Initial: 0.0720 M Changes: x M +x M +x M Equil: 0.0720 x M xM x M b g 796 Chapter 17: Additional Aspects of AcidBase Equilibria Kb = NH 3 H 3O + NH 4
+ Kw 1.0 1014 xx x2 = 5.6 1010 = = = Kb 1.8 10 5 0.0720 x 0.0720 x 0.0720 5.6 1010 6.3 10 6 M=[H 3 O + ] The assumption is valid (x << 0.0720). pH = log 6.3 106 = 5.20
9A (M) The acidity of the solution is principally the result of the hydrolysis of the carbonate ion, which is considered first. c h Equation: Initial: Changes: Equil: + H 2 O(l) HCO3 aq + OH aq 0M 0 M 1.0 M x M +x M +x M 1.0 x M x M x M CO3 2 aq b g HCO3 OH xx x2 1.0 1014 Kw 4 Kb = = = 2.110 = 11 1.0 x 1.0 CO32 K a (HCO3 ) 4.7 10 x 1.0 2.1 104 1.5 102 M 0.015 M [OH ] The assumption is valid (x 1.0 M ). Now we consider the hydrolysis of the bicarbonate ion. Equation: Initial: Changes: Equil: HCO3 aq + H 2 O(l) 0.015 M y M 0.015 y M H 2CO3 aq + OH aq 0M +y M y M 0.015 M +y M 0.015 + y M b g b g H 2 CO 3 OH y 0.015 + y 0.015 y 1.0 1014 Kw 8 =y = 2.3 10 = = = Kb = F 7 0.015 x 0.015 Ka H H 2 CO 3 I 4.4 10 HCO 3 K b g The assumption is valid (y << 0.015) and y = H 2 CO 3 = 2.3 108 M. Clearly, the second hydrolysis makes a negligible contribution to the acidity of the solution. For the entire solution, then pOH = log OH = log 0.015 = 1.82
9B b g pH = 14.00 1.82 = 12.18 (M) The acidity of the solution is principally the result of hydrolysis of the sulfite ion. 2 Equation: SO3 aq + H 2 O(l) HSO3 aq + OH aq Initial: Changes: Equil: 0.500 M x M 0.500 x M b g 0M +x M xM 0 M +x M x M 797 Chapter 17: Additional Aspects of AcidBase Equilibria HSO3 OH Kw x x x2 1.0 1014 7 Kb = = = 1.6 10 2 6.2 108 [SO3 ] 0.500 x 0.500 K a HSO3
x 0.500 1.6 107 2.8 104 M = 0.00028 M = [OH ] The assumption is valid (x << 0.500). Next we consider the hydrolysis of the bisulfite ion. Equation: HSO3 aq + H 2O(l) H 2SO3 aq + OH aq Initial: 0.00028 M 0M 0.00028 M +y M +y M Changes: y M Equil: 0.00028 y M y M 0.00028 + y M b g b g Kb = Kw 1.0 1014 = = 7.7 1013 2 K a H 2SO3 1.3 10
13 K b = 7.7 10 = H 2SO3 OH HSO3 = y 0.00028 + y 0.00028 y =y 0.00028 y 0.00028 The assumption is valid (y << 0.00028) and y = H 2 SO3 = 7.7 1013 M. Clearly, the second hydrolysis makes a negligible contribution to the acidity of the solution. For the entire solution, then pOH = log OH = log 0.00028 = 3.55 b g pH = 14.00 3.55 = 10.45 INTEGRATIVE EXAMPLE
A. (D) From the given information, the following can be calculated: pH of the solution = 2.716 therefore, [H+] = 1.92103 pH at the halfway point = pKa pH = 4.602 = pKa pKa = log Ka therefore Ka = 2.50105
FP = 0 C + Tf Tf = i K f m Tf 1 1.86 C / m molality 798 Chapter 17: Additional Aspects of AcidBase Equilibria molality = # moles solute # moles solute = kg solvent 0.500 kg  0.00750 kg To determine the number of moles of solute, convert 7.50 g of unknown acid to moles by using its molar mass. The molar mass can be calculated as follows: pH = 2.716
HA 7.50 g MM [H+] = 1.92103 H+ + A Initial Change Equilibrium 0.500 L 0 0 x 7.50 g MM 1.92 103 0.500 L 2.50 105 = 2.50 105 [H + ][A  ] [HA] (1.92 103 ) 2 x
1.92103 x
1.92103 7.50 MM 1.92 103 0.500 MM =100.4 g/mol 1 mol 0.0747 mol 100.4 g # moles of solute = 7.50 g Molality = # moles solute 0.0747 mol 0.152 m = Kg solvent 0.500 Kg  0.00750 Kg Tf = i K f m Tf 1 1.86 C / m 0.152 m Tf 0.283 C FP = 0C + Tf = 0C 0.283 C = 0.283 C 799 Chapter 17: Additional Aspects of AcidBase Equilibria B. (M) By looking at the titration curve provided, one can deduce that the titrant was a strong acid. The pH before titrant was added was basic, which means that the substance that was titrated was a base. The pH at the end of the titration after excess titrant was added was acidic, which means that the titrant was an acid. Based on the titration curve provided, the equivalence point is at approximately 50 mL of titrant added. At the halfway point, of approximately 25 mL, the pH = pKa. A pH ~8 is obtained by extrapolation a the halfway point. pH = 8 = pKa Ka = 108 = 1 108 Kb = Kw/Ka = 1 106 ~50 mL of 0.2 M strong acid (1102 mol) was needed to reach the equivalence point. This means that the unknown contained 1102 mol of weak base. The molar mass of the unknown can be determined as follows:
0.800 g 80 g/mol 1 102 mol EXERCISES
The CommonIon Effect
1. (M) (a) Note that HI is a strong acid and thus the initial H 3O + = HI = 0.0892 M
Equation : Initial : Changes : Equil : HC3 H 5 O 2 + 0.275 M x M 0.275 x M H 2 O C3 H 5 O 2 + H 3 O + 0M 0.0892M +x M +x M xM 0.0892 + x M C3 H 5 O 2 H 3O + x 0.0892 + x 0.0892 x Ka = = 1.3 105 = 0.275 x 0.275 HC3 H 5 O 2 The assumption that x 0.0892 M is correct. H 3O + = 0.0892 M
(b) (c) x 4.0 105 M OH = Kw 1.0 1014 = = 1.1 1013 M + 0.0892 H 3O C3 H 5O 2 = x = 4.0 105 M 800 Chapter 17: Additional Aspects of AcidBase Equilibria (d) 2. I = HI int = 0.0892 M (M) (a) The NH 4 Cl dissociates completely, and thus, NH + = Cl = 0.102 M 4 int int Equation: NH 3 (aq) + H 2 O(l) NH 4 (aq) + OH (aq) Initial: 0.164 M Changes: x M Equil: 0.164 x M b g b 0.102 M +x M 0.102 + x M g 0 M +x M x M NH + OH 0.102 + x x 0.102 x 4 Kb = = = 1.8 105 ; x = 2.9 105 M 0.164 x 0.164 NH 3 Assumed x 0.102 M , a clearly valid assumption. OH = x = 2.9 10 5 M
(b) (c) (d) 3. NH 4 = 0.102 + x = 0.102 M Cl = 0.102 M H 3O
+ 1.0 1014 = = 3.4 1010 M 5 2.9 10 (M) (a) We first determine the pH of 0.100 M HNO 2 . Equation HNO 2 (aq) + H 2 O(l) NO 2 (aq) + H 3 O + (aq) 0M + xM xM 0.100 M xM Changes : Equil : Initial : 0M +x M xM 0.100 x M NO 2 H 3 O + x2 K a = = 7.2 104 = 0.100 x HNO 2 Via the quadratic equation roots formula or via successive approximations, x = 8.1 103 M = H 3O + . Thus pH = log 8.1 103 = 2.09 When 0.100 mol NaNO 2 is added to 1.00 L of a 0.100 M HNO 2 , a solution with NO 2 = 0.100M = HNO 2 is produced. The answer obtained with the HendersonHasselbalch equation, is pH = pKa = log 7.2 104 = 3.14 . Thus, the addition has caused a pH change of 1.05 units. c h 801 Chapter 17: Additional Aspects of AcidBase Equilibria (b) NaNO 3 contributes nitrate ion, NO 3 , to the solution. Since, however, there is no molecular HNO 3 aq in equilibrium with hydrogen and nitrate ions, there is no equilibrium to be shifted by the addition of nitrate ions. The [H3O+] and the pH are thus unaffected by the addition of NaNO 3 to a solution of nitric acid. The pH changes are not the same because there is an equilibrium system to be shifted in the first solution, whereas there is no equilibrium, just a change in total ionic strength, for the second solution. b g 4. (M) The explanation for the different result is that each of these solutions has acetate ion present, C 2 H 3O 2 , which is produced in the ionization of acetic acid. The presence of this ion suppresses the ionization of acetic acid, thus minimizing the increase in H 3O + . All three solutions are buffer solutions and their pH can be found with the aid of the HendersonHasselbalch equation.
(a)
pH = pKa + log C 2 H 3O 2 HC 2 H 3O 2 = 4.74 + log 0.10 = 3.74 1.0 H 3O + = 103.74 = 1.8 104 M % ionization = H 3O HC 2 H 3O 2 100% 1.8 104 M 100% 0.018% 1.0M (b) C 2 H 3O 2 = 4.74 + log 0.10 = 4.74 pH = pK a + log 0.10 HC 2 H 3 O 2 H 3O + = 104.74 = 1.8 105 M 1.8 105 M % ionization = 100% 100% 0.018% HC 2 H 3O 2 0.10M
(c) H 3O C 2 H 3O 2 = 4.74 + log 0.10 = 5.74 pH = pK a + log 0.010 HC 2 H 3O 2 H 3O + = 105.74 = 1.8 106 M % ionization = H 3O HC 2 H 3O 2 100% 1.8 106 M 100% 0.018% 0.010M 5. (M) (a) The strong acid HCl suppresses the ionization of the weak acid HOCl to such an extent that a negligible concentration of H 3O + is contributed to the solution by HOCl. Thus, H 3O + = HCl = 0.035 M 802 Chapter 17: Additional Aspects of AcidBase Equilibria (b) This is a buffer solution. Consequently, we can use the HendersonHasselbalch equation to determine its pH. pK a = log 7.2 104 = 3.14; NO 2 = 3.14 + log 0.100 M = 3.40 pH = pK a + log 0.0550 M HNO 2 H 3O + = 103.40 = 4.0 104 M
(c) This also is a buffer solution, as we see by an analysis of the reaction between the components. H 3O + aq, from HCl + C 2 H 3O 2 aq, from NaC 2 H 3 O 2 HC2 H 3O 2 aq + H 2 O(l) Equation: In soln: Produce HAc: Initial: 0.0525 M 0.0525 M 0M 0.0768 M 0.0525 M 0.0243 M 0M +0.0525 M 0.0525 M Now the HendersonHasselbalch equation can be used to find the pH. pKa = log 1.8 105 = 4.74
C2 H3 O2 = 4.74 + log 0.0243M = 4.41 pH = pK a + log 0.0525 M HC 2 H 3 O 2 H 3 O + = 104.41 = 3.9 105 M 6. (M) (a) Neither Ba 2+ aq nor Cl aq hydrolyzes to a measurable extent and hence they c h b g
2 b g have no effect on the solution pH. Consequently, OH is determined entirely by the Ba OH OH = b g solute. 2 mol OH = 0.012 M 1mol Ba OH 2 0.0062 mol Ba OH 2 1 L soln (b) We use the HendersonHasselbalch equation to find the pH for this buffer solution.
2 mol NH 4 NH 4 + = 0.315 M NH 4 SO 4 = 0.630 M 2 1 mol NH 4 2 SO 4
+ pK a = 9.26 for NH 4 . + pH = pK a + log NH3 NH 4 + = 9.26 + log 0.486 M = 9.15 pOH = 14.00 9.15 = 4.85 0.630 M OH = 104.85 = 1.4 105 M 803 Chapter 17: Additional Aspects of AcidBase Equilibria (c) This solution also is a buffer, as analysis of the reaction between its components shows. Equation: NH 4 aq, from NH 4 Cl + OH aq, from NaOH NH 3 aq + H 2 O(l) In soln: Form NH 3 : Initial: 0.264 M 0.196 M 0.068 M 0.196 M 0.196 M 0M 0M +0.196 M 0.196 M pH = pK a + log NH3 = 9.26 + log 0.196 M = 9.72
NH + 4 0.068 M pOH = 14.00 9.72 = 4.28 Buffer Solutions
7. (M) H 3O + = 104.06 = 8.7 105 M . We let S = CHO 2 int Equation: HCHO 2 aq + H 2 O(l) Initial : 0.366 M Changes : 8.7 10 M Equil : 0.366 M 5 CHO 2 aq SM
5 5 + H 3 O + aq 0M +8.7 105 M 8.7 105 M + 8.7 10 M S + 8.7 10 M
4 Ka = H 3O + CHO 2 LM HCHO 2 OP Q N = 1.8 10 cS + 8.7 10 h8.7 10 =
5 5 0.366 8.7 105 S ; S = 0.76 M 0.366 To determine S , we assumed S 8.7 105 M , which is clearly a valid assumption. Or, we could have used the HendersonHasselbalch equation (see below). pKa = log 1.8 104 = 3.74 c h CHO 2 ; 4.06 = 3.74 + log HCHO 2 CHO 2 = 2.1; HCHO 2 CHO 2 = 2.1 0.366 = 0.77 M The difference in the two answers is due simply to rounding.
8. (E) We use the HendersonHasselbalch equation to find the required [NH3]. pKb = log 1.8 105 = 4.74 pKa = 14.00 pKb = 14.00 4.74 = 9.26 pH = 9.12 = 9.26 + log NH 3 NH 4
+ c h NH3 NH 4 + = 100.14 = 0.72 NH3 = 0.72 NH 4+ = 0.72 0.732 M = 0.53M 804 Chapter 17: Additional Aspects of AcidBase Equilibria 9. (M) (a) Equation: HC7 H 5O 2 aq + H 2 O(l) C7 H 5O 2 (aq) + 0.033 M Initial: 0.012 M +x M Changes: x M 0.033 + x M Equil: 0.012 x M H 3O + (aq) 0 M +x M x M x = 2.3 105 M b g H 3O + C7 H 5 O x 0.033 + x 0.033x 2 Ka = = 6.3 105 = 0.012 x 0.012 HC 7 H 5 O 2 To determine the value of x , we assumed x 0.012 M, which is an assumption that clearly is correct. H 3O + = 2.3 105 M pH = log 2.3 105 = 4.64 c h (b) Equation: Initial : Changes : Equil :
Kb = NH 4
+ NH 3 aq + H 2 O(l) 0.408 M x M NH 4 (aq) 0.153 M +x M + OH (aq) 0M +xM xM
x = 4.8 105 M 0.408 x M
OH 0.153 + x M LM NH 3 OP N Q = 1.8 105 = x 0.153 + x 0.153x 0.408 x 0.408 b g To determine the value of x , we assumed x 0.153 , which clearly is a valid assumption. OH = 4.8 105 M; pOH = log 4.8 105 = 4.32; 10. pH = 14.00 4.32 = 9.68 (M) Since the mixture is a buffer, we can use the HendersonHasselbalch equation to determine Ka of lactic acid. C3 H 5 O3 = = 0.0892 M 100.0 mL soln 1 L soln 112.1 g NaC3 H 5 O 3 1 mol NaC3 H 5 O 3 1.00 g NaC3 H 5 O3 1000 mL 1 mol NaC3 H 5 O3 1 mol C3 H 5 O 3 C3 H 5 O 3 = pK + log 0.0892 M = pK + 0.251 pH = 4.11 = pK a + log a a 0.0500 M HC3 H 5 O3 pK a = 4.11 0.251 = 3.86; K a = 103.86 = 1.4 104
11. (M) (a) 0.100 M NaCl is not a buffer solution. Neither ion reacts with water to a detectable extent. (b) 0.100 M NaCl0.100 M NH 4 Cl is not a buffer solution. Although a weak acid, + NH 4 , is present, its conjugate base, NH3, is not. 805 Chapter 17: Additional Aspects of AcidBase Equilibria (c) + 0.100 M CH 3 NH 2 and 0.150 M CH 3 NH 3 Cl is a buffer solution. Both the weak + base, CH 3 NH 2 , and its conjugate acid, CH 3NH 3 , are present in approximately equal concentrations. 0.100 M HCl0.050 M NaNO 2 is not a buffer solution. All the NO 2 has converted to HNO 2 and thus the solution is a mixture of a strong acid and a weak acid. (d) (e) 0.100 M HCl0.200 M NaC2 H 3O 2 is a buffer solution. All of the HCl reacts with half of the C 2 H 3O 2 to form a solution with 0.100 M HC2 H 3O 2 , a weak acid, and 0.100 M C 2 H 3O 2 , its conjugate base. 0.100 M HC 2 H 3O 2 and 0.125 M NaC 3 H 5O 2 is not a buffer in the strict sense because it does not contain a weak acid and its conjugate base, but rather the conjugate base of another weak acid. These two weak acids (acetic, Ka = 1.8 10 5 , and propionic, Ka = 1.35 105 ) have approximately the same strength, however, this solution would resist changes in its pH on the addition of strong acid or strong base, consequently, it could be argued that this system should also be called a buffer. (f) 12. (M) 2 (a) Reaction with added acid: HPO 4 + H 3O + H 2 PO 4 + H 2O Reaction with added base: H 2 PO 4 + OH HPO 4 (b) 2 + H2O We assume initially that the buffer has equal concentrations of the two ions, H 2 PO 4 = HPO 4 2 HPO 4 2 = 7.20 + 0.00 = 7.20 (pH at which the buffer is most effective). pH = pK a 2 + log H 2 PO4 (c) HPO 4 2 = 7.20 + log 0.150 M = 7.20 + 0.48 = 7.68 pH = 7.20 + log 0.050 M H 2 PO 4 13. 1 mol C6 H 5 NH 3+ Cl 1 mol C6 H 5 NH 3+ 1g (M) moles of solute = 1.15 mg 1000 mg 129.6 g 1mol C6 H 5 NH 3+ Cl = 8.87 106 mol C6 H 5 NH 3+ C 6 H 5 NH 3 Equation: Initial: Changes: Equil: + 8.87 106 mol C 6 H 5 NH 3 + = = 2.79 10 6 M 3.18 L soln
+ C6 H 5 NH 2 aq + H 2 O(l) C6 H 5 NH 3 aq + OH aq 0M +x M xM b 0.105 M x M 0.105 x M g c
806 2.79 106 M +x M 2.79 106 + x M h Chapter 17: Additional Aspects of AcidBase Equilibria Kb = C 6 H 5 NH 3 + OH LMC6 H 5 NH 2 OP Q N = 7.4 10 10 c2.79 10 = 6 +x x h 0.105 x 7.4 1010 0.105 x = 2.79 106 + x x; 7.8 1011 7.4 1010 x = 2.79 106 x + x 2 x 2 + 2.79 106 + 7.4 1010 x 7.8 1011 = 0;
x= x 2 + 2.79 106 x 7.8 1011 0 b b 2 4ac 2.79 106 7.78 1012 + 3.1 1010 = = 7.5 106 M = OH 2a 2 pOH = log 7.5 106 = 5.12
14. c h pH = 14.00 5.12 = 8.88 (M) We determine the concentration of the cation of the weak base. 1 mmol C6 H 5 NH 3 Cl 1 mmol C6 H 5 NH 3 + 129.6 g 1 mmol C6 H 5 NH 3 Cl + [C6 H 5 NH 3 ] 0.0874 M 1L 750 mL 1000 mL In order to be an effective buffer, each concentration must exceed the ionization constant Kb = 7.4 1010 by a factor of at least 100, which clearly is true. Also, the
+ + 8.50 g c h ratio of the two concentrations must fall between 0.1 and 10: + C 6 H 5 NH 3 0.0874 M LMC6 H 5 NH 2 OP = 0.215M = 0.407 . Q N Since both criteria are met, this solution will be an effective buffer.
15. (M) (a) First use the HendersonHasselbalch equation. pK b = log 1.8 105 = 4.74, pK a = 14.00 4.74 = 9.26 to determine NH 4 + in the buffer solution.
pH = 9.45 = pK a + log NH 3 NH3 NH 4 + NH + 4 = 100.19 = 1.55 NH 4 NH 4 + NH 3 0.258M NH 4 + = 1.55 = 1.55 = 0.17M
+ = 9.26 + log NH 3 ; log NH 3 = 9.45 9.26 = +0.19 We now assume that the volume of the solution does not change significantly when the solid is added. + 1 L soln 0.17 mol NH 4 1 mol NH 4 2 SO 4 mass NH 4 2 SO 4 = 425 mL + 1000 mL 1 L soln 2 mol NH 4 132.1 g NH 4 2 SO 4 1 mol NH 4 2 SO 4 = 4.8 g NH 4 2 SO 4 807 Chapter 17: Additional Aspects of AcidBase Equilibria (b) We can use the HendersonHasselbalch equation to determine the ratio of concentrations of cation and weak base in the altered solution.
pH = 9.30 = pK a + log NH 3 + NH3 NH + 4 = 100.04 NH 4 0.258 = 1.1 = 0.17 M + x M = 9.26 + log NH 3 NH 4 + log NH 3 NH 4 + = 9.30 9.26 = +0.04 0.19 1.1x 0.258 x 0.062 M The reason we decided to add x to the denominator follows. (Notice we cannot remove a component.) A pH of 9.30 is more acidic than a pH of 9.45 and therefore the conjugate acid's NH 4 + concentration must increase. Additionally, d i mathematics tells us that for the concentration ratio to decrease from 1.55 to 1.1, its denominator must increase. We solve this expression for x to find a value of 0.062 M. We need to add NH4+ to increase its concentration by 0.062 M in 100 mL of solution. NH 4 2 SO 4 mass = 0.100 L + 0.062 mol NH 4 1L 1mol NH 4 2 SO 4 2 mol NH + 4 132.1g 1mol NH 4 2 SO 4 NH 4 2 SO 4 = 0.41g NH 4 2 SO 4 Hence, we need to add 0.4 g 16. (M) (a) nHC7 H5O2 = 2.00 g HC7 H 5O 2 nC H O = 2.00 g NaC7 H 5O 2 7 5 2 1mol HC7 H 5O 2 = 0.0164 mol HC7 H 5O 2 122.1g HC7 H 5O 2
1mol NaC7 H 5O 2 1mol C7 H 5O 2 144.1g NaC7 H 5O 2 1mol NaC7 H 5O 2 = 0.0139 mol C7 H 5O 2 C 7 H 5 O 2 = log 6.3 10 5 + log 0.0139 mol C7 H 5 O 2 /0.7500 L pH = pK a + log HC H O 0.0164 mol HC 7 H 5 O 2 /0.7500 L 7 5 2 = 4.20 0.0718 = 4.13 (b) To lower the pH of this buffer solution, that is, to make it more acidic, benzoic acid must be added. The quantity is determined as follows. We use moles rather than concentrations because all components are present in the same volume of solution. 4.00 = 4.20 + log 0.0139 mol C 7 H 5O 2 x mol HC 7 H 5O 2 x= log 0.0139 mol C 7 H 5O 2 = 0.20 x mol HC 7 H 5O 2 0.0139 mol C7 H 5O 2 = 100.20 = 0.63 x mol HC7 H 5O 2 0.0139 = 0.022 mol HC7 H 5O 2 (required) 0.63 808 Chapter 17: Additional Aspects of AcidBase Equilibria HC7 H 5O 2 that must be added = amount required amount already in solution HC7 H 5O 2 that must be added = 0.022 mol HC7 H 5O 2 0.0164 mol HC7 H 5O 2 HC7 H 5O 2 that must be added = 0.006 mol HC7 H 5O 2 122.1g HC 7 H 5O 2 added mass HC 7 H 5O 2 = 0.006 mol HC 7 H 5O 2 = 0.7g HC 7 H 5O 2 1 mol HC 7 H 5O 2
17. (M) The added HCl will react with the ammonia, and the pH of the buffer solution will decrease. The original buffer solution has NH 3 = 0.258 M and NH 4 = 0.17 M . We first calculate the [HCl] in solution, reduced from 12 M because of dilution. [HCl] 0.55 mL added = 12 M = 0.066 M We then determine pKa for ammonium ion: 100.6 mL pKb = log 1.8 105 = 4.74 pKa = 14.00 4.74 = 9.26
+ H 3O + aq NH 4 aq + H 2 O l Buffer: 0.258 M 0M 0.17 M Added: +0.066 M Changes: 0.066 M 0.066 M +0.066 M Final: 0.192 M 0M 0.24 M NH 3 0.192 pH = pKa + log L = 9.26 + log = 9.16 +O 0.24 MN NH 4 PQ c h Equation: NH 3 aq + 18. (M) The added NH 3 will react with the benzoic acid, and the pH of the buffer solution will increase. Original buffer solution has [C7H5O2] = 0.0139 mol C7H5O2/0.750 L = 0.0185 M and HC7 H 5O 2 = 0.0164 mol HC7 H 5O 2 /0.7500 L = 0.0219 M . We first calculate the NH 3 in solution, reduced from 15 M because of dilution. NH3 added = 15 M Equation: NH 3 aq Buffer: 0M Added: 0.0070 M Changes: 0.0070M Final: 0.000 M 0.35 mL = 0.0070 M 750.35 mL For benzoic acid, pKa = log 6.3 105 = 4.20 + HC7 H 5O 2 aq NH 4+ aq + C7 H 5O 2 aq c h 0.0219 M 0.0070 M 0.0149 M 0M + 0.0070 M 0.0070 M 0.0185 M + 0.0070 M 0.0255 M C7 H 5 O 2 = 4.20 + log 0.0255 = 4.43 pH = pK a + log 0.0149 HC 7 H 5 O 2 809 Chapter 17: Additional Aspects of AcidBase Equilibria 19. (M) The pKa 's of the acids help us choose the one to be used in the buffer. It is the acid with a pKa within 1.00 pH unit of 3.50 that will do the trick. pKa = 3.74 for HCHO 2 , pKa = 4.74 for HC2 H 3O 2 , and pK a1 = 2.15 for H 3 PO 4 . Thus, we choose HCHO 2 and NaCHO 2 to prepare a buffer with pH = 3.50 . The HendersonHasselbalch equation is used to determine the relative amount of each component present in the buffer solution. CHO 2 pH = 3.50 = 3.74 + log HCHO2 CHO 2 = 100.24 = 0.58 HCHO 2 CHO 2 = 3.50 3.74 = 0.24 log HCHO 2 This ratio of concentrations is also the ratio of the number of moles of each component in the buffer solution, since both concentrations are a number of moles in a certain volume, and the volumes are the same (the two solutes are in the same solution). This ratio also is the ratio of the volumes of the two solutions, since both solutions being mixed contain the same concentration of solute. If we assume 100. mL of acid solution, Vacid = 100. mL . Then the volume of salt solution is Vsalt = 0.58 100. mL = 58 mL 0.100 M NaCHO2
20. (D) We can lower the pH of the 0.250 M HC 2 H 3O 2  0.560 M C 2 H 3O 2 buffer solution by increasing HC 2 H 3O 2 or lowering C 2 H 3O 2 . Small volumes of NaCl solutions will have no effect, and the addition of NaOH(aq) or NaC 2 H 3O 2 aq will raise the pH. The addition of 0.150 M HCl will raise HC 2 H 3O 2 and lower C 2 H 3O 2 through the reaction H 3O + (aq) + C 2 H 3O 2 (aq) HC 2 H 3O 2 (aq) + H 2 O(l) and bring about the desired lowering of the pH. We first use the HendersonHasselbalch equation to determine the ratio of the concentrations of acetate ion and acetic acid. C 2 H 3O 2 pH = 5.00 = 4.74 + log L O NM HC 2 H 3O 2 QP C 2 H 3O 2 = 5.00 4.74 = 0.26; C2 H 3O 2 = 100.26 = 1.8 log HC 2 H 3O 2 HC 2 H 3O 2 Now we compute the amount of each component in the original buffer solution. b g amount of C2 H 3O 2 = 300. mL amount of HC2 H 3O 2 = 300. mL 0.560 mmol C 2 H 3O 2 1 mL soln = 168 mmol C 2 H 3O 2 0.250 mmol HC2 H 3O 2 = 75.0 mmol HC2 H 3O 2 1mL soln 810 Chapter 17: Additional Aspects of AcidBase Equilibria Now let x represent the amount of H 3O + added in mmol. 1.8 = 168 x ; 168 x = 1.8 75 + x = 135 +1.8 x 75.0 + x 168 135 = 2.8 x x= 168 135 = 12 mmol H 3O + 2.7
1mmol HCl 1mL soln + 1mmol H 3 O 0.150 mmol HCl Volume of 0.150 M HCl = 12 mmol H 3 O + = 80 mL 0.150 M HCl solution
21. (M) (a) The pH of the buffer is determined via the HendersonHasselbalch equation.
pH = pKa + log L 0.100M = 4.89 + log = 4.89 O 0.100M NM HC 3 H 5O 2 QP C 3 H 5O 2 The effective pH range is the same for every propionate buffer: from pH = 3.89 to pH = 5.89 , one pH unit on either side of pKa for propionic acid, which is 4.89.
(b) To each liter of 0.100 M HC3 H 5O 2  0.100M NaC 3 H 5O 2 we can add 0.100 mol OH before all of the HC3 H 5O 2 is consumed, and we can add 0.100 mol H 3O + before all of the C 3 H 5O 2 is consumed. The buffer capacity thus is 100. millimoles (0.100 mol) of acid or base per liter of buffer solution. 22. (M) (a) The solution will be an effective buffer one pH uniton either side of the pKa of methylammonium ion, CH 3 NH 3 , K b = 4.2 104 for methylamine, pKb = log 4.2 104 = 3.38 . For methylammonium cation, pKa = 14.00 3.38 = 10.62 . Thus, this buffer will be effective from a pH of 9.62 to a pH of 11.62. The capacity of the buffer is reached when all of the weak base or all of the conjugate acid has been neutralized by the added strong acid or strong base. Because their concentrations are the same, the number of moles of base is equal to the number of moles of conjugate acid in the same volume of solution. 0.0500 mmol amount of weak base = 125 mL = 6.25 mmol CH 3 NH 2 or CH 3 NH 3 1 mL Thus, the buffer capacity is 6.25 millimoles of acid or base per 125 mL buffer solution. c h (b) 811 Chapter 17: Additional Aspects of AcidBase Equilibria 23. (M) (a) The pH of this buffer solution is determined with the HendersonHasselbalch equation. CHO 2 = log 1.8 104 + log 8.5 mmol/75.0 mL pH = pK a + log 15.5 mmol/75.0 mL HCHO 2 = 3.74 0.26 = 3.48 [Note: the solution is not a good buffer, as CHO 2 = 1.1 101 , which is only ~ 600 times Ka ]
(b) Amount of added OH = 0.25 mmol Ba OH 2 b g 2 mmol OH 1 mmol Ba OH b g = 0.50 mmol OH 2 The OH added reacts with the formic acid and produces formate ion. Equation: HCHO 2 (aq) + OH (aq) CHO 2 (aq) + H 2 O(l) Buffer: Add base: React: Final: 0.50 mmol 15.0 mmol 15.5 mmol 0M +0.50 mmol 0.50 mmol 0 mmol +0.50 mmol 9.0 mmol 8.5 mmol CHO 2 = log 1.8 104 + log 9.0 mmol/75.0 mL pH = pK a + log 15.0 mmol/75.0 mL HCHO 2 = 3.74 0.22 = 3.52
(c)
12 mmol HCl 1 mmol H 3 O + = 13 mmol H 3O + Amount of added H 3O = 1.05 mL acid 1 mL acid 1 mmol HCl
+ The H 3O added reacts with the formate ion and produces formic acid.
Equation: Buffer : Add acid : React : 8.5 mmol CHO 2 (aq) 8.5 mmol + H 3 O + (aq) 0 mmol 13mmol 8.5 mmol +8.5 mmol HCHO 2 (aq) + H 2 O(l) 15.5 mmol Final : 0 mmol 4.5 mmol 24.0 mmol The buffer's capacity has been exceeded. The pH of the solution is determined by the excess strong acid present. 4.5 mmol H 3O + = 75.0 mL +1.05 mL = 0.059 M; pH = log 0.059 = 1.23 812 Chapter 17: Additional Aspects of AcidBase Equilibria 24. (D) For NH 3 , pKb = log 1.8 105 For NH 4 , pKa = 14.00 pKb = 14.00 4.74 = 9.26 (a) c h NH3 = 1.68g NH 3 1 mol NH 3 = 0.197 M 0.500 L 17.03 g NH 3
4.05 g NH 4 2 SO 4 0.500 L NH 4 = 2 mol NH 4 = 0.123 M 132.1 g NH 4 2 SO 4 1mol NH 4 2 SO 4 1 mol NH 4 2 SO 4 NH 3 0.197 M pH = pKa + log L = 9.26 + log = 9.46 O 0.123 M MN NH 4 PQ
(b) The OH aq reacts with the NH 4 aq to produce an equivalent amount of NH 3 aq . OH =
i b g b g b g 0.88 g NaOH 1mol NaOH 1mol OH = 0.044 M 0.500 L 40.00 g NaOH 1mol NaOH NH 4 aq + OH aq 0.123M 0.044 M 0.079 M 0M 0.044 M 0.044 M 0.0000 M Equation: Initial : Add NaOH : React : Final : NH 3 aq + 0.197 M +0.044 M 0.241M H 2 O(l) NH 3 0.241 M pH = pKa + log L = 9.26 + log = 9.74 0.079 M MN NH 4 OPQ
(c) Equation: Initial : Add HCl : React : Final : NH 3 aq + H 3O + aq NH 4 aq 0.197 M 0.123M 0M +x M + xM x M x M + H 2 O(l) 0.197 x M
NH 3 1109 M 0 0.123 + x M 0.197 x = 9.26 + log 0.123 + x MN NH 4 OPQ 0.197 x 0.197 x M = 9.00 9.26 = 0.26 log 0.123 + x 0.123 + x M pH = 9.00 = pKa + log L b b g g 0.197 x = 0.55 0.123 + x = 0.068 + 0.55 x b g b b b b gM gM g M = 10 gM 0.26 = 0.55 1.55 x = 0.197 0.068 = 0.129 x= 0.129 = 0.0832 M 1.55 0.0832 mol H 3O + 1mol HCl 1000 mL HCl = 3.5 mL 1L soln 1mol H 3O + 12 mol HCl volume HCl = 0.500 L 813 Chapter 17: Additional Aspects of AcidBase Equilibria 25. (D) (a) We use the HendersonHasselbalch equation to determine the pH of the solution. The total solution volume is
36.00 mL + 64.00 mL = 100.00 mL. pK a = 14.00 pK b = 14.00 + log 1.8 105 = 9.26 NH 3 = 36.00 mL 0.200 M NH 3 7.20 mmol NH 3 = 0.0720 M = 100.0 mL 100.00 mL 64.00 mL 0.200 M NH 4 12.8 mmol NH 4 = = 0.128 M 100.00 mL 100.0 mL
[NH 3 ] 0.0720 M 9.26 log 9.01 9.00 0.128 M [NH 4 ] NH 4 = pH = pK a log (b) The solution has OH = 104.99 = 1.0 105 M The HendersonHasselbalch equation depends on the assumption that: NH3 1.8 105 M NH 4 If the solution is diluted to 1.00 L, NH 3 = 7.20 103 M , and NH 4 = 1.28 102 M . These concentrations are consistent with the assumption. However, if the solution is diluted to 1000. L, NH 3 = 7.2 106 M , and NH 3 = 1.28 105 M , and these two concentrations are not consistent with the assumption. Thus, in 1000. L of solution, the given quantities of NH 3 and NH 4 will not produce a solution with pH = 9.00 . With sufficient dilution, the solution will become indistinguishable from pure water (i.e.; its pH will equal 7.00).
(c) The 0.20 mL of added 1.00 M HCl does not significantly affect the volume of the solution, but it does add 0.20 mL 1.00 M HCl = 0.20 mmol H 3O + . This added H 3O + reacts with NH 3 , decreasing its amount from 7.20 mmol NH 3 to 7.00 mmol NH 3 , and increasing the amount of NH 4 from 12.8 mmol NH 4 to 13.0 mmol NH 4 , as the reaction: NH 3 + H 3O + NH 4 + H 2 O pH = 9.26 + log 7.00 mmol NH 3 / 100.20 mL = 8.99 13.0 mmol NH 4 / 100.20 mL (d) We see in the calculation of part (c) that the total volume of the solution does not affect the pOH of the solution, at least as long as the HendersonHasselbalch equation is obeyed. We let x represent the number of millimoles of H 3O + added, 814 Chapter 17: Additional Aspects of AcidBase Equilibria through 1.00 M HCl. This increases the amount of NH 4 and decreases the amount of NH 3 , through the reaction NH 3 + H 3O + NH 4 + H 2 O 7.20 x 7.20 x pH = 8.90 = 9.26 + log ; log = 8.90 9.26 = 0.36 12.8 + x 12.8 + x Inverting, we have: 12.8 + x = 100.36 = 2.29; 12.8 + x = 2.29 7.20 x = 16.5 2.29 x 7.20 x x= 16.5 12.8 = 1.1 mmol H 3O + 1.00 + 2.29
1mmol HCl 1mmol H 3O
+ vol 1.00 M HCl = 1.1mmol H 3O + 26. (D) (a)
C2 H 3 O 2 = 1mL soln 1.00 mmol HCl = 1.1mL 1.00 M HCl 12.0 g NaC 2 H 3 O 2 1mol NaC 2 H 3 O 2 1mol C 2 H 3 O 2 = 0.488 M C2 H 3 O 2 0.300 L soln 82.03g NaC 2 H 3 O 2 1mol NaC 2 H 3 O 2 Equation : Initial : Add C2 H 3 O 2 Buffer : HC2 H 3 O 2 + H 2 O C2 H 3 O 2 + 0M 0M 0M + 0.200 M 0.200 M 0.488 M 0.200 M 0.288 M H3O+ 0.200 M 0M 0.200 M 0M Consume H 3 O + Then use the HendersonHasselbalch equation to find the pH.
C 2 H 3O 2 pH = pKa + log L = 4.74 + log = 4.74 + 0.16 = 4.90 0.200 M MN HC2 H 3O 2 OPQ 0.288 M (b) We first calculate the initial OH due to the added Ba OH 2 . OH = 1.00 g Ba OH 0.300 L b g
2 b g 2 b g 171.3 g BabOH g
1 mol Ba OH
2 2 2 mol OH 1 mol Ba OH b g = 0.0389 M Then HC 2 H 3O 2 is consumed in the neutralization reaction shown directly below. HC2 H 3O 2 + OH C2 H 3O 2 + H 2 O Initial: 0.200 M 0.0389 M 0.288 M  Consume OH : 0.0389 M 0.0389 M +0.0389 M  Buffer: 0.161 M ~ 0M 0.327 M  Equation: 815 Chapter 17: Additional Aspects of AcidBase Equilibria Then use the HendersonHasselbalch equation.
pH = pKa + log L 0.327 M = 4.74 + log = 4.74 + 0.31 = 5.05 O 0.161 M NM HC 2 H 3O 2 QP
2 3 2 C 2 H 3O 2 (c) b g can be added up until all of the HC H O is consumed. BabOH g + 2 HC H O BabC H O g + 2 H O Ba OH
2
2 2 3 2 2 3 2 2 2 moles of Ba OH 2 = 0.300 L 0.200 mol HC 2 H 3 O 2 1L soln 1mol Ba OH 2 2 mol HC 2 H 3 O 2 = 0.0300 mol Ba OH 2 mass of Ba OH 2 = 0.0300 mol Ba OH 2 171.3g Ba OH 2 1mol Ba OH 2 = 5.14 g Ba OH 2 (d) 0.36 g Ba OH b g 2 is too much for the buffer to handle and it is the excess of OH originating from the Ba(OH)2 that determines the pOH of the solution. OH = 0.36 g Ba OH b g 2 0.300 L soln b g 171.3 g BabOH g
1 mol Ba OH
2 2 2 mol OH 1 mol Ba OH b g = 1.4 102 M OH 2 pOH = log 1.4 102 = 1.85 c h pH = 14.00 1.85 = 12.15 AcidBase Indicators
27. (E) (a) The pH color change range is 1.00 pH unit on either side of pKHln . If the pH color change range is below pH = 7.00 , the indicator changes color in acidic solution. If it is above pH = 7.00 , the indicator changes color in alkaline solution. If pH = 7.00 falls within the pH color change range, the indicator changes color near the neutral point. Indicator K HIn Bromphenol blue 1.4 104 Bromcresol green 2.1 105 Bromthymol blue 7.9 108 2,4Dinitrophenol 1.3 104 Chlorophenol red 1.0 106 Thymolphthalein 1.0 1010
pK HIn 3.85 4.68 7.10 3.89 6.00 10.00 pH Color Change Range Changes Color in? 2.9 (yellow) to 4.9 (blue) acidic solution 3.7 (yellow) to 5.7 (blue) acidic solution 6.1 (yellow) to 8.1 (blue) neutral solution 2.9 (colorless) to 4.9 (yellow) acidic solution 5.0 (yellow) to 7.0 (red) acidic solution 9.0 (colorless) to 11.0 (blue) basic solution (b) If bromcresol green is green, the pH is between 3.7 and 5.7, probably about pH = 4.7 . If chlorophenol red is orange, the pH is between 5.0 and 7.0, probably about pH = 6.0 . 816 Chapter 17: Additional Aspects of AcidBase Equilibria 28. (M) We first determine the pH of each solution, and then use the answer in Exercise 27 (a) to predict the color of the indicator. (The weakly acidic or basic character of the indicator does not affect the pH of the solution, since very little indicator is added.) (a) 1 mol H 3O + H 3O + = 0.100 M HCl = 0.100 M; pH = log 0.100 M = 1.000 1 mol HCl 2,4dinitrophenol is colorless. Solutions of NaCl(aq) are pH neutral, with pH = 7.000 . Chlorphenol red assumes its neutral color in such a solution; the solution is red/orange.
+ Equation: NH 3 (aq) + H 2 O(l) NH 4 (aq) + OH (aq) Initial: 1.00 M  0M 0 M  +x M +x M Changes: x M x M x M Equil: 1.00 x M  (b) (c) x2 x2 LM NH 3 OP 1.00 x 1.00 N Q (x << 1.00 M, thus the approximation was valid). Kb =
= 1.8 105 = NH 4 + OH x = 4.2 10 3 M = OH pOH = log 4.2 103 = 2.38 c h pH = 14.00 2.38 = 11.62 Thus, thymolphthalein assumes its blue color in solutions with pH 11.62 .
(d) From Figure 178, seawater has pH = 7.00 to 8.50. Bromcresol green solution is blue. 29. (M) (a) In an AcidBase titration, the pH of the solution changes sharply at a definite pH that is known prior to titration. (This pH change occurs during the addition of a very small volume of titrant.) Determining the pH of a solution, on the other hand, is more difficult because the pH of the solution is not known precisely in advance. Since each indicator only serves to fix the pH over a quite small region, often less than 2.0 pH units, several indicatorscarefully chosen to span the entire range of 14 pH unitsmust be employed to narrow the pH to 1 pH unit or possibly lower. An indicator is, after all, a weak acid. Its addition to a solution will affect the acidity of that solution. Thus, one adds only enough indicator to show a color change and not enough to affect solution acidity. (b) 817 Chapter 17: Additional Aspects of AcidBase Equilibria 30. (E) (a) We use an equation similar to the HendersonHasselbalch equation to determine the relative concentrations of HIn, and its anion, In , in this solution. In ; 4.55 = 4.95 + log In ; log In = 4.55 4.95 = 0.40 pH = pK HIn + log HIn HIn HIn In x = 100.40 = 0.40 = 100 x HIn x = 40 0.40 x x= 40 = 29% In and 71% HIn 1.40 (b) When the indicator is in a solution whose pH equals its pKa (4.95), the ratio In / HIn = 1.00 . And yet, at the midpoint of its color change range (about pH = 5.3 ), the ratio In /[HIn] is greater than 1.00. Even though HIn In at this midpoint, the contribution of HIn to establishing the color of the solution is about the same as the contribution of In . This must mean that HIn (red) is more strongly colored than In (yellow). 31. (E) (a) (b) (c) (d) (e) (f) 0.10 M KOH is an alkaline solution and phenol red will display its basic color in such a solution; the solution will be red. 0.10 M HC 2 H 3O 2 is an acidic solution, although that of a weak acid, and phenol red will display its acidic color in such a solution; the solution will be yellow. 0.10 M NH 4 NO 3 is an acidic solution due to the hydrolysis of the ammonium ion. Phenol red will display its acidic color, that is, yellow, in this solution. 0.10 M HBr is an acidic solution, the aqueous solution of a strong acid. Phenol red will display its acidic color in this solution; the solution will be yellow. 0.10 M NaCN is an alkaline solution because of the hydrolysis of the cyanide ion. Phenol red will display its basic color, red, in this solution. An equimolar acetic acidpotassium acetate buffer has pH = pKa = 4.74 for acetic acid. In this solution phenol red will display its acid color, namely, yellow. 32. (M) (a) pH = log 0.205 = 0.688 The indicator is red in this solution. b g (b) The total volume of the solution is 600.0 mL. We compute the amount of each solute. amount H 3O + = 350.0 mL 0.205 M = 71.8 mmol H 3O + amount NO 2 = 250.0 mL 0.500 M = 125 mmol NO 2 71.8 mmol 125 mmol H 3O + = = 0.120 M NO 2 = = 0.208 M 600.0 mL 600.0 mL The H 3O + and NO 2 react to produce a buffer solution in which HNO 2 = 0.120 M and NO 2 = 0.208 0.120 = 0.088 M . We use the 818 Chapter 17: Additional Aspects of AcidBase Equilibria HendersonHasselbalch equation to determine the pH of this solution. pKa = log 7.2 104 = 3.14 c h NO 2 = 3.14 + log 0.088 M = 3.01 The indicator is yellow in this pH = pK a + log HNO 0.120 M 2 solution.
(c) The total volume of the solution is 750. mL. We compute the amount and then the concentration of each solute. Amount OH = 150 mL 0.100 M = 15.0 mmol OH This OH reacts with HNO 2 in the buffer solution to neutralize some of it and leave 56.8 mmol ( = 71.8 15.0 ) unneutralized. 125 +15 mmol 56.8 mmol = 0.187 M HNO 2 = = 0.0757 M NO 2 = 750. mL 750. mL We use the HendersonHasselbalch equation to determine the pH of this solution. NO 2 0.187 M pH = pKa + log L = 3.14 + log = 3.53 O 0.0757 M NM HNO 2 QP The indicator is yellow in this solution. b g (d) We determine the OH due to the added Ba OH 2 . OH = 5.00 g Ba OH 0.750 L b g 2 b g 171.34 g BabOH g
1 mol Ba OH
2 b g 2 2 mol OH 1 mol Ba OH b g = 0.0778 M
2 This is sufficient OH to react with the existing HNO 2 and leave an excess OH = 0.0778 M 0.0757 M = 0.0021M. pOH = log 0.0021 = 2.68. pH = 14.00 2.68 = 11.32 The indicator is blue in this solution.
33. (M) Moles of HCl = C V = 0.04050 M 0.01000 L = 4.050 104 moles Moles of Ba(OH)2 at endpoint = C V = 0.01120 M 0.01790 L = 2.005 104 moles. Moles of HCl that react with Ba(OH)2 = 2 moles Ba(OH)2 Moles of HCl in excess 4.050 104 moles 4.010 104 moles = 4.05 106 moles Total volume at the equivalence point = (10.00 mL + 17.90 mL) = 27.90 mL [HCl]excess =
(a) (b) 4.05 106 mole HCl = 1.45 104 M; pH = log(1.45 104) = 3.84 0.02790 L The approximate pKHIn = 3.84 (generally + 1 pH unit) This is a relatively good indicator (with 1 % of the equivalence point volume), however, pKHin is not very close to the theoretical pH at the equivalence point (pH = 7.000) For very accurate work, a better indicator is needed (i.e., bromthymol blue (pKHin = 7.1). Note: 2,4dinitrophenol works relatively well here because the pH near the equivalence point of a strong acid/strong base 819 Chapter 17: Additional Aspects of AcidBase Equilibria titration rises very sharply ( 6 pH units for an addition of only 2 drops (0.10 mL)).
34. Solution (a): 100.0 mL of 0.100 M HCl, [H3O+] = 0.100 M and pH = 1.000 (yellow) Solution (b): 150 mL of 0.100 M NaC2H3O2 Ka of HC2H3O3 = 1.8 105 Kb of C2H3O2 = 5.6 1010 C2H3O2(aq) + H2O(l) Initial Change Equil. 0.100 M x 0.100 x   
K b = 5.6 10 10 HC2H3O2(aq) + OH(aq) 0M +x x ~0M +x x Assume x is small: 5.6 1011 = x2; x = 7.48 106 M (assumption valid by inspection) [OH] = x = 7.48 106 M, pOH = 5.13 and pH = 8.87 (greenblue) Mixture of solution (a) and (b). Total volume = 250.0 mL nHCl = C V = 0.1000 L 0.100 M = 0.0100 mol HCl nC2H3O2 = C V = 0.1500 L 0.100 M = 0.0150 mol C2H3O2 HCl is the limiting reagent. Assume 100% reaction. Therefore, 0.0050 mole C2H3O2 is left unreacted, and 0.0100 moles of HC2H3O2 form. [C2H3O2] = n 0.0050 mol n 0.0100 mol = = = 0.020 M [HC2H3O2] = = 0.0400 M V V 0.250 L 0.250 L + K a = 1.8 105 HC2H3O2(aq) + H2O(l) C2H3O2 (aq) + H3O (aq) Initial 0.0400 M  0.020 M ~0M  +x +x Change x Equil.  0.020 + x x 0.0400 x x(0.020 x) x(0.020) 0.0400 x 0.0400 x = 3.6 105 1.8 105 = (proof 0.18 % < 5%, the assumption was valid) [H3O+] = 3.6 105 pH = 4.44 Color of thymol blue at various pHs:
pH 1.2 Red Orange pH 2.8 Yellow Solution (c) YELLOW pH 8.0 Green Solution (b) GREEN pH 9.8 Blue Solution (a) RED 820 Chapter 17: Additional Aspects of AcidBase Equilibria Neutralization Reactions
35. (E) The reaction (to second equiv. pt.) is: H 3PO 4 aq + 2 KOH aq K 2 HPO 4 aq + 2H 2O(l) . The molarity of the H 3 PO 4 solution is determined in the following manner. 31.15 mL KOH soln H 3PO 4 molarity =
36. 0.2420 mmol KOH 1mmol H 3 PO 4 1mL KOH soln 2 mmol KOH = 0.1508 M 25.00 mL H 3PO 4 soln (E) The reaction (first to second equiv. pt.) is:
NaH 2 PO 4 aq + NaOH aq Na 2 HPO 4 aq + H 2 O(l) . The molarity of the H 3 PO 4 solution is determined in the following manner.
18.67 mL NaOH soln H 3 PO 4 molarity = 0.1885mmol NaOH 1 mmol H 3 PO 4 1mL NaOH soln 1 mmol NaOH = 0.1760 M 20.00 mL H 3 PO 4 soln 37. (M) Here we must determine the amount of H 3O + or OH in each solution, and the amount of excess reagent. 0.0150 mmol H 2SO 4 2 mmol H 3O + + amount H 3O = 50.00 mL = 1.50 mmol H 3O + 1 mL soln 1 mmol H 2SO 4 (assuming complete ionization of H2SO4 and HSO4 in the presence of OH) amount OH = 50.00 mL 0.0385 mmol NaOH 1mmol OH = 1.93 mmol OH 1mL soln 1mmol NaOH OH aq + 1.93mmol 0.43mmol Result: Titration reaction : Initial amounts : After reaction : H 3 O + aq 2H 2 O(l) 1.50 mmol 0 mmol 0.43mmol OH 3 OH = 100.0 mL soln = 4.3 10 M pOH = log 4.3 103 = 2.37 pH = 14.00 2.37 = 11.63 821 Chapter 17: Additional Aspects of AcidBase Equilibria 38. (M) Here we must determine the amount of solute in each solution, followed by the amount of excess reagent. H 3O + = 102.50 = 0.0032 M mmol HCl = 100.0 mL 0.0032 mmol H 3O + 1 mmol HCl = 0.32 mmol HCl 1 mL soln 1 mmol H 3O + OH = 103.00 = 1.0 103 M 0.0010 mmol OH 1 mmol NaOH mmol NaOH = 100.0 mL = 0.10 mmol NaOH 1 mL soln 1 mmol OH Result: pOH = 14.00 11.00 = 3.00
Titration reaction : Initial amounts : After reaction : NaOH aq + HCl aq NaCl aq 0.10 mmol 0.00 mmol 0.32 mmol 0.22 mmol 0 mmol 0.10 mmol + H 2 O(l) 0.22 mmol HCl 1mmol H 3O + 3 H 3O + = 200.0 mL soln 1mmol HCl = 1.110 M pH = log 1.1103 = 2.96 Titration Curves
39. (M) First we calculate the amount of HCl. The relevant titration reaction is
HCl aq + KOH aq KCl aq + H 2O(l) amount HCl = 25.00 mL 0.160 mmol HCl = 4.00 mmol HCl = 4.00 mmol H 3O + present 1 mL soln Then, in each case, we calculate the amount of OH that has been added, determine which ion, OH aq or H 3O + aq , is in excess, compute the concentration of that ion, and determine the pH. 0.242 mmol OH (a) amount OH = 10.00 mL = 2.42 mmol OH ; H 3O + is in excess. 1 mL soln b g b g 4.00 mmol H 3O 2.42 mmol OH [H 3O + ] = 1 mmol H 3O + 1 mmol OH = 0.0451 M 25.00 mL originally +10.00 mL titrant FG H IJ K pH = log 0.0451 = 1.346 b g 822 Chapter 17: Additional Aspects of AcidBase Equilibria (b) amount OH = 15.00 mL 0.242 mmol OH = 3.63 mmol OH ; H3O+ is in excess. 1 mL soln 4.00 mmol H 3O 3.63 mmol OH [H 3O + ] = 1 mmol H 3O + 1 mmol OH = 0.00925 M 25.00 mL originally +15.00 mL titrant FG H IJ K pH = log 0.00925 = 2.034
40. (M) The relevant titration reaction is KOH aq + HCl aq KCl aq + H 2O(l) b g mmol of KOH = 20.00 mL (a) 0.275 mmol KOH = 5.50 mmol KOH 1 mL soln The total volume of the solution is V = 20.00 mL +15.00 mL = 35.00 mL mmol HCl = 15.00 mL 0.350 mmol HCl = 5.25 mmol HCl 1 mL soln
1 mmol OH 1 mmol KOH = 0.25 mmol OH  mmol excess OH  = 5.50 mmol KOH5.25 mmol HCl 0.25 mmol OH OH = = 0.0071M 35.00 mL soln pH = 14.00 2.15 = 11.85
(b) pOH = log 0.0071 = 2.15 The total volume of solution is V = 20.00 mL + 20.00 mL = 40.00 mL mmol HCl = 20.00 mL 0.350 mmol HCl = 7.00 mmol HCl 1 mL soln 1 mmol H 3O + 1 mmol HCl mmol excess H 3O + = 7.00 mmol HCl 5.50 mmol KOH = 1.50 mmol H 3O + H 3O + =
41. 1.50 mmol H 3O + = 0.0375 M 40.00 mL pH = log 0.0375 = 1.426 b g (M) The relevant titration reaction is HNO 2 aq + NaOH aq NaNO 2 aq + H 2 O(l) amount HNO 2 = 25.00 mL 0.132 mmol HNO 2 = 3.30 mmol HNO 2 1 mL soln 823 Chapter 17: Additional Aspects of AcidBase Equilibria (a) The volume of the solution is 25.00 mL +10.00 mL = 35.00 mL amount NaOH = 10.00 mL 0.116 mmol NaOH = 1.16 mmol NaOH 1 mL soln 1.16 mmol NaNO 2 are formed in this reaction and there is an excess of (3.30 mmol HNO 2 1.16 mmol NaOH) = 2.14 mmol HNO 2 . We can use the HendersonHasselbalch equation to determine the pH of the solution. pKa = log 7.2 104 = 3.14
pH = pKa + log L 1.16 mmol NO 2 / 35.00 mL OP = 3.14 + log 2.14 mmol HNO 2 / 35.00 mL = 2.87 MN HNO 2 Q NO 2 c h (b) The volume of the solution is 25.00 mL + 20.00 mL = 45.00 mL amount NaOH = 20.00 mL 0.116 mmol NaOH = 2.32 mmol NaOH 1 mL soln 2.32 mmol NaNO 2 are formed in this reaction and there is an excess of (3.30 mmol HNO 2 2.32 mmol NaOH =) 0.98 mmol HNO 2 .
pH = pKa + log L 2.32 mmol NO 2 / 45.00 mL = 3.14 + log = 3.51 O 0.98 mmol HNO 2 / 45.00 mL NM HNO 2 QP NO 2 42. (M) In this case the titration reaction is NH 3 aq + HCl aq NH 4 Cl aq + H 2 O l b g b g b g bg amount NH 3 = 20.00 mL (a) 0.318 mmol NH 3 = 6.36 mmol NH 3 1 mL soln The volume of the solution is 20.00 mL +10.00 mL = 30.00 mL 0.475 mmol NaOH amount HCl = 10.00 mL = 4.75 mmol HCl 1 mL soln 4.75 mmol NH 4 Cl is formed in this reaction and there is an excess of (6.36 mmol NH3 4.75 mmol HCl =) 1.61 mmol NH3. We can use the HendersonHasselbalch equation to determine the pH of the solution. pK b = log 1.8 105 = 4.74; pH = pKa + log NH 3 NH 4
+ pK a = 14.00 pK b = 14.00 4.74 = 9.26 1.61 mmol NH 3 / 30.00 mL = 8.79 + 4.75 mmol NH 4 / 30.00 mL = 9.26 + log (b) The volume of the solution is 20.00 mL +15.00 mL = 35.00 mL amount HCl = 15.00 mL 0.475 mmol NaOH = 7.13 mmol HCl 1 mL soln 824 Chapter 17: Additional Aspects of AcidBase Equilibria 6.36 mmol NH 4 Cl is formed in this reaction and there is an excess of (7.13 mmol HCl 6.36 mmol NH3 =) 0.77 mmol HCl; this excess HCl determines the pH of the solution. 0.77 mmol HCl 1mmol H 3O + H 3O + = 35.00 mL soln 1mmol HCl = 0.022 M pH = log 0.022 = 1.66
43. (E) In each case, the volume of acid and its molarity are the same. Thus, the amount of acid is the same in each case. The volume of titrant needed to reach the equivalence point will also be the same in both cases, since the titrant has the same concentration in each case, and it is the same amount of base that reacts with a given amount (in moles) of acid. Realize that, as the titration of a weak acid proceeds, the weak acid will ionize, replenishing the H 3O + in solution. This will occur until all of the weak acid has ionized and all of the released H+ has subsequently reacted with the strong base. At the equivalence point in the titration of a strong acid with a strong base, the solution contains ions that do not hydrolyze. But the equivalence point solution of the titration of a weak acid with a strong base contains the anion of a weak acid, which will hydrolyze to produce a basic (alkaline) solution. (Don't forget, however, that the inert cation of a strong base is also present.) (E) (a) 44. This equivalence point solution is the result of the titration of a weak acid with a 2 strong base. The CO 3 in this solution, through its hydrolysis, will form an alkaline, or basic solution. The other ionic species in solution, Na + , will not hydrolyze. Thus, pH > 7.0. This is the titration of a strong acid with a weak base. The NH 4 present in the equivalence point solution hydrolyzes to form an acidic solution. Cl does not hydrolyze. Thus, pH < 7.0. This is the titration of a strong acid with a strong base. Two ions are present in the solution at the equivalence point, namely, K + and Cl , neither of which hydrolyzes. Thus the solution will have a pH of 7.00. Initial OH = 0.100 M OH pOH = log 0.100 = 1.000
+ (b) (c) 45. (D) (a) b g pH = 13.00 Since this is the titration of a strong base with a strong acid, KI is the solute present at the equivalence point, and since KI is a neutral salt, the pH = 7.00 . The titration reaction is:
KOH aq + HI aq KI aq + H 2 O(l) VHI = 25.0 mL KOH soln = 12.5 mL HI soln 0.100 mmol KOH soln 1mmol HI 1mL HI soln 1mL soln 1mmol KOH 0.200 mmol HI 825 Chapter 17: Additional Aspects of AcidBase Equilibria Initial amount of KOH present = 25.0 mL KOH soln 0.100 M = 2.50 mmol KOH At the 40% titration point: 5.00 mL HI soln 0.200 M HI = 1.00 mmol HI excess KOH = 2.50 mmol KOH 1.00 mmol HI = 1.50 mmol KOH 1.50 mmol KOH 1mmol OH OH = = 0.0500 M pOH = log 0.0500 = 1.30 30.0 mL total 1mmol KOH pH = 14.00 1.30 = 12.70 At the 80% titration point: 10.00 mL HI soln 0.200 M HI = 2.00 mmol HI excess KOH = 2.50 mmol KOH 2.00 mmol HI = 0.50 mmol KOH 0.50 mmol KOH 1mmol OH OH = = 0.0143M 35.0 mL total 1mmol KOH pOH = log 0.0143 = 1.84 pH = 14.00 1.84 = 12.16 At the 110% titration point:13.75 mL HI soln 0.200 M HI = 2.75 mmol HI excess HI = 2.75 mmol HI 2.50 mmol HI = 0.25 mmol HI [H 3O + ] 0.25 mmol HI 1mmol H 3O 0.0064 M; pH= log(0.0064) 2.19 38.8 mL total 1mmol HI Since the pH changes very rapidly at the equivalence point, from about pH = 10 to about pH = 4 , most of the indicators in Figure 178 can be used. The main exceptions are alizarin yellow R, bromphenol blue, thymol blue (in its acid range), and methyl violet.
(b) Initial pH: Since this is the titration of a weak base with a strong acid, NH4Cl is the solute present at the equivalence point, and since NH4+ is a slightly acidic cation, the pH should be slightly lower than 7. The titration reaction is:
Equation: Initial : Changes : Equil : NH 3 aq + H 2 O(l) NH 4 + aq + OH aq 0M 1.00 M 0M x M +x M + xM 1.00 x M xM xM NH 4 + OH x2 x2 5 Kb = = 1.8 10 = 1.00 x 1.00 NH 3 (x << 1.0, thus the approximation is valid) x = 4.2 103 M = OH , pOH = log 4.2 103 = 2.38, pH = 14.00 2.38 = 11.62 = initial pH Volume of titrant : NH 3 + HCl NH 4 Cl + H 2 O 826 Chapter 17: Additional Aspects of AcidBase Equilibria VHCl = 10.0mL 1.00 mmol NH 3 1mmol HCl 1mL HCl soln = 40.0 mL HCl soln 1mL soln 1mmol NH 3 0.250 mmol HCl pH at equivalence point: The total solution volume at the equivalence point is 10.0 + 40.0 mL = 50.0 mL
Also at the equivalence point, all of the NH 3 has reacted to form NH 4 . It is this NH 4 that hydrolyzes to determine the pH of the solution.
10.0 mL NH 4
+ + + = 1.00 mmol NH 3 1 mmol NH 4+ 1 mL soln 1 mmol NH 3 = 0.200 M 50.0mL total solution Equation : NH 4 + aq + H 2 O(l) NH 3 aq + H 3 O + aq Initial : 0.200 M 0M 0M Changes : x M + xM + xM Equil : 0.200 x M xM xM + K w 1.0 1014 NH 3 H 3O x2 x2 = Ka = = = Kb 1.8 105 0.200 x 0.200 NH 3 (x << 0.200, thus the approximation is valid) x = 1.1105 M; H 3O + = 1.1105 M; pH = log 1.1105 = 4.96 Of the indicators in Figure 178, one that has the pH of the equivalence point within its pH color change range is methyl red (yellow at pH = 6.2 and red at pH = 4.5); Bromcresol green would be another choice. At the 50% titration point, NH 3 = NH 4 + and pOH = pK b = 4.74 pH = 14.00 4.74 = 9.26 The titration curves for parts (a) and (b) follow. 827 Chapter 17: Additional Aspects of AcidBase Equilibria 46. (M) (a) This part simply involves calculating the pH of a 0.275 M NH 3 solution. Equation: NH 3 aq + H 2 O(l) NH 4 + aq + OH aq Initial: 0.275 M 0M 0M Changes: x M +xM +xM Equil: (0.275 x) M xM xM
x2 x2 LM NH 3 OP 0.275 x 0.275 N Q (x << 0.275, thus the approximation is valid) pOH = log 2.2 103 = 2.66 pH = 11.34 Kb =
= 1.8 105 = NH 4
+ OH x = 2.2 10 3 M = OH (b) This is the volume of titrant needed to reach the equivalence point. The relevant titration reaction is NH 3 aq + HI aq NH 4 I aq b g b g b g VHI = 20.00 mL NH 3 aq VHI = 16.9 mL HI soln
(c) 0.275 mmol NH 3 1mmol HI 1mL HI soln 1mL NH3 soln 1mmol NH 3 0.325 mmol HI The pOH at the halfequivalence point of the titration of a weak base with a strong acid is equal to the pKb of the weak base. pOH = pK b = 4.74; pH = 14.00 4.74 = 9.26 NH 4 is formed during the titration, and its hydrolysis determines the pH of the solution. Total volume of solution = 20.00 mL +16.9 mL = 36.9 mL
mmol NH 4
+ (d) + 0.275 mmol NH 3 1mmol NH 4 + = 20.00 mL NH 3 aq = 5.50 mmol NH 4 1mL NH 3 soln 1mmol NH 3 + NH 4 + 5.50 mmol NH 4 + = = 0.149 M 36.9 mL soln Equation: NH 4 + aq + H 2 O(l) NH 3 aq + H 3 O + aq Initial: 0.149 M 0M 0M Changes: +x M +x M xM Equil: (0.149 x) M xM xM 828 Chapter 17: Additional Aspects of AcidBase Equilibria
+ K w 1.0 1014 NH 3 H 3 O x2 x2 = Ka = = = Kb 0.149 x 0.149 1.8 105 NH 4 + (x << 0.149, thus the approximation is valid) x = 9.1 106 M = H 3 O + 47. pH = log 9.1 106 = 5.04 (M) A pH greater than 7.00 in the titration of a strong base with a strong acid means that the base is not completely titrated. A pH less than 7.00 means that excess acid has been added. (a) We can determine OH of the solution from the pH. OH is also the quotient of the amount of hydroxide ion in excess divided by the volume of the solution: 20.00 mL base +x mL added acid. pOH = 14.00 pH = 14.00 12.55 = 1.45 0.175 mmol OH 20.00 mL base 1mL base [OH ] OH = 10 pOH = 101.45 = 0.035 M 20.00 mL + x mL 0.200 mmol H 3 O + x mL acid 1mL acid 0.035 M 3.50 0.200 x = 0.70 + 0.035 x; 3.50 0.70 = 0.035 x + 0.200 x; 2.80 = 0.235 x 2.80 x= = 11.9 mL acid added. 0.235
(b) The setup here is the same as for part (a). pOH = 14.00 pH = 14.00 10.80 = 3.20 OH = 10 pOH = 103.20 = 0.00063 M 0.200 mmol H 3O + 0.175 mmol OH 20.00 mL base x mL acid 1mL base 1mL acid [OH ] 20.00 mL + x mL
[OH ] 0.00063 mmol = 0.00063M mL 3.50 0.200 x = 0.0126 + 0.00063x; 3.50 0.0126 = 0.00063 x + 0.200 x; 3.49 = 0.201 x 3.49 x= = 17.4 mL acid added. This is close to the equivalence point at 17.5 mL. 0.201 829 Chapter 17: Additional Aspects of AcidBase Equilibria (c) Here the acid is in excess, so we reverse the setup of part (a). We are just slightly beyond the equivalence point. This is close to the "mirror image" of part (b). H3O + = 10 pH = 104.25 = 0.000056 M FG x mLacid 0.200 mmol H O IJ FG 20.00 mL base 0.175 mmol OH IJ 1 mL acid 1mL base H K H K [H O ] =
+ 3 + 3 20.00 mL+ x mL = 5.6 10 M 5 0.200 x 3.50 = 0.0011+ 5.6 105 x; 3.50 + 0.0011 = 5.6 105 x + 0.200 x; 3.50 = 0.200 x 3.50 = 17.5 mL acid added, which is the equivalence point for this titration. 0.200 (D) In the titration of a weak acid with a strong base, the middle range of the titration, with the pH within one unit of pKa ( = 4.74 for acetic acid), is known as the buffer region. The HendersonHasselbalch equation can be used to determine the ratio of weak acid and anion concentrations. The amount of weak acid then is used in these calculations to determine the amount of base to be added. x=
(a) C 2 H 3O 2 = 3.85 = 4.74 + log C 2 H 3O 2 pH = pK a + log 48. HC H O 2 3 2 HC H O 2 3 2 C 2 H 3O 2 = 3.85 4.74 log HC H O 2 3 2 HC H O 2 3 2 C 2 H 3O 2 = 10 0.89 = 0.13;
= 2.50 mmol nHC2 H3O2 = 25.00 mL 0.100 mmol HC 2 H 3 O 2 1mL acid Since acetate ion and acetic acid are in the same solution, we can use their amounts in millimoles in place of their concentrations. The amount of acetate ion is the amount created by the addition of strong base, one millimole of acetate ion for each millimole of strong based added. The amount of acetic acid is reduced by the same number of millimoles. HC 2 H 3O 2 + OH C 2 H 3O 2 + H 2 O
0.200 mmol OH 1 mmol C 2 H 3O 2 x mL base 0.200 x mL base 1 mmol OH 013 . 2.50 0.200 x 0.200 mmol OH 2.50 mmol HC 2 H 3O 2 x mL base mL base 0.200 x = 0.13 2.50 0.200 x = 0.33 0.026 x x= 0.33 = 1.5 mL of base 0.226 b g FG H IJ K 0.33 = 0.200 x + 0.026 x = 0.226 x 830 Chapter 17: Additional Aspects of AcidBase Equilibria (b) This is the same setup as part (a), except for a different ratio of concentrations. C 2 H 3O 2 C 2 H 3O 2 pH = 5.25 = 4.74 + log L log L O O = 5.25 4.74 = 0.51 NM HC 2 H 3O 2 QP NM HC2 H 3O 2 QP
0.51 LM HC2 H 3O 2 OP = 10 = 3.2 N Q C 2 H 3O 2 3.2 = 0.200 x = 3.2 2.50 0.200 x = 8.0 0.64 x
x= b g 0.200 x 2.50 0.200 x 8.0 = 0.200 x + 0.64 x = 0.84 x 8.0 = 9.5 mL base. This is close to the equivalence point, which is reached 0.84 by adding 12.5 mL of base. (c) This is after the equivalence point, where the pH is determined by the excess added base. pOH = 14.00 pH = 14.00 11.10 = 2.90 OH = 10 pOH = 102.90 = 0.0013 M
0.200 mmol OH 0.200 x 1 mL base OH = 0.0013 M = = x mL + 12.50 mL + 25.00 mL 37.50 + x x mL b g 0.200 x = 0.0013 37.50 + x = 0.049 + 0.0013 x 0.049 = 0.25 mL excess 0.200 0.0013 Total base added = 12.5 mL to equivalence point + 0.25 mL excess = 12.8 mL x= b g 49. (D) For each of the titrations, the pH at the halfequivalence point equals the pKa of the acid. x2 x [H 3 O ] The initial pH is that of 0.1000 M weak acid: K a 0.1000 x x must be found using the quadratic formula roots equation unless the approximation is valid. One method of determining if the approximation will be valid is to consider the ratio Ca/Ka. If the value of Ca/Ka is greater than 1000, the assumption should be valid, however, if the value of Ca/Ka is less than 1000, the quadratic should be solved exactly (i.e., the 5% rule will not be satisfied). The pH at the equivalence point is that of 0.05000 M anion of the weak acid, for which the OH is determined as follows.
Kw Kw x2 x= 0.05000 [OH ] K a 0.05000 Ka We can determine the pH at the quarter and three quarter of the equivalence point by using the HendersonHasselbalch equation (effectively + 0.48 pH unit added to the pKa). Kb = 831 Chapter 17: Additional Aspects of AcidBase Equilibria And finally, when 0.100 mL of base has been added beyond the equivalence point, the pH is determined by the excess added base, as follows (for all three titrations). 0100 mL . OH = 0.1000 mmol NaOH 1 mmol OH 1mL NaOH soln 1 mmol NaOH = 4.98 104 M 20.1 mL soln total pOH = log 4.98 104 = 3.303
(a) c h pH = 14.000 3.303 = 10.697 Ca/Ka = 14.3; thus, the approximation is not valid and the full quadratic equation must be solved. Initial: From the roots equation x = [H 3 O ] 0.023M pH=1.63 Half equivalence point: pH = pKa = 2.15 pH at quarter equivalence point = 2.15  0.48 = 1.67 pH at three quarter equivalence point = 2.15 + 0.48 = 2.63 pOH 6.57 1.0 1014 0.05000 2.7 107 Equiv: x = OH = 3 pH 14.00 6.57 7.43 7.0 10 Indicator: bromthymol blue, yellow at pH = 6.2 and blue at pH = 7.8 (b) Ca/Ka = 333; thus, the approximation is not valid and the full quadratic equation must be solved. Initial: From the roots equation x = [H 3 O ] 0.0053M Half equivalence point: pH = pKa = 3.52 pH at quarter equivalence point = 3.52  0.48 = 3.04 pH at three quarter equivalence point = 3.52 + 0.48 = 4.00 Equiv: x = OH = 1.0 1014 0.05000 1.3 106 3.0 104 pH=2.28 pOH 5.89 pH = 14.00 5.89 811 . Indicator: thymol blue, yellow at pH = 8.0 and blue at pH = 9.6
(c) Ca/Ka = 5106; thus, the approximation is valid.. Initial: [ H 3O ] 01000 2.0 108 0.000045 M pH = 4.35 . pH at quarter equivalence point = 4.35  0.48 = 3.87 pH at three quarter equivalence point = 4.35 + 0.48 = 4.83 Half equivalence point: pH = pKa = 7.70 Equiv: x = OH 10 1014 . = 0.0500 16 104 . 8 2.0 10 . pOH 380 pH 14.00 380 10.20 . 832 Chapter 17: Additional Aspects of AcidBase Equilibria Indicator: alizarin yellow R, yellow at pH = 10.0 and violet at pH = 12.0 The three titration curves are drawn with respect to the same axes in the diagram below.
12 10 8 pH 6 4 2 0 0 2 4 6 8 10 12 Volume of 0.10 M NaOH added (mL) 50. (D) For each of the titrations, the pOH at the halfequivalence point equals the pKb of the base. The initial pOH is that of 0.1000 M weak base, determined as follows. x2 Kb ; x 0.1000 K a [OH ] if Cb/Kb > 1000. For those cases 0.1000 x where this is not the case, the approximation is invalid and the complete quadratic equation must be solved. The pH at the equivalence point is that of 0.05000 M cation of the weak base, for which the H 3O + is determined as follows.
Ka Kw x2 K b 0.05000 x Kw 0.05000 [H 3O + ] Kb And finally, when 0.100 mL of acid has been added beyond the equivalence point, the pH for all three titrations is determined by the excess added acid, as follows. 0.100 mL HCl H 3O + = 0.1000 mmol HCl 1 mmol H 3O + 1 mL HCl soln 1 mmol HCl = 4.98 104 M 20.1 mL soln total pH = log 4.98 104 = 3.303
(a) c h Cb/Kb = 100; thus, the approximation is not valid and the full quadratic equation must be solved. Initial: From the roots equation x = 0.0095. Therefore, [OH ] 0.0095 M pOH=2.02 pH=11.98 Halfequiv: Equiv: pOH = log 1 103 = 3.0 1.0 1014 0.05000 7 107 1 103 c h pH = 11.0 pH = 6.2 x = [H 3 O + ] 833 Chapter 17: Additional Aspects of AcidBase Equilibria Indicator: methyl red, yellow at pH = 6.3 and red at pH = 4.5
(b) Cb/Kb = 33,000; thus, the approximation is valid. Initial: [OH ] 0.1000 3 106 5.5 104 M pOH=3.3 pH=10.7 pH = 8.5 Halfequiv: pOH = log 3 106 = 5.5 Equiv: x = [H 3 O + ] pH = 14 pOH 1.0 1014 0.05000 1 10 5 6 3 10 pH log(1 10 5 ) = 5.0 Indicator: methyl red, yellow at pH = 6.3 and red at pH = 4.5
(c) Cb/Kb = 1.4106; thus, the approximation is valid. Initial: [OH ] 01000 7 108 8 105 M . pOH = 4.1 pH = 9.9 pH = 6.8 pH=4.1 Halfequiv: pOH = log 7 108 = 7.2 Equiv:
+ pH = 14 pOH 1.0 1014 x = [H 3O ] 0.05000 8.5 105 8 7 10 Indicator: bromcresol green, blue at pH = 5.5 and yellow at pH = 4.0 The titration curves are drawn with respect to the same axes in the diagram below. 51. (D) 25.00 mL of 0.100 M NaOH is titrated with 0.100 M HCl (i) Initial pOH for 0.100 M NaOH: [OH] = 0.100 M, pOH = 1.000 or pH = 13.000 25.00 mL = 0.0510 M (ii) After addition of 24 mL: [NaOH] = 0.100 M 49.00 mL 24.00 mL [HCl] = 0.100 M = 0.0490 M 49.00 mL 834 Chapter 17: Additional Aspects of AcidBase Equilibria NaOH is in excess by 0.0020 M = [OH] pOH = 2.70 (iii) At the equivalence point (25.00 mL), the pOH should be 7.000 and pH = 7.000 25.00 = 0.0490 M 51.00 26.00 mL [HCl] = 0.100 M = 0.0510 M 51.00 mL HCl is in excess by 0.0020 M = [H3O+] pH = 2.70 or pOH = 11.30 (iv) After addition of 26 mL: [NaOH] = 0.100M (v) After addition of 33.00 mL HCl( xs) [NaOH] = 0.100 M [HCl] = 0.100 M 33.00 mL = 0.0569 M 58.00 mL 25.00 mL = 0.0431 M 58.00 mL [HCl]excess= 0.0138 M pH = 1.860, pOH = 12.140 The graphs look to be mirror images of one another. In reality, one must reflect about a horizontal line centered at pH or pOH = 7 to obtain the other curve.
14 12 10 pOH 8 6 4 2 0 0 10 20 30 40 Volume of HCl (mL) 50 60 0.100 M NaOH(25.00 mL) + 0.100 M HCl
14 12 10 pH 8 6 4 2 0 0 0.100 M NaOH(25.00 mL) + 0.100 M HCl 10 20 30 40 Volume of HCl (mL) 50 60 52. (D) 25.00 mL of 0.100 M NH3 is titrated with 0.100 M HCl Ka = (i) Kw 1 1014 = = 5.6 1010 5 Kb 1.8 10 For initial pOH, use I.C.E.(initial, change, equilibrium) table. NH3(aq) + H2O(l) K b = 1.8 10 5 NH4+(aq) + OH(aq) 0M +x x Initial 0.100 M 0M Change +x x Equil. 0.100 x x 2 2 x x (Assume x ~0) x = 1.3 103 1.8 105 = 0.100 x 0.100 (x < 5% of 0.100, thus, the assumption is valid). Hence, x = [OH] = 1.3 103 pOH = 2.89, pH = 11.11 (ii) After 2 mL of HCl is added: [HCl] = 0.100 M 2.00 mL = 0.00741 M (after 27.00 mL dilution) [NH3] = 0.100 M 25.00 mL = 0.0926 M (after dilution) 27.00 mL 835 Chapter 17: Additional Aspects of AcidBase Equilibria The equilibrium constant for the neutralization reaction is large (Kneut = Kb/Kw = 1.9105) and thus the reaction goes to nearly 100% completion. Assume that the limiting reagent is used up (100% reaction in the reverse direction) and reestablish the equilibrium by a shift in the forward direction. Here H3O+ (HCl) is the limiting reagent. NH4+(aq) + H2O(l) Initial 0M Change +x 100% rxn 0.00741 Change y Equil 0.00741y 5.6 1010 =
(0.00741 y ) Ka = 5.6 10 10 NH3(aq) + H3O+(aq) 0.00741 M x 0 +y y 0.0926 M x 0.0852 +y reestablish equilibrium 0.0852 + y
x = 0.00741 y (0.0852 y ) = y (0.0852) (0.00741) (set y ~ 0) y = 4.8 1011 (the approximation is clearly valid) y = [H3O+] = 4.8 1011; pH = 10.32 and pOH = 3.68 (iii) pH at 1/2 equivalence point = pKa = log 5.6 1010 = 9.25 and pOH = 4.75 (iv) After addition of 24 mL of HCl: [HCl] = 0.100 M 24.00 mL 25.00 mL = 0.0490 M; [NH3] = 0.100 M = 0.0510 M 49.00 mL 49.00 mL The equilibrium constant for the neutralization reaction is large (see above), and thus the reaction goes to nearly 100% completion. Assume that the limiting reagent is used up (100% reaction in the reverse direction) and reestablish the equilibrium in the reverse direction. Here H3O+ (HCl) is the limiting reagent. NH4+(aq) + H2O(l) Initial 0M Change +x 100% rxn 0.0490 Change y Equil 0.0490y 5.6 1010 = Ka = 5.6 10 10 NH3(aq) + H3O+(aq) 0.0490 M x 0 +y y 0.0541 M x = 0.0490 x 0.0020 +y reestablish equilibrium 0.0020 + y y (0.0020 y ) y (0.0020) = (Assume y ~ 0) y = 1.3 108 (valid) (0.0490 y ) (0.0490) + 8 y = [H3O ] = 1.3 10 ; pH = 7.89 and pOH = 6.11 Equiv. point: 100% reaction of NH3 NH4+: [NH4+] = 0.100 NH4+(aq) Initial 0.0500 M Change x Equil 0.0500x + H2O(l) Ka = 5.6 10 10 (v) 25.00 mL =0.0500 M 50.00 mL NH3(aq) + H3O+(aq) 0M +x x ~0M +x x 836 Chapter 17: Additional Aspects of AcidBase Equilibria 5.6 1010= x2 x2 = (Assume x ~ 0) x = 5.3 106 (0.0500 x) 0.0500 (the approximation is clearly valid) x = [H3O+] = 5.3 106 ; pH = 5.28 and pOH = 8.72 (vi) After addition of 26 mL of HCl, HCl is in excess. The pH and pOH should be the same as those in Exercise 51. pH = 2.70 and pOH = 11.30 (vii) After addition of 33 mL of HCl, HCl is in excess. The pH and pOH should be the same as those in Exercise 51. pH = 1.860 and pOH = 12.140. This time the curves are not mirror images of one another, but rather they are related through a reflection in a horizontal line centered at pH or pOH = 7.
0.100 M NH3 (25mL) + 0.100 HCl
14 12 10 pOH 8 6 4 2 0 0 10 20 30 Volume of HCl (mL) 40 50 14 12 10 pH 8 6 4 2 0 0 10 20 30 Volume of HCl (mL) 40 50 0.100 M NH3 (25mL) + 0.100 HCl Salts of Polyprotic Acids
53. (E) We expect a solution of Na 2S to be alkaline, or basic. This alkalinity is created by the hydrolysis of the sulfide ion, the anion of a very weak acid ( K2 = 1 1019 for H 2S ). S2 aq + H 2 O(l) HS aq + OH aq (E) We expect the pH of a solution of sodium dihydrogen citrate, NaH 2 Cit , to be acidic because the pKa values of first and second ionization constants of polyprotic acids are reasonably large. The pH of a solution of the salt is the average of pK1 and pK2 . For citric acid, in fact, this average is 3.13 + 4.77 2 = 3.95 . Thus, NaH2Cit affords acidic 54. solutions.
55. (M) 2 (a) H 3 PO 4 aq + CO3 aq H 2 PO 4 aq + HCO3 aq H 2 PO 4 aq + CO3 2 aq HPO4 2 aq + HCO3 aq HPO 4 2 aq + OH aq PO43 aq + H 2O(l) 837 Chapter 17: Additional Aspects of AcidBase Equilibria (b) The pH values of 1.00 M solutions of the three ions are; 2 1.0 M OH pH = 14.00 1.0 M CO 3 pH = 12.16
2 1.0 M PO 4 pH = 13.15 3 Thus, we see that CO 3 is not a strong enough base to remove the third proton from H 3 PO 4 . As an alternative method of solving this problem, we can compute the equilibrium constants of the reactions of carbonate ion with H 3 PO 4 , H2PO4and HPO42.
H 3 PO 4 + CO 3
2 H 2 PO 4 + HCO 3 K K a1{H 3 PO 4 } K a 2 {H 2 CO 3 } K a 2 {H 3 PO 4 } K a 2 {H 2 CO3 } K a 3 {H 3 PO 4 } K a 2 {H 2 CO3 } 7.1 103 4.7 10
11 1.5 108 2 H 2 PO 4 + CO3 HPO 2 + HCO3 K 4 6.3 108 4.7 10 11 4.2 10 13 4.7 10
11 1.3 103 HPO 4 + CO3 2 2 3 PO 4 + HCO3 K 8.9 10 3 Since the equilibrium constant for the third reaction is much smaller than 1.00, we conclude that it proceeds to the right to only a negligible extent and thus is not a 3 practical method of producing PO 4 . The other two reactions have large equilibrium constants, and products are expected to strongly predominate. They have the advantage of involving an inexpensive base and, even if they do not go to completion, they will be drawn to completion by reaction with OH in the last step of the process.
56. (M) We expect CO32 to hydrolyze and the hydrolysis products to determine the pH of the solution. Equation: Initial Changes: Equil: HCO3 (aq) 1.00 M x M (1.00 x) M H 2 O(l) "H 2 CO3 "(aq) 0M +x M xM + OH (aq) 0M +x M xM H 2CO3 OH x x x 2 K w 1.00 1014 = 8 = = 2.3 10 = Kb = 7 4.4 10 1.00 x 1.00 K1 HCO3 (Cb/Kb = a very large number; thus, the approximation is valid). x 100 2.3 108 15 104 M [OH ] ; pOH = log(15 104 ) 382 pH = 10.18 . . . . For 1.00 M NaOH, OH = 1.00 pOH = log 1.00 = 0.00 pH = 14.00 Both 1.00 M NaHCO3 and 1.00 M NaOH have an equal capacity to neutralize acids since one mole of each neutralizes one mole of strong acid. NaOH aq + H 3O + aq Na + aq + 2H 2 O(l) 838 Chapter 17: Additional Aspects of AcidBase Equilibria NaHCO 3 aq + H 3O + aq Na + aq + CO 2 g + 2H 2 O(l) .
But on a per gram basis, the one with the smaller molar mass is the more effective. Because the molar mass of NaOH is 40.0 g/mol, while that of NaHCO 3 is 84.0 g/mol, NaOH(s) is more than twice as effective as NaHCO3(s) on a per gram basis. NaHCO3 is preferred in laboratories for safety and expense reasons. NaOH is not a good choice because it can cause severe burns. NaHCO3, baking soda, by comparison, is relatively nonhazardous. It also is much cheaper than NaOH.
57. (M) Malonic acid has the formula H2C3H2O4 MM = 104.06 g/mol 1 mol = 0.187 mol Moles of H2C3H2O4 = 19.5 g 104.06 g moles 0.187 moles Concentration of H2C3H2O4 = = = 0.748 M V 0.250 L The second proton that can dissociate has a negligible effect on pH ( K a 2 is very small). Thus the pH is determined almost entirely by the first proton loss. H2A(aq) + H2O(l) Initial 0.748 M Change x Equil 0.748x So, x = + HA(aq) + H3O (aq) 0M 0M +x +x x x x2 = Ka ; pH = 1.47, therefore, [H3O+] = 0.034 M = x , 0.748 x (0.034) 2 = 1.6 103 K a1 = 0.748 0.034 (1.5 103 in tables, difference owing to ionization of the second proton) HA(aq) + H2O(l) Initial 0.300 M Change x Equil 0.300x + A2(aq) + H3O (aq) 0M 0M +x +x x x (5.5 105 ) 2 = 1.0 108 pH = 4.26, therefore, [H3O+] = 5.5 105 M = x, K a 2 = 0.300 5.5 105 58. (M) Orthophthalic acid. K a1 = 1.1 103, K a 2 = 3.9 106 (a) We have a solution of HA. HA(aq) + H2O(l) Initial 0.350 M Change x Equil 0.350x + A2(aq) + H3O (aq) 0M ~0M +x +x x x 839 Chapter 17: Additional Aspects of AcidBase Equilibria x2 x2 , x = 1.2 103 0.350 x 0.350 0.350, thus, the approximation is valid) (x x = [H3O+] = 1.2 103, pH = 2.92 3.9 106 =
(b) 36.35 g of potassium orthophthalate (MM = 242.314 g mol1) 1 mol = 0.150 mol in 1 L moles of potassium orthophthalate = 36.35 g 242.314 g A2(aq) + H2O(l) Initial 0.150 M Change x Equil 0.150 x
K b,A2 HA(aq) + 0M +x x OH(aq) 0M +x x K w 1.0 1014 x2 x2 2.6 109 Ka2 0.150 x 0.150 3.9 106 x 2.0 105 (x << 0.150, so the approximation is valid) = [OH] pOH = log 2.0105 = 4.70; pH = 9.30 General AcidBase Equilibria
59. (E) (a) Ba OH b g 2 is a strong base. pOH = 14.00 11.88 = 2.12 Ba OH b g 2 OH = 10 2.12 = 0.0076 M 1 mol Ba OH 2 0.0076 mol OH = = 0.0038 M 1L 2 mol OH b g (b) C 2 H 3O 2 = 4.74 + log 0.294 M pH = 4.52 = pK a + log HC 2 H 3O 2 HC 2 H 3O 2 log 0.294 M = 4.52 4.74 HC 2 H 3O 2 0.294 M = 100.22 = 0.60 HC2 H 3O 2 HC2 H3O2 = 0.294 M = 0.49 M 0.60 840 Chapter 17: Additional Aspects of AcidBase Equilibria 60. (M) (a) pOH = 14.00 8.95 = 5.05 OH = 105.05 = 8.9 106 M H 2 O(l) C 6 H 5 NH 3 (aq) Equation: Initial Changes: Equil: C6 H 5 NH 2 (aq) xM 8.9 106 M ( x 8.9 106 ) M OH (aq) 0 M 8.9 106 M 8.9 106 M 10 0M 8.9 106 M 8.9 106 M Kb = C 6 H 5 NH 3 + OH LMC6 H 5 NH 2 OP N Q
6 = 7.4 10
6 2 c8.9 10 h = 6 2 x 8.9 106
x = 0.11 M = molarity of aniline x 8.9 10 c8.9 10 h =
7.4 1010 = 0.11 M (b) H 3O + = 105.12 = 7.6 106 M Equation: NH 4 + (aq) H 2 O(l) NH 3 (aq) 0M 7.6 106 M 7.6 106 M H 3O (aq) 0 M 7.6 106 M 7.6 106 M Initial xM Changes: 7.6 106 M Equil: ( x 7.6 106 ) M Ka = NH 3 H 3O + NH 4 +
6 7.6 106 Kw 1.0 1014 10 = = = 5.6 10 = Kb for NH 3 1.8 105 x 7.6 106
6 2 10 c h 2 x 7.6 10 c7.6 10 h =
5.6 10 = 0.10 M x = NH 4 + = NH 4 Cl = 0.10 M 61. (M) (a) A solution can be prepared with equal concentrations of weak acid and conjugate base (it would be a buffer, with a buffer ratio of 1.00, where the pH = pKa = 9.26). Clearly, this solution can be prepared, however, it would not have a pH of 6.07. (b) These solutes can be added to the same solution, but the final solution will have an appreciable HC 2 H 3O 2 because of the reaction of H 3O + aq with C 2 H 3O 2 aq b g b g H 3O + (aq) + C2 H3O 2 (aq) HC 2 H 3O 2 (aq) + H 2 O(l) Initial 0.058 M 0.10 M 0M Changes: 0.058 M 0.058 M +0.058 M Equil: 0.000 M 0.04 M 0.058 M + Of course, some H 3O will exist in the final solution, but not equivalent to 0.058 M HI.
Equation: 841 Chapter 17: Additional Aspects of AcidBase Equilibria (c) Both 0.10 M KNO 2 and 0.25 M KNO 3 can exist together. Some hydrolysis of the (d) b g b g BabOH g is a strong base and will react as much as possible with the weak conjugate acid NH , to form NH baq g .We will end up with a solution of BaCl baq g, NH baq g , and unreacted NH Clbaq g .
NO 2 aq ion will occur, forming HNO 2 aq and OH(aq).
2 4 3 2 3 4 (e) This will be a benzoic acidbenzoate ion buffer solution. Since the two components have the same concentration, the buffer solution will have pH = pKa = log 6.3 105 = 4.20 . This solution can indeed exist. c h (f) The first three components contain no ions that will hydrolyze. But C2 H3O 2 is the anion of a weak acid and will hydrolyze to form a slightly basic solution. Since pH = 6.4 is an acidic solution, the solution described cannot exist. 62. (M) (a) When H 3O + and HC 2 H 3O 2 are high and C2 H3O 2 is very low, a common ion H 3O + has been added to a solution of acetic acid, suppressing its ionization.
(b) When C2 H3O 2 is high and H 3O + and HC 2 H 3O 2 are very low, we are dealing with a solution of acetate ion, which hydrolyzes to produce a small concentration of HC2 H 3O 2 .
(c) When HC 2 H 3O 2 is high and both H 3O + and C2 H3O 2 are low, the solution is an acetic acid solution, in which the solute is partially ionized. (d) When both HC 2 H 3O 2 and C2 H3O 2 are high while H 3O + is low, the solution is a buffer solution, in which the presence of acetate ion suppresses the ionization of acetic acid. INTEGRATIVE AND EXERCISES
63. (M) (a) NaHSO 4 (aq) NaOH(aq) Na 2 SO 4 (aq) H 2 O(l) HSO 4 (aq) OH (aq) SO 42 (aq) H 2 O(l) (b) We first determine the mass of NaHSO4. mass NaHSO 4 36.56 mL 1L 0.225 mol NaOH 1 mol NaHSO 4 1000 mL 1 L 1 mol NaOH 120.06 g NaHSO 4 0.988 g NaHSO 4 1 mol NaHSO 4 842 Chapter 17: Additional Aspects of AcidBase Equilibria 1.016 g sample 0.988 g NaHSO 4 100% 2.8% NaCl 1.016 g sample (c) At the endpoint of this titration the solution is one of predominantly SO42, from which the pH is determined by hydrolysis. Since Ka for HSO4 is relatively large (1.1 102), base hydrolysis of SO42 should not occur to a very great extent. The pH of a neutralized solution should be very nearly 7, and most of the indicators represented in Figure 178 would be suitable. A more exact solution follows. % NaCl 1 mol SO 4 2 0.988 g NaHSO 4 1 mol NaHSO 4 [SO ] 0.225 M 0.03656 L 120.06 g NaHSO 4 1 mol NaHSO 4
24 Equation: Initial: Changes: Equil: SO 4 2 (aq) H 2 O(l) HSO 4  (aq) 0.225 M 0M OH (aq) 0M xM xM x M (0.225 x)M xM xM Kb [HSO 4 ] [OH ] [SO 4 2 ] K w 1.0 10 14 xx x2 9.1 10 13 0.225 x 0.225 K a2 1.1 10 2 [OH ] 9.1 1013 0.225 4.5 107 M (the approximation was valid since x << 0.225 M) pOH log(4.5 107 ) 6.35 pH 14.00 6.35 7.65 Thus, either bromthymol blue (pH color change range from pH = 6.1 to pH = 7.9) or phenol red (pH color change range from pH = 6.4 to pH = 8.0) would be a suitable indicator, since either changes color at pH = 7.65.
64. (D) The original solution contains 250.0 mL 0.100 mmol HC3 H 5O 2 25.0 mmol HC3 H 5O 2 1 mL soln pK a log(1.35 105 ) 4.87 We let V be the volume added to the solution, in mL. (a) Since we add V mL of HCl solution (and each mL adds 1.00 mmol H3O+ to the solution), we have added V mmol H3O+ to the solution. Now the final [H3O+] = 101.00 = 0.100 M.
V mmol H 3 O 0.100 M (250.0 V ) mL 25.0 27.8 mL added V 25.0 0.100 V , therefore, V 0.900 [H 3 O ] Now, we check our assumptions. The total solution volume is 250.0 mL+ 27.8 mL = 277.8 mL There are 25.0 mmol HC2H3O2 present before equilibrium is established, and 27.8 mmol H3O+ also. 843 Chapter 17: Additional Aspects of AcidBase Equilibria [HC 2 H 3 O 2 ] Equation: Initial: Changes: Equil: 25.0 mmol 27.8 mmol [H 3 O ] 0.0900 M 0.100 M 277.8 mL 277.8 mL HC3 H 5O 2 (aq) H 2 O(l) C3H 5O 2 ( aq) H 3O ( aq) 0.0900 M x M (0.08999 x)M 0 M xM xM 0.100 M x M (0.100 x)M Ka [H 3O ][C3 H 5O 2 ] [HC3 H 5O 2 ] 1.35 105 x (0.100 x) 0.100 x 0.0900 0.0900 x x 1.22 105 M The assumption used in solving this equilibrium situation, that x << 0.0900 clearly is correct. In addition, the tacit assumption that virtually all of the H3O+ comes from the HCl also is correct.
(b) The pH desired is within 1.00 pH unit of the pKa. We use the HendersonHasselbalch equation to find the required buffer ratio.
pH pK a log nAnHA [A ] pK a log [HA] nAnHA nA / V nHA / V PK a log nA25.00 nAnHA 4.00 4.87 log nA 3.3 nAnHA 0.87 100.87 0.13 Vsoln 3.3 mmol C3 H5 O2 1 mmol NaC3 H5 O 2 1 mL soln 3.3 mL added 1.00 mmol NaC3 H 5 O 2 1 mmol C3 H 5 O 2 We have assumed that all of the C3H5O2 is obtained from the NaC3H5O2 solution, since the addition of that ion in the solution should suppress the ionization of HC3H5O2.
(c) We let V be the final volume of the solution. Equation: Initial: Changes: Equil: Ka HC3 H 5 O 2 ( aq) H 2 O(l) C3 H 5 O 2 ( aq) H 3 O ( aq)
25.0/V 0M x/V M x/V M 0M x/V M x/V M x/V M (25.0/V x/V) M [ H 3 O ] [C 3 H 5 O 2 ] [HC 3 H 5 O 2 ] 1.35 10 5 (x /V )2 x2 /V 25.0 / V x / V 25.0 [H 3 O ] 0.29 mmol 1.2 10 3 M 250.0 mL When V = 250.0 mL, x = 0.29 mmol H3O+ pH = log(1.2 103) = 2.92 An increase of 0.15 pH unit gives pH = 2.92 + 0.15 = 3.07 844 Chapter 17: Additional Aspects of AcidBase Equilibria [H3O+] = 103.07 = 8.5 104 M 1.3 105 (8.5 104 ) 2 25.0 / V 8.5 104 This is the value of x/V. Now solve for V. 25.0 / V 8.5 104 V (8.5 104 ) 2 0.056 1.3 105 25.0 4.4 102 mL 0.057 On the other hand, if we had used [H3O+] = 1.16 103 M (rather than 1.2 103 M), we would obtain V = 4.8 102 mL. The answer to the problem thus is sensitive to the last significant figure that is retained. We obtain V = 4.6 102 mL, requiring the addition of 2.1 102 mL of H2O. 25.0 / V 0.056 0.00085 0.057 Another possibility is to recognize that [H3O+] K a Ca for a weak acid with ionization constant Ka and initial concentration Ca if the approximation is valid. If Ca is changed to Ca/2, [H3O+] K a C a 2 2. Since, pH = log [H3O+], the change in pH given by: pH log 2 log 21 / 2 0.5 log 2 0.5 0.30103 0.15 . This corresponds to doubling the solution volume, that is, to adding 250 mL water. Diluted by half with water, [H3O+] goes down and pH rises, then (pH1 pH2) < 0.
65. (E) Carbonic acid is unstable in aqueous solution, decomposing to CO2(aq) and H2O. The CO2(aq), in turn, escapes from the solution, to a degree determined in large part by the partial pressure of CO2(g) in the atmosphere. H2CO3(aq) H2O(l) + CO2(aq) CO2(g) + H2O(l) Thus, a solution of carbonic acid in the laboratory soon will reach a low [H2CO3], since the partial pressure of CO2(g) in the atmosphere is quite low. Thus, such a solution would be unreliable as a buffer. In the body, however, the [H2CO3] is regulated in part by the process of respiration. Respiration rates increase when it is necessary to decrease [H2CO3] and respiration rates decrease when it is necessary to increase [H2CO3].
66. (E) (a) At a pH = 2.00, in Figure 179 the pH is changing gradually with added NaOH. There would be no sudden change in color with the addition of a small volume of NaOH. (b) At pH = 2.0 in Figure 179, approximately 20.5 mL have been added. Since, equivalence required the addition of 25.0 mL, there are 4.5 mL left to add. Therefore, % HCl unneutralized 4.5 100% 18% 25.0 845 Chapter 17: Additional Aspects of AcidBase Equilibria 67. (D) Let us begin the derivation with the definition of [H3O+]. amount of excess H 3 O volume of titrant volume of solution being titrated Vb volume of base (titrant) Let Va volume of acid ( solution being titrated) [H 3 O ] M a molarity of acid [H 3 O ] Va M a Vb M b Va Vb
Vb ([H 3 O ] M b ) Va ( M a [H 3 O ]) Vb Va ( M a 10 pH ) 10 pH M b M b molarity of base Now we solve this equation for Vb V a M a Vb M b [H 3 O ] (Va Vb )
Vb Va ( M a [H 3 O ]) [H 3 O ] M b (a) 10 pH 102.00 1.0 102 M (b) 10 pH 103.50 3.2 104 M (c) 10 pH 105.00 1.0 105 M Vb Vb Vb 20.00 mL (0.1500 0.010) 25.45 mL 0.010 0.1000 20.00 mL (0.1500 0.0003) 29.85 mL 0.0003 0.1000 20.00 mL (0.1500 0.00001) 30.00 mL 0.00001 0.1000 Beyond the equivalence point, the situation is different. amount of excess OH Vb M b V a M a [OH ] total solution volume V a Vb Solve this equation for Vb. [OH ](Va + Vb) = Vb Mb Va Ma V ([OH ] M a ) Va ([OH ] M a ) Vb ( M b [OH ]) Vb a M b [OH ]
(d) [OH ] 10 pOH 1014.00 pH 103.50 0.00032 M; 0.00032 0.1500 Vb 20.00 mL 30.16 mL 0.1000 0.00032 [OH ] 10 pOH 1014.00 pH 102.00 0.010 M; (e) 0.010 0.1500 Vb 20.00 mL 35.56 mL 0.1000 0.010 The initial pH = log(0.150) = 0.824. The titration curve is sketched below. 846 Chapter 17: Additional Aspects of AcidBase Equilibria 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 25 30
Volume NaOH (ml) pH 35 68. (D) (a) The expressions that we obtained in Exercise 67 were for an acid being titrated by a base. For this titration, we need to switch the a and b subscripts and exchange [OH] and [H3O+].
Before the equivalence point : After the equivalence point : Va Vb M b [OH ] [OH ] M a [H 3 O ] M b Va Vb M a [H 3 O ] pOH 14.00 13.00 1.00 [OH ] 101.00 1.0 101 0.10 M 0.250 0.10 9.38 mL Va 25.00 mL 0.10 0.300 pH 12.00 pOH 14.00 12.00 2.00 [OH ] 102.00 1.0 102 0.010 M pH 13.00 Va 25.00 mL 0.250 0.010 19.4 mL 0.010 0.300 pOH 14.00 10.00 4.00 pH 10.00 [OH ] 104.00 1.0 104 0.00010 M Va 25.00 mL pH 4.00 Va 25.00 0.250 0.00010 20.8 mL 0.00010 0.300 [H 3 O ] 104.00 1.0 104 0.00010 M 0.00010 0.250 20.8 mL 0.300 0.00010 pH 3.00 [H 3 O ] 103.00 1.0 103 0.0010 M 0.0010 0.250 21.0 mL 0.300 0.0010 Va 25.00 847 Chapter 17: Additional Aspects of AcidBase Equilibria (b) Our expression does not include the equilibrium constant, Ka. But Ka is not needed after the equivalence point. pOH = 14.00 11.50 = 2.50 [OH] = 102.50 = 3.2 103 = 0.0032 M
Vb Va ( M a [OH ]) 50.00 mL (0.0100 0.0032) 14.1 mL 0.0500 0.0032 M b [OH ] Let us use the HendersonHasselbalch equation as a basis to derive an expression that incorporates Ka. Note that because the numerator and denominator of that expression are concentrations of substances present in the same volume of solution, those concentrations can be replaced by numbers of moles. Amount of anion = VbMb since the added OH reacts 1:1 with the weak acid. Amount of acid = VaMa VbMb the acid left unreacted pH pK a log Vb M b [A ] pK a log V a M a Vb M b [HA]
Vb M b V a M a Vb M b Rearrange and solve for Vb 10 pH pK (V a M a Vb M b ) 10
pH pK Vb M b pH 4.50, 10pH pK 104.50 4.20 pH 5.50, 10pH pK 105.50 4.20 M b (1 10 pH pK ) 50.00 mL 0.0100 M 2.0 2.0 Vb 6.7 mL 0.0500(1 2.0) 50.00 mL 0.0100 M 20 20. Vb 9.5 mL 0.0500(1 20.) Vb V a M a 10 pH pK 69. (M) (a) We concentrate on the ratio of concentrations of which the logarithm is taken. f initial amt. weak acid f equil. amount conj. base [weak acid]eq equil. amount weak acid (1 f ) initial amt. weak acid 1 f The first transformation is the result of realizing that the volume in which the weak acid and its conjugate base are dissolved is the same volume, and therefore the ratio of equilibrium amounts is the same as the ratio of concentrations. The second transformation is the result of realizing, for instance, that if 0.40 of the weak acid has been titrated, 0.40 of the original amount of weak acid now is in the form of its conjugate base, and 0.60 of that amount remains as weak acid. Equation 17.2 then is: pH = pKa + log (f /(1 f )) 0.27 (b) We use the equation just derived. pH 10.00 log 9.56 1 0.23 [conjugate base]eq 848 Chapter 17: Additional Aspects of AcidBase Equilibria 70. (M)
22[HPO 4 ] [HPO 2 ] [HPO 4 ] 4 7.20 log 7.40 100.20 1.6 [H 2 PO 4 ] [H 2 PO 4 ] [H 2 PO 4 ] (b) In order for the solution to be isotonic, it must have the same concentration of ions as does the isotonic NaCl solution. 9.2 g NaCl 1 mol NaCl 2 mol ions [ions] 0.31 M 1 L soln 58.44 g NaCl 1 mol NaCl Thus, 1.00 L of the buffer must contain 0.31 moles of ions. The two solutes that are used to formulate the buffer both ionize: KH2PO4 produces 2 mol of ions (K+ and H2PO4) per mole of solute, while Na 2 HPO 4 12 H 2 O produces 3 mol of ions (2 Na+ and HPO42 ) per mole of solute. We let x = amount of KH2PO4 and y = amount of Na 2 HPO 4 12 H 2 O . y 1.6 or y 1.6 x 2x 3 y 0.31 2 x 3(1.6 x) 6.8 x x 0.31 x 0.046 mol KH 2 PO 4 y 1.6 0.046 0.074 mol Na 2 HPO 4 12 H 2O 6.8 136.08 g KH 2 PO 4 mass KH 2 PO 4 0.046 mol KH 2 PO 4 6.3 g KH 2 PO 4 1 mol KH 2 PO 4 358.1 g Na 2 HPO 4 12H 2 O mass of Na 2 HPO 4 12H 2O 0.074 mol Na 2 HPO 4 12H 2O 1 mol Na 2 HPO 4 12H 2O (a) pH pK a 2 log 26 g Na 2 HPO 4 12H 2O 71. (M) A solution of NH4Cl should be acidic, hence, we should add an alkaline solution to make it pH neutral. We base our calculation on the ionization equation for NH3(aq), and assume that little NH4+(aq) is transformed to NH3(aq) because of the inhibition of that reaction by the added NH3(aq), and because the added volume of NH3(aq) does not significantly alter the Vtotal. Equation: NH 3 (aq) H 2 O(l) NH 4 (aq) OH (aq)
Initial: xM 0.500 M 1.0 10 7 M [NH 4 ][OH ] (0.500 M)(1.0 107 M) 1.8 105 Kb [NH 3 ] xM (0.500 M)(1.0 107 M) 2.8 103 M [NH 3 ] 5 1.8 10 M 2.8 103 mol NH 3 1 L conc. soln 1 drop V 500 mL = 2.8 drops 3 drops 1 L final soln 10.0 mol NH 3 0.05 mL xM 849 Chapter 17: Additional Aspects of AcidBase Equilibria 72. (D) (a) In order to sketch the titration curve, we need the pH at the following points. i ) Initial point. That is the pH of 0.0100 M pnitrophenol, which we represent by the general formula for an indicator that also is a weak acid, HIn.
Equation: Initial: Changes: Equil: HIn(aq) 0.0100 M x M (0.0100x) M H 2 O(l) H 3O (aq) 0M xM x M In (aq) 0M x M xM K HIn 107.15 7.1 108 [H 3O ] [In ] xx x2 [HIn] 0.0100 x 0.0100 Ca / K a 1105 ; thus, the approximation is valid [H 3O ] 0.0100 7.1108 2.7 105 M pH log(2.7 10 5 ) 4.57 ii) At the halfequivalence point, the pH = pKHIn = 7.15 iii) At the equiv point, pH is that of In. nIn = 25.00 mL 0.0100 M = 0.250 mmol In
Vtitrant = 0.25 mmol HIn [In ] 1 mmol NaOH 1 mL titrant 12.5 mL titrant 1 mmol HIn 0.0200 mmol NaOH 0.25 mmol In 6.67 103 M 0.00667 M 25.00 mL 12.5 mL
In (aq) 0.00667 M H2O(l) HIn(aq) 0M x M xM OH (aq) 0M x M xM Equation: Initial: Changes: x M Equil: (0.00667 x) M [HIn][OH ] K w 1.0 104 xx x x Kb 1.4 107 8 [In ] K a 7.1 10 0.00667 x 0.00667 5 (Ca/Ka = 1.410 ; thus, the approximation is valid) [OH ] 0.00667 1.4 107 3.1105 pOH log(3.1105 ) 4.51; therefore, pH 14.00 pOH 9.49 iv) Beyond the equivalence point, the pH is determined by the amount of excess OH in solution. After 13.0 mL of 0.0200 M NaOH is added, 0.50 mL is in excess, and the total volume is 38.0 mL. 0.50 mL 0.0200 M [OH ] 2.63 10 4 M pOH 3.58 pH 10.42 38.0 mL After 14.0 mL of 0.0200 M NaOH is added, 1.50 mL is in excess, and Vtotal = 39.0 mL. 1.50 mL 0.0200 M 7.69 10 4 M pOH 3.11 pH 10.89 [OH ] 39.0 mL v) In the buffer region, the pH is determined with the use of the HendersonHasselbalch equation. 850 Chapter 17: Additional Aspects of AcidBase Equilibria 0.10 0.90 6.20 At f 0.90, pH 7.15 log 8.10 0.90 0.10 f = 0.10 occurs with 0.10 12.5 mL = 1.25 mL added titrant; f = 0.90 with 11.25 mL added. The titration curve plotted from these points follows. At f 0.10, pH 7.15 log
12 11 10 9
pH 8 7 6 5 4 0 5 10
Volume NaOH (ml) 15 20 (b) The pH color change range of the indicator is shown on the titration curve. (c) The equivalence point of the titration occurs at a pH of 9.49, far above the pH at which pnitrophenol has turned yellow. In fact, the color of the indicator changes gradually during the course of the titration, making it unsuitable as an indicator for this titration. Possible indicators are as follows. Phenolphthalein: pH color change range from colorless at pH = 8.0 to red at pH = 10.0 Thymol blue: pH color change range from yellow at pH = 8.0 to blue at pH = 9.8 Thymolphthalein: pH color change range from colorless at pH 9 to blue at pH 11 The red tint of phenolphthalein will appear orange in the titrated pnitrophenol solution. The blue of thymol blue or thymolphthalein will appear green in the titrated pnitrophenol solution, producing a somewhat better yellow end point than the orange phenolphthalein endpoint. 73. (M) (a) Equation (1) is the reverse of the equation for the autoionization of water. Thus, its equilibrium constant is simply the inverse of Kw. 1 1 K 1.00 1014 K w 1.0 10 14 Equation (2) is the reverse of the hydrolysis reaction for NH4+. Thus, its equilibrium constant is simply the inverse of the acid ionization constant for NH4+, Ka = 5.6 1010 1 1 K' 1.8 10 9 10 K a 5.6 10 851 Chapter 17: Additional Aspects of AcidBase Equilibria (b) The extremely large size of each equilibrium constant indicates that each reaction goes essentially to completion. In fact, a general rule of thumb suggests that a reaction is considered essentially complete if Keq > 1000 for the reaction. 74. (D) The initial pH is that of 0.100 M HC2H3O2. Equation: HC2 H 3 O 2 (aq) H 2 O(l) C2 H 3 O 2 (aq) H3 O (aq) 0M x M xM Initial: Changes: Equil: Ka 0.100 M xM (0.100 x) M 0M x M xM [H 3 O ][C2 H 3 O 2 ] xx x2 1.8 105 [HC2 H 3 O 2 ] 0.100 x 0.100 3 (Ca/Ka = 5.510 ; thus, the approximation is valid) [H 3 O ] 0.100 1.8 105 1.3 103 M pH log(1.3 103 ) 2.89 At the equivalence point, we have a solution of NH4C2H3O2 which has a pH = 7.00, because both NH4+ and C2H3O2 hydrolyze to an equivalent extent, since K a (HC 2 H 3 O 2 ) K b (NH 3 ) and their hydrolysis constants also are virtually equal. Total volume of titrant = 10.00 mL, since both acid and base have the same concentrations. At the half equivalence point, which occurs when 5.00 mL of titrant have been added, pH = pKa = 4.74. When the solution has been 90% titrated, 9.00 mL of 0.100 M NH3 has been added. We use the HendersonHasselbalch equation to find the pH after 90% of the acid has been titrated [C 2 H 3 O 2 ] 9 pH pK a log 4.74 log 5.69 [HC 2 H 3 O 2 ] 1 When the solution is 110% titrated, 11.00 mL of 0.100 M NH3 have been added. amount NH3 added = 11.00 mL 0.100 M = 1.10 mmol NH3 amount NH4+ produced = amount HC2H3O2 consumed = 1.00 mmol NH4+ amount NH3 unreacted = 1.10 mmol NH3 1.00 mmol NH4+ = 0.10 mmol NH3 We use the HendersonHasselbalch equation to determine the pH of the solution. [NH 4 ] 1.00 mmol NH 4 pOH pK b log 4.74 log 5.74 pH 8.26 [ NH 3 ] 0.10 mmol NH 3 When the solution is 150% titrated, 15.00 mL of 0.100 M NH3 have been added. amount NH3 added = 15.00 mL 0.100 M = 1.50 mmol NH3 amount NH4+ produced = amount HC2H3O2 consumed = 1.00 mmol NH4+ amount NH3 unreacted = 1.50 mmol NH3 1.00 mmol NH4+ = 0.50 mmol NH3 [NH 4 ] 1.00 mmol NH 4 pOH pK b log 4.74 log 5.04 pH 8.96 [ NH 3 ] 0.50 mmol NH 3 The titration curve based on these points is sketched next. We note that the equivalence point is not particularly sharp and thus, satisfactory results are not obtained from acetic acidammonia titrations. 852 Chapter 17: Additional Aspects of AcidBase Equilibria 10 9 8 7 6 5 4 3 2 1 0 0 2 4 6 8 10 Volume of NH3 (mL) 12 14 16 pH 75. (D) C6H5NH3+ is a weak acid, whose acid ionization constant is determined from Kb(C6H5NH2) = 7.4 1010. 1.0 10 14 Ka 1.4 10 5 and pK a 4.85 . We first determine the initial pH. 7.4 10 10 Equation: C6 H 5 NH 3 (aq) H 2 O(l) C6 H 5 NH 2 (aq) H 3O (aq)
Initial: Changes: Equil: 0.0500 M x M (0.0500 x) M
5 0 M x M x M 0 M x M x M K a 1.4 10 [C6 H 5 NH 2 ] [H 3 O ] xx x2 0.0500 x 0.0500 [C6 H 5 NH 3 ] pH log (8.4 10 4 ) 3.08 (Since Ca /K a = 3.6103 ; thus, the approximation is valid) [H 3 O ] 0.0500 1.4 105 8.4 104 M At the equivalence point, we have a solution of C6H5NH2(aq). Now find the volume of titrant.
V 10.00 mL 0.0500 mmol C6 H 5 NH 3 1 mmol NaOH 1 mL titrant 5.00 mL 1 mL 1 mmol C6 H 5 NH 3 0.100 mmol NaOH amount C6 H 5 NH 2 10.00 mL 0.0500 M 0.500 mmol C6 H 5 NH 2 [C6 H 5 NH 2 ] 0.500 mmol C6 H 5 NH 2 0.0333 M 10.00 mL 5.00 mL 853 Chapter 17: Additional Aspects of AcidBase Equilibria Equation: Initial: Changes: Equil: C6 H 5 NH 2 (aq) 0.0333 M x M (0.0333 x) M H 2 O(l) C6 H 5 NH 3 (aq) 0 M OH (aq) 0M x M xM x M xM
2 K b 7.4 1010 [C6 H 5 NH 3 ][OH ] xx x [C6 H 5 NH 2 ] 0.0333 x 0.0333 pOH 5.30 pH 8.70 (Cb /K b = very large number; thus, the approximation is valid) [OH ] 0.0333 7.4 1010 5.0 106 At the half equivalence point, when 5.00 mL of 0.1000 M NaOH has been added, pH = pKa = 4.85. At points in the buffer region, we use the expression derived in Exercise 69.
After 4.50 mL of titrant has been added, f = 0.90, pH = 4.85 + log 0.90 = 5.80 0.10 0.95 After 4.75 mL of titrant has been added, f = 0.95, pH = 4.85 + log = 6.13 0.05 After the equivalence point has been reached, with 5.00 mL of titrant added, the pH of the solution is determined by the amount of excess OH that has been added. 0.25 mL 0.100 M 0.0016 M 15.25 mL 1.00 mL 0.100 M At 120% titrated, [OH ] 0.00625 M 16.00 mL At 105% titrated, [OH ] pOH 2.80 pH 11.20 pOH 2.20 pH 11.80 Of course, one could sketch a suitable titration curve by calculating fewer points. Below is the titration curve.
12 10
pH 8 6 4 2 0
Volume NaOH (ml) 5 854 Chapter 17: Additional Aspects of AcidBase Equilibria 76. In order for a diprotic acid to be titrated to two distinct equivalence points, the acid must initially start out with two undissociated protons and the Ka values for the first and second protons must differ by >1000. This certainly is not the case for H2SO4, which is a strong acid (first proton is 100% dissociated). Effectively, this situation is very similar to the titration of a monoprotic acid that has added strong acid (i.e., HCl or HNO3). With the leveling effect of water, Ka1 = 1 and Ka2 = 0.011, there is a difference of only 100 between Ka1 and Ka2 for H2SO4.
(D) For both titration curves, we assume 10.00 mL of solution is being titrated and the concentration of the solute is 1.00 M, the same as the concentration of the titrant. Thus, 10.00 mL of titrant is required in each case. (You may be able to sketch titration curves based on fewer calculated points. Your titration curve will look slightly different if you make different initial assumptions.) (a) The initial pH is that of a solution of HCO3(aq). This is an anion that can ionize to CO32 (aq) or be hydrolyzed to H2CO3(aq). Thus, pH 1 (pK a1 pK a2 ) 1 (6.35 10.33) 8.34 2 2 77. The final pH is that of 0.500 M H2CO3(aq). All of the NaOH(aq) has been neutralized, as well as all of the HCO3(aq), by the added HCl(aq). Equation: Initial: Changes: Equil: H 2 CO3 (aq) H 2 O(l) 0.500 M xM (0.500 x ) M HCO3 (aq) H 3 O (aq) 0M xM xM 0M xM xM Kb [HCO3 ][H 3 O ] xx x2 4.43 107 [H 2 CO3 ] 0.500 x 0.500 pH 3.33 (Ca /K a = 1.1106 ; thus, the approximation is valid) [OH ] 0.500 4.4 107 4.7 104 M During the course of the titration, the pH is determined by the HendersonHasselbalch equation, with the numerator being the percent of bicarbonate ion remaining, and as the denominator being the percent of bicarbonate ion that has been transformed to H2CO3.
90% titrated: pH = 6.35 + log 10% 5.40 90% 10% titrated: pH = 6.35 + log 90% 7.30 10% 95% titrated: pH = 6.35 + log 5% 95% 5.07 5% titrated: pH = 6.35 + log 7.63 95% 5% After the equivalence point, the pH is determined by the excess H3O+. 0.50 mL 1.00 M At 105% titrated, [H 3 O ] 0.024 pH 1.61 20.5 mL 855 Chapter 17: Additional Aspects of AcidBase Equilibria 2.00 mL 1.00 M 0.091 M pH 1.04 22.0 mL The titration curve derived from these data is sketched below. At 120% titrated, [H 3 O ] 9 8 7 6 5 4 3 2 1 0 0 2 4 6 8 Volume 1.00 M HCl 10 12 (b) The final pH of the first step of the titration is that of a solution of HCO3(aq). This is an anion that can be ionized to CO32(aq) or hydrolyzed to H2CO3(aq). Thus, pH 1 (pK a1 pK a 2 ) 1 (6.35 10.33) 8.34 . The initial pH is that of 1.000 M 2 2 CO32(aq), which pH is the result of the hydrolysis of the anion.
Equation: Initial: Changes: Equil: Kb CO32 (aq) H 2 O(l) HCO3 (aq) OH (aq) 0M x M xM 0M x M xM pH 1.000 M x M (1.000 x) M [HCO3 ][OH ] K w 1.0 1014 xx x2 2.1 104 [CO32 ] K a2 4.7 1011 1.000 x 1.000 pOH 1.85 pH 12.15 [OH ] 1.000 2.1104 1.4 102 M During the course of the first step of the titration, the pH is determined by the HendersonHasselbalch equation, modified as in Exercise 69, but using as the numerator the percent of carbonate ion remaining, and as the denominator the percent of carbonate ion that has been transformed to HCO3.
10% 9.38 90% 5% 95% titrated: pH 10.33 log 9.05 95% 90% titrated: pH 10.33 log 10% titrated: pH 10.33 log 90% 11.28 10% 95% 5% titrated: pH 10.33 log 11.61 5% During the course of the second step of the titration, the values of pH are precisely as they are for the titration of NaHCO3, except the titrant volume is 10.00 mL more (the volume needed to reach the first equivalence point). The solution at the second 856 Chapter 17: Additional Aspects of AcidBase Equilibria equivalence point is 0.333 M H2CO3, for which the setup is similar to that for 0.500 M H2CO3.
[H 3 O ] 0.333 4.4 10 7 3.8 10 4 M pH 3.42 After the equivalence point, the pH is determined by the excess H3O+. 0.50 mL 1.00 M 0.0164 M pH 1.8 At 105% titrated, [H3O ] 30.5 mL 2.00 mL 1.00 M 0.0625 M pH 1.20 At 120% titrated, [H3O ] 32.0 mL The titration curve from these data is sketched below.
12 10 8 pH 6 4 2 0 3 2 7 12 Volume 1.00 M HCl, mL 17 22 (c) VHCl 1.00 g NaHCO3 119 mL 0.100 M HCl 1 mol NaHCO3 1 mol HCl 1000 mL 84.01 g NaHCO3 1 mol NaHCO3 0.100 mol HCl 1 mol Na 2 CO3 2 mol HCl 1000 mL 105.99 g Na 2 CO3 1 mol Na 2 CO3 0.100 mol HCl (d) VHCl 1.00 g Na 2 CO3 189 mL 0.100 M HCl (e) The phenolphthalein endpoint occurs at pH = 8.00 and signifies that the NaOH has been neutralized, and that Na2CO3 has been half neutralized. The methyl orange endpoint occurs at about pH = 3.3 and is the result of the second equivalence point of Na2CO3. The mass of Na2CO3 can be determined as follows: 857 Chapter 17: Additional Aspects of AcidBase Equilibria 1L 0.1000 mol HCl 1 mol HCO3 1 mol Na 2 CO3 mass Na 2 CO3 0.78 mL 1000 mL 1 L soln 1 mol HCl 1 mol HCO3 105.99 g Na 2 CO3 0.0083 g Na 2 CO3 1 mol Na 2 CO3 0.0083g 100 8.3% Na 2 CO3 by mass 0.1000g % Na 2 CO3 78. (M) We shall represent piperazine as Pip in what follows. The cation resulting from the first ionization is HPiP+, and that resulting from the second ionization is H2Pip2+. 1.00 g C4 H10 N 2 6 H 2 O 1mol C4 H10 N 2 6 H 2 O (a) [Pip] = 0.0515M 0.100 L 194.22 g C 4 H10 N 2 6 H 2 O H 2O(l) HPip (aq) OH (aq) Initial: 0.0515 M 0M 0 M Changes: x M x M xM Equil: (0.0515 x) M xM xM Ca/Ka = 858; thus, the approximation is not valid. The full quadratic equation must be solved. [HPip ][OH ] xx K b1 104.22 6.0 105 [ Pip ] 0.0515 x Equation: Pip (aq) From the roots of the equation, x = [OH ] 1.7 103 M pOH 2.77 pH 11.23 (b) At the halfequivalence point of the first step in the titration, pOH = pKb1 = 4.22 pH = 14.00 pOH = 14.00 4.22 = 9.78 (c) Volume of HCl = 100. mL 0.0515 mmol Pip 1mmol HCl 1mL titrant = 10.3mL 1mL 1mmol Pip 0.500 mmol HCl (d) At the first equivalence point we have a solution of HPip+. This ion can react as a base with H2O to form H2Pip2+ or it can react as an acid with water, forming Pip (i.e. HPip+ is amphoteric). The solution's pH is determined as follows (base hydrolysis predominates): pOH = (pKb1 + pKb2) = (4.22 + 8.67) = 6.45, hence pH = 14.00 6.45 = 7.55 (e) The pOH at the halfequivalence point in the second step of the titration equals pKb2. pH = 14.00 8.67 = 5.33 pOH = pKb2 = 8.67 (f) The volume needed to reach the second equivalence point is twice the volume needed to reach the first equivalence point, that is 2 10.3 mL = 20.6 mL. (g) The pH at the second equivalence point is determined by the hydrolysis of the H2Pip2+ cation, of which there is 5.15 mmol in solution, resulting from the reaction 858 Chapter 17: Additional Aspects of AcidBase Equilibria of HPip+ with HCl. The total solution volume is 100. mL + 20.6 mL = 120.6 mL 5.15 mmol 0.0427 M [H2Pip2+] = K b 2 108.67 2.1 109 120.6 mL Equation: Initial: Changes: Equil: H 2 Pip 2+ (aq) H 2 O(l) HPip (aq) H 3O (aq) 0.0427 M 0M 0 M x M x M x M xM xM (0.0427 x) M K w 1.00 1014 [HPip ][H 3O ] x x x2 6 Ka 4.8 10 Kb 2 2.1 109 [H 2 Pip 2+ ] 0.0427 x 0.0427 x [H 3O ] 0.0427 4.8 106 4.5 104 M (x << 0.0427; thus, the approximation is valid). 79. (M) Consider the two equilibria shown below. H2PO4 (aq) + H2O(l) HPO42(aq) + H3O+(aq)
 pH 3.347 [HPO 4 2 ][H 3 O + ] Ka2 = [H 2 PO 4 ] H2PO4 (aq) + H2O(l) H3PO4(aq) + OH(aq)
 [H 3 PO 4 ][OH  ] Kb = Kw/Ka1 = [H 2 PO 4 ] All of the phosphorus containing species must add up to the initial molarity M. Hence, mass balance: [HPO42 ] + [H2PO4 ] + [H3PO4] = M charge balance: [H3O+] + [Na+] = [H2PO4 ] + 2[HPO42 ] Note: [Na+] for a solution of NaH2PO4 = M [H3O+ ] + [Na+ ] = [H2PO4 ] + 2[HPO42 ] = [H3O+ ] + M (substitute mass balance equation) [H3O+ ] + [HPO42 ] + [H2PO4 ] + [H3PO4] = [H2PO4 ] + 2[HPO42 ] (cancel terms) [H3O+ ] = [HPO42 ]  [H3PO4] (Note: [HPO42 ] = initial [H3O+ ] and [H3PO4] = initial [OH ]) Thus: [H3O+ ]equil = [H3O+ ]initial ]initial (excess H3O+ reacts with OH to form H2O) Thus, Rearrange the expression for Ka2 to solve for [HPO42], and the expression for Kb to solve for [H3PO4].
[H 3 O + ] = [HPO 4 2 ]  [H 3 PO 4 ] =
 K a2 [H 2 PO 4  ] K b [H 2 PO 4 ] [H 3 O + ] [OH  ] Kw [H 2 PO 4  ] K a2 [H 2 PO 4 ] K a1 K [H PO  ] [H O + ][H 2 PO 4  ] [H 3 O + ] a2 2 + 4 3 Kw K a1 [H 3 O + ] [H 3 O ] + [H 3 O ] Multiply through by [H3O+] Ka1 and solve for [H3O+]. 859 Chapter 17: Additional Aspects of AcidBase Equilibria K a1 [H 3 O + ][H 3 O + ] K a1 K a2 [H 2 PO 4 ] [H 3 O + ][H 3 O + ][H 2 PO 4 ] K a1 [H 3 O + ]2 K a1 K a2 [H 2 PO 4  ] [H 3 O + ]2 [H 2 PO 4  ] K a1 [H 3 O + ]2 [H 3 O + ]2 [H 2 PO 4 ] K a1 K a2 [H 2 PO 4 ] = [H 3 O + ]2 (K a1 [H 2 PO 4 ]) K a1 [H 3 O + ]2 [H 3 O + ]2 [H 2 PO 4 ] K a1 K a2 [H 2 PO 4 ] = [H 3 O + ]2 (K a1 [H 2 PO 4 ]) [H 3 O + ]2 = K a1 K a2 [H 2 PO 4  ] For moderate concentrations of [H 2 PO 4  ], K a1 << [H 2 PO4  ] (K a1 [H 2 PO 4 ]) This simplifies our expression to: [H 3O + ]2 = K a1K a2 [H 2 PO 4  ] = K a1K a2 [H 2 PO 4  ] Take the square root of both sides: [H 3O + ]2 = K a1K a2 [H3O+ ] = K a1K a2 (K a1K a2 )1 2 Take the log of both sides and simplify: log[H 3O + ] = log(K a1K a2 )1 2 1 2 log(K a1K a2 ) 1 2 (log K a1 +logK a2 ) log[H 3O + ] 1 2 ( log K a1 logK a2 ) Use log[H 3O + ] = pH and log K a1 pK a1 logK a2 pK a2 Hence, log[H 3O + ] = 1 2 ( log K a1 logK a2 ) becomes pH =1 2 (pK a1 +pK a2 ) (Equation 17.5) Equation 17.6 can be similarly answered. 80. (D) H2PO4 can react with H2O by both ionization and hydrolysis. H 2 PO 4  (aq) H 2 O(l) HPO 4 2 (aq) H 3 O (aq) HPO 4 2 (aq) H 2 O(l) PO 4 3 (aq) H3 O (aq) The solution cannot have a large concentration of both H3O+ and OH (cannot be simultaneously an acidic and a basic solution). Since the solution is acidic, some of the H3O+ produced in the first reaction reacts with virtually all of the OH produced in the second. In the first reaction [H3O+] = [HPO42] and in the second reaction [OH] = [H3PO4]. Thus, following the neutralization of OH by H3O+, we have the following. Kw [H 2 PO4 ] K a2 [H 2 PO ] K b [H 2 PO4 ] K a2 [H 2 PO ] K a1 2 [H 3O ] [HPO 4 ] [H 2 PO 4 ] Kw [H 3O ] [OH ] [H 3O ] [H 3O ]
4 4 K a2 [H 2 PO4 ] [H 3O ][H 2 PO4 ] K a1 [H 3O ]
4 Then multiply through by [H3O ]Ka 1 and solve for [H3O ]. 2 4 [H 3 O ] K a1 K a1 K a2 [H 2 PO ] [H 3 O ] [H 2 PO ] 2 [H 3 O ] K a1 K a2 [H 2 PO 4 ]
[ K a1 [H 2 PO4 ] But, for a moderate [H2PO 4 ], from Table 166, K a1 = 7.1 103 < [H2PO 4 ] and thus K a1 [H 2 PO 4 ] [H 2 PO 4 ] . Then we have the following expressions for [H3O+] and pH. 860 Chapter 17: Additional Aspects of AcidBase Equilibria [H 3O ] K a1 K a2 pH log[H 3O ] log(K a1 K a2 )1/2 1 [ log(K a1 K a2 )] 1 ( log K a1 log K a2 ) 2 2 pH 1 2 (pK a1 pK a2 ) Notice that we assumed Ka < [H2PO4]. This assumption is not valid in quite dilute solutions because Ka = 0.0071.
81. (D) (a) A buffer solution is able to react with small amounts of added acid or base. When strong acid is added, it reacts with formate ion.
CHO 2 (aq) H 3 O (aq) HCHO 2 (aq) H 2 O Added strong base reacts with acetic acid.
HC 2 H 3 O 2 (aq) OH (aq) H 2 O(l) C 2 H 3 O 2 (aq) Therefore neither added strong acid nor added strong base alters the pH of the solution very much. Mixtures of this type are referred to as buffer solutions.
(b) We begin with the two ionization reactions. HCHO 2 (aq) H 2 O(l) CHO 2 (aq) H 3 O (aq) HC2 H 3 O 2 (aq) H 2 O(l) C2 H 3 O 2 (aq) H 3 O (aq) [Na ] 0.250 [OH ] 0 [H 3O ] x [C2 H 3O 2 ] y [CHO 2 ] z 0.150 [HC 2 H 3 O 2 ] [C 2 H 3 O 2 ] [HC 2 H 3 O 2 ] 0.150 [C 2 H 3 O 2 ] 0.150 y 0.250 [HCHO 2 ] [CHO 2 ] [HCHO 2 ] 0.250 [CHO 2 ] 0.250 z [ Na ] [H 3 O ] [C 2 H 3 O 2 ] [CHO 2 ] [OH ] (electroneutrality) 0.250 x [C 2 H 3 O 2 ] [CHO 2 ] y z (1) (2) [H 3 O ][C 2 H 3 O 2 ] x y K A 1.8 10 5 [HC 2 H 3 O 2 ] 0.150 y [H 3 O ][CHO 2 ] xz K F 1.8 10 4 [HCHO 2 ] 0.250 z (3) There now are three equations(1), (2), and (3)in three unknownsx, y, and z. We solve equations (2) and (3), respectively, for y and z in terms of x. 861 Chapter 17: Additional Aspects of AcidBase Equilibria 0.150 K A y K A xy 0.250 K F z K F xz y z 0.150 K A KA x 0.250 K F KF x Then we substitute these expressions into equation (1) and solve for x.
0.250 x 0.150 K A 0.250 K F 0.250 KA x KF x since x 0.250 0.250 ( K F x) ( K A x) 0.150 K A ( K F x) 0.250 K F ( K A x) K A K F ( K A K F ) x x 2 1.60 K A K F x(0.600 K A 1.00 K F ) x 2 0.400 K A x 0.600 K A K F 0 x 2 7.2 10 6 x 1.9 10 9 7.2 10 6 5.2 10 11 7.6 10 9 4.0 10 5 M [H 3 O ] 2 pH 4.40 (c) Adding 1.00 L of 0.100 M HCl to 1.00 L of buffer of course dilutes the concentrations of all components by a factor of 2. Thus, [Na+] = 0.125 M; total acetate concentration = 0.0750 M; total formate concentration = 0.125 M. Also, a new ion is added to the solution, namely, [Cl] = 0.0500 M. x [ Na ] 0.125 [OH ] 0 [H 3 O ] x [Cl ] 0.0500 M [C 2 H 3 O 2 ] y [CHO 2 ] z 0.0750 [HC 2 H 3 O 2 ] [C 2 H 3 O 2 ] [HC 2 H 3 O 2 ] 0.0750 [C 2 H 3 O 2 ] 0.0750 y 0.125 [HCHO 2 ] [CHO 2 ] [HCHO 2 ] 0.125 [CHO 2 ] 0.125 z [ Na ] [H 3 O ] [C 2 H 3 O 2 ] [OH ] [CHO 2 ] [Cl ] (electroneutrality)
0.125 x [C 2 H 3 O 2 ] [CHO 2 ] [Cl ] y z 0 0.0500 0.075 x y z [H 3 O ][C 2 H 3 O 2 ] x y K A 1.8 10 5 [ HC 2 H 3 O 2 ] 0.0750 y
[H 3 O ][CHO 2 ] [HCHO 2 ] K F 1.8 10 4 xz 0.125 z Again, we solve the last two equations for y and z in terms of x. 0.0750 K A y K A xy 0.125 K F z K F xz y 0.0750 K A KA x 0.125 K F KF x z 862 Chapter 17: Additional Aspects of AcidBase Equilibria Then we substitute these expressions into equation (1) and solve for x.
0.0750 x 0.0750 K A KA x 0.125 K F KF x 0.0750 since x 0.0750 0.0750 (K F x) (K A x) 0.0750 K A (K F x) 0.125 K F (K A x) K A K F (K A K F )x x 2 2.67 K A K F x(1.67 K F 1.00 K A ) x 2 0.67 K F x 1.67 K A K F 0 x 2 1.2 10 4 x 5.4 10 9 x 1.2 104 1.4 108 2.2 10 8 2 1.55 10 4 M [H 3O ] pH 3.81 3.8 As expected, the addition of HCl(aq), a strong acid, caused the pH to drop. The decrease in pH was relatively small, nonetheless, because the H3O+(aq) was converted to the much weaker acid HCHO2 via the neutralization reaction: CHO2(aq) + H3O+(aq) HCHO2(aq) + H2O(l)
82. (M) First we find the pH at the equivalence point: CH 3 CH(OH)COOH(aq) OH ( aq) CH 3 CH(OH)COO  (aq) H 2 O(l)
1 mmol 1 mmol 1 mmol The concentration of the salt is 1 103 mol/0.1 L = 0.01 M The lactate anion undergoes hydrolysis thus: Initial Change Equilibrium (buffering action) CH 3 CH(OH)COO  ( aq ) H 2 O(l) CH 3 CH(OH)COOH(aq) OH (aq ) 0 M 0.01 M 0M
x (0.01 x) M +x x
 +x x Where x is the [hydrolyzed lactate ion], as well as that of the [OH ] produced by hydrolysis K for the above reaction so [CH 3 CH(OH)COOH][OH ] K w 1.0 1014 7.2 1011 [CH 3 CH(OH)COO ] Ka 103.86 x2 x2 7.2 1011 and x [OH  ] 8.49 107 M 0.01 x 0.01 x 0.01, thus, the assumption is valid pOH log (8.49 107 M ) 6.07 pH 14 pOH 14.00 6.07 7.93 (a) Bromthymol blue or phenol red would be good indicators for this titration since they change color over this pH range. 863 Chapter 17: Additional Aspects of AcidBase Equilibria (b) The H2PO4/HPO42 system would be suitable because the pKa for the acid (namely, H2PO4) is close to the equivalence point pH of 7.93. An acetate buffer would be too acidic, an ammonia buffer too basic. (c) H 2 PO 4  H 2 O H 3 O HPO 4 2 K a2 6.3 10 8 K a2 [HPO 4 2 ] 6.3 108 5.4 (buffer ratio required) Solving for [H 2 PO 4  ] [H 3 O ] 107.93 83. (M) H 2 O 2 (aq) H 2 O(l) H 3 O (aq) HO 2 (aq) K a Data taken from experiments 1 and 2:
[H 2 O 2 ] [HO 2 ] 0.259 M [HO 2 ] 0.235 M [H 3 O ] 10 ( pKw pOH ) [H 3 O ][HO 2 ] [H 2 O 2 ] (6.78) (0.00357 M) [HO 2 ] 0.259 M [H 2 O 2 ] (6.78)(0.00357 M) 0.0242 M log 0.250  0.235) M NaOH log (0.015 M NaOH) = 1.824 [H 3 O ][HO 2 ] (7.7 1014 M)(0.235 M) 7.4 1013 [H 2 O 2 ] (0.0242 M) [H 3 O ] 10 (14.941.824) 7.7 1014 Ka pK a 12.14 From data taken from experiments 1 and 3: [H 2 O 2 ] [HO 2 ] 0.123 M (6.78)(0.00198 M) [HO 2 ] 0.123 M
[HO 2 ] 0.1096 M [H 2 O 2 ] (6.78)(0.00198 M) 0.0134 M [H 3 O ] 10 ( pKw pOH ) , pOH = log[(0.125 0.1096 ) M NaOH] 1.81 [H 3 O ] 10 (14.941.81) 7.41 1014 [H 3 O ][HO 2 ] (7.41 1014 M)(0.1096 M) Ka 6.06 1013 [H 2 O 2 ] (0.0134 M) pK a = log (6.06 1013 ) = 12.22 Average value for pK a = 12.17 864 Chapter 17: Additional Aspects of AcidBase Equilibria 84. (D) Let's consider some of the important processes occurring in the solution. (1) HPO 4 2 (aq) H 2 O(l) H 2 PO 4 (aq) OH (aq) 2 K (1) = K a2 4.2 1013 1.0 1014 K (2) = K b (2) HPO 4 1.6 107 8 6.3 10 1.0 1014 K (3) = K a (3) NH 4 (aq) H 2 O(l) H 3 O (aq) NH 3 (aq) 5.6 1010 1.8 105 1 K (4) (4) NH 4 (aq) OH  (aq) H 2 O(l) NH 3 (aq) 5.6 104 1.8 105 May have interaction between NH 4 and OH  formed from the hydrolysis of HPO 4 2 . (aq) H 2 O(l) H 3 O (aq) PO 43 (aq) NH 4 (aq) HPO 4 2 (aq) H 2 PO 4  (aq) NH 3 (aq) Initial 0.10M 0.10M 0 M 0 M x x x Change x Equil. 0.100 x 0.100 x x x (where x is the molar concentration of NH 4 that hydrolyzes)
K K (2) K (4) (1.6 107 ) (5.6 104 ) 9.0 103 K [H 2 PO 4  ] [NH 3 ] [x]2 9.0 103 and x 8.7 103M 2 2 [NH 4 ] [HPO 4 ] [0.10x] Finding the pH of this buffer system: pH pK a log 0.1008.7 103 M [HPO 4 2 ] log(6.3 108 ) log 8.2 3 [H 2 PO 4  ] 8.7 10 M As expected, we get the same result using the NH 3 /NH 4 buffer system. 85. Consider the two weak acids and the equilibrium for water (autodissociation) HA(aq) + H 2 O(l) H3 O+ (aq) + A (aq) K HA = HB(aq) + H 2 O(l) H 3 O + (aq) + B (aq) K HB = [H 3 O + ][A  ] [HA] [H 3 O + ][ B ] [HB] [H 3 O + ][A  ] [HA]= K HA [HB]= [H 3 O + ][ B ] K HB Kw [H 3 O + ] H 2 O(l) + H 2 O(l) H3 O+ (aq) + OH  (aq) K w =[H 3 O + ][OH  ] [OH  ] = 865 Chapter 17: Additional Aspects of AcidBase Equilibria Mass Balance: [HA]initial =[HA]+[A  ] = From which: [A  ] = [HA]initial [H 3O + ] +1 K HA [H O + ] [H 3O + ][A  ] +[A  ] = [A  ] 3 +1 K HA K HA [HB]initial = [HB] + [B ] = From which: [B ] = [H O + ] [H 3O + ][B ] + [B ] = [B ] 3 +1 K HB K HB [HB]initial [H 3O + ] +1 K HB Charge Balance: [H3O+ ] = [A  ] + [B ] + [OH  ] (substitute above expressions) [H 3O + ] = [HA]initial [HB]initial Kw + + [H3O ] [H 3O ] [H 3O + ] +1 +1 K HA K HB 86. (D) We are told that the solution is 0.050 M in acetic acid (Ka = 1.8 105) and 0.010 M in phenyl acetic acid (Ka = 4.9 105). Because the Ka values are close, both equilibria must be satisfied simultaneously. Note: [H3O+] is common to both equilibria (assume z is the concentration of [H3O+]). Reaction: Initial: Change: Equilibrium HC2H3O2(aq) + H2O(l) C2H3O2(aq) + H3O+(aq) 0.050 M  0M 0M x M  +xM +zM (0.050 x)M  xM zM + [H3 O ][C2 H 3 O 2 ] 1.8 105 (0.050 x) xz 1.8 105 or x For acetic acid: K a = [HC2 H3 O 2 ] (0.050 x) z HC8H7O2(aq) 0.010 M y M (0.010 y)M    + H2O(l) Reaction: H3O+(aq) Initial: Change: Equilibrium 0M +yM yM C8H7O2(aq) 0M +zM zM + [H3O + ][C8 H 7 O 2 ] yz For phenylacetic acid: K a = 4.9 105 [HC8 H 7 O 2 ] (0.010 y ) or y 4.9 105 (0.010 y ) z 866 Chapter 17: Additional Aspects of AcidBase Equilibria There are now three variables: x = [C2H3O2 ], y = [C8H7O2 ], and z = [H3O+ ] These are the only charged species, hence, x + y = z. (This equation neglects contribution from the [H3O+] from water.) Next we substitute in the values of x and y from the rearranged Ka expressions above. 1.8 105 (0.050 x) 4.9 105 (0.010 y ) z z x+y=z= + Since these are weak acids, we can simplify this expression by assuming that x << 0.050 and y << 0.010. 1.8 105 (0.050) 4.9 105 (0.010) z z z= + z2 = 9.0107 + 4.9107 = 1.4 106 z = 1.18103 = [H3O+] From which we find that x = 7.6 104 and y = 4.2104 (as a quick check, we do see that x + y = z). We finish up by checking to see if the approximation is valid (5% rule).. For x : 7.6 104 100% 1.5% 0.050 For y : 4.2 104 100% 4.2% 0.010 Simplify even further by multiplying through by z. Since both are less than 5%, we can be assured that the assumption is valid. The pH of the solution = log(1.18103) = 2.93. (If we simply plug the appropriate values into the equation developed in the previous question we get the exact answer 2.933715 (via the method of successive approximations), but the final answers can only be reported to 2 significant figures. Hence the best we can do is say that the pH is expected to be 2.93 (which is the same level of precision as that for the result obtained following the 5% rule).
87. (D) (a) By using dilution, there are an infinite number of ways of preparing the pH = 7.79 buffer. We will consider a method that does not use dilution. Let TRIS = weak base and TRISH+ be the conjugate acid. pKb = 5.91, pKa = 14 pKb = 8.09 Use the Henderson Hasselbalch equation. pH = pKa + log (TRIS/TRISH+) = 7.79 = 8.09+ log (TRIS/TRISH+) log (TRIS/TRISH+) = 7.79 8.09 = 0.30 Take antilog of both sides (TRIS/TRISH+) = 100.30 = 0.50; Therefore, nTRIS = 0.50(nTRISH+) 867 Chapter 17: Additional Aspects of AcidBase Equilibria Since there is no guidance given on the capacity of the buffer, we will choose an arbitrary starting point. We start with 1.00 L of 0.200 M TRIS (0.200 moles). We need to convert 2/3 of this to the corresponding acid (TRISH+) using 10.0 M HCl, in order for the TRIS/TRISH+ ratio to be 0.5. In all, we need (2/3)0.200 mol of HCl, or a total of 0.133 moles of HCl. Volume of HCl required = 0.133 mol 10.0 mol/L = 0.0133 L or 13.3 mL. This would give a total volume of 1013.3 mL, which is almost a liter (within 1.3%). If we wish to make up exactly one liter, we should only use 987 mL of 0.200 M TRIS. This would require 13.2 mL of HCl, resulting in a final volume of 1.0002 L. Let's do a quick double check of our calculations. nHCl = 0.0132 L10.0 M = 0.132 mol = nTRISH+ nTRIS = ninitial nreacted = 0.200 M0.987 L 0.132 mol = 0.0654 mol pH = pKa + log (TRIS/TRISH+) = 8.09+ log (0.0654 mol/0.132 mol) = 7.785 We have prepared the desired buffer (realize that this is just one way of preparing the buffer).
(b) To 500 mL of the buffer prepared above, is added 0.030 mol H3O+. In the 500 mL of solution we have 0.132 mol 2 mol TRISH+ = 0.0660 mol TRISH+ and 0.0654 mol 2 mol TRIS = 0.0327 mol TRIS (we will assume no change in volume). HCl will completely react ( 100%) with TRIS, converting it to TRISH+. After complete reaction, there will be no excess HCl, 0.0327 mol 0.0300 mol = 0.0027 mol TRIS and 0.0660 mol + 0.0300 mol = 0.0960 mol TRISH+. By employing the Henderson Hasselbalch equation, we can estimate the pH of the resulting solution: pH = pKa + log (TRIS/TRISH+) = 8.09 + log (0.0027 mol/0.0960 mol) = 6.54 This represents a pH change of 1.25 units. The buffer is nearly exhausted owing to the fact that almost all of the TRIS has been converted to TRISH+. Generally, a pH change of 1 unit suggests that the capacity of the buffer has been pretty much completely expended. This is the case here. (Alternatively, you may solve this question using an I.C.E. table).
(c) Addition of 20.0 mL of 10.0 M HCl will complete exhaust the buffer (in part (b) we saw that addition of 3 mL of HCl used up most of the TRIS in solution). The buffer may be regenerated by adding 20.0 mL of 10.0 M NaOH. The buffer will be only slightly diluted after the addition of HCl and NaOH (500 mL 540 mL). Through the slow and careful addition of NaOH, one can regenerate the pH = 7.79 buffer in this way (if a pH meter is used to monitor the addition). 868 Chapter 17: Additional Aspects of AcidBase Equilibria 88. (M) (a) The formula given needs to be rearranged to isolate , as shown below. 1 pH pK a log 1 H 3O log 1 1 pH pKa log Ka H 3O 1 1 Ka = Ka K pH a H 3O K a 10 K a Now, use the given Ka (6.3108) and calculate for a range of pH values.
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 6 7 (c)
8 9 10 11 12 13 14 pH (b) When pH = pKa, = 0.5 or 50%. (c) At a pH of 6.0, = Ka 6.3 108 6.0 0.059 or 5.9%. 10 pH K a 10 6.3 108 869 Chapter 17: Additional Aspects of AcidBase Equilibria 89. (D) (a) We start by writing the equilibrium expression for all reactions: CO 2 aq K1 CO 2 g HCO3 K3 2 H 3O CO3 2 K 2 Ca 2 CO3 K4 CO 2 aq HCO3 H 3O First, we have to express [H3O+] using the available expressions: HCO3 H 3O K CO 2 3 3 CO aq H 3O 2 K HCO 4 3 HCO3 CO 2 aq CO 2 aq H 3O 2 2 K 3 CO3 K 4 HCO3 K 3 K 4 CO3 2 From the expression for K1, we know that [CO2(aq)] = K1[CO2(g)]. Therefore,
K CO g 2 H 3O 1 2 2 K 3 K 4 CO3 Now, the expression for K2 can be plugged into the above expression as follows: 2 K1 CO 2 g K1 CO 2 g Ca 2 H 3O 2 K 2 K3 K 4 K 3 K 4 CO3 H 3O K1 CO 2 g Ca 2 K 2 K3 K 4 (b) The K values for the reactions are as follows: K1 = 0.8317 (given in the problem) K2 = Ksp (CaCO3) = 2.8109 K3 = 1/Ka of HCO3 = 1/(4.71011) = 2.131010 K4 = 1/Ka of H2CO3 = 1/(4.4107) = 2.27106 280 106 L CO 2 1 mol CO 2 mol CO 2 1.145 105 CO2 (g) L air 24.45 L CO 2 L air 870 Chapter 17: Additional Aspects of AcidBase Equilibria H 3O K1 CO 2 g Ca 2 K 2 K3 K 4 0.8317 1.145 105 10.24 103 2.8 10 2.13 10 2.27 10 9 10 6 2.684 108 M pH log 2.684 108 7.57
90. (M) To determine pH, we must first determine what is the final charge balance of the solution without adding any H3O+ or OH ions. As such, we have to calculate the number of moles of each ion (assume 1 L): Moles of Na+: 23.0 g Na+ (1 mol/23.0 g) = 1.00 mol Na+ Moles of Ca2+: 10.0 g Ca2+ (1 mol/40.0 g) = 0.250 mol Ca2+ Moles of CO32: 40.02 g CO32 (1 mol/60.01 g) = 0.670 mol CO32Moles of SO42: 9.6 g SO42 (1 mol/96.056 g) = 0.100 mol Looking at the list of ions and consulting the solubility guide, we note that Ca2+ will precipitate with both SO42 and CO32. Since both of these anions have a 2 charge, it doesn't matter with which anion the precipitation occurs (in reality, it does a little, but the effects are small compared to the effects of adding H3O+ or OH ions). Moles of anions left = (0.670 + 0.100) 0.250 = 0.520 moles Since we have 0.520 moles of 2 ions, we must have 1.04 moles of 1+ ions to balance. However, we only have 1 mole of Na+. The difference in charge is: 1.04() 1.0(+) = 0.04() moles of ions To balance, we need 0.04 moles of H3O+. The pH = log (0.04) = 1.40.
91. (M) (a) We note that the pH is 5.0. Therefore, H 3O 105.0 1.0 105 M 2 5 C K a H 3O 2.0 10 1.8 10 1.0 105 4.6 103 2 5 5 2 K a H 3O 1.8 10 1.0 10 (b) dCA d pH d pH dCA d pH 1.0 103 4.6 103 0.22 pH 5 0.22 4.78 (c) At an acetic acid concentration of 0.1 M, C is also 0.1, because C is the total concentration of the acetic acid and acetate. The maximum buffer index (2.50102) happens at a pH of 4.75, 871 Chapter 17: Additional Aspects of AcidBase Equilibria where [HAc] = [Ac]. The minima are located at pH values of 8.87 (which correspond to pH of a solution of 0.1 M acetic acid and 0.1 M acetate, respectively).
3.0E02 2.5E02 2.0E02 1 .5E02 1 .0E02 5.0E03 0.0E+00 0 1 2 3 4 5 6 7 8 9 pH FEATURE PROBLEMS
92. (D) (a) The two curves cross the point at which half of the total acetate is present as acetic acid and half is present as acetate ion. This is the half equivalence point in a titration, where pH = pKa = 4.74 . For carbonic acid, there are three carbonate containing species: " H 2 CO 3 " which predominates at low pH, HCO3 , and CO 2 , which predominates in alkaline 3 solution. The points of intersection should occur at the halfequivalence points in each stepwise titration: at pH = pKa1 = log 4.4 107 = 6.36 and at pH = pKa 2 = log 4.7 10 11 = 10.33 . The following graph was computercalculated (and then drawn) from these equations. f in each instance represents the fraction of the species whose formula is in parentheses.
1 f H2A = 1+ K1 (b) c h c h H + + + K1 K 2 H+ 2 f HA 1 H K2 = +1+ K1 H+ 2 1 f A 2 H+ H+ = + +1
K1 K 2 K2 872 Chapter 17: Additional Aspects of AcidBase Equilibria f H 2 CO3 f HCO3 f CO3 2 (c) For phosphoric acid, there are four phosphate containing species: H 3 PO 4 under acidic conditions, H 2 PO 4 , HPO 4 , and PO 4 , which predominates in alkaline solution. The points of intersection should occur at pH = pKa1 = log 7.1 103 = 2.15, pH = pKa2 = log 6.3 10 8 = 7.20 , and pH = pKa3 2 3 c h c h = logc4.2 10 h = 12.38 , a quite alkaline
13 solution. The graph that follows was computercalculated and drawn. f PO4 f H 3(H3PO4)
 f (H PO ) f H 2 PO2 4 4
f HPO4 2 3f PO43 f (HPO42) f (PO4 ) 873 Chapter 17: Additional Aspects of AcidBase Equilibria 93. (D) (a) This is exactly the same titration curve we would obtain for the titration of 25.00 mL of 0.200 M HCl with 0.200 M NaOH, because the acid species being titrated is H 3O + . Both acids are strong acids and have ionized completely before titration begins. The initial pH is that of 0.200 M H3O+ = [HCl] + [HNO3]; pH = log 0.200 = 0.70. At the equivalence point, pH = 7.000 . We treat this problem as we would for the titration of a single strong acid with a strong base.
14 12 10 pH 8 6 4 2 0 0 5 10 15 20 25 Volume of 0.200 M NaOH (ml) (b) In Figure 179, we note that the equivalence point of the titration of a strong acid occurs at pH = 7.00 , but that the strong acid is essentially completely neutralized at pH = 4. In Figure 1713, we see that the first equivalence point of H 3 PO 4 occurs at about pH = 4.6. Thus, the first equivalence point represents the complete neutralization of HCl and the neutralization of H 3 PO 4 to H 2 PO 4 . Then, the second equivalence point represents the neutralization of H 2 PO 4 to HPO 4 . To reach the first equivalence point requires about 20.0 mL of 0.216 M NaOH, while to reach the second one requires a total of 30.0 mL of 0.216 M NaOH, or an additional 10.0 mL of base beyond the first equivalence point. The equations for the two titration reactions are as follows.
NaOH + H 3PO 4 NaH 2 PO 4 + H 2 O To the first equivalence point: NaOH + HCl NaCl + H 2 O 2 To the second equivalence point: NaOH + NaH 2 PO 4 Na 2 HPO 4 + H 2O There is a third equivalence point, not shown in the figure, which would require an additional 10.0 mL of base to reach. Its titration reaction is represented by the following equation. To the third equivalence point:
NaOH + Na 2 HPO 4 Na 3 PO 4 + H 2 O 874 Chapter 17: Additional Aspects of AcidBase Equilibria We determine the molar concentration of H 3 PO 4 and then of HCl. Notice that only 10.0 mL of the NaOH needed to reach the first equivalence point reacts with the HCl(aq); the rest reacts with H 3 PO 4 .
(30.0 20.0) mL NaOH(aq) 0.216 mmol NaOH 1 mmol H 3 PO 4 1 mL NaOH soln 1 mmol NaOH 10.00 mL acid soln (20.0 10.0) mL NaOH(aq) 0.216 mmol NaOH 1 mL NaOH soln 0.216 M H 3 PO 4 1 mmol HCl 1 mmol NaOH 10.00 mL acid soln 0.216 M HCl (c) We start with a phosphoric aciddihydrogen phosphate buffer solution and titrate until all of the H 3 PO 4 is consumed. We begin with 10.00 mL 0.0400 mmol H 3PO 4 = 0.400 mmol H 3PO 4 1 mL and the diprotic anion, 0.0150 mmol H 2 PO 4 2 10.00 mL = 0.150 mmol H 2 PO 4 1 mL . The volume of 0.0200 M NaOH needed is: 0.400 mmol H 3 PO 4 1 mmol NaOH 1 mmol H 3 PO 4 1 mL NaOH 0.0200 mmol NaOH = 20.00 mL To reach the first equivalence point. The pH value of points during this titration are computed with the HendersonHasselbalch equation. H 2 PO 4 = log 7.110 3 + log 0.0150 = 2.15 0.43 = 1.72 Initially : pH = pK1 + log 0.0400 H 3 PO 4 0.150 + 0.100 At 5.00 mL : pH = 2.15 + log = 2.15 0.08 2.07 0.400 0.100 At 10.0 mL, pH = 2.15 log 0.150 0.200 2.15 0.24 2.39 0.400 0.200 0.150 0.300 2.15 0.65 2.80 0.400 0.300 At 15.0 mL, pH = 2.15 log This is the first equivalence point, a solution of 30.00 mL (= 10.00 mL originally + 20.00 mL titrant), containing 0.400 mmol H 2 PO 4 from the titration and the 0.150 mmol H 2 PO 4 2 originally present. This is a solution with
875 Chapter 17: Additional Aspects of AcidBase Equilibria H 2 PO 4 = (0.400 + 0.150) mmol H 2 PO 4 = 0.0183 M , which has 30.00 mL pH = 1 pK1 + pK 2 = 0.50 2.15 log 6.3 108 = 0.50 2.15 + 7.20 = 4.68 2 To reach the second equivalence point means titrating 0.550 mmol H 2 PO 4 , which requires an additional volume of titrant given by
0.550 mmol H 2 PO 4 1 mmol NaOH 1 mL NaOH = 27.5 mL . 1 mmol H 2 PO 4 0.0200 mmol NaOH To determine pH during this titration, we divide the region into five equal portions of 5.5 mL and use the HendersonHasselbalch equation. At 20.0 + 5.5 mL,
2 HPO 4 2 = 7.20 + log 0.20 0.550 mmol HPO 4 formed pH = pK 2 + log H 2 PO 4 0.80 0.550 mmol H 2 PO4 remaining pH = 7.20 0.60 = 6.60 At 20.0 +11.0 mL = 31.0 mL, pH = 7.20 + log At 36.5 mL, pH = 7.38 0.40 0.550 = 7.02 0.60 0.550 At 42.0 mL, pH = 7.80 The pH at the second equivalence point is given by pH = 1 pK 2 + pK3 = 0.50 7.20 log 4.2 1013 = 0.50 7.20 + 12.38 = 9.79 . 2 b g e c hj b g Another 27.50 mL of 0.020 M NaOH would be required to reach the third equivalence point. pH values at each of four equally spaced volumes of 5.50 mL additional 0.0200 M NaOH are computed as before, assuming the HendersonHasselbalch equation is valid. PO 43 = 12.38 + log 0.20 0.550 At 47.50 + 5.50 mL = 53.00 mL, pH = pK 3 + log 2 0.80 0.550 HPO 4 = 12.38 0.60 = 11.78 876 Chapter 17: Additional Aspects of AcidBase Equilibria At 58.50 mL, pH = 12.20 At 64.50 mL, pH = 12.56 At 70.00 mL, pH = 12.98 But at infinite dilution with 0.0200 M NaOH, the pH = 12.30, so this point can't be reached.
3 At the last equivalence point, the solution will contain 0.550 mmol PO 4 in a total of 10.00 + 20.00 + 27.50 + 27.50 mL = 85.00 mL of solution, with 0.550 mmol PO 43 = = 0.00647 M. But we can never reach this point, because the 85.00 mL pH of the 0.0200 M NaOH titrant is 12.30. Moreover, the titrant is diluted by its addition to the solution. Thus, our titration will cease sometime shortly after the second equivalence point. We never will see the third equivalence point, largely because the titrant is too dilute. Our results are plotted below. 877 Chapter 17: Additional Aspects of AcidBase Equilibria 94. (D) p K a1 = 2.34; K a1 = 4.6 103 and p K a 2 = 9.69; K a 2 = 2.0 1010 (a) Since the Ka values are so different, we can treat alanine (H2A+) as a monoprotic acid with K a1 = 4.6 103. Hence: H2A+(aq) + H2O(l) Initial 0.500 M  Change  x M  Equilibrium (0.500 x) M
K a1 = HA(aq) + H3O+(aq) 0M 0M +x M +x M xM xM [HA][H 3O + ] ( x)( x) x2 = 4.6 103 = 0.500 (0.500 x) [H 2 A + ] + x = 0.048 M = [H3O ] (x = 0.0457, solving the quadratic equation) pH = log[H3O+] = log(0.046) = 1.34
(b) At the first halfneutralization point a buffer made up of H2A+/HA is formed, where [H2A+] = [HA]. The HendersonHasselbalch equation gives pH = pKa = 2.34. At the first equivalence point, all of the H2A+(aq) is converted to HA(aq). HA(aq) is involved in both K a1 and K a 2 , so both ionizations must be considered. If we assume that the solution is converted to 100% HA, we must consider two reactions. HA may act as a weak acid (HAA + H+ ) or HA may act as a base (HA + H+ H2A+). See the following diagram.
acid A as
HA a s (c) Some HA unreacted 100 % HA amphoteric nature (reactions occur)
Some HA reacts as an acid HA reacts as a base H A + + H3O base H2A+ Using the diagram above, we see that the following relations must hold true. [A] = [H3O+] + [H2A+]
Ka2 = K a [HA] [HA][H 3O + ] [A ][H 3O + ] [H O + ][HA] or [A] = 2 + & K a1 = or [H2A+] = 3 [H3O ] [HA] K a1 [H 2 A + ] Substitute for [A] and [H2A+] in [A] = [H3O+] + [H2A+] 878 Chapter 17: Additional Aspects of AcidBase Equilibria K a 2 [HA] [H 3O + ] = [H3O+] + [H 3O + ][HA] (multiply both sides by K a1 [H3O+]) K a1 K a1 K a 2 [HA] = K a1 [H3O+][H3O+] + [H 3O + ][H 3O + ][HA] K a1 K a 2 [HA] = [H3O+]2( K a1 + [HA]) [H3O+]2 = K a1 K a 2 [HA]
(K a1 + [HA]) Usually, [HA] K a1 (Here, 0.500 4.6 103) Make the assumption that K a1 + [HA] [HA] [H3O+]2 =
K a1 K a 2 [HA]
[HA] = K a1 K a 2 Take log of both sides log[H3O+]2 = 2log [H3O+] = 2(pH) = log K a1 K a 2 = log K a1 log K a 2 = p K a1 + p K a 2 2(pH) = p K a1 + p K a 2 pH =
(d) pK a1 + pK a 2 2 = 2.34 + 9.69 = 6.02 2 Halfway between the first and second equivalence points, half of the HA(aq) is converted to A(aq). We have a HA/A buffer solution where [HA] = [A]. The HendersonHasselbalch equation yields pH = p K a 2 = 9.69. At the second equivalence point, all of the H2A+(aq) is converted to A(aq). We can treat this simply as a weak base in water having: K 11014 Kb = w = = 5.0 105 10 Ka2 2.0 10 Note: There has been a 1:3 dilution, hence the [A] = 0.500 M + H2O(l) A(aq) Initial 0.167 M   Change x M  Equilibrium (0.167 x) M
Kb = (e) 1V = 0.167 M 3V HA(aq) + OH(aq) 0M 0M +x M +x M xM xM [HA][OH ] ( x)( x) x2 = = 5.0 105 [A ] 0.167 (0.167 x) x = 0.0029 M = [OH]; pOH = log[OH] = 2.54; pH = 14.00 pOH = 14.00 2.54 = 11.46 879 Chapter 17: Additional Aspects of AcidBase Equilibria (f) All of the points required in (f) can be obtained using the HendersonHasselbalch equation (the chart below shows that the buffer ratio for each point is within the acceptable range (0.25 to 4.0))
mL NaOH 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 110.0 %H2A+ %HA %Abuffer base = acid ratio 100 0 0 0 80 20 0 0.25 60 40 0 0.67 40 60 0 1.5 20 80 0 4.0 0 100 0 8 0 80 20 0.25 0 60 40 0.67 0 40 60 1.5 0 20 80 4.0 0 0 100 8 0 0 100 8 May use HendersonHasselbalch equation May use HendersonHasselbalch equation (i) After 10.0 mL Here we will show how to obtain the answer using both the HendersonHasselbalch equation and setting up the I. C. E. (Initial, Change, Equilibrium) table. The results will differ within accepted experimental limitation of the experiment (+ 0.01 pH units) nH2A+ = (C V) = (0.500 M)(0.0500 L) = 0.0250 moles H2A
nOH = (C V) = (0.500 M)(0.0100 L) = 0.00500 moles OH Vtotal = (50.0 + 10.0) mL = 60.0 mL or 0.0600 L nH A+ 0.0250 mol n 0.00500 mol = 0.417 M [H2A+] = 2 = [OH] = OH = = 0.0833M Vtotal 0.0600 L Vtotal 0.0600 L K a1 4.6 103 1 1 = = 4.6 1011 Keq for titration reaction = 14 K b(HA) K w K w 1.00 10 Ka 1 H2A+(aq) + OH(aq) HA(aq) + H2O(l) Initial: 0.417 M 100% rxn: 0.0833 New initial: 0.334 M Change: +x M Equilibrium: 0.334 M 4.6 1011 = 0.0833 M 0.0833 M 0M +x M xM 0M +0.0833 M 0.0833 M x M 0.0833 M      (0.0833) (0.0833) ; x= = 5.4 1013 (valid assumption) 11 (0.334)( x) (0.334)(4.6 10 ) x = 5.4 1013 = [OH]; pOH = log(5.4 1013) = 12.27; pH = 14.00 pOH = 14.00 12.27 = 1.73 Alternative method using the HendersonHasselbalch equation: 880 Chapter 17: Additional Aspects of AcidBase Equilibria (i) After 10.0 mL, 20% of H2A+ reacts, forming the conjugate base HA. Hence the buffer solution is 80% H2A+ (acid) and 20% HA (base). pK a1 base 20.0 + log acid = 2.34 + log 80.0 = 2.34 + (0.602) = 1.74 (within + 0.01) pH = For the remainder of the calculations we will employ the HendersonHasselbalch equation with the understanding that using the method that employs the I.C.E. table gives the same result within the limitation of the data. (ii) After 20.0 mL, 40% of H2A+ reacts, forming the conjugate base HA. Hence the buffer solution is 60% H2A+ (acid) and 40% HA (base). pK a1 base 40.0 + log acid = 2.34 + log 60.0 = 2.34 + (0.176) = 2.16 pH = (iii) After 30.0 mL, 60% of H2A+ reacts, forming the conjugate base HA. Hence the buffer solution is 40% H2A+ (acid) and 60% HA (base). pK a1 base 60.0 + log acid = 2.34 + log 40.0 = 2.34 + (+0.176) = 2.52 pH = (iv) After 40.0 mL, 80% of H2A+ reacts, forming the conjugate base HA. Hence the buffer solution is 20% H2A+ (acid) and 80% HA (base). base 80.0 = 2.34 + log = 2.34 + (0.602) = 2.94 pH = pK a1 + log acid 20.0 (v) After 50 mL, all of the H2A+(aq) has reacted, and we begin with essentially 100% HA(aq), which is a weak acid. Addition of base results in the formation of the conjugate base (buffer system) A(aq). We employ a similar solution, however, now we must use p K a 2 = 9.69. (vi) After 60.0 mL, 20% of HA reacts, forming the conjugate base A. Hence the buffer solution is 80% HA (acid) and 20% A (base) base 20.0 = 9.69 + log = 9.69 + (0.602) = 9.09 pH = pK a 2 + log acid 80.0 (vii) After 70.0 mL, 40% of HA reacts, forming the conjugate base A. Hence the buffer solution is 60% HA (acid) and 40% A (base). base 40.0 = 9.69 + log = 9.69 + (0.176) = 9.51 pH = pK a 2 + log acid 60.0 881 Chapter 17: Additional Aspects of AcidBase Equilibria (viii) After 80.0 mL, 60% of HA reacts, forming the conjugate base A. Hence the buffer solution is 40% HA (acid) and 60% A (base). base 60.0 = 9.69 + log = 9.69 + (+0.176) = 9.87 pH = pK a 2 + log acid 40.0 (ix) After 90.0 mL, 80% of HA reacts, forming the conjugate base A. Hence the buffer solution is 20% HA (acid) and 80% A (base). base 80.0 = 9.69 + log = 9.69 + (0.602) = 10.29 pH = pK a 2 + log acid 20.0 (x) After the addition of 110.0 mL, NaOH is in excess. (10.0 mL of 0.500 M NaOH is in excess, or, 0.00500 moles of NaOH remains unreacted). The pH of a solution that has NaOH in excess is determined by the [OH] that is in excess. (For a diprotic acid, this occurs after the second equivalence point.) n 0.00500 mol = 0.0312 M ; pOH = log(0.03125) = 1.51 [OH]excess = OH = Vtotal 0.1600 L pH = 14.00 pOH = 14.00 1.51 = 12.49
(g) A sketch of the titration curve for the 0.500 M solution of alanine hydrochloride, with some significant points labeled on the plot, is shown below.
14 12 10 8
pH Plot of pH versus Volume of NaOH Added
HA/Abuffer region NaOH in excess pH = pKa2 isoelectric point H2A /HA buffer region
+ 6 4 2 0 0
pH = pKa1 20 40 60
NaOH volume (mL) 80 100 120 882 Chapter 17: Additional Aspects of AcidBase Equilibria SELFASSESSMENT EXERCISES
95. (E) (a) mmol: millimoles, or 1103 mol (b) HIn: An indicator, which is a weak acid (c) Equivalence point of a titration: When the moles of titrant equals the moles of the substance being titrated (d) Titration curve: A curve of pH of the solution being titrated versus the pH of the titrating solution (E) (a) The commonion effect: A process by which ionization of a compound is suppressed by having present one of the product ions (from another source) in the solution (b) use of buffer to maintain constant pH: A buffer is a solution of a weak acid and its conjugate base, and it resists large changes in pH when small amounts of an acid or base are added. (c) determination of pKa from titration curve: At the halfway point (when half of the species being titrated is consumed, the concentration of that species and its conjugate is the same, and the equilibrium expression simplifies to pKa = pH (d) measurement of pH with an indicator: An approximate method of measuring the pH, where the color of an ionizable organic dye changes based on the pH (E) (a) Buffer capacity and buffer range: Buffer capacity is a measure of how much acid or base can be added to the buffer without an appreciable change in the pH, and is determined by the concentration of the weak acid and conjugate base in the buffer solution. The buffer range, however, refers to the range over which a buffer effectively neutralizes added acids and bases and maintains a fairly constant pH. (b) Hydrolysis and neutralization: Hydrolysis is reaction of an acid or base with water molecules, which causes water to split into hydronium and hydroxide ions. Neutralization is the reaction of H3O+ and OH together to make water. (c) First and second equivalence points in the titration of a weak diprotic acid: First equivalence point is when the first proton of a weak diprotic acid is completely abstracted and the resulting acid salt and base have been consumed and neither is in excess. Second equivalence point is the equivalence point at which all protons are abstracted from the acid and what remains is the (2) anion. (d) Equivalence point of a titration and end point of an indicator: Equivalence point of a titration is when the moles of titrant added are the same as the moles of acid or base in the solution being titrated. The endpoint of an indicator is when the indicator changes color because of abstraction or gain of a proton, and is ideally as close as possible to the pH of the equivalence point. 96. 97. 883 Chapter 17: Additional Aspects of AcidBase Equilibria 98. (E) (a) HCHO2 + OH CHO2 + H2O CHO2 + H3O+ HCHO2 + H2O (b) C6H5NH3+ + OH C6H5NH2 + H2O C6H5NH2 + H3O+ C6H5NH3+ + H2O (c) H2PO4 + OH HPO42 + H2O HPO42 + H3O+ H2PO4 + H2O 99. (M) (a) The pH at theequivalence point is 7. Use bromthymol blue.
14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 pH Vol of Titrant (mL) (b) The pH at the equivalence point is ~5.3 for a 0.1 M solution. Use methyl red.
14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 pH Vol of Titrant (mL) 884 Chapter 17: Additional Aspects of AcidBase Equilibria (c) The pH at the equivalence point is ~8.7 for a 0.1 M solution. Use phenolphthalein, because it just begins to get from clear to pink around the equivalence point.
14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 pH Vol of Titrant (mL) (d) The pH for the first equivalence point (NaH2PO4 to Na2HPO42) for a 0.1 M solution is right around ~7, so use bromthymol blue.
14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 pH Vol of Titrant (mL) 885 Chapter 17: Additional Aspects of AcidBase Equilibria 100. (D) (a) This is the initial equilibrium before any base has reacted with the acid. The reaction that dominates, along with changes in concentration, is shown below: HC7H5O2 0.0100 x 0.0100  x + H2O C7H5O2 0 +x x + H3O+ 0 +x x H 3O + C7 H 5O2 Ka HC7 H5O2 0.0100  x Solving for x using the quadratic formula, x = 7.63 104 M. pH = log H3O+ log 7.63 104 3.12 (b) In this case, we titrate the base with 0.00625 L of Ba(OH)2. Therefore, we have to calculate the final moles of the base and the total volume to determine the concentration. mol HC7 H 5O 2 0.02500 L 0.0100 M HC7 H 5O 2 2.5 104 mol 6.3 105 x x 2 mol OH mol OH 0.00625 L 0.0100 M Ba OH 2 = 1.25 104 mol 1 mol Ba OH 2 Since the amount of OH is half of the initial amount of HC7H5O2, the moles of HC7H5O2 and C7H5O2 are equal. Therefore, Ka = [H3O+], and pH = log(6.3105) = 4.20.
(c) At the equivalence point, there is initially no HC7H5O2. The equilibrium is dominated by C7H5O2 hydrolyzing water. The concentration of C7H5O2 is: Total moles of HC7H5O2 = 2.5104 mol as shown previously. At the equivalence point, the moles of acid equal moles of OH. 1 mol Ba OH 2 mol Ba OH 2 2.5 104 mol OH 1.25 104 mol 2 mol OH Vol of Ba(OH)2 = 1.25104 mol / 0.0100 M = 0.0125 L Total volume of the solution at the equivalence point is the sum of the initial volume plus the volume of Ba(OH)2 added. That is, VTOT = 0.02500 L + 0.0125 L = 0.0375 L Therefore, the concentration of C7H5O2 = 2.5104/0.0375 L = 0.00667 M. 886 Chapter 17: Additional Aspects of AcidBase Equilibria C7H5O2 0.00667 x 0.00667  x + H2O HC7H5O2 0 +x x + OH 0 +x x Since this is a base reaction, K b K w K a 1.00 104 6.3 105 1.587 1010.
OH  HC7 H 5O 2 Kb = C7 H 5O2 1.59 1010 x x 0.00667 x solving for x (by simplifying the formula above) yields x = 1.03 10 6 M. pH = 14 pOH 14 log 1.03 106 14 6.00 8.00 (d) In this part, we have an excess of a strong base. As such, we have to determine how much excess base there is and what is the final volume of the solution. mol Ba OH 2 2 mol OH mol OH 0.01500 L 0.0100 L 1 mol Ba OH 2 3.000 104 mol OH Excess mol OH mol HC7 H5 O 2 mol OH 3.000 104 2.500 104 5.000 105 mol 5.0 105 mol OH 0.02500 L+0.01500 L 0.00125 M pH = 14 pOH 14 log 0.00125 14 2.903 11.1 101. (E) The answer is (c); because of the common ioneffect, the presence of HCO2 will repress ionization of formic acid. 102. (E) The answer is (d), because NaHCO3 is a weak base and will react with protons in water, shifting the formic acid ionization equilibrium to the right. 103. (E) The answer is (b), raise the pH. NH4+ is an acid, and to be converted to its conjugate base, it must react with a base to abstract its proton. 104. (E) The answer is (b), because at that point, the number of moles of weak base remaining is the same as its conjugate acid, and the equilibrium expression simplifies to Ka = [H3O+]. 887 Chapter 17: Additional Aspects of AcidBase Equilibria 105. (M) The base, C2H5NH2, is reacted with HClO4. The reaction is: C2 H 5 NH 2 HClO 4 C2 H 5 NH 3 ClO 4 Assuming a volume of 1 L for each solution, C2 H5 NH 2 1.49 mol 1.001 mol 0.489 mol 0.2445 M 2L 2L 1.001 mol C2 H 5 NH 3 0.5005 M 2L C2 H 5 NH 3 OH 0.5005 +x x 4.3 104 0.2445 x C2 H5 NH 2 x 2.10 104 pOH log 2.10 104 3.68 pH 14 3.68 10.32
106. (D) We assume that all of Ca(HSe)2 dissociates in water. The concentration of HSe is therefore: 2 mol HSe 1.0 M HSe 1 mol Ca HSe 2 We note that HSe is amphoteric; that is, it can act either as an acid or a base. The acid reaction of HSe and the concentration of [H3O+] generated, are as follows: 0.5 M Ca HSe 2 HSe + H2O Se2 + H3O+
1.00 10
11 Se 2 H 3O x x HSe 1.00 x x = 1.00 1011 3.16 106 The basic reaction of HSe and the concentration of [OH] generated, are as follows: HSe + H2O H2Se + OH Kb = 1.001014/1.3104 = 7.691011
7.69 10
11 H 2Se OH HSe 1.00 x x x x = 7.69 1011 8.77 106 888 Chapter 17: Additional Aspects of AcidBase Equilibria Therefore, we have [H3O+] = 3.16106 and [OH] = 8.77106. Since these two react to give H2O, the result is 8.77106 3.16106 = 5.61106 M [OH]. The pH of the solution is: pH = 14 pOH = 14 5.25 = 8.75 107. (E) The answer is (a). The solution system described is a buffer, and will resist large changes in pH. Adding KOH should raise the pH slightly. 108. (E) The answer is (b), because HSO3 is a much stronger acid (Ka = 1.3102) than H2PO4 (Ka = 6.3108). 109. (E) The answer is (b). The pKa of the acid is 9, which puts it squarely in the middle of the 810 pH range for the equivalence point. 110. (E) (a) NaHCO3 titrated with NaOH: pH > 7, because HCO3 is itself slightly basic, and is being titrated with NaOH to yield CO32 at the equivalence point, which is even more basic. (b) HCl titrated with NH3: pH < 7, because the resulting NH4+ at the equivalence point is acidic. (c) KOH titrated with HI: pH = 7, because a strong base is being titrated by a strong acid, and the resulting anions and cations are all nonbasic and nonacidic. 111. (M) The concepts that define Sections 172, 173, and 174 are buffers, indicators, and titrations. For buffers, after definition, there are a number of concepts, such as composition, application, the equilibrium expression (the HendersonHasselbalch equation is a subtopic of the equilibrium expression). Under composition, there are other subtopics such as buffer range and capacity. For indicators, after defining it, there are subtopics such as equilibrium expression and usage. With titration, topics such as form (weak acid titrated by strong base, and weak base titrated by strong acid), the titration curve, and the equivalence point. Look at these sections to find interrelated terminology and other concepts. 889 CHAPTER 18 SOLUBILITY AND COMPLEXION EQUILIBRIA
PRACTICE EXAMPLES
1A (E) In each case, we first write the balanced equation for the solubility equilibrium and then the equilibrium constant expression for that equilibrium, the Ksp expression: (a) (b) 1B (E) (a) MgCO3 s Mg 2+ aq + CO3 Ag 3 PO 4 s 3Ag + aq + PO 4 2 aq aq Ksp = [Mg2+][CO32 ] Ksp = [Ag+]3[PO43 ] 3 Provided the [OH] is not too high, the hydrogen phosphate ion is not expected to ionize in aqueous solution to a significant extent because of the quite small values for the second and third ionization constants of phosphoric acid. 2 CaHPO 4 s Ca 2+ aq + HPO 4 aq The solubility product constant is written in the manner of a Kc expression: Ksp = [Ca2+][HPO42] = 1. 107 (b) 2A (M) We calculate the solubility of silver cyanate, s , as a molarity. We then use the solubility equation to (1) relate the concentrations of the ions and (2) write the Ksp expression. s= 7 mg AgOCN 1000 mL 1g 1 mol AgOCN = 5 104 moles/L 100 mL 1L 1000 mg 149.9 g AgOCN AgOCN s Ag + aq + OCN aq Equation : Solubility Product : s s
2 K sp = Ag + OCN = s s = s 2 = 5 104 = 3 107 2B (E) We calculate the solubility of lithium phosphate, s , as a molarity. We then use the solubility equation to (1) relate the concentrations of the ions and (2) write the Ksp expression. s= 0.034 g Li3 PO 4 1000 mL 1 mol Li3 PO 4 = 0.0029 moles/L 100 mL soln 1L 115.79 g Li3 PO 4
3 Equation: Li3 PO 4 s 3Li + aq + PO 4 aq Solubility Product:
3 (3 s)3 s 3 3 K sp = Li + PO 4 = 3s s = 27 s 4 27 (0.0029) 4 1.9 109 890 Chapter 18: Solubility and ComplexIon Equilibria 3A (E) We use the solubility equilibrium to write the Ksp expression, which we then solve to obtain the molar solubility, s , of Cu3(AsO4)2. Cu 3 (AsO 4 ) 2 . s 3 Cu 2+ aq + 2 AsO 4 aq K sp = Cu 2+ AsO 4 = 3s 2 s = 108s 5 = 7.6 1036 3 2 3 2 Solubility :
3B s 5 7.6 1036 3.7 108 M 108 (E) First we determine the solubility of BaSO 4 , and then find the mass dissolved.
2 BaSO 4 aq Ba 2+ aq + SO 4 aq Ksp = Ba 2+ SO 4 2 = s 2 The last relationship is true because Ba 2+ = SO 4 2 in a solution produced by dissolving BaSO 4 in pure water. Thus, s = K sp 1.1 1010 1.05 105 M. mass BaSO 4 = 225 mL 1.05 105 mmol BaSO 4 233.39 mg BaSO 4 = 0.55 mg BaSO 4 1 mL sat'd soln 1mmol BaSO4
2 4A (M) For PbI 2 , Ksp = Pb 2+ I = 7.1 109 . The solubility equilibrium is the basis of the
+ Pb 2+ aq 0.10 M +s M (0.10 + s) M calculation. Equation: Initial: Changes: Equil:
Ksp = Pb
2+ PbI 2 s
   I 2 bg b g 2I aq 0M +2s M 2s M
2 b g
7.1 109 . s 13 104 M 0.40
2 = 7.1 10 = 0.10 + s 2 s 0.40 s
2 9 b gb g (assumption 0.10 s is valid) This value of s is the solubility of PbI 2 in 0.10 M Pb NO 3
4B (E) We find pOH from the given pH: b g baqg . pOH = 14.00 8.20 = 5.80; [OH] = 10pOH = 105.80 = 1.6106 M. We assume that pOH remains constant, and use the Ksp expression for Fe OH 3 . b g K sp = Fe3+ OH = 4 1038 = Fe3+ 1.6 106 3 3 Fe3+ = 1.6 10 4 1038 6 3 = 9.8 1021 M Therefore, the molar solubility of Fe(OH)3 is 9.81021 M. The dissolved Fe OH b g 3 does not significantly affect OH . 891 Chapter 18: Solubility and ComplexIon Equilibria 5A (M) First determine I as altered by dilution. We then compute Qsp and compare it with Ksp . 3drops I = 0.05 mL 0.20 mmol KI 1mmol I 1drop 1mL 1mmol KI = 3 104 M 100.0 mL soln Qsp = Ag + I = 0.010 3 104 = 3 106 Qsp 8.5 1017 = K sp Thus, precipitation should occur.
5B (M) We first use the solubility product constant expression for PbI 2 to determine the I needed in solution to just form a precipitate when Pb 2+ = 0.010 M. We assume that the volume of solution added is small and that Pb 2+ remains at 0.010 M throughout. Ksp = Pb 2+ I 2 = 7.1 10 = 0.010 I 9 b g 2 I 7.1 109 = 8.4 104 M = 0.010 We determine the volume of 0.20 M KI needed.
volume of KI aq = 100.0 mL 8.4 104 mmol I 1 mmol KI 1 mL KI aq 1 drop 1 mL 1 mmol I 0.20 mmol KI 0.050 mL = 8.4 drops = 9 drops Since one additional drop is needed, 10 drops will be required. This is an insignificant volume compared to the original solution, so Pb 2+ remains constant. b g b g 6A (E) Here we must find the maximum Ca 2+ that can coexist with OH = 0.040 M. 6 2 5.5 10 2 K sp = 5.5 106 = Ca 2+ OH = Ca 2+ 0.040 ; Ca 2+ = = 3.4 103 M 2 0.040 For precipitation to be considered complete, Ca 2+ should be less than 0.1% of its original value. 3.4 103 M is 34% of 0.010 M and therefore precipitation of Ca OH under these conditions.
6B (E) We begin by finding Mg 2+ that corresponds to1 g Mg 2+ / L . b g 2 is not complete 1 g Mg 2+ 1g 1 mol Mg 2+ Mg = 6 = 4 108 M 2+ 1 L soln 10 g 24.3 g Mg
2+ Now we use the Ksp expression for Mg OH Ksp = 1.8 10 11 = Mg 2+ OH 2 b g to determine OH . 18 10 . = c4 10 h OH OH = 4 10 2
8 2 11 8 0.02 M 892 Chapter 18: Solubility and ComplexIon Equilibria 7A (M) Let us first determine Ag + when AgCl(s) just begins to precipitate. At this point, Qsp and Ksp are equal.
Ksp = 1.8 1010 = Ag + Cl = Qsp = Ag + 0.115M
Ag + = 1.8 1010 = 1.6 109 M 0.115 Now let us determine the maximum Br that can coexist with this Ag + . Ksp = 5.0 1013 = Ag + Br = 1.6 109 M Br ; Br = 5.0 1013 = 3.1 10 4 M 1.6 109 The remaining bromide ion has precipitated as AgBr(s) with the addition of AgNO 3 aq . Percent of Br remaining 7B b g [Br ]final 3.1 104 M 100% 100% 0.12% [Br ]initial 0.264 M (M) Since the ions have the same charge and the same concentrations, we look for two Ksp values for the salt with the same anion that are as far apart as possible. The Ksp values for the carbonates are very close, while those for the sulfates and fluorides are quite different. However, the difference in the Ksp values is greatest for the chromates; Ksp for
2 precipitate first and SrCrO 4 will begin to precipitate when CrO 4 has the value: BaCrO 4 1.2 1010 is so much smaller than Ksp for SrCrO 4 2.2 105 , BaCrO 4 will K sp 2.2 105 CrO 4 2 = = = 2.2 104 M . Sr 2+ 0.10 At this point Ba 2+ is found as follows. K sp 1.2 1010 Ba 2+ = = = 5.5 107 M ; 4 2 2.2 10 CrO 4 [Ba2+] has dropped to 0.00055% of its initial value and therefore is considered to be completely precipitated, before SrCrO 4 begins to precipitate. The two ions are thus effectively separated as chromates. The best precipitating agent is a group 1 chromate salt.
8A (M) First determine OH resulting from the hydrolysis of acetate ion. Equation: Initial: Changes: Equil: C2 H 3 O2 aq + H 2 O(l) HC 2 H 3O 2 aq 0M +x M xM b g + b 0.10 M x M 0.10 x M g    OH aq 0M +x M xM b g 893 Chapter 18: Solubility and ComplexIon Equilibria
2 HC2 H3O2 OH 1.0 1014 xx x 5.6 1010 = 1.8 105 0.10 x 0.10 C 2 H 3O 2 Kb x [OH ] 010 5.6 1010 7.5 106 M (the assumption x 0.10 was valid) . Now compute the value of the ion product in this solution and compare it with the value of Ksp for Mg OH 2 . b g Qsp = Mg 2+ OH will not occur.
8B 2 = 0.010M 7.5 106 M b gc Because Qsp is smaller than Ksp b g , this solution is unsaturated and precipitation of MgbOH g bsg h
2 = 5.6 1013 1.8 1011 = Ksp Mg OH 2 2 (M) Here we can use the HendersonHasselbalch equation to determine the pH of the buffer.
pH = pKa + log L = log 1.8 10 MN HC2 H 3O 2 OPQ C 2 H 3O 2 c 5 h + log 0.250 M = 4.74 + 0.22 = 4.96 0.150 M
OH = 10 pOH = 10 9.04 = 9.1 1010 M pOH = 14.00 pH = 14.00 4.96 = 9.04 Now we determine Qsp to see if precipitation will occur. Qsp = Fe3+ OH = 0.013M 9.1 1010 = 9.8 1030 38 Qsp 4 10 = K sp ; Thus, Fe(OH)3 precipitation should occur.
3 3 9A (M) Determine OH , and then the pH necessary to prevent the precipitation of Mn OH 2 . b g K sp = 1.9 10 13 = Mn 2+ OH = 0.0050M OH OH = 2 2 1.9 1013 6.2 106 M 0.0050 pOH = log 6.2 106 = 5.21 c h pH = 14.00 5.21 = 8.79 + We will use this pH in the HendersonHasselbalch equation to determine NH 4 . pK b = 4.74 for NH 3 .
pH = pK a + log NH3 + NH 4 = 8.79 = 14.00 4.74 + log 0.025 M NH 4 + log 0.025 M = 8.79 14.00 4.74 = 0.47 NH 4 + 0.025 NH 4 + = 100.47 = 0.34 0.025 NH 4 + = 0.34 = 0.074M 894 Chapter 18: Solubility and ComplexIon Equilibria 9B (M) First we must calculate the [H3O+] in the buffer solution that is being employed to dissolve the magnesium hydroxide: NH3(aq) + H2O(l) NH4+(aq) + OH(aq) ; Kb = 1.8 105 Kb [NH 4 ][OH ] [0.100M][OH ] 1.8 105 [NH 3 ] [0.250M] 1.00 1014 M 2 2.22 1010 M 4.5 105 M Now we can employ Equation 18.4 to calculate the molar solubility of Mg(OH)2 in the buffer solution; molar solubility Mg(OH)2 = [Mg2+]equil [OH ] 4.5 105 M ; [H 3O ] K = 1.8 10 Mg(OH)2(s) + 2 H3O+(aq) Mg2+(aq) + 4H2O(l) ; 10 Equilibrium 2.2 10 x 17 K x [Mg 2 ] 1.8 1017 2 10 2 [H 3O ] [2.2 10 M] x 8.7 103 M [Mg 2 ]equil So, the molar solubility for Mg(OH) 2 = 8.7 103 M. 10A (M) (a) In solution are Cu 2+ aq , SO 4 2 aq , Na + aq , and OH aq .
Cu
2+ aq + 2 OH aq Cu OH 2 s b g b g b g (b) In the solution above, Cu OH Cu OH 2 s Cu 2+ aq + 2 OH aq b g bsg is Cu baqg :
2+ 2 This Cu 2+ aq reacts with the added NH 3 aq : Cu 2+ aq + 4 NH 3 aq Cu NH 3 4 2+ b g b g aq 2+ The overall result is: Cu OH 2 s + 4NH 3 aq Cu NH 3 4 (c) aq + 2 OH aq HNO 3 aq (a strong acid), forms H 3O + aq , which reacts with OH aq and NH 3 aq .
OH aq + H 3O + aq 2 H 2 O(l); b g b g b g b g As NH 3 aq is consumed, the reaction below shifts to the left. Cu OH 2 s + 4 NH 3 aq Cu NH 3 4 2+ b g NH 3 aq + H 3O + aq NH 4 aq + H 2 O(l)
+ aq + 2 OH aq But as OH aq is consumed, the dissociation reactions shift to the side with the dissolved ions: Cu OH s Cu 2+ aq + 2 OH aq 2 b g The species in solution at the end of all this are 895 Chapter 18: Solubility and ComplexIon Equilibria Cu 2+ aq , NO3 aq , NH 4 + aq , excess H 3O + aq , Na + aq , and SO 4 2 aq (probably HSO4(aq) as well).
10B (M) 2 (a) In solution are Zn 2+ (aq), SO 4 (aq), and NH 3 (aq), Zn 2+ aq + 4 NH 3 aq Zn NH 3 4 (b)
2+ aq HNO 3 aq , a strong acid, produces H 3O + aq , which reacts with NH 3 aq . NH 3 aq + H 3O + aq NH 4 aq + H 2 O(l) As NH 3 aq is consumed, the tetrammine
+ b g b g b g b g complex ion is destroyed. Zn NH 3 4 (c)
2+ aq + 4H3O+ aq Zn H 2O 4 aq + 4NH 4 + aq 2+ NaOH(aq) is a strong base that produces OH aq , forming a hydroxide precipitate. Zn H 2 O 4 aq + 2 OH aq Zn OH 2 s + 4 H 2O l Another possibility is a reversal of the reaction of part (b). 2+ 2+ Zn H 2 O 4 aq + 4 NH 4 + aq + 4 OH aq Zn NH 3 4 aq + 8 H 2O l 2+ b g (d) The precipitate dissolves in excess base. 2 Zn OH 2 s + 2 OH aq Zn OH 4 aq 11A (M) We first determine Ag + in a solution that is 0.100 M Ag + (aq) (from AgNO 3 ) and 0.225 M NH 3 aq . Because of the large value of Kf = 1.6 107 , we start by having the reagents form as much complex ion as possible, and approach equilibrium from this point. Equation: Ag + aq b g b g + 2 NH 3 aq b g Ag NH 3 b g baqg
+ 2 In soln 0.100 M Form complex 0.100 M Initial 0M Changes +x M Equil xM
Kf = 1.6 107 = Ag NH 3
+ 0.225 M 0.200 M 0.025 M +2x M (0.025 + x) M
+ 2 2 0M +0.100 M 0.100 M x M (0.100 x) M b g Ag L NH 3 O NM QP
7 = 0.100 x x 0.025 + 2 x 2 0.100 x 0.025 2 x= b0.025g 1.6 10
2 0.100 = 1.0 105 M = Ag + = concentration of free silver ion (x << 0.025 M, so the approximation was valid) The Cl is diluted:
1.00 mLinitial Cl = Cl final initial 1,500 mL = 3.50 M 1500 = 0.00233M
final 896 Chapter 18: Solubility and ComplexIon Equilibria Finally we compare Qsp with Ksp to determine if precipitation of AgCl(s) will occur. Qsp = Ag + Cl = 1.0 105 M 0.00233 M = 2.3 108 1.8 10 10 = Ksp
Because the value of the Qsp is larger than the value of the Ksp, precipitation of AgCl(s) should occur.
11B (M) We organize the solution around the balanced equation of the formation reaction. c hb g Equation: Pb 2+ aq b g + EDTA 4 aq b g g PbEDTA 2 baqg Initial 0.100 M Form Complex: 0.100 M Equil xM 0.250 M (0.250 0.100) M 0.150 + x M b b 0M 0.100 M 0.100 x M g PbEDTA 2 = 2 1018 = 0.100 x 0.100 K f = 2+ 4 x 0.150 + x 0.150 x Pb EDTA x= 0.100 = 3 1019 M 0.150 2 1018
2 (x << 0.100 M, thus the approximation was valid.) We calculate Qsp and compare it to Ksp to determine if precipitation will occur. Qsp = Pb 2+ I = 3 1019 M 0.10 M = 3 1021 . Qsp 7.1109 = K sp Thus precipitation will not occur.
2 12A (M) We first determine the maximum concentration of free Ag + . 1.8 1010 = 2.4 108 M. 0.0075 + This is so small that we assume that all the Ag in solution is present as complex ion: K sp = Ag + Cl = 1.8 1010 + Ag + = Ag NH 3 Kf = b g = 0.13 M. We use K to determine the concentration of free NH . Agb NH g 0.13M
2 f 3 + 3 2 2 Ag L NH 3 O NM QP + = 1.6 107 = 2.4 108 L NH 3 O NM QP 2 NH 3 = 013 . = 0.58 M. 2.4 10 16 107 .
8 If we combine this with the ammonia present in the complex ion, the total ammonia concentration is 0.58 M + 2 0.13 M = 0.84 M . Thus, the minimum concentration of ammonia necessary to keep AgCl(s) from forming is 0.84 M. b g 897 Chapter 18: Solubility and ComplexIon Equilibria 12B (M) We use the solubility product constant expression to determine the maximum Ag + that can be present in 0.010 M Cl without precipitation occurring. 1.8 1010 Ag + = K sp = 1.8 1010 = Ag + Cl = Ag + 0.010 M = 1.8 108 M 0.010 This is also the concentration of free silver ion in the Kf expression. Because of the large value of Kf , practically all of the silver ion in solution is present as the complex ion, [Ag(S2O3)2]3]. We solve the expression for [S2O32] and then add the [S2O32] "tied up" in the complex ion. Ag S O 3 2 3 2 0.10 M K f 1.7 1013 2 2 2 Ag + S2 O32 1.8 108 M S2 O3 0.10 2 S2 O32 = = 5.7 104 M = concentration of free S2O3 8 13 1.8 10 1.7 10 total S2 O32 = 5.7 104 M + 0.10 M Ag S2 O3 2 = 0.20 M + 0.00057 M = 0.20 M 3 2 mol S2 O32 1mol Ag S2 O3 2 3 13A (M) We must combine the two equilibrium expressions, for Kf and for Ksp, to find Koverall. Fe OH 3 s Fe3+ aq + 3OH aq Fe3+ aq + 3C2 O 4 Fe OH 3 s + 3C2 O 4
2 2 Ksp = 4 1038
3 aq Fe C2O4 3 aq 3 Kf = 2 1020 aq Fe C2O4 3 aq + 3OH aq K overall = 8 1018 Initial Changes Equil 0.100 M 3x M 0.100 3x M b g 0M +x M xM
3 0M +3x M 3x M K overall 3 4 Fe C 2 O 4 3 OH = x 3 x = 8 10 18 27 x = 3 3 3 2 0.100 3 x 0.100 C 2 O 4 3 (3x << 0.100 M, thus the approximation was valid.) (0100) 3 8 1018 . x 4 106 M The assumption is valid. 27 2 Thus the solubility of Fe OH 3 in 0.100 M C 2 O 4 is 4 10 6 M .
4 b g 898 Chapter 18: Solubility and ComplexIon Equilibria 13B (M) In Example 1813 we saw that the expression for the solubility, s , of a silver halide in an aqueous ammonia solution, where NH 3 is the concentration of aqueous ammonia, is given by: s K sp K f = NH 3 2 s 2 or K sp K f s [NH 3 ] 2 s For all scenarios, the NH 3 stays fixed at 0.1000 M and K f is always 1.6 107 . We see that s will decrease as does Ksp . The relevant values are:
K sp AgCl = 1.8 1010 , K sp AgBr = 5.0 1013 , K sp AgI = 8.5 1017 . Thus, the order of decreasing solubility must be: AgI > AgBr > AgCl.
14A (M) For FeS, we know that Kspa = 6 102 ; for Ag 2S, Kspa = 6 1030 . We compute the value of Qspa in each case, with H 2S = 0.10 M and H 3O + = 0.30 M .
For FeS, Qspa = 0.020 0.10 0.30 2 = 0.022 6 102 = K spa Thus, precipitation of FeS should not occur. 2 0.010 0.10 = 1.1104 For Ag 2S, Qspa = 2 0.30 Qspa 6 1030 = K spa ; thus, precipitation of Ag 2S should occur. 14B (M) The H 3O + needed to just form a precipitate can be obtained by direct substitution of the provided concentrations into the Kspa expression. When that expression is satisfied, a precipitate will just form. Fe 2+ H 2S 0.015 M Fe 2+ 0.10 M H 2S = 6 10 2 = , H 3O + = K spa = + 2 + 2 H 3O H 3O 0.015 0.10 6 102 = 0.002 M pH = log H 3O + = log 0.002 = 2.7 b g 899 Chapter 18: Solubility and ComplexIon Equilibria INTEGRATIVE EXAMPLES
A. (D) To determine the amount of Ca(NO3)2 needed, one has to calculate the amount of Ca2+ that will result in only 1.001012 M of PO43Ca 3 PO 4 2 3 Ca 2+ +2 PO34
K sp 3s 2s 3 2 Using the commonion effect, we will try to determine what concentration of Ca2+ ions forces the equilibrium in the lake to have only 1.001012 M of phosphate, noting that (2s) is the equilibrium concentration of phosphate.
1.30 1032 Ca 2 1.00 1012 2+ Solving for [Ca ] yields a concentration of 0.00235 M.
3 2 The volume of the lake: V = 300 m 150 m 5 m = 225000 m3 or 2.25108 L. Mass of Ca(NO3)2 is determined as follows: 0.00235 mol Ca 2 1 mol Ca NO3 2 164.1 g Ca NO3 2 mass Ca NO3 2 2.25 108 L L 1 mol Ca 2 1 mol Ca NO3 2 87 106 g Ca NO3 2
B. (M) The reaction of AgNO3 and Na2SO4 is as follows: 2AgNO3 Na 2SO 4 2NaNO3 Ag 2SO 4 mol Ag 2SO 4 0.350 L 0.200 M AgNO3 mol NaNO3 0.250 L 0.240 M Na 2SO 4 1 mol Ag 2SO 4 3.5 102 mol 2 mol AgNO3 2 mol NaNO3 0.12 mol 1 mol Na 2SO 4 Ag2SO4 is the precipitate. Since it is also the limiting reagent, there are 3.5102 moles of Ag2SO4 produced. The reaction of Ag2SO4 with Na2S2O3 is as follows:
mol Na 2S2 O3 0.400 L 0.500 M Na 2S2 O3 = 0.200 mol Na 2S2 O3 Ag 2SO 4 Na 2SO 4 Ag 2S2 O3 Ag2SO4 is the limiting reagent, so no Ag2SO4 precipitate is left. 900 Chapter 18: Solubility and ComplexIon Equilibria EXERCISES
Ksp and Solubility
1. (E) (a) (b) (c) (d) 2. (E) (a) (b) (c) (d) 3. (E) (a) (b) (c) (d) 4.
2 Ag 2SO 4 s 2 Ag + aq + SO 4 aq 2 K sp = Ag + SO 4 2 Ra IO3 2 s Ra 2+ aq + 2 IO3 aq K sp = Ra 2+ IO3 3 2 3 Ni3 PO 4 2 s 3 Ni 2+ aq + 2 PO 4 aq 2+ 2 PuO 2 CO3 s PuO 2 aq + CO3 aq 3 K sp = Ni 2+ PO 4 2 2+ 2 K sp = PuO 2 CO3 K sp = Fe3+ OH 3 Fe OH 3 s Fe3+ aq + 3OH aq BiOOH s BiO + aq + OH aq 2+ Hg 2 I 2 s Hg 2 aq + 2 I aq 2 3 Pb3 AsO 4 2 s 3Pb 2+ aq + 2 AsO 4 aq K sp = BiO + OH 2+ K sp = Hg 2 I 3 2 3 K sp = Pb 2+ AsO 4 CrF3 s Cr 3+ aq + 3F aq K sp = Cr 3+ F = 6.6 1011 2 3 3 2 2 Au 2 C2O 4 3 s 2 Au 3+ aq + 3C 2O 4 aq K sp = Au 3+ C 2 O 4 = 11010 Cd 3 PO 4 2 s 3Cd 2+ aq + 2 PO 4 SrF2 s Sr 2+ aq + 2 F aq 3 aq Ksp = Cd 2+ 3 PO 4
2 3 2 = 2.1 1033 Ksp = Sr 2+ F = 2.5 109 (E) Let s = solubility of each compound in moles of compound per liter of solution. (a) (b) (c) (d) Ksp = Ba 2+ CrO 4 2 = s s = s 2 = 1.2 1010 Ksp = Pb 2+ Ksp = Ce 3+
Ksp = Mg 2+ 2 2 3 b gb g Br = b sgb2 sg = 4 s = 4.0 10 F = b sgb3sg = 27 s = 8 10 AsO = b3sg b2 sg = 108s = 2.1 10
5 3 3 4 16 3 3 2 3 2 5 4 s = 1.1 105 M s = 2.2 102 M s = 7 105 M
20 s = 4.5 105 M 901 Chapter 18: Solubility and ComplexIon Equilibria 5. (E) We use the value of Ksp for each compound in determining the molar solubility in a saturated solution. In each case, s represents the molar solubility of the compound. AgCN K sp = Ag + CN = s s = s 2 = 1.2 1016 s = 1.1108 M AgIO3 AgI AgNO 2 Ag 2SO 4 K sp = Ag + IO3 + K sp = Ag I = s s = s 2 = 3.0 108 = s s = s 2 = 8.5 1017 s = 1.7 104 M s = 9.2 109 M s = 2.4 102 M K sp = Ag + NO 2 = s s = s 2 = 6.0 104 2 2 2 K sp = Ag + SO 4 = 2s s = 4s 3 = 1.4 105 s = 1.5 102 M Thus, in order of increasing molar solubility, from smallest to largest: AgI AgCN AgIO 3 Ag 2SO 4 AgNO 2
6. (E) We use the value of Ksp for each compound in determining Mg 2+ in its saturated solution. In each case, s represents the molar solubility of the compound.
(a) MgCO 3 Ksp = Mg 2+ CO 32 = s s = s2 = 3.5 108 s = 1.9 104 M Mg 2+ = 1.9 104 M
2 b gb g
2 (b) MgF2 Ksp = Mg 2+ F s = 2.1 103 M
3 = s 2 s = 4 s 3 = 3.7 108 Mg 2+ = 2.1 103 M
2 b gb g (c) Mg 3 (PO 4 ) 2 3 2 3 K sp = Mg 2+ PO 4 = 3s 2 s = 108s 5 = 2.11025 Mg 2+ = 1.4 105 M s = 4.5 106 M Thus a saturated solution of MgF2 has the highest Mg 2+ .
7. (M) We determine F in saturated CaF2 , and from that value the concentration of F in ppm. For CaF2 Ksp = Ca 2+ F 2 = s 2 s = 4 s 3 = 5.3 109
2 b gb g s = 1.1 103 M The solubility in ppm is the number of grams of CaF2 in 106 g of solution. We assume a solution density of 1.00 g/mL. mass of F 106 g soln 1mL 1L soln 1.1103 mol CaF2 1.00 g soln 1000 mL 1L soln 2 mol F 19.0 g F 42 g F 1mol CaF2 1mol F This is 42 times more concentrated than the optimum concentration of fluoride ion for fluoridation. CaF2 is, in fact, more soluble than is necessary. Its uncontrolled use might lead to excessive F in solution. 902 Chapter 18: Solubility and ComplexIon Equilibria 8. (E) We determine OH in a saturated solution. From this OH we determine the pH. Ksp = BiO + OH = 4 1010 = s 2 pOH = log 2 105 = 4.7
9. s = 2 105 M = OH = BiO + pH = 9.3 c h (M) We first assume that the volume of the solution does not change appreciably when its temperature is lowered. Then we determine the mass of Mg C16 H 31O 2 2 dissolved in each solution, recognizing that the molar solubility of Mg C16 H 31O 2 Ksp 4 s 3
3 b b g g 2 equals the cube root of one fourth of its solubility product constant, since it is the only solute in the solution. s 3 Ksp / 4
3 At 50 C : s = 4.8 1012 / 4 1.1 104 M; At 25C: s = 3.3 1012 / 4 9.4 105 M amount of Mg C16 H 31O 2 2 (50C) = 0.965 L b b g g 1.1104 mol Mg C16 H 31O 2 2 1L soln 9.4 105 mol Mg C16 H 31O 2 2 1L soln = 1.1104 mol amount of Mg C16 H 31O 2 2 (25C) = 0.965 L mass of Mg C16 H 31O 2 2 precipitated: = 1.1 0.91 104 mol = 0.91104 mol 10. 535.15g Mg C16 H 31O 2 2 1000 mg = 11mg 1mol Mg C16 H 31O 2 2 1g (M) We first assume that the volume of the solution does not change appreciably when its temperature is lowered. Then we determine the mass of CaC 2 O 4 dissolved in each solution, recognizing that the molar solubility of CaC 2 O 4 equals the square root of its solubility product constant, since it is the only solute in the solution. At 95 C s = 1.2 108 1.1 104 M; At 13 C: mass of CaC 2 O 4 (95 C) = 0.725 L mass of CaC 2 O 4 (13 C) = 0.725 L 1 L soln 5.2 10 5 mol PbSO 4 1 L soln s 2.7 109 5.2 105 M 128.1g CaC 2 O 4 1 mol CaC 2 O 4 1mol CaC 2 O 4 = 0.010 g CaC 2 O 4 = 0.0048 g CaC 2 O 4 1.1 10 4 mol CaC 2 O 4 128.1g CaC 2 O 4 mass of CaC2 O 4 precipitated = 0.010 g 0.0048 g 1000 mg = 5.2 mg 1g 903 Chapter 18: Solubility and ComplexIon Equilibria 11. (M) First we determine I in the saturated solution. Ksp = Pb 2+ I 2 = 7.1 109 = s 2 s = 4 s 3
2 b gb g s = 1.2 103 M The AgNO 3 reacts with the I in this saturated solution in the titration. Ag + aq + I aq AgI s We determine the amount of Ag + needed for this titration, and b g b g bg then AgNO 3 in the titrant. moles Ag + = 0.02500 L AgNO 3 molarity = 1.2 103 mol PbI 2 2 mol I 1 mol Ag + = 6.0 105 mol Ag + 1 L soln 1 mol PbI 2 1 mol I 6.0 105 mol Ag + 1 mol AgNO 3 = 4.5 103 M + 0.0133 L soln 1 mol Ag
2 12. (M) We first determine C 2 O 4
2 = s, the solubility of the saturated solution.
2 C2O4 0.00134 mmol KMnO 4 5 mmol C 2 O 4 4.8 mL 1 mL soln 2 mmol MnO 4 = = 6.4 105 M = s = Ca 2+ 250.0 mL
2 Ksp = Ca 2+ C 2 O 4 = s s = s2 = 6.4 105 b gb g c h 2 = 4.1 109 13. (M) We first use the ideal gas law to determine the moles of H 2S gas used. FG 748 mmHg 1atm IJ FG 30.4 mL 1 L IJ 1000 mL K 760 mmHg K H PV H . 123 10 n RT
0.08206 L atm mol K (23 273) K
1 1 3 moles If we assume that all the H 2S is consumed in forming Ag 2S , we can compute the Ag + in the AgBrO 3 solution. This assumption is valid if the equilibrium constant for the cited reaction is large, which is the case, as shown below: 2Ag + aq + HS aq + OH aq Ag 2S s + H 2 O(l) H 2S aq + H 2 O(l) HS aq + H 3O + aq 2H 2 O(l) H 3O + aq + OH aq K a 2 /K sp = 1.0 1019 3.8 1031 51 2.6 10 K1 = 1.0 107 K w = 1.0 1014 2Ag + aq + H 2 S aq + 2H 2 O(l) Ag 2 S s + 2H 3 O + aq K overall = ( K a 2 / K sp )( K1 )( K w ) (3.8 1031 )(1.0 107 )(1.0 1014 ) = 3.8 1010
Ag + = 1.23 103 mol H 2S 1000 mL 2 mol Ag + = 7.28 103 M 338 mL soln 1 L soln 1 mol H 2S
2 Then, for AgBrO3 , K sp = Ag + BrO3 = 7.28 103 = 5.30 105 904 Chapter 18: Solubility and ComplexIon Equilibria 14. (M) The titration reaction is Ca OH 2 aq + 2HCl aq CaCl2 aq + 2H 2 O(l) 1L 0.1032 mol HCl 1 mol OH 10.7 mL HCl 1000 mL 1L 1 mol HCl OH 0.0221 M 1L 50.00 mL Ca(OH) 2 soln 1000 mL In a saturated solution of Ca OH 2 , Ca 2+ = OH 2 K sp = Ca 2+ OH = 0.0221 2 0.0221 = 5.40 106 ( 5.5 106 in Appendix D). 2 2 The CommonIon Effect
15. (a) (b) (E) We let s = molar solubility of Mg OH Ksp = Mg 2+ OH Equation : Initial : Changes : Equil : 2 = s 2 s = 4 s 3 = 1.8 1011
2 b gb g b g 2 in moles solute per liter of solution. s = 1.7 104 M 2OH aq 0M +2s M 2s M
2 Mg OH 2 s Mg 2+ aq + 0.0862 M +s M 0.0862 + s M
2 K sp = 0.0862 + s 2 s = 1.8 1011 0.0862 2s = 0.34 s 2 s = 7.3 106 M (s << 0.0802 M, thus, the approximation was valid.)
(c) OH = KOH = 0.0355 M Equation : Initial : Changes : Equil : Mg OH 2 s Mg 2+ aq 0M + sM sM
2 2 + 2OH aq 0.0355 M + 2s M 0.0355 + 2s M s = 1.4 108 M
2 K sp = s 0.0355 + 2s = 1.8 1011 s 0.0355 = 0.0013 s
16. (E) The solubility equilibrium is (a) CaCO3 s Ca 2+ aq + CO3 aq The addition of Na 2 CO 3 aq produces CO 32 aq in solution. This common ion suppresses the dissolution of CaCO 3 s . b g bg b g (b) HCl(aq) is a strong acid that reacts with carbonate ion: 2 CO3 aq + 2H 3O + aq CO 2 g + 3H 2 O(l) . This decreases CO 32 in the solution, allowing more CaCO 3 s to dissolve. bg 905 Chapter 18: Solubility and ComplexIon Equilibria (c) HSO 4 aq is a moderately weak acid. It is strong enough to protonate appreciable b g concentrations of carbonate ion, thereby decreasing CO 32 and enhancing the solubility of CaCO 3 s , as the value of Kc indicates. HSO 4 aq + CO3 bg 2 aq SO4 2 aq + HCO3 aq Kc K a (HSO 4 ) 0.011 2.3 108 11 K a (HCO3 ) 4.7 10 17. (E) The presence of KI in a solution produces a significant I in the solution. Not as much AgI can dissolve in such a solution as in pure water, since the ion product, Ag + I , cannot exceed the value of K sp (i.e., the I from the KI that dissolves represses the dissociation of AgI(s). In similar fashion, AgNO 3 produces a significant Ag + in solution, again influencing the value of the ion product; not as much AgI can dissolve as in pure water.
18. (E) If the solution contains KNO 3 , more AgI will end up dissolving than in pure water, because the activity of each ion will be less than its molarity. On an ionic level, the reason is that ion pairs, such as Ag+NO3 (aq) and K+I (aq) form in the solution, preventing Ag+ and I ions from coming together and precipitating. (E) Equation: 19. Ag 2SO 4 s    Original: Add solid: Equil: + 2Ag aq 0M +2x M 2x M + SO 2 aq 4 0.150 M +x M 0.150 + x M 2 x = Ag + = 9.7 103 M; x = 0.00485 M 2 2 K sp = Ag + SO 4 = 2 x 0.150 + x = 9.7 103 0.150 + 0.00485 = 1.5 105 2 2 20. (M) Equation: CaSO 4 s Ca 2+ aq + 0M +x M xM SO 4 2 aq . Soln: Add CaSO 4 s Equil:    0.0025 M +x M 0.0025 + x M If we use the approximation that x << 0.0025, we find x = 0.0036. Clearly, x is larger than 0.0025 and thus the approximation is not valid. The quadratic equation must be solved.
2 K sp = Ca 2+ SO 4 = 9.1106 = x 0.0025 + x = 0.0025 x + x 2 2 x + 0.0025 x 9.1106 = 0 906 Chapter 18: Solubility and ComplexIon Equilibria x= b b 2 4ac 0.0025 6.3 106 + 3.6 105 = 2.0 10 3 M = CaSO 4 = 2 2a mass CaSO 4 = 0.1000 L 2.0 10 3 mol CaSO 4 136.1 g CaSO 4 = 0.027 g CaSO 4 1 L soln 1 mol CaSO 4
2 21. (M) For PbI 2 , Ksp = 7.1 109 = Pb 2+ I In a solution where 1.5 104 mol PbI 2 is dissolved, Pb 2 1.5 104 M , and I = 2 Pb 2+ = 3.0 104 M Equation: Initial: Add lead(II): Equil: PbI 2 s    2+ Pb aq + 2I aq 3.0 104 M 0.00030 M 1.5 104 M +x M 0.00015 + x M
2 K sp = 7.1109 = 0.00015 + x 0.00030 ; 0.00015 + x = 0.079 ; x = 0.079 M = Pb 2+ 22. (M) For PbI 2 , Ksp = 7.1 109 = Pb 2+ I 2 In a solution where 1.5 105 mol PbI 2 /L is dissolved, Pb 2 1.5 105 M , and 2+ 5 I = 2 Pb = 3.0 10 M Equation: Initial: Add KI: Equil: PbI 2 s Pb 2+ aq + 2I aq    1.5 105 M
1.5 105 M 3.0 105 M +x M (3.0 105 x) M Ksp = 7.1 109 = (1.5 105) (3.0 105 + x)2 (3.0 105 + x) = (7.1 109 1.5 105)1/2 x = (2.2 102) (3.0 105) = 2.1 102 M = [I]
23.
2 (D) For Ag 2 CrO 4 , K sp = 1.11012 = Ag + CrO 4 8 2 In a 5.0 10 M solution of Ag 2 CrO 4 , CrO 4 = 5.0 108 M and Ag + = 1.0 107 M 2 Equation: Initial: Add chromate: Equil: Ag 2 CrO 4 s    + 2Ag aq + CrO 4 2 aq 1.0 107 M
1.0 107 M
2 5.0 108 M +x M 5.0 108 + x M c h K sp = 1.1 1012 = 1.0 107 5.0 108 + x ; 5.0 10 8 + x = 1.1 102 = CrO 4 2 . 907 Chapter 18: Solubility and ComplexIon Equilibria This is an impossibly high concentration to reach. Thus, we cannot lower the solubility of 2 Ag 2 CrO 4 to 5.0 108 M with CrO 4 as the common ion. Let's consider using Ag+ as the common ion. 2 Ag 2CrO 4 s 2Ag + aq + CrO 4 aq Equation: Initial: Add silver(I) ion: Equil:   
2 1.0 107 M +x M 1.0 107 + x M 5.0 108 M 5.0 108 M K sp = 1.1 1012 = 1.0 + x 5.0 108 1.1 1012 = 4.7 103 8 5.0 10 3 7 3 x = 4.7 10 1.0 10 = 4.7 10 M = I ; this is an easytoreach concentration. Thus, 1.0 107 + x = the solubility can be lowered to 5.0108 M by carefully adding Ag+(aq).
24. (M) Even though BaCO 3 is more soluble than BaSO 4 , it will still precipitate when 0.50 M Na 2 CO 3 aq is added to a saturated solution of BaSO 4 because there is a sufficient Ba 2+ in such a solution for the ion product Ba 2+ CO 32 to exceed the value of Ksp for the compound. An example will demonstrate this phenomenon. Let us assume that the two solutions being mixed are of equal volume and, to make the situation even more unfavorable, that the saturated BaSO 4 solution is not in contact with solid BaSO 4 , meaning that it does not maintain its saturation when it is diluted. First we determine Ba 2+ in saturated BaSO 4 (aq) .
Ksp = Ba 2+ SO 4
2 b g = 1.1 1010 = s 2 s = 11 1010 10 105 M . . Mixing solutions of equal volumes means that the concentrations of solutes not common to the two solutions are halved by dilution. 1 1.0 105 mol BaSO 4 1 mol Ba 2+ Ba 2+ = = 5.0 106 M 2 1L 1 mol BaSO 4 1 0.50 mol Na 2 CO3 1 mol CO3 CO32 = = 0.25 M 2 1L 1 mol Na 2 CO3
2 2 Qsp {BaCO3 } = Ba 2+ CO3 = 5.0 106 0.25 = 1.3 106 5.0 109 = K sp {BaCO3} Thus, precipitation of BaCO 3 indeed should occur under the conditions described.
25. 115g Ca 2+ 1 mol Ca 2+ 1000 g soln 3 (E) Ca 2+ = 6 10 g soln 40.08 g Ca 2+ 1 L soln = 2.87 10 M
9 3 3 2+ Ca F = K sp = 5.3 10 = 2.87 10 F F = 1.4 10 M 1.4 103 mol F 19.00 g F 1 L soln ppm F = 106 g soln = 27 ppm 1 L soln 1 mol F 1000 g
2 2 908 Chapter 18: Solubility and ComplexIon Equilibria 26. (M) We first calculate the Ag + and the Cl in the saturated solution. Ksp = Ag + Cl = 1.8 1010 = s s = s 2 Ag + = Cl = 1.3 105 M b gb g s = 1.3 105 M = Ag + = Cl Both of these concentrations are marginally diluted by the addition of 1 mL of NaCl(aq) 100.0 mL = 1.3 105 M 100.0 mL +1.0 mL 1.0 mL = 9.9 103 M The Cl in the NaCl(aq) also is diluted. Cl = 1.0 M 100.0 mL +1.0 mL Let us use this Cl to determine the Ag + that can exist in this solution. 1.8 1010 8 Ag + = 9.9 103 = 1.8 10 M We compute the amount of AgCl in this final solution, and in the initial solution. Ag + Cl = 1.8 1010 = Ag + 9.9 103 M mmol AgCl final = 101.0 mL 1.8 108 mol Ag + 1 mmol AgCl = 1.8 106 mmol AgCl + 1 L soln 1 mmol Ag 1.3 105 mol Ag + 1 mmol AgCl = 1.3 103 mmol AgCl + 1 L soln 1 mmol Ag The difference between these two amounts is the amount of AgCl that precipitates. Next we compute its mass. 143.3 mg AgCl mass AgCl = 1.3 103 1.8 106 mmol AgCl = 0.19 mg 1 mmol AgCl We conclude that the precipitate will not be visible to the unaided eye, since its mass is less than 1 mg. mmol AgCl initial = 100.0 mL Criteria for Precipitation from Solution
27. (E) We first determine Mg 2+ , and then the value of Qsp in order to compare it to the value of Ksp . We express molarity in millimoles per milliliter, entirely equivalent to moles per liter.
[Mg 2+ ] 22.5 mg MgCl2 1mmol MgCl2 6H 2O 1mmol Mg 2+ 3.41 104 M 325 mL soln 203.3mg MgCl2 6H 2O 1mmol MgCl2 Qsp [Mg 2+ ][F ]2 (3.41 104 )(0.035) 2 4.2 107 3.7 108 K sp Thus, precipitation of MgF2 s should occur from this solution.
28. (E) The solutions mutually dilute each other. bg 155 mL Cl = 0.016 M = 6.2 103 M 155 mL + 245 mL 245 mL Pb 2+ = 0.175 M = 0.107 M 245 mL +155 mL 909 Chapter 18: Solubility and ComplexIon Equilibria Then we compute the value of the ion product and compare it to the solubility product constant value. Qsp = Pb 2+ Cl = 0.107 6.2 103 = 4.1106 1.6 105 = K sp 2 2 Thus, precipitation of PbCl 2 s will not occur from these mixed solutions.
29. (E) We determine the OH needed to just initiate precipitation of Cd(OH)2.
K sp = Cd 2+ OH 2 bg = 2.5 10 14 = 0.0055M OH b g 2 OH = 2.5 10 14 = 2.1 10 6 M 0.0055 pOH = log 2.1 106 = 5.68 Thus, Cd OH
30. b g c h pH = 14.00 5.68 = 8.32 2 will precipitate from a solution with pH 8.32. (E) We determine the OH needed to just initiate precipitation of Cr(OH)3.
K sp [Cr 3+ ][OH ]3 6.3 1031 (0.086 M) [OH ]3 ; [OH ] 6.3 1031 1.9 1010 M 0.086 c Thus, Cr bOH g
31. (D) (a) pOH = log 1.9 1010 = 9.72
3 h pH = 14.00 9.72 = 4.28 will precipitate from a solution with pH > 4.28. First we determine Cl due to the added NaCl. Cl = 0.10 mg NaCl 1g 1 mol NaCl 1 mol Cl = 1.7 106 M 1.0 L soln 1000 mg 58.4 g NaCl 1 mol NaCl Then we determine the value of the ion product and compare it to the solubility product constant value. Qsp = Ag + Cl = 0.10 1.7 106 = 1.7 107 1.8 1010 = K sp for AgCl Thus, precipitation of AgCl(s) should occur.
(b) The KBr(aq) is diluted on mixing, but the Ag + and Cl are barely affected by dilution. Br = 0.10 M 0.05 mL = 2 105 M 0.05 mL + 250 mL Now we determine Ag + in a saturated AgCl solution. Ksp = Ag + Cl = s s = s 2 = 1.8 1010 b gb g s = 1.3 105 M 910 Chapter 18: Solubility and ComplexIon Equilibria Then we determine the value of the ion product for AgBr and compare it to the solubility product constant value. Qsp = Ag + Br = 1.3 105 2 105 = 3 1010 5.0 1013 = K sp for AgBr Thus, precipitation of AgBr(s) should occur.
(c) The hydroxide ion is diluted by mixing the two solutions. 0.05 mL OH = 0.0150 M = 2.5 107 M 0.05 mL + 3000 mL But the Mg 2+ does not change significantly. 2.0 mg Mg 2+ 1g 1 mol Mg 2+ Mg 2+ = = 8.2 105 M 1.0 L soln 1000 mg 24.3 g Mg Then we determine the value of the ion product and compare it to the solubility product constant value. Qsp = Mg 2+ OH = 2.5 107 8.2 105 = 5.1 1018 2 2 Qsp 1.8 1011 = K sp for Mg OH 2 Thus, no precipitate forms. 32. First we determine the moles of H 2 produced during the electrolysis, then determine [OH]. 1 atm 752 mmHg 0.652 L PV 760 mmHg moles H 2 = = = 0.0267 mol H 2 RT 0.08206 L atm mol1 K 1 295 K 0.0267 mol H 2 2 mol OH 1 mol H 2 = 0.170 M 0.315 L sample
2 2 OH = Qsp = Mg 2+ OH = 0.185 0.170 = 5.35 103 1.8 1011 = K sp Thus, yes, precipitation of Mg OH
33.
2 b g bsg should occur during the electrolysis. (D) First we must calculate the initial [H2C2O4] upon dissolution of the solid acid: 1 mol H 2 C 2 O 4 1 [H2C2O4]initial = 1.50 g H2C2O4 = 0.0833 M 90.036 g H 2 C2 O 4 0.200 L (We assume the volume of the solution stays at 0.200 L.) Next we need to determine the [C2O42] in solution after the hydrolysis reaction between oxalic acid and water reaches equilibrium. To accomplish this we will need to solve two I.C.E. tables: 911 Chapter 18: Solubility and ComplexIon Equilibria
K 5.2 102 Table 1: Initial: Change: Equilibrium: a1 H2C2O4(aq) + H2O(l) HC2O4(aq) + H3O+(aq) 0.0833 M  0M 0M x  +x M +x M 0.0833 x M  xM xM Since Ca/Ka = 1.6, the approximation cannot be used, and thus the full quadratic equation must be solved: x2/(0.0833 x) = 5.2102; x2 + 5.2102x 4.33103
2.7103 + 0.0173 x = 0.045 M = [HC 2 O 4 ] [H 3 O ] 2 Now we can solve the I.C.E. table for the second proton loss: x= 5.2102 Table 2: Initial: Change: Equilibrium: a1 HC2O4(aq) + H2O(l) C2O42(aq) + H3O+(aq) 0.045 M  0M 0.045 M y  +y M +yM (0.045 y) M  yM (0.045 +y) M K 5.4 105 Since Ca/Ka = 833, the approximation may not be valid and we yet again should solve the full quadratic equation: y (0.045 y ) = 5.4 105 ; y2 + 0.045y = 2.43106 5.4105y (0.045 y ) 2.03103 + 9.72106 y = 8.2 105 M = [C2 O 4 2 ] 2 Now we can calculate the Qsp for the calcium oxalate system: Qsp = [Ca2+]initial[ C2O42]initial = (0.150)(8.2105) = 1.2105 > 1.3 109 (Ksp for CaC2O4) Thus, CaC2O4 should precipitate from this solution. y=
34. (D) The solutions mutually dilute each other. We first determine the solubility of each compound in its saturated solution and then its concentration after dilution.
2 2 K sp = Ag + SO 4 = 1.4 105 = 2 s s = 4s 3 2 3 0.045 s= 1.4 105 1.5 102 M 4 Ag + = 0.0086 M 100.0 mL SO 4 2 = 0.015 M = 0.0043 M 100.0 mL + 250.0 mL
Ksp [ Pb 2+ ][CrO 2 ] 2.8 1013 ( s)( s) s2 4 s 2.8 1013 5.3 107 M 250.0 mL Pb 2+ = CrO 4 2 = 5.3 107 = 3.8 107 M 250.0 mL +100.0 mL From the balanced chemical equation, we see that the two possible precipitates are PbSO 4 and Ag 2 CrO 4 . (Neither PbCrO 4 nor Ag 2SO 4 can precipitate because they have been diluted below their saturated concentrations.) PbCrO 4 + Ag 2SO 4 PbSO 4 + Ag 2CrO 4 Thus, we compute the value of Qsp for each of these compounds and compare those values with the solubility constant product value. 912 Chapter 18: Solubility and ComplexIon Equilibria
2 Qsp = Pb 2+ SO 4 = 3.8 107 0.0043 = 1.6 109 1.6 108 = K sp for PbSO 4 Thus, PbSO 4 (s) will not precipitate.
2 2 Qsp = Ag + CrO 4 = 0.0086 3.8 107 = 2.8 1011 1.11012 = K sp for Ag 2 CrO 4 2 Thus, Ag 2 CrO 4 s should precipitate. bg Completeness of Precipitation
35. (M) First determine that a precipitate forms. The solutions mutually dilute each other. CrO 4 2 = 0.350 M Ag + = 0.0100 M 200.0 mL = 0.175 M 200.0 mL + 200.0 mL 200.0 mL = 0.00500 M 200.0 mL + 200.0 mL We determine the value of the ion product and compare it to the solubility product constant value.
2 2 Qsp = Ag + CrO 4 = 0.00500 0.175 = 4.4 106 1.110 12 = K sp for Ag 2 CrO 4 2 Ag 2 CrO 4 should precipitate. Now, we assume that as much solid forms as possible, and then we approach equilibrium by dissolving that solid in a solution that contains the ion in excess.
1.1 10 Equation: Ag 2 CrO 4 s 2Ag + aq + CrO 4 2 aq Orig. soln : 0.00500 M 0.175 M Form solid : 0.00500 M 0.00250 M 0M 0.173 M Not at equilibrium Changes : +2 x M +x M Equil : 2x M 0.173 + x M
12 K sp Ag CrO 4 2 1.1 1012 2 x 0.173 x 4 x 2 0.173 2 x = 1.11012 4 0.173 = 1.3 106 M % Ag + unprecipitated 36. Ag + = 2 x = 2.6 106 M 2.6 106 M final 100% 0.052% unprecipitated 0.00500 M initial
425 mL (M) [Ag+]diluted = 0.0208 M 175 mL = 0.008565 M
[CrO42]diluted = 0.0380 M 250 mL = 0.02235 M
425 mL Qsp Ag CrO 4 1.6 106 K sp 2 2 913 Chapter 18: Solubility and ComplexIon Equilibria Because Qsp > Ksp, then more Ag2Cr2O4 precipitates out. Assume that the limiting reagent is used up (100% reaction in the reverse direction) and reestablish the equilibrium in the reverse direction. Here Ag+ is the limiting reagent. Ag2CrO4(s) Initial Change 100% rxn Change Equil 2CrO4 ) Ksp(Ag =1.1 1012 2 Ag+(aq) 0.008565 M 2x 0 +2y 2y + CrO42(aq) 0.02235 M x 0.0181 +y 0.0181 + y x = 0.00428 M reestablish equil (assume y ~ 0) 1.1 1012 = (2y)2(0.0181+ y) (2y)2(0.0181) y = 3.9 106 M y << 0.0181, so this assumption is valid. 2y = [Ag+] = 7.8 106 M after precipitation is complete. 7.8 106 + 100% = 0.091% (precipitation is essentially quantitative) % [Ag ]unprecipitated = 0.00856
37. (M) We first use the solubility product constant expression to determine Pb 2+ in a solution with 0.100 M Cl . K sp = Pb 2+ Cl = 1.6 105 = Pb 2+ 0.100 2 2 Pb 2+ = 1.6 105 0.100 2 = 1.6 103 M Thus, % unprecipitated 1.6 10 M 100% 2.5% 0.065 M 3 Now, we want to determine what [Cl] must be maintained to keep [Pb 2+ ]final 1%; [Pb 2+ ]initial 0.010 0.065 M = 6.5 104 M
Ksp = Pb 2+ Cl 2 = 1.6 105 = 6.5 10 4 Cl c h 2 Cl = 6 105 = 0.16 M 6.5 10 4 38. (M) Let's start by assuming that the concentration of Pb2+ in the untreated wine is no higher than 1.5 104M (this assumption is not unreasonable.) As long as the Pb2+ concentration is less than 1.5 104 M, then the final sulfate ion concentration in the CaSO4 treated wine should be virtually the same as the sulfate ion concentration in a saturated solution of CaSO4 formed by dissolving solid CaSO4 in pure water ( i.e., with [Pb2+] less than or equal to 1.5 104 M, the [SO42] will not drop significantly below that for a saturated solution, 3.0 103 M.) Thus, the addition of CaSO4 to the wine would result in the precipitation of solid PbSO4, which would continue until the concentration of Pb2+ was equal to the Ksp for PbSO4 divided by the concentration of dissolved sulfate ion, i.e., [Pb2+]max = 1.6 108M2/3.0 103 M = 5.3 106 M. 914 Chapter 18: Solubility and ComplexIon Equilibria Fractional Precipitation
39. (M) First, assemble all of the data. Ksp for Ca(OH)2 = 5.5106, Ksp for Mg(OH)2 = 1.81011 440 g Ca 2+ 1 mol Ca 2+ 1 kg seawater 1.03 kg seawater 2+ [Ca ] 0.0113 M 2+ 1000 kg seawater 40.078 g Ca 1000 g seawater 1 L seawater [Mg2+] = 0.059 M, obtained from Example 186. [OH] = 0.0020 M (maintained) (a) (b) Qsp = [Ca2+][OH]2 = (0.0113)( 0.0020)2 = 4.5108 Qsp < Ksp no precipitate forms.
2+ For the separation to be complete, >>99.9 % of the Mg must be removed before Ca2+ begins to precipitate. We have already shown that Ca2+ will not precipitate if the [OH] = 0.0020 M and is maintained at this level. Let us determine how much of the 0.059 M Mg2+ will still be in solution when [OH] = 0.0020 M. Ksp = [Mg2+][OH]2 = (x)( 0.0020)2 = 1.81011 x = 4.5106 M 4.5 106 100% = 0.0076 % The percent Mg2+ ion left in solution = 0.059 This means that 100%  0.0076 % Mg = 99.992 % has precipitated. Clearly, the magnesium ion has been separated from the calcium ion (i.e., >> 99.9% of the Mg2+ ions have precipitated and virtually all of the Ca2+ ions are still in solution.) 40. (D) (a) (b) 0.10 M NaCl will not work at all, since both BaCl 2 and CaCl 2 are soluble in water.
Ksp = 1.1 1010 for BaSO 4 and Ksp = 9.1 106 for CaSO 4 . Since these values differ by more than 1000, 0.05 M Na 2SO 4 would effectively separate Ba 2+ from Ca 2+ . We first compute SO 4 2 when BaSO 4 begins to precipitate. Ba 2+ SO 4 2 = 0.050 SO 4 2 = 1.11010 ; SO 4 2 = 1.11010 = 2.2 109 M 0.050 And then we calculate SO 4 2 when Ba 2+ has decreased to 0.1% of its initial value, that is, to 5.0 105 M. 1.11010 6 Ba 2+ SO 4 2 = 5.0 105 SO 4 2 = 1.11010 ; SO 4 2 = 5.0 105 = 2.2 10 M And finally, SO 4 2 when CaSO 4 begins to precipitate. 9.1 106 Ca 2+ SO 4 2 = 0.050 SO 4 2 = 9.1106 ; SO 4 2 = = 1.8 104 M 0.050 915 Chapter 18: Solubility and ComplexIon Equilibria (c) Now, Ksp = 5 103 for Ba OH b g 2 and Ksp = 5.5 106 for Ca OH 2 . The fact that these two Ksp values differ by almost a factor of 1000 does not tell the entire story, because OH is squared in both Ksp expressions. We compute OH when Ca OH to precipitate. . 55 106 = 1.0 102 M 0.050 Precipitation will not proceed, as we only have 0.001 M NaOH, which has OH = 1 103 M. Ca 2+ OH 2 b g 2 begins = 5.5 106 = 0.050 OH b g 2 OH = (d) Ksp = 5.1 109 for BaCO 3 and Ksp = 2.8 109 for CaCO3 . Since these two values differ by less than a factor of 2, 0.50 M Na 2 CO 3 would not effectively separate Ba 2+ from Ca 2+ .
41. (M) (a) Here we need to determine I when AgI just begins to precipitate, and I when PbI 2 just begins to precipitate. Ksp = Ag + I = 8.5 1017 = 0.10 I K sp = Pb 2+ I = 7.1 109 = 0.10 I 2 2 b g I = 8.5 1016 M I 7.1 109 = 2.7 104 M 0.10 Since 8.5 1016 M is less than 2.7 104 M, AgI will precipitate before PbI 2 .
(b) (c) (d) I must be equal to 2.7 104 M before the second cation, Pb 2+ , begins to precipitate. Ksp = Ag + I = 8.5 1017 = Ag + 2.7 104 c h Ag + = 3.1 10 13 M Since Ag + has decreased to much less than 0.1% of its initial value before any PbI 2 begins to precipitate, we conclude that Ag + and Pb 2+ can be separated by precipitation with iodide ion. 42. (D) Normally we would worry about the mutual dilution of the two solutions, but the values of the solubility product constants are so small that only a very small volume of 0.50 M Pb NO 3 2 solution needs to be added, as we shall see. b g (a) Since the two anions are present at the same concentration and they have the same type of formula (one anion per cation), the one forming the compound with the smallest Ksp value will precipitate first. Thus, CrO 4
2 is the first anion to precipitate. 916 Chapter 18: Solubility and ComplexIon Equilibria (b) At the point where SO 4
2+ 2 begins to precipitate, we have 1.6 108 SO 4 2 = 1.6 108 = Pb 2+ 0.010M ; Pb 2+ = K sp = Pb = 1.6 106 M 0.010 Now we can test our original assumption, that only a very small volume of 0.50 M Pb NO 3 2 solution has been added. We assume that we have 1.00 L of the original
2+
6
3 2 b g solution, the one with the two anions dissolved in it, and compute the volume of 0.50 M Pbb NO g that has to be added to achieve Pb = 1.6 10 M.
1.6 106 mol Pb 2+ 1 mol Pb NO3 2 1 L Pb 2+soln = 1.00 L 1 L soln 1 mol Pb 2+ 0.50 mol Pb NO3 2 Vadded Vadded = 3.2 105 L Pb 2+soln = 0.0032mL Pb 2+soln This is less than one drop (0.05 mL) of the Pb 2+ solution, clearly a very small volume.
(c) The two anions are effectively separated if Pb 2+ has not reached 1.6 106 M when CrO 4 2 is reduced to 0.1% of its original value, that is, to
2 0.010 103 M = 1.0 105 M = CrO 4 Ksp = Pb 2+ CrO 4 2 = 2.8 10 13 = Pb 2+ 1.0 105 c h 2.8 1013 8 Pb 2+ = 1.0 105 = 2.8 10 M Thus, the two anions can be effectively separated by fractional precipitation.
43. (M) First, let's assemble all of the data. Ksp for AgCl = 1.81010 [Ag+] = 2.00 M [Cl] = 0.0100 M [I] = 0.250 M (a) (b) Ksp for AgI = 8.51017 AgI(s) will be the first to precipitate by virtue of the fact that the Ksp value for AgI is about 2 million times smaller than that for AgCl. AgCl(s) will begin to precipitate when the Qsp for AgCl(s) > Ksp for AgCl(s). The concentration of Ag+ required is: Ksp = [Ag+][Cl] = 1.81010 = (0.0100)(x) 1.8108 M Using this data, we can determine the remaining concentration of I using the Ksp. Ksp = [Ag+][I] = 8.51017 = (x)( 1.8108) x = 4.7109 M
x= (c) In part (b) we saw that the [I] drops from 0.250 M 4.7109 M. Only a small 4.7 109 100% = 0.0000019 % percentage of the ion remains in solution. 0.250 This means that 99.999998% of the I ion has been precipitated before any of the Cl ion has precipitated. Clearly, the fractional separation of Cl from I is feasible. 917 Chapter 18: Solubility and ComplexIon Equilibria 44. (M) (a) We first determine the Ag + needed to initiate precipitation of each compound.
AgCl : K sp = Ag + Cl = 1.8 1010 = Ag + 0.250 ; AgBr : Ksp = Ag + Br = 5.0 1013 = Ag + 0.0022 ; Ag + = 1.8 1010 = 7.2 1010 M 0.250 5.0 1013 Ag + = = 2.3 1010 M 0.0022 Thus, Br precipitates first, as AgBr, because it requires a lower Ag + .
(b) (c) Ag + = 7.2 1010 M when chloride ion, the second anion, begins to precipitate. Cl and Br cannot be separated by this fractional precipitation. Ag + will have to rise to 1000 times its initial value, to 2.3 107 M, before AgBr is completely precipitated. But as soon as Ag + reaches 7.2 1010 M, AgCl will begin to precipitate. Solubility and pH
45. (E) In each case we indicate whether the compound is more soluble in water. We write the net ionic equation for the reaction in which the solid dissolves in acid. Substances are more soluble in acid if either (1) an acidbase reaction occurs or (2) a gas is produced, since escape of the gas from the reaction mixture causes the reaction to shift to the right. Same: KCl (K+ and Cl do not react appreciably with H2O) Acid: MgCO3 s + 2H + aq Mg 2+ aq + H 2O(l) + CO 2 g Acid: FeS s + 2H + aq Fe 2+ aq + H 2S g Acid: Ca OH 2 s + 2H + aq Ca 2+ aq + 2H 2 O(l) Water:
46. C6 H 5COOH is less soluble in acid, because of the H 3O + common ion. (E) In each case we indicate whether the compound is more soluble in base than in water. We write the net ionic equation for the reaction in which the solid dissolves in base. Substances are soluble in base if either (1) acidbase reaction occurs [as in (b)] or (2) a gas is produced, since escape of the gas from the reaction mixture causes the reaction to shift to the right. Water: BaSO 4 is less soluble in base, hydrolysis of SO 4
2 2 will be repressed. Base: H 2C 2 O 4 s + 2OH aq C2 O 4 Water: Same: Fe OH aq + 2H 2O(l) b g 3 is less soluble in base because of the OH common ion. NaNO 3 (neither Na+ nor NO3 react with H2O to a measurable extent). Water: MnS is less soluble in base because hydrolysis of S2 will be repressed. 918 Chapter 18: Solubility and ComplexIon Equilibria 47. (E) We determine Mg 2+ in the solution. Mg 2+ = 0.65 g Mg OH 1 L soln b g 2 b g 58.3 g MgbOH g
1 mol Mg OH 2 2 1 mol Mg 2+ 1 mol Mg OH b g = 0.011 M
2 Then we determine OH in the solution, and its pH.
K sp Mg 2 OH 1.8 1011 0.011 OH ; OH 2 2 1.8 1011 4.0 105 M 0.011 pOH = log 4.0 105 = 4.40
48. pH = 14.00 4.40 = 9.60 (M) First we determine the [Mg2+] and [NH3] that result from dilution to a total volume of 0.500 L. 0.150 Linitial Mg 2+ = 0.100 M = 0.0300 M; 0.500 Lfinal NH3 = 0.150 M 0.350 Linitial = 0.105 M 0.500 L final Then determine the OH that will allow Mg 2+ = 0.0300 M in this solution.
K sp 1.8 10 11 = Mg 2+ OH = 0.0300 OH ; OH = 2 2 1.8 10 11 0.0300 = 2.4 10 5 M This OH is maintained by the NH3 / NH + buffer; since it is a buffer, we can use the 4 HendersonHasselbalch equation to find the [NH4+].
pH = 14.00 pOH = 14.00 + log 2.4 105 = 9.38 = pK a + log NH3 = 9.38 9.26 = +0.12; log NH 4 + NH3 = 10+0.12 = 1.3; NH 4 + + NH3 NH 4 + = 9.26 + log NH3 NH 4 + NH 4 + = 0.105 M NH3 = 0.081 M 1.3 = 2.7 g mass NH 4 2 SO 4 = 0.500 L 0.081 mol NH 4 L soln 1 mol NH 4 2 SO 4 2 mol NH 4
+ 132.1 g NH 4 2 SO 4 1 mol NH 4 2 SO 4 49. (M) (a) Here we calculate OH needed for precipitation. K sp [Al3+ ][OH ]3 1.3 1033 (0.075 M)[OH ]3 [OH ] 1.3 1033 2.6 1011 pOH = log 2.6 1011 = 10.59 0.075 pH = 14.00 10.59 = 3.41
3 919 Chapter 18: Solubility and ComplexIon Equilibria (b) We can use the HendersonHasselbalch equation to determine C 2 H 3O 2 .
pH = 3.41 = pKa + log L O = 4.74 + log 1.00 M NM HC2 H 3O 2 QP C 2 H 3O 2 C 2 H 3O 2 C2 H3O2 C H O = 3.41 4.74 = 1.33; 2 3 2 = 101.33 = 0.047; C H O = 0.047 M log 2 3 2 1.00 M 1.00 M This situation does not quite obey the guideline that the ratio of concentrations must fall in the range 0.10 to 10.0, but the resulting error is a small one in this circumstance.
mass NaC2 H3O2 = 0.2500 L 0.047 mol C2 H3O 2 1 mol NaC2 H3O 2 82.03 g NaC2 H3O2 1 L soln 1 mol NaC2 H3O2 1 mol C2 H3O2 0.96 g NaC2 H3O2 50. (D) (a) Since HI is a strong acid, I = 1.05 103 M +1.05 103 M = 2.10 103 M. We determine the value of the ion product and compare it to the solubility product constant value.
Qsp = Pb 2+ I = 1.1103 2.10 103 = 4.9 109 7.1109 = K sp for PbI 2 2 2 Thus a precipitate of PbI 2 will not form under these conditions.
(b) We compute the OH needed for precipitation. Ksp Mg 2 OH 1.8 1011 0.0150 OH ; OH 2 2 1.8 1011 3.5 105 M 0.0150 Then we compute OH in this solution, resulting from the ionization of NH3. 0.05 103 L = 1.2 104 M NH3 = 6.00 M 2.50 L Even though NH 3 is a weak base, the OH produced from the NH 3 hydrolysis reaction will approximate 4 105 M (3.85 if you solve the quadratic) in this very dilute solution. (Recall that degree of ionization is high in dilute solution.) And since OH = 3.5 105 M is needed for precipitation to occur, we conclude that Mg OH this solution (note: not much of a precipitate is expected).
(c) b g 2 will precipitate from 0.010 M HC2 H 3O 2 and 0.010 M NaC2 H 3O 2 is a buffer solution with pH = pKa of acetic acid (the acid and its anion are present in equal concentrations.) From this, we determine the OH . 920 Chapter 18: Solubility and ComplexIon Equilibria pH = 4.74 pOH = 14.00 4.74 = 9.26
3 3 OH = 109.26 = 5.5 1010 Q = Al3+ OH = 0.010 5.5 10 10 = 1.7 10 30 1.3 10 33 = K sp of Al OH 3 Thus, Al OH b g bsg should precipitate from this solution.
3 ComplexIon Equilibria
51. (E) Lead(II) ion forms a complex ion with chloride ion. It forms no such complex ion with nitrate ion. The formation of this complex ion decreases the concentrations of free Pb 2+ aq and free Cl aq . Thus, PbCl 2 will dissolve in the HCl(aq) up until the value of the solubility product is exceeded. Pb 2+ aq + 3Cl aq PbCl3 aq b g b g 52. (E) Zn 2+ aq + 4NH 3 aq Zn NH 3 4 2+ NH 3 aq will be least effective in reducing the concentration of the complex ion. In fact, the addition of NH 3 aq will increase the concentration of the complex ion by favoring a shift of b g aq K f = 4.1108 b g the equilibrium to the right. NH 4 + aq will have a similar effect, but not as direct. NH 3 aq is formed by the hydrolysis of NH 4 +
+ b g cause the greatest decrease in the concentration of the complex ion. HCl(aq) will react with NH baq g to decrease its concentration (by forming NH ) and this will cause the complex ion NH 3 aq : NH 4 aq + H 2 O(l) NH 3 aq + H 3O + aq . The addition of HCl(aq) will
+ 3 4 b g baqg and, thus, increasing b g NH 4 + will eventually increase equilibrium reaction to shift left toward free aqueous ammonia and Zn2+(aq).
53. (E) We substitute the given concentrations directly into the Kf expression to calculate Kf.
Kf [[Cu(CN) 43] 0.0500 2.0 1030 4 + [Cu ][CN ] (6.1 1032 )(0.80) 4 54. (M) The solution to this problem is organized around the balanced chemical equation. Free NH 3 is 6.0 M at equilibrium. The size of the equilibrium constant indicates that most copper(II) is present as the complex ion at equilibrium. 2+ Equation: Cu 2+ aq + 4NH 3 aq Cu NH 3 4 aq b g b g Initial: 0.10 M Change(100 % rxn): 0.10 M Completion: 0M Changes: +x M Equil: x M 6.00 M 0.40 M 5.60 M +4x M 5.60 + 4x M b 0M +0.10 M 0.10 M x M 0.10 x M g 921 Chapter 18: Solubility and ComplexIon Equilibria Let's assume (5.60 + 4x) M 5.60 M and (0.10 x) M 0.10 M Cu NH 2+ 3 4 0.10 x 0.10 0.10 = 1.1 1013 Kf = 4 4 4 2+ (5.60) x 983.4x Cu NH 3 x 5.60 4 x x 0.10 9.2 1018 M Cu 2 983.4 (1.1 1013 ) ( x << 0.10, thus the approximation was valid) 55. (M) We first find the concentration of free metal ion. Then we determine the value of Qsp for the precipitation reaction, and compare its value with the value of Ksp to determine whether precipitation should occur. Equation: Initial: Changes: Equil: Ag + aq + 0M +x M x M b g 2 S2 O3 2 aq Ag S2 O 3 b g baqg
3 2 0.76 M +2 x M 0.76 + 2 x M 0.048 M x M 0.048 x M Ag S O 3 2 3 2 = 1.7 1013 = 0.048 x 0.048 ; x = 4.9 1015 M = Ag + Kf = 2 2 2 (0.76) 2 x x 0.76 + 2 x Ag + S2 O3 (x << 0.048 M, thus the approximation was valid.) Qsp = Ag + I = 4.9 1015 2.0 = 9.8 1015 8.5 1017 = K sp . Because Qsp K sp , precipitation of AgI(s) should occur.
56. (M) We need to determine OH in this solution, and also the free Cu 2+ .
pH = pK a + log NH3 NH 4 + = 9.26 + log 0.10 M = 9.26 pOH = 14.00 9.26 = 4.74 0.10 M OH = 104.74 = 1.8 105 M Cu 2+ aq + 4NH 3 aq Cu NH 3 4 2+ aq Cu NH 3 2+ 4 0.015 0.015 = 1.1 1013 = 11 Kf = ; Cu 2+ = 4 2+ 4 1.1 1013 0.10 4 = 1.4 10 M 2+ Cu 0.10 Cu NH 3 Now we determine the value of Qsp and compare it with the value of Ksp for Cu OH 2 .
2 5 2 b g Qsp = Cu 2+ OH = 1.41011 1.810 = 4.51021 < 2.21020 (K sp for Cu OH 2 ) Precipitation of Cu OH 2 s from this solution should not occur. b gbg 922 Chapter 18: Solubility and ComplexIon Equilibria 57. (M) We first compute the free Ag + in the original solution. The size of the complex ion formation equilibrium constant indicates that the reaction lies far to the right, so we form as much complex ion as possible stoichiometrically. Equation: In soln: Form complex: Changes: Equil:
7 Ag + aq + 0.10 M 0.10 M 0M +x M x M
+ b g 2 NH 3 aq b g g Ag NH 3 b g baqg
+ 2 b 1.00 M 0.20 M 0.80 M +2x M 0.80 + 2x M
x= b 0M +0.10 M 0.10 M x M 0.10 x M g Ag NH 3 2 0.10 x 0.10 = K f = 1.6 10 = 2 2 + 2 Ag NH3 x 0.80 + 2 x x 0.80 0.10 = 9.8 109 M . 2 7 1.6 10 0.80 (x << 0.80 M, thus the approximation was valid.) Thus, Ag + = 9.8 109 M. We next determine the I that can coexist in this solution without precipitation. 8.5 1017 + 17 9 9 I = K sp = Ag I = 8.5 10 = 9.8 10 I ; 9.8 109 = 8.7 10 M Finally, we determine the mass of KI needed to produce this I mass KI = 1.00 L soln 8.7 109 mol I 1mol KI 166.0 g KI = 1.4 106 g KI 1 L soln 1mol I 1mol KI 58. (M) First we determine Ag + that can exist with this Cl . We know that Cl will be unchanged because precipitation will not be allowed to occur. 1.8 1010 = 1.8 109 M 0.100 We now consider the complex ion equilibrium. If the complex ion's final concentration is x , then the decrease in NH 3 is 2x , because 2 mol NH 3 react to form each mole of complex K sp = Ag + Cl = 1.8 1010 = Ag + 0.100 M; Ag + = ion, as follows. Ag + aq + 2NH 3 aq Ag NH 3 2 aq We can solve the Kf + expression for x.
Kf = 1.6 107 = Ag NH 3 b g
hb + 2 2 x = 1.6 107 1.8 109 1.00 2 x c hc Ag + L NH 3 O PQ MN = 1.8 109 1.00 2 x g b x 2 = 0.029 1.00 4.00 x + 4.00 x 2 = 0.029 0.12 x + 0.12 x 2 c g 2 h 0 = 0.029 1.12 x + 0.12 x 2 We use the quadratic formula roots equation to solve for x. x= b b 2 4ac 1.12 (1.12) 2 4 0.029 0.12 1.12 1.114 = = = 9.3, 0.025 2a 2 0.12 0.24 923 Chapter 18: Solubility and ComplexIon Equilibria Thus, we can add 0.025 mol AgNO 3 (~ 4.4 g AgNO 3 ) to this solution before we see a precipitate of AgCl(s) form. Precipitation and Solubilities of Metal Sulfides
59. (M) We know that Kspa = 3 107 for MnS and Kspa = 6 102 for FeS. The metal sulfide will begin to precipitate when Qspa = Kspa . Let us determine the H 3O + just necessary to form each precipitate. We assume that the solution is saturated with H 2S, H 2S = 0.10 M . K spa [M 2 ][H 2S] [H 3O + ]2 [M 2 ][H 2S] (0.10 M)(0.10 M) 1.8 105 M for MnS [H 3O ] 7 3 10 K spa
+ H 3O + = (0.10 M)(0.10 M) 4.1103 M for FeS 2 6 10 Thus, if the [H3O+] is maintained just a bit higher than 1.8 105 M , FeS will precipitate and Mn 2+ aq will remain in solution. To determine if the separation is complete, we see whether b g Fe 2+ has decreased to 0.1% or less of its original value when the solution is held at the aforementioned acidity. Let H 3O + = 2.0 105 M and calculate Fe 2+ . K spa Fe 2+ H 2S Fe 2+ 0.10 M 6 102 2.0 105 = 2.4 106 M = = 6 102 = ; Fe2+ = 2 2 0.10 H 3O + 2.0 105 M 2 2.4 106 M % Fe aq remaining = 100% = 0.0024% 0.10 M
2+ Separation is complete. 60. (M) Since the cation concentrations are identical, the value of Qspa is the same for each one. It is this value of Qspa that we compare with Kspa to determine if precipitation occurs.
Qspa M 2+ H 2S 0.05 M 0.10 M = = = 5 101 2 + 2 0.010 M H 3O If Qspa Kspa , precipitation of the metal sulfide should occur. But, if Qspa Kspa , precipitation will not occur. For CuS, Kspa = 6 10 16 Qspa = 5 101 For HgS, Kspa = 2 1032 Qspa = 5 101 For MnS, Kspa = 3 107 Qspa = 5 101 Precipitation of CuS(s) should occur. Precipitation of HgS(s) should occur. Precipitation of MnS(s) will not occur. 924 Chapter 18: Solubility and ComplexIon Equilibria 61. (M) (a) We can calculate H 3O + in the buffer with the HendersonHasselbalch equation. C2 H 3O 2 = 4.74 + log 0.15 M = 4.52 H O + = 104.52 = 3.0 10 5 M pH = pK a + log 3 HC H O 0.25 M 2 3 2 We use this information to calculate a value of Qspa for MnS in this solution and then comparison of Qspa with Kspa will allow us to decide if a precipitate will form.
Qspa Mn 2+ H 2S 0.15 0.10 = = = 1.7 107 3 107 = K spa for MnS + 2 5 2 H 3O 3.0 10 Thus, precipitation of MnS(s) will not occur.
(b) We need to change H 3O + so that
Qspa = 3 107 = 0.15 0.10 H3 O + 2 ; H3 O+ = (0.15) 0.10 3 10
7 [H 3 O + ] = 2.2 105 M pH = 4.66 This is a more basic solution, which we can produce by increasing the basic component of the buffer solution, namely, the acetate ion. We can find out the necessary acetate ion concentration with the HendersonHasselbalch equation. [C 2 H 3O ] [C2 H 3O 2 ] 2 4.66 4.74 log pH pK a log [HC2 H 3O 2 ] 0.25 M log [C2 H 3O ] 2 4.66 4.74 0.08 0.25 M
C 2 H 3O 2 = 0.83 0.25 M = 0.21M C 2 H 3O 2 = 100.08 = 0.83 0.25 M 62. (M) (a) CuS is in the hydrogen sulfide group of qualitative analysis. Its precipitation occurs when 0.3 M HCl is saturated with H 2S. It will certainly precipitate from a (nonacidic) saturated solution of H2S which has a much higher S2 .
Cu 2+ aq + H 2S satd aq CuS s + 2H + aq This reaction proceeds to an essentially quantitative extent in the forward direction.
(b) (c) MgS is soluble, according to the solubility rules listed in Chapter 5. 0.3 M HCl Mg 2+ aq + H 2S satd aq no reaction As in part (a), PbS is in the qualitative analysis hydrogen sulfide group, which precipitates from a 0.3 M HCl solution saturated with H 2S. Therefore, PbS does not dissolve appreciably in 0.3 M HCl. PbS s + HCl 0.3M no reaction 925 Chapter 18: Solubility and ComplexIon Equilibria (d) Since ZnS(s) does not precipitate in the hydrogen sulfide group, we conclude that it is soluble in acidic solution. ZnS s + 2HNO3 aq Zn NO3 2 aq + H 2S g Qualitative Cation Analysis
63. (E) The purpose of adding hot water is to separate Pb 2+ from AgCl and Hg 2 Cl2 . Thus, the most important consequence would be the absence of a valid test for the presence or absence of Pb 2+ . In addition, if we add NH 3 first, PbCl 2 may form Pb OH 2 . If Pb OH 2 does form, it will be present with Hg 2 Cl 2 in the solid, although Pb OH added NH 3 . Thus, we might falsely conclude that Ag is present.
64.
+ b g b g 2 will not darken with (M) For PbCl2 aq , 2 Pb 2+ = Cl where s = molar solubility of PbCl 2 . 2+ Thus s = Pb . Ksp [Pb 2+ ][Cl ]2 ( s )(2s ) 2 4s 3 1.6 105 ; s 1.6 105 4 1.6 102 M = [Pb 2+ ]
3 Both Pb 2+ and CrO 4 2 are diluted by mixing the two solutions. 1.00 mL Pb 2+ = 0.016 M = 0.015 M 1.05 mL 0.05 mL CrO 4 2 = 1.0 M = 0.048 M 1.05 mL 2 Qsp = Pb 2+ CrO 4 = 0.015 M 0.048 M = 7.2 104 2.8 1013 = K sp Thus, precipitation should occur from the solution described. 65. (E) (a) (b) (c) (d) Ag + and/or Hg 2 are probably present. Both of these cations form chloride precipitates from acidic solutions of chloride ion. We cannot tell whether Mg 2+ is present or not. Both MgS and MgCl 2 are water soluble. Pb 2+ possibly is absent; it is the only cation of those given which forms a precipitate in an acidic solution that is treated with H 2S, and no sulfide precipitate was formed. We cannot tell whether Fe2+ is present. FeS will not precipitate from an acidic solution that is treated with H 2S; the solution must be alkaline for a FeS precipitate to form. 2+ (a) and (c) are the valid conclusions. 66. (E) (a) (b) (c)
Pb 2+ aq + 2Cl aq PbCl2 s Zn OH 2 s + 2OH aq Zn OH 4 2 aq 3+ Fe OH 3 s + 3H 3O + aq Fe3+ aq + 6H 2 O(l) or Fe H 2O 6 aq 926 Chapter 18: Solubility and ComplexIon Equilibria (d) Cu 2+ aq + H 2S aq CuS s + 2H + aq INTEGRATIVE AND ADVANCED EXERCISES
67. (M) We determine s, the solubility of CaSO4 in a saturated solution, and then the concentration of CaSO4 in ppm in this saturated solution, assuming that the solution's density is 1.00 g/mL. Ksp [Ca 2 ][SO 4 2 ] ( s )( s ) s 2 9.1 106 s 3.0 103 M
ppm CaSO 4 106 g soln 1 mL 1 L soln 0.0030 mol CaSO4 136.1 g CaSO 4 4.1 102 ppm 1.00 g 1000 mL 1 L soln 1 mol CaSO 4 Now we determine the volume of solution remaining after we evaporate the 131 ppm CaSO4 down to a saturated solution (assuming that both solutions have a density of 1.00 g/mL.) 10 6 g sat' d soln 1 mL volume sat' d soln 131 g CaSO 4 3.2 10 5 mL 2 4.1 10 g CaSO 4 1.00 g Thus, we must evaporate 6.8 105 mL of the original 1.000 106 mL of solution, or 68% of the water sample. 68. (M) (a) First we compute the molar solubility, s, of CaHPO42H2O. 0.32 g CaHPO 4 2H 2 O 1 mol CaHPO 4 2H 2 O s 1.9 103 M 1 L soln 172.1 g CaHPO 4 2H 2 O K sp [Ca 2 ][HPO 4 2 ] ( s )( s ) s 2 (1.9 103 ) 2 3.6 106 Thus the values are not quite consistent. (b) The value of Ksp given in the problem is consistent with a smaller value of the molar solubility. The reason is that not all of the solute ends up in solution as HPO42 ions. HPO42 can act as either an acid or a base in water (see below), but it is base hydrolysis that predominates. . HPO 4 2 (aq) H 2 O(l) H 3O (aq) PO43 (aq) K a 3 4.2 1013 HPO 4 2 (aq) H 2 O(l) OH (aq) H 2 PO 4 (aq) Kb Kw 1.6 10 7 K2 69. (M) The solutions mutually dilute each other and, because the volumes are equal, the concentrations are halved in the final solution: [Ca2+] = 0.00625 M, [SO42] = 0.00760 M. We cannot assume that either concentration remains constant during the precipitation. Instead, we assume that precipitation proceeds until all of one reagent is used up. Equilibrium is reached from that point. 927 Chapter 18: Solubility and ComplexIon Equilibria Equation: CaSO 4 (s) In soln Form ppt Not at equil Changes Equil: Ca 2 (aq) 0.00625 M 0.00625 M 0M xM xM SO 4 2 (aq) K sp 9.1 10 6 0.00760 M K sp [Ca 2 ][SO 4 2 ] K sp 0.00135 x 0.00135 {not a reasonable assumption!} x 9.1 106 6.7 103 M 0.00625 M K sp ( x )(0.00135 x) 0.00135 M xM (0.00135 x ) M Solving the quadratic 0 = x2 + (1.35 102)x 9.1 106 yields x = 2.4 103. 2.4103 M final % unprecipitated Ca = 100% = 38% unprecipitated 0.00625 M initial
2+ 70. (M) If equal volumes are mixed, each concentration is reduced to onehalf of its initial value. We assume that all of the limiting ion forms a precipitate, and then equilibrium is achieved by the reaction proceeding in the forward direction to a small extent. BaCO3 (s) Ba 2 (aq) Orig. soln: 0.00050 M Form ppt: 0M
Equation: Changes: Equil: CO32 (aq) 0.0010 M 0.0005 M xM xM xM
(0.0005 x)M K sp [Ba 2 ][CO32 ] 5.1109 (x)(0.0005 x) 0.0005 x 5.1 109 1 105 M The [Ba 2+ ] decreases from 5 104 M to 1 105 M. 0.0005 (x < 0.0005 M, thus the approximation was valid.) x % unprecipitated 1 10 5 M 100% 2% 5 10 4 M Thus, 98% of the Ba2+ is precipitated as BaCO3(s). 71. (M) The pH of the buffer establishes [H3O+] = 10pH = 103.00 = 1.0 103 M. Now we combine the two equilibrium expressions, and solve the resulting expression for the molar solubility of Pb(N3)2 in the buffer solution. 928 Chapter 18: Solubility and ComplexIon Equilibria Pb(N 3 ) 2 (s) Pb 2 (aq) 2 N 3 (aq) 2 H 3O (aq) 2 N 3 (aq) 2 HN 3 (aq) 2 H 2O(l)
1/K a 2 K sp 2.5 10 9 1 2.8 109 5 2 (1.9 10 ) Pb(N 3 ) 2 (s) 2 H 3O (aq) Pb 2 (aq) 2 HN 3 (aq) 2 H 2 O (l) Pb(N 3 ) 2 (s) 2 H 3O (aq) Buffer: Dissolving: Equilibrium: K [H 3O ]2 0.0010 M (buffer) 0.0010 M 7.0 (x) (2 x) 2 (0.0010) 2 K K sp (1/ K a 2 ) 7.0 Pb 2 (aq) 2 HN 3 (aq) 2 H 2 O (l) 0M x M xM 0M 2x M 2x M x
3 0.012 M [Pb 2 ][HN 3 ]2 4x 3 7.0 (0.0010) 2 7.0(0.0010) 2 4 Thus, the molar solubility of Pb(N3)2 in a pH = 3.00 buffer is 0.012 M.
72. (M) We first determine the equilibrium constant of the suggested reaction. Mg(OH)2 (s) Mg 2 (aq) 2 OH (aq) K sp 1.8 1011 2 NH 4 (aq) 2 OH (aq) 2 NH3 (aq) 2 H 2 O(l) 1/K b 2 1/(1.8 105 )2 Mg(OH)2 (s) 2 NH 4 (aq) Mg 2 (aq) 2 NH3 (aq) 2 H 2 O(l) Initial: 1.00 M 0M 0M Changes: Equil: K x
3 2 x M (1.00 2 x) M x M xM 2x M 2x M K sp Kb2 [Mg 2 ][NH 3 ]2 x(2 x) 2 1.8 1011 4 x3 0.056 (1.8 105 ) 2 [NH 4 ]2 (1.00 2 x) 2 1.002 0.056 0.24 M 4 Take this as a first approximation and cycle through again. 2x = 0.48
4x 3 K 0.056 (1.00 0.48) 2
x
3 0.056 (1.00 0.48) 2 0.16 M 4 Yet another cycle gives a somewhat more consistent value. 2x = 0.32
K 4x3 0.056 (1.00 0.32) 2 x 3 0.056 (1.00 0.38) 2 0.19 M 4 Another cycle gives more consistency. 2x = 0.38 929 Chapter 18: Solubility and ComplexIon Equilibria K 4x3 0.056 (1.00 0.38) 2 x 3 0.056 (1.00 0.38) 2 0.18 M [Mg 2 ] 4 The molar solubility of Mg(OH)2 in a 1.00 M NH4Cl solution is 0.18 M.
73. (D) First we determine the value of the equilibrium constant for the cited reaction. CaCO3 (s) Ca 2 (aq) CO32 (aq) H 3 O (aq) CO32 (aq) H 2 O(l) HCO3 (aq) CaCO3 (aq) H 3 O (aq) Ca 2 (aq) HCO3 (aq) H 2 O(l) Ksp 2.8 109 1/K a 2 1/ 4.7 1011 K overall K sp 1/K a 2 K overall 2.8 109 60 4.7 1011 Equation: CaCO3 (s) H 3O (aq) Ca 2 (aq) HCO3 (aq) H 2 O(l) Initial: Change: Equil: 10 pH (buffer ) 10 pH 0M x M xM 0M x M xM In the above setup, we have assumed that the pH of the solution did not change because of the dissolving of the CaCO3(s). That is, we have treated the rainwater as if it were a buffer solution.
(a) K [Ca 2 ][HCO 3 ] [H 3 O ] 60. x2 3 10 6 x 60 3 10 6 1 10 2 M [Ca 2 ] [Ca 2 ][HCO3 ] x2 (b) K 60. [H 3 O ] 6.3 105 x 60 6.3 105 6.1 102 M [Ca 2 ] 74. We use the HendersonHasselbalch equation to determine [H3O+] in this solution, and then use the Ksp expression for MnS to determine [Mn2+] that can exist in this solution without precipitation occurring. 930 Chapter 18: Solubility and ComplexIon Equilibria [C2 H 3 O 2 ] 0.500 M pH pK a log 4.74 log 4.74 0.70 5.44 [HC2 H 3 O 2 ] 0.100 M [H 3 O ] 105.44 3.6 106 M MnS(s) 2 H 3 O (aq) Mn 2 (aq) H 2 S(aq) 2 H 2 O(l) Note that [H 2 S] [Mn 2 ] s, the molar solubility of MnS. K spa [Mn 2 ][H 2 S] s2 7 3 10 [H 3 O ]2 (3.6 106 M) 2 s 0.02 M K spa 3 107 mass MnS/L 0.02 mol Mn 2 1 mol MnS 87 g MnS 2 g MnS/L 1 L soln 1 mol Mn 2 1 mol MnS 75. (M) (a) The precipitate is likely Ca3(PO4)2. Let us determine the %Ca of this compound (by mass) %Ca 3 40.078 g Ca 100% 38.763% Ca 310.18 g Ca 3 (PO 4 ) 2
2 3 Ca 2 (aq) 2 HPO 4 (aq) Ca 3 (PO 4 ) 2 (s) 2 H (aq) (b) The bubbles are CO 2 (g) : CO 2 (g) H 2 O(l) "H 2 CO 3 (aq)" precipitation redissolving Ca 2 (aq) H 2 CO3 (aq) CaCO3 (s) 2 H (aq) CaCO3 (s) + H 2 CO3 (aq) Ca 2+ (aq) + 2 HCO3 (aq) 76. (D) (a) A solution of CO2(aq) has [CO32 ] = Ka[H2CO3] = 5.6 1011. Since Ksp = 2.8 109 for CaCO3, the [Ca2+] needed to form a precipitate from this solution can be computed. 2.8109 [Ca ] = = = 50.M [CO32 ] 5.61011
2+ K sp This is too high to reach by dissolving CaCl2 in solution. The reason why the technique works is because the OH(aq) produced by Ca(OH)2 neutralizes some of the HCO3(aq) from the ionization of CO2(aq), thereby increasing the [CO32] above a value of 5.6 1011.
(b) The equation for redissolving can be obtained by combining several equations. 931 Chapter 18: Solubility and ComplexIon Equilibria CaCO3 (s) Ca 2+ (aq) + CO32 (aq) CO32 (aq) + H 3O + (aq) HCO3 (aq) + H 2O(l) CO 2 (aq) + 2 H 2O(l) HCO3 (aq) +H 3O + (aq) CaCO3 (s) + CO 2 (aq) + H 2 O(l) Ca 2+ (aq) + 2 HCO3 (aq)
9 7 2+  2 K sp = 2.8109 1 1 = K a 2 5.61011 K a1 = 4.2107 K= K sp K a1 Ka 2 [Ca ][HCO3 ] (2.810 )( 4.210 ) = 2.1105 = 11 5.610 [CO 2 ] 2+ If CaCO3 is precipitated from 0.005 M Ca (aq) and then redissolved, [Ca2+] = 0.005 M and [HCO32] = 20.005 M = 0.010 M. We use these values in the above expression to compute [CO2]. [Ca 2+ ][HCO3 ]2 (0.005)(0.010) 2 [CO 2 ]= = = 0.02 M 2.1105 2.1105 We repeat the calculation for saturated Ca(OH)2, in which [OH] = 2 [Ca2+], after first determining [Ca2+] in this solution. K= Ksp = [Ca2+][OH]2 = 4 [Ca2+] = 5.5106 [Ca2+]
3 5.5 106 = 0.011 4 [Ca 2+ ][HCO3 ]2 (0.011)(0.022) 2 = = 0.25 M M [CO 2 ]= 2.1105 2.1105 Thus, to redissolve the CaCO3 requires that [CO2] = 0.25 M if the solution initially is saturated Ca(OH)2, but only 0.02 M CO2 if the solution initially is 0.005 M Ca(OH)2(aq). A handbook lists the solubility of CO2 as 0.034 M. Clearly the CaCO3 produced cannot redissolve if the solution was initially saturated with Ca(OH)2(aq).
77. (M) (a) BaSO 4 (s) Ba 2 (aq) SO 4 2 (aq) Ba 2 (aq) CO 3 2 (aq) BaCO 3 (s) BaCO 3 (s) + SO 4 2 (aq) Sum BaSO 4 (s) +CO 3 2 (aq) Initial Equil.
K
K sp 1.1 1010 1 1 = Ksp 5.1 109
K overall 1.1 1010 5.1 109 0.0216 3M (3x )M 0 x
Since 0.063 M 0.050 M, the response is yes. (where x is the carbonate used up in the reaction)
[SO 4 2 ] x 0.0216 and x 0.063 2[CO 3 ] 3 x 932 Chapter 18: Solubility and ComplexIon Equilibria (b) 2AgCl(s) 2Ag + (aq) + 2 Cl (aq) 2 Ag + (aq) + CO32 (aq) Ag 2 CO3 (s) Sum 2AgCl(s) +CO32 (aq) Ag 2 CO3 (s) + 2 Cl (aq) K sp = (1.8 1010 ) 2 1/ K sp = K overall = 1 8.5 1012 (1.8 1010 )2 =3.8 109 8.5 1012 Initial 3M 0M Equil. (3x )M 2x (where x is the carbonate used up in the reaction) K overall = [Cl ]2 (2x) 2 = = 3.8 109 and x = 5.3 105 M 2[CO3 ] 3x Since 2x , (2(5.35 105 M)) , << 0.050 M, the response is no.
(c) MgF2 (s) Mg 2+ (aq) 2 F (aq) Mg 2+ (aq) CO32 (aq) MgCO3 (s) sum initial equil. MgF2 (s) CO32 (aq) MgCO3 (s) 2 F (aq) 3M 0M 2x K sp 3.7 108 1/ K sp K overall 1 3.5 108 3.7 108 1.1 3.5 108 (3x )M (where x is the carbonate used up in the reaction) K overall [F ]2 (2x) 2 1.1 and x 0.769M [CO32 ] 3 x Since 2x , (2(0.769M)) 0.050 M, the response is yes. 78. (D) For ease of calculation, let us assume that 100 mL, that is, 0.100 L of solution, is to be titrated. We also assume that [Ag+] = 0.10 M in the titrant. In order to precipitate 99.9% of the Br as AgBr, the following volume of titrant must be added.
volume titrant 0.999 100 mL 0.010 mmol Br 1 mmol Ag 1 mL sample 1 mmol Br 1 mL titrant 9.99 mL 0.10 mmol Ag We compute [Ag+] when precipitation is complete.
[Ag ] K sp [Br ] f 5.0 10 13 5.0 10 8 M 0.001 0.01 M Thus, during the course of the precipitation of AgBr, while 9.99 mL of titrant is added, [Ag+] increases from zero to 5.0 108 M. [Ag+] increases from 5.0 108 M to 1.0 105 M from the point where AgBr is completely precipitated to the point where Ag2CrO4 begins to precipitate. Let us assume, for the purpose of making an initial estimate, that this increase in [Ag+] occurs while the total volume of the solution is 110 mL (the original 100 mL plus 10 mL 933 Chapter 18: Solubility and ComplexIon Equilibria of titrant needed to precipitate all the AgBr.) The amount of Ag+ added during this increase is the difference between the amount present just after AgBr is completely precipitated and the amount present when Ag2CrO4 begins to precipitate. 1.0 10 5 mmol Ag 5.0 10 8 mmol Ag 110 mL amount Ag added 110 mL 1 mL soln 1 mL soln 3 1.1 10 mmol Ag 1 mL titrant volume added titrant 1.1 10 3 mmol Ag 1.1 10 2 mL titrant 0.10 mmol Ag We see that the total volume of solution remains essentially constant. Note that [Ag+] has increased by more than two powers of ten while the volume of solution increased a very small degree; this is indeed a very rapid rise in [Ag+], as stated.
79. (D) The chemistry of aluminum is complex, however, we can make the following assumptions. Consider the following reactions at various pH: pH > 7 overall f sp Al(OH)3(s) + OH Al(OH)4(aq) K K K = 1.43 sp pH < 7 Al(OH)3(s) Al3+(aq) + 3 OH K = 1.31033 For the solubilities in basic solutions, consider the following equilibrium at pH = 13.00 and 11.00. At pH = 13.00: Initial : Equilibrium: Kf = 1.43 =
ov f sp Al(OH)3(s) + OH Al(OH)4(aq) K K K = 1.43 0.10 x 0.10 x 0 +x x Change: x x (buffered at pH = 13.00) x = 0.143 M 0.10 x 0.10 Kf = 1.43 =
x x (buffered at pH = 11) x = 0.00143 M 0.0010 x 0.0010 At pH = 11.00: Thus, in these two calculations we see that the formation of the aluminate ion [Al(OH)4] is favored at high pH values. Note that the concentration of aluminum(III) increases by a factor of onehundred when the pH is increased from 11.00 to 13.00, and it will increase further still above pH 13. Now consider the solubility equilibrium established at pH = 3.00, 4.00, and 5.00. These correspond to [OH] = 1.0 1011 M, 1.0 1010 M, and 1.0 109 M, respectively.
sp Al(OH)3(s) Al3+(aq) + 3 OH(aq) K = 1.31033 At these pH values, [Al3+] in saturated aqueous solutions of Al(OH)3 are [Al3+] = 1.3 1033 /(1.0 1011)3 = 1.3 M at pH = 3.00 934 Chapter 18: Solubility and ComplexIon Equilibria [Al3+] = 1.3 1033 /(1.0 1010)3 = 1.3 103 M at pH = 4.00 [Al3+] = 1.3 1033 /(1.0 109)3 = 1.3 106 M at pH = 5.00 These three calculations show that in strongly acidic solutions the concentration of aluminum(III) can be very high and is certainly not limited by the precipitation of Al(OH)3(s). However, this solubility decreases rapidly as the pH increases beyond pH 4. At neutral pH 7, a number of equilibria would need to be considered simultaneously, including the selfionization of water itself. However, the results of just these five calculations do indicate that concentration of aluminum(III) increases at both low and high pH values, with a very low concentration at intermediate pH values around pH 7, thus aluminum(III) concentration as a function of pH is a Ushaped (or Vshaped) curve.
80. (M) We combine the solubility product expression for AgCN(s) with the formation expression for [Ag(NH3)2]+(aq). Solubility: Formation: AgCN(s) Ag (aq) CN (aq) Ag (aq) 2NH 3 (aq) [Ag(NH 3 ) 2 ] (aq) K sp ? K f 1.6 107 Net reaction: AgCN(s) 2 NH3 (aq) [Ag(NH3 )2 ] CN (aq) K overall K sp K f K overall K sp [[Ag(NH 3 ) 2 ] ][CN ] (8.8 106 ) 2 1.9 109 K sp 1.6 107 [NH 3 ]2 (0.200) 2 1.9 109 1.2 1016 1.6 107 Because of the extremely low solubility of AgCN in the solution, we assumed that [NH3] was not altered by the formation of the complex ion. 81. We combine the solubility product constant expression for CdCO3(s) with the formation expression for [CdI4]2(aq).
Solubility: CdCO3 (s) Cd 2 (aq) CO32 (aq) K sp 5.2 1012 Kf ? K overall K sp K f Formation: Cd 2 (aq) 4 I (aq) [CdI 4 ]2 (aq) Net reaction: CdCO3 (s) 4 I (aq) [CdI4 ]2 (aq) CO32 (aq) K overall Kf [[CdI 4 ]2 ][CO32 ] (1.2 103 ) 2 1.4 106 5.2 1012 K f 4 4 [I ] (1.00) 1.4 106 2.7 105 12 5.2 10 935 Chapter 18: Solubility and ComplexIon Equilibria Because of the low solubility of CdCO3 in this solution, we assumed that [I] was not appreciably lowered by the formation of the complex ion.
82. (M) We first determine [Pb2+] in this solution, which has [Cl] = 0.10 M. K sp 1.6 105 [Pb 2 ][Cl ]2 [Pb 2 ] K sp [Cl ]2 1.6 105 1.6 103 M 2 (0.10) Then we use this value of [Pb2+] in the Kf expression to determine [[PbCl3]].
Kf [[PbCl 3 ] ] [Pb 2 ][Cl ] 3 24 [[PbCl 3 ] ] (1.6 10 3 )(0.10) 3 [[PbCl 3 ] ] 1.6 10 6 [[PbCl 3 ] ] 24 1.6 10 6 3.8 10 5 M The solubility of PbCl2 in 0.10 M HCl is the following sum. solubility = [Pb2+] + [[PbCl3]] = 1.6 103 M + 3.8 105 M = 1.6 103 M 83. (M) Two of the three relationships needed to answer this question are the two solubility product 4.0 107 = [Pb2+][S2O32] expressions. 1.6 108 = [Pb2+][SO42] The third expression required is the electroneutrality equation, which states that the total positive charge concentration must equal the total negative charge concentration: [Pb2+] = [SO42] + [S2O32], provided [H3O+] = [OH]. Or, put another way, there is one Pb2+ ion in solution for each SO42 ion and for each S2O32 ion. We solve each of the first two expressions for the concentration of each anion, substitute these expressions into the electroneutrality expression, and then solve for [Pb2+].
[SO 4
2 1.6 108 ] [Pb 2 ] [S2 O3 2 4.0 107 ] [Pb 2 ] 1.6 108 4.0 107 [Pb ] [Pb 2 ] [Pb 2 ]
2 [Pb 2 ]2 1.6 108 4.0 107 4.2 107 [Pb 2 ] 4.2 107 6.5 104 M 84. (D) The chemical equations are: PbCl 2 (s) Pb 2 (aq) 2 Cl (aq) and PbBr2 (s) Pb 2 (aq) 2 Br (aq) The two solubility constant expressions, [Pb2+][Cl]2 = 1.6 105, [Pb2+][Br]2 = 4.0 105 and the condition of electroneutrality, 2 [Pb2+] = [Cl] + [Br], must be solved simultaneously to determine [Pb2+]. (Rearranging the electroneutrality relationship, one obtains: [Pb 2 ] 1 [Cl ] 1 [Br ] .) First we solve each of the solubility constant expressions for the 2 2 concentration of the anion. Then we substitute these values into the electroneutrality expression and solve the resulting equation for [Pb2+].
[Cl ] 1.6 10 5 [Pb 2 ] [Br ] 4.0 10 5 [Pb 2 ] 2 [Pb 2 ] 1.6 10 5 4.0 10 5 [Pb 2 ] [Pb 2 ] 2 [Pb 2 ]3 1.6 10 5 4.0 10 5 4.0 10 3 6.3 10 3 1.03 10 2 [Pb 2 ]3 5.2 10 3 [Pb 2 ]3 2.7 10 5 [Pb 2 ] 3.0 10 2 M 936 Chapter 18: Solubility and ComplexIon Equilibria 85. (D) (a) First let us determine if there is sufficient Ag2SO4 to produce a saturated solution, in which [Ag+] = 2 [SO42].
K sp 1.4 10 5 [Ag ] 2 [SO 4 ] 4 [SO 4 ]3 [SO 4 ] 3 [SO 4 ] 2 2 2 2 2 1.4 10 5 0.015 M 4 2.50 g Ag 2 SO 4 1 mol Ag 2 SO 4 1 mol SO 4 0.0535 M 0.150 L 311.8 g Ag 2 SO 4 1 mol Ag 2 SO 4 Thus, there is more than enough Ag2SO4 present to form a saturated solution. Let us now see if AgCl or BaSO4 will precipitate under these circumstances. [SO42] = 0.015 M and [Ag+] = 0.030 M. Q = [Ag+][Cl] = 0.030 0.050 = 1.5 103 > 1.8 1010 = Ksp, thus AgCl should precipitate. Q = [Ba2+][SO42] = 0.025 0.015 = 3.8 104 > 1.1 1010 = Ksp, thus BaSO4 should precipitate.
Thus, the net ionic equation for the reaction that will occur is as follows. Ag 2 SO 4 (s) Ba 2 (aq) 2 Cl (aq) BaSO 4 (s) 2 AgCl(s)
(b) Let us first determine if any Ag2SO4(s) remains or if it is all converted to BaSO4(s) and AgCl(s). Thus, we have to solve a limiting reagent problem.
amount Ag 2 SO 4 2.50 g Ag 2 SO 4 amount BaCl 2 0.150 L 1 mol Ag 2 SO 4 8.02 10 3 mol Ag 2 SO 4 311.8 g Ag 2 SO 4 0.025 mol BaCl 2 3.75 10 3 mol aCl 2 1 L soln Since the two reactants combine in a 1 mole to 1 mole stoichiometric ratio, BaCl2 is the limiting reagent. Since there must be some Ag2SO4 present and because Ag2SO4 is so much more soluble than either BaSO4 or AgCl, we assume that [Ag+] and [SO42] are determined by the solubility of Ag2SO4. They will have the same values as in a saturated solution of Ag2SO4. [SO42] = 0.015 M [Ag+] = 0.030 M We use these values and the appropriate Ksp values to determine [Ba2+] and [Cl]. [Ba 2 ] 1.1 10 10 7.3 10 9 M 0.015 [Cl ] 1.8 10 10 6.0 10 9 M 0.030 Since BaCl2 is the limiting reagent, we can use its amount to determine the masses of BaSO4 and AgCl. mass BaSO 4 0.00375 mol BaCl 2 mass AgCl 0.00375 mol BaCl 2 1 mol BaSO 4 233.4 g BaSO 4 0.875 g BaSO 4 1 mol BaCl 2 1 mol BaSO 4 2 mol AgCl 143.3 g AgCl 1.07 g AgCl 1 mol BaCl 2 1 mol AgCl 937 Chapter 18: Solubility and ComplexIon Equilibria The mass of unreacted Ag2SO4 is determined from the initial amount and the amount that reacts with BaCl2. 1 mol Ag 2 SO 4 mass Ag 2 SO4 0.00802 mol Ag 2 SO4 0.00375 mol BaCl2 1 mol BaCl2 mass Ag 2 SO4 1.33 g Ag 2 SO4 unreacted 311.8 g Ag 2 SO4 1 mol Ag 2 SO 4 Of course, there is some Ag2SO4 dissolved in solution. We compute its mass. mass dissolved Ag 2SO 4 0.150 L 0.0150 mol Ag 2SO 4 311.8 g Ag 2SO 4 0.702 g dissolved 1 L soln 1 mol Ag 2SO 4 mass Ag 2SO 4 (s) 1.33 g 0.702 g 0.63 g Ag 2SO 4 (s)
86. (M) To determine how much NaF must be added, determine what the solubility of BaF2 is considering that the final concentration of F must be 0.030 M. Ksp = 1106. BaF2 (s) Ba2+ (aq) + 0 +s s
2 2F (aq) 0 +2s 2s (0.030 M) K sp 1 106. s 0.030 s 9.0 104 ; Therefore, s 0.0011 M . Since s is 0.0011 M, the F contribution from BaF2 considering the common ion effect exerted by NaF is 0.0022 M. Therefore, the total number of moles of NaF in a 1 L solution required is therefore 0.0278 moles. To determine how much BaF2 precipitates, we must first determine what the value of s is for a saturated solution in absence of the common ion effect. That is, 2 K sp 1 106. s 2 s 4s 3 ; Therefore, s 0.0063 M . Since in absence of the common ion effect from NaF the solubility of BaF2 is 0.0063 M, and with NaF the solubility is 0.0011, the difference between the two (0.0063 0.0011 = 0.0052 M) has to be precipitated. The mass of the precipitate is therefore: mass of BaF2 0.0052 mol BaF2 175.3 g BaF2 0.91 g 1 L solution 1 mol BaF2 938 Chapter 18: Solubility and ComplexIon Equilibria FEATURE PROBLEMS
87. (M) Ca 2+ = SO 4
2 in the saturated solution. Let us first determine the amount of H 3O + in
8.25 mL base 0.0105 mmol NaOH 1 mmol H 3O + 10.00 mL sample 1 mL base 1 mmol NaOH the 100.0 mL diluted effluent. H 3O + aq + NaOH aq 2H 2O(l) + Na + aq mmol H 3O + = 100.0 mL = 0.866 mmol H 3O + aq Now we determine Ca 2+ in the original 25.00 mL sample, remembering that 2 H 3O + were produced for each Ca 2+ . Ca 2+ = 0.866 mmol H 3O + (aq) 25.00 mL
2 1mmol Ca 2 2 mmol H 3O = 0.0173 M 88. Ksp = Ca 2+ SO4 2 = 0.0173 = 3.0 104 ; the K sp for CaSO4 is 9.1106 in Appendix D. (D) (a) We assume that there is little of each ion present in solution at equilibrium (that this is a simple stoichiometric calculation). This is true because the K value for the titration reaction is very large. Ktitration = 1/Ksp(AgCl) = 5.6 109. We stop the titration when just enough silver ion has been added. Ag + aq + Cl aq AgCl s V = 100.0 mL = 8.32 mL
(b) 29.5 mg Cl 1 mmol Cl 1 mmol Ag + 1mL 1000 mL 35.45 mg Cl 1 mmol Cl 0.01000 mmol AgNO3 We first calculate the concentration of each ion as the consequence of dilution. Then we determine the [Ag+] from the value of Ksp. 8.32 mL added initial Ag + = 0.01000 M = 7.68 104 M 108.3mL final volume 29.5 mg Cl initial Cl = 1mmol Cl 100.0 mL taken 35.45 mg Cl = 7.68 104 M 1000 mL 108.3 mL final volume The slight excess of each ion will precipitate until the solubility constant is satisfied. Ag + = Cl = Ksp 18 10 10 13 105 M . .
(c) If we want Ag 2 CrO 4 to appear just when AgCl has completed precipitation, Ag + = 1.3 105 M . We determine CrO 4 2 from the Ksp expression. 939 Chapter 18: Solubility and ComplexIon Equilibria Ksp = Ag + CrO42 = 1.11012 = 1.3105 2 2 CrO 2 ; 4 12 CrO 2 = 1.110 = 0.0065 M 4 2 5 1.310 (d) If CrO 4 2 were greater than the answer just computed for part (c), Ag 2 CrO 4 would appear before all Cl had precipitated, leading to a false early endpoint. We would calculate a falsely low Cl for the original solution. If CrO 4 2 were less than computed in part 3, Ag 2 CrO 4 would appear somewhat after all Cl had precipitated, leading one to conclude there was more Cl in solution than actually was the case. (e) If it was Ag + that was being titrated, it would react immediately with the CrO 4 in the sample, forming a redorange precipitate. This precipitate would not likely dissolve, and thus, very little if any AgCl would form. There would be no visual indication of the endpoint. 2 89. (D) (a) We need to calculate the Mg 2+ in a solution that is saturated with Mg(OH)2. K sp = 1.8 1011 = Mg 2+ OH = s 2s = 4 s 3 2 2 3 s=
(b) 1.8 1011 1.7 104 M = [Mg 2+ ] 4 Even though water has been added to the original solution, it remains saturated (it is in equilibrium with the undissolved solid Mg OH 2 ). Mg 2+ = 1.7 104 M. Although HCl(aq) reacts with OH , it will not react with Mg 2+ . The solution is simply a more dilute solution of Mg2+. 100.0 mL initial volume Mg 2+ = 1.7 104 M = 2.8 105 M 100.0 + 500. mL final volume In this instance, we have a dual dilution to a 275.0 mL total volume, followed by a commonion scenario. 1.7 104 mmol Mg 2+ 0.065 mmol Mg 2+ 25.00 mL 250.0 mL 1 mL 1 mL initial Mg 2+ = 275.0 mL total volume 0.059 M (c) (d) 25.00 mL initial OH = 1.7 104 mmol Mg 2+ 2 mmol OH 1 mL 1 mmol Mg 2+ 3.1105 M 275.0 mL total volume 940 Chapter 18: Solubility and ComplexIon Equilibria Let's see if precipitation occurs. Qsp = Mg 2+ OH 2 = 0.059 3.1 105 b gc h 2 = 5.7 1011 1.8 1011 = Ksp Thus, precipitation does occur, but very little precipitate forms. If OH goes down by 1.4 105 M (which means that Mg 2+ drops by 0.7 105 M ), then OH = 1.7 105 M and Mg 2+ = 0.059 M 0.7 105 M = 0.059 M , then Qsp Ksp and precipitation will stop. Thus, Mg 2+ = 0.059 M . (e) Again we have a dual dilution, now to a 200.0 mL final volume, followed by a commonion scenario. 50.00 mL initial Mg 2+ = 1.7 104 mmol Mg 2+ 1 mL 4.3 105 M 200.0 mL total volume 150.0 mL initial volume 0.113 M initial OH = 0.150 M 200.0 mL total volume Now it is evident that precipitation will occur. Next we determine the Mg 2+ that can exist in solution with 0.113 M OH . It is clear that Mg 2+ will drop dramatically to satisfy the Ksp expression but the larger value of OH will scarcely be affected. K sp = Mg 2+ OH = 1.8 1011 = Mg 2+ 0.0113 M 11 1.8 10 Mg 2+ = = 1.4 109 M 2 0.113
2 2 941 Chapter 18: Solubility and ComplexIon Equilibria SELFASSESSMENT EXERCISES
90. (E) (a) Ksp: The solubility product constant, which is the constant for the equilibrium established between a solid solute and its ions in a saturated solution (b) Kf: The formation constant of a complex ion, which is the equilibrium constant describing the formation of a complex ion from a central ion (typically a metal cation) and the ligands that attach to it (c) Qsp: The ion product, which is the product of concentrations of the constituent ions of a compound, and is used to determine if they meet the precipitation criteria or not. (d) Complex ion: A polyatomic cation or anion composed of a central metal ion to which other groups (molecules or ions) called ligands are coordinated/attached to. 91. (E) (a) Commonion effect in solubility equilibrium: Where the solubility of a sparingly soluble compound is suppressed by the presence of a quantity (often large) of one of the compound's ions from another source (b) Fractional precipitation: A technique in which two or more ions in solution, each capable of being precipitated by the same reagent, are separated by the proper use of that reagent: One ion is precipitated, while the other(s) remains in solution. (c) Ionpair formation: Pair of ions held loosely together by electrostatic interactions in water (d) Qualitative cation analysis: A qualitative method that is used to identify various the cations present in a solution by stepwise addition of various anions in order to selectively precipitate out various cations in steps. (E) (a) Solubility and solubility product constant: Solubility is the number of moles of a precipitate dissolved in a certain mass of water (or some other solvent.) The solubility product constant is the constant for the equilibrium established between a solid solute and its ions in a saturated solution. (b) Commonion effect and salt effect: The common ion effect is a consequence of Le Chtelier's principle, where the equilibrium concentration of all ions resulting from the dissolution of a sparingly soluble precipitate is reduced because a large amount of one (or more) of the ions from another source is present. The salt effect is the presence of a large quantity of ions from a very soluble salt which slightly increases the solubility of a sparingly soluble precipitate. (c) Ion pair and ion product: An ion pair comprises two oppositely charged ions held together by electrostatic forces. The ion product (Qsp) is the product of concentrations of the constituent ions of a compound, and is used to determine if they meet the precipitation criteria or not. 92. 942 Chapter 18: Solubility and ComplexIon Equilibria 93. (E) The answer is (d). See the reasoning below. (a) Wrong, because the stoichiometry is wrong (b) Wrong, because Ksp = [Pb2+][I]2 (c) Wrong, because [Pb2+] = Ksp/[ I]2 (d) Correct because of the [Pb2+]:2[I] stoichiometry
2 (E) The answer is (a). BaSO 4 s Ba 2 SO 4 . If Na2SO4 is added, the commonion effect 94. forces the equilibrium to the left and reduces [Ba2+].
95. (E) The answer is (c). Choices (b) and (d) reduce solubility because of the commonion effect. In the case of choice (c), the diverse noncommonion effect or the "salt effect" causes more Ag2CrO4 to dissolve. (E) The answer is (b). The sulfate salt of Cu is soluble, whereas the Pb salt is insoluble. 96. 97. (M) The answers are (c) and (d). Adding NH3 causes the solution to become basic (adding OH). Mg, Fe, Cu, and Al all have insoluble hydroxides. However, only Cu can form a complex ion with NH3, which is soluble. In the case of (NH4)2SO4, it is slightly acidic and dissolved readily in a base. 98. (M) The answer is (a). CaCO3 is slightly basic, so it is more soluble in an acid. The only option for an acid given is NH4Cl. (M) The answer is (c). Referring to Figure 187, it is seen that ammonia is added to an aqueous H2S solution to precipitate more metal ions. Since ammonia is a base, increasing the pH should cause more precipitation. 99. 100. (M) (a) H2C2O4 is a moderately strong acid, so it is more soluble in a basic solution. (b) MgCO3 is slightly basic, so it is more soluble in an acidic solution. (c) CdS is more soluble in acidic solutions, but the solubility is still so small that it is essentially insoluble even in acidic solutions. (d) KCl is a neutral salt, and therefore its solubility is independent of pH. (e) NaNO3 is a neutral salt, and therefore its solubility is independent of pH. (f) Ca(OH)2, a strong base, is more soluble in an acidic solution. 101. (E) The answer is NH3. NaOH(aq) precipitates both, and HCl(aq) precipitates neither. Mg(OH)2 precipitates from an NH3(aq) solution but forms the soluble complex Cu(NH3)4(OH)2. 102. (D) Al(OH)3 will precipitate. To demonstrate this, the pH of the acetate buffer needs to be determined first, from which the OH can be determined. The OH concentration can be used to calculate Qsp, which can then be compared to Ksp to see if any Al(OH)3 will precipitate. This is shown below: 943 Chapter 18: Solubility and ComplexIon Equilibria Ac log 1.8 105 log 0.35 M 4.65 pH pK a log 0.45 M HAc 14 pH 4.467 1010 M [OH ] 10 Al OH 3 Al3+ +3OH  Qsp ( s )(3s )3 Qsp 0.275 4.467 1010 2.45 1029
3 Qsp K sp , therefore there will be precipitation.
103. (E) The answer is (b). Based on solubility rules, Cu3(PO4)2 is the only species that is sparingly soluble in water. 104. (M) The answer is (d). The abbreviated work shown for each part calculates the molar solubility (s) for all the salts. They all follow the basic outlined below: M x A y xM + yA K sp x s y s x y and we solve for s.
(a) MgF2 Mg 2 2F
3.7 108 s (2s) 2 4s3 s 2.1 103 M
2 (b) MgCO3 Mg 2 CO3 (c) Mg 3 PO 4 2 3Mg 2 2PO3 4
1 1025 3s 2s 108s5
3 2 s 3.9 106 M (d) Li3PO 4 3Li PO3 4
3.2 109 3s s 27s 4
3 3.5 108 s s s 1.9 104 M s 3.3 103 M 105. (E) The answer is (b). This is due to the "salt effect." The more moles of salt there are available, the greater the solubility. For (b), there are 0.300 moles of ions (30.100 M Na2S2O3). 106. (E) It will decrease the amount of precipitate. Since all salts of NO3 are highly soluble and Ag+ and Hg2+ salts of I are not, anything that forms a soluble complex ion with Ag+ and Hg2+ will reduce the amount of those ions available for precipitation with I and therefore will reduce the amount of precipitate. 944 Chapter 18: Solubility and ComplexIon Equilibria 107. (M) No precipitate will form. To demonstrate this, one has to calculate [Ag+], and then, using [I], determine the Qsp and compare it to Ksp of AgI. Since Kf is for the formation of the complex ion Ag(CN)2, its inverse is for the dissociation of Ag(CN)2 to Ag+ and CN. K dis 1 K f 1 5.6 1018 1.786 1019 The dissociation of Ag(CN)2 is as follows: Ag CN 2 (aq) Ag + +2CN  x(1.05 x) 1.786 1019 (0.012 x) We can simplify the calculations by noting that x is very small in relation to [Ag+] and [CN]. Therefore, x = 1.701019 M. K dis Dissociation of AgI is as follows: AgI Ag + + I Qsp 1.7 1019 2.0 3.4 1019
Since Qsp < Ksp, no precipitate will form.
108. (M) In both cases, the dissolution reaction is similar to reaction (18.5), for which K = Kf Ksp. This is an exceedingly small quantity for (a) but large for (b). CuCO3 is soluble in NH3(aq) and CuS(s) is not. 109. (M) In this chapter, the concept of solubility of sparingly soluble precipitates is the overarching concept. Deriving from this overall concept is the dissociation constant for the precipitate, Ksp and molar solubility. Also emanating from the solubility concept are factors that affect solubility: the common ioneffect, the salt effect, the effects of pH on solubility, and formation of complex ions. Take a look at the subsection headings and problems for more refining of the general and specific concepts. 945 CHAPTER 19 SPONTANEOUS CHANGE: ENTROPY AND GIBBS ENERGY
PRACTICE EXAMPLES
1A (E) In general, S 0 if ngas 0 . This is because gases are very dispersed compared to liquids or solids; (gases possess large entropies). Recall that ngas is the difference between the sum of the stoichiometric coefficients of the gaseous products and a similar sum for the reactants.
(a) (b) ngas = 2 + 0 2 +1 = 1 . One mole of gas is consumed here. We predict S 0 . ngas = 1+ 0 0 = +1. Since one mole of gas is produced, we predict S 0 . 1B (E) (a) The outcome is uncertain in the reaction between ZnS(s) and Ag 2O s . We have used ngas to estimate the sign of entropy change. There is no gas involved in this reaction and thus our prediction is uncertain.
(b) In the chloralkali process the entropy increases because two moles of gas have formed where none were originally present ( ngas (1 1 0) (0 0) 2 2A (E) For a vaporization, Gvap = 0 = Hvap TS vap . Thus, S vap = Hvap / Tvap . H vap We substitute the given values. Svap = Tvap = 20.2 kJ mol1 = 83.0 J mol1 K 1 29.79 + 273.15 K 2B (E) For a phase change, Gtr = 0 = Htr TStr . Thus, Htr = TS tr . We substitute in the given values. H tr = T Str = 95.5 + 273.2 K 1.09 J mol1 K 1 = 402 J/mol 3A (M) The entropy change for the reaction is expressed in terms of the standard entropies of the reagents. S = 2S NH 3 g S N 2 g 3S H 2 g = 2 192.5 J mol1K 1 191.6 J mol1 K 1 3 130.7 J mol1 K 1 = 198.7 J mol1 K 1 Thus to form one mole of NH 3 g , the standard entropy change is 99.4 J mol 1 K 1 946 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 3B (M) The entropy change for the reaction is expressed in terms of the standard entropies of the reagents.
S o = S o NO g + S o NO 2 g S o N 2 O 3 g 138.5 J mol1 K 1 = 210.8 J mol1 K 1 + 240.1 J mol1 K 1 S N 2 O3 g = 450.9 J mol1 K 1 S N 2 O3 g S N 2 O3 g = 450.9 J mol1 K 1 138.5 J mol1 K 1 = 312.4 J mol1 K 4A (E) (a) Because ngas = 2 1 + 3 = 2 for the synthesis of ammonia, we would predict S 0 for the reaction. We already know that H 0 . Thus, the reaction falls into case 2, namely, a reaction that is spontaneous at low temperatures and nonspontaneous at high temperatures. (b) For the formation of ethylene ngas = 1 2 + 0 = 1 and thus S 0 . We are given that H 0 and, thus, this reaction corresponds to case 4, namely, a reaction that is nonspontaneous at all temperatures. 4B (E) (a) Because ngas = +1 for the decomposition of calcium carbonate, we would predict (b) S 0 for the reaction, favoring the reaction at high temperatures. High temperatures also favor this endothermic H o 0 reaction. The "roasting" of ZnS(s) has ngas = 2 3 = 1 and, thus, S 0 . We are given that H 0 ; thus, this reaction corresponds to case 2, namely, a reaction that is spontaneous at low temperatures, and nonspontaneous at high ones.
5A (E) The expression G = H TS is used with T = 298.15 K . G o = H o T S o = 1648 kJ 298.15 K 549.3 J K 1 1 kJ / 1000 J 5B = 1648 kJ +163.8 kJ = 1484 kJ (M) We just need to substitute values from Appendix D into the supplied expression. G = 2Gf NO 2 g 2Gf NO g Gf O 2 g = 2 51.31 kJ mol1 2 86.55 kJ mol1 0.00 kJ mol1 = 70.48 kJ mol1
6A (M) Pressures of gases and molarities of solutes in aqueous solution appear in thermodynamic equilibrium constant expressions. Pure solids and liquids (including solvents) do not appear. (a) K= PSiCl
2 PCl 4 Kp (b) K= 2 HOCl H + Cl PCl
2 K = K p for (a) because all terms in the K expression are gas pressures. 947 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 6B (M) We need the balanced chemical equation in order to write the equilibrium constant expression. We start by translating names into formulas. PbS s + HNO 3 aq Pb NO 3 2 aq + S s + NO g The equation then is balanced with the ionelectron method. oxidation : {PbS s Pb 2+ aq + S s + 2e } 3 reduction :{NO3 aq + 4H + aq + 3e NO g + 2H 2O(l) } 2 net ionic : 3 PbS s + 2NO3 aq + 8H + aq 3Pb 2+ aq + 3S s + 2NO g + 4H 2 O(l) In writing the thermodynamic equilibrium constant, recall that neither pure solids (PbS(s) and S(s)) nor pure liquids H 2 O(l) appear in the thermodynamic equilibrium constant expression. Note also that we have written H + aq here for brevity even though we
2 [Pb 2 ]3 pNO [NO3 ]2 [H ]8 understand that H 3O + aq is the acidic species in aqueous solution. K 7A (M) Since the reaction is taking place at 298.15 K, we can use standard free energies of formation to calculate the standard free energy change for the reaction: N 2 O 4 (g) 2 NO 2 (g) G = 2Gf NO 2 g Gf N 2 O 4 g 2 51.31 kJ/mol 97.89 kJ/mol 4.73kJ Gorxn = +4.73 kJ . Thus, the forward reaction is nonspontaneous as written at 298.15 K.
7B (M) In order to answer this question we must calculate the reaction quotient and compare it to the Kp value for the reaction:
N 2 O 4 (g) 2 NO 2 (g) (0.5) 2 Qp 0.5 0.5 0.5 bar 0.5 bar o G rxn = +4.73 kJ = RTlnKp; 4.73 kJ/mol = (8.3145 10 3 kJ/Kmol)(298.15 K)lnKp Therefore, Kp = 0.148. Since Qp is greater than Kp, we can conclude that the reverse reaction will proceed spontaneously, i.e. NO2 will spontaneously convert into N2O4.
8A (D) We first determine the value of G and then set G = RT ln K to determine K . G Gf Ag + aq + Gf [I (aq)] Gf AgI s [(77.11 51.57) (66.19)] kJ/mol 91.73 948 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 91.73 kJ / mol 1000 J G ln K 37.00 1 1 RT 8.3145 J mol K 298.15 K 1 kJ K e 37.00 8.5 1017 This is precisely equal to the value for the Ksp of AgI listed in Appendix D.
8B (D) We begin by translating names into formulas. MnO 2 s + HCl aq Mn 2 + aq + Cl2 aq Then we produce a balanced net ionic equation with the ionelectron method. oxidation : 2 Cl aq Cl2 g + 2e
net ionic : MnO 2 s + 4H + aq + 2Cl aq Mn 2+ aq + Cl2 g + 2H 2 O(l) reduction : MnO 2 s + 4H + aq + 2e Mn 2+ aq + 2H 2O(l) Next we determine the value of G for the reaction and then the value of K.
G = Gf Mn 2+ aq + Gf Cl2 g + 2Gf H 2 O l Gf MnO 2 s 4Gf H + aq 2Gf Cl aq = 228.1 kJ + 0.0 kJ + 2 237.1 kJ 465.1 kJ 4 0.0 kJ 2 131.2 kJ = +25.2 kJ ln K (25.2 103 J mol1 ) G 10.17 RT 8.3145 J mol1 K 1 298.15 K K e 10.2 4 105 Because the value of K is so much smaller than unity, we do not expect an appreciable forward reaction.
9A (M) We set equal the two expressions for G and solve for the absolute temperature. G = H T S = RT ln K T= H = T S RT ln K = T S R ln K H o 114.1 103 J/mol = = 607 K S o R ln K 146.4 8.3145 ln 150 J mol1 K 1 9B (D) We expect the value of the equilibrium constant to increase as the temperature decreases since this is an exothermic reaction and exothermic reactions will have a larger equilibrium constant (shift right to form more products), as the temperature decreases. Thus, we expect K to be larger than 1000, which is its value at 4.3 102 K. (a) The value of the equilibrium constant at 25 C is obtained directly from the value of G o , since that value is also for 25 C . Note: G o = H o T S o = 77.1 kJ/mol 298.15 K 0.1213 kJ/mol K = 40.9 kJ/mol ln K (40.9 103 J mol1 ) G 16.5 RT 8.3145 J mol1 K 1 298.15 K K e +16.5 1.5 107 949 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (b) First, we solve for G o at 75 C = 348 K kJ 1000 J J G o = H o T S o = 77.1 348.15 K 121.3 mol 1 kJ mol K = 34.87 103 J/mol Then we use this value to obtain the value of the equilibrium constant, as in part (a). (34.87 103 J mol1 ) G ln K 12.05 K e+12.05 1.7 105 1 1 RT 8.3145 J mol K 348.15 K As expected, K for this exothermic reaction decreases with increasing temperature.
10A (M) We use the value of Kp = 9.1 102 at 800 K and H o = 1.8 105 J / mol , for the appropriate terms, in the van't Hoff equation. 5.8 102 1 1 1 9.66 8.3145 1.8 105 J/mol 1 ln = = = 9.66; 2 1 1 9.1 10 8.3145 J mol K 800 K T K T 800 1.8 105 1/T = 1.25 103 4.5 104 = 8.0 104 T = 1240 K 970 C This temperature is an estimate because it is an extrapolated point beyond the range of the data supplied.
10B (M) The temperature we are considering is 235 C = 508 K. We substitute the value of Kp = 9.1 102 at 800 K and H o = 1.8 105 J/mol, for the appropriate terms, in the van't Hoff equation. Kp Kp 1 1.8 105 J/mol 1 ln = = e +15.6 = 6 106 = +15.6 ; 2 2 1 1 9.1 10 8.3145 J mol K 800 K 508 K 9.110
K p = 6 106 9.1 102 = 5 109 INTEGRATIVE EXAMPLE
11A (D) The value of G can be calculated by finding the value of the equilibrium constant Kp at 25 oC. The equilibrium constant for the reaction is simply given by K p p{N 2O5 (g)} .
o The vapor pressure of N2O5(g) can be determined from the ClausiusClapeyron eqution, which is a specialized version of the van't Hoff equation. Stepwise approach: We first determine the value of H sub .
ln 1 p2 H sub 1 1 H sub 1 760 mmHg ln 1 1 T T 7.5 273.15 32.4 273.15 p1 R 1 100 mmHg 8.314Jmol K 2 2.028 5.81 10 4 J/mol 5 3.49 10 Using the same formula, we can now calculate the vapor pressure of N2O5 at 25 oC. H sub 950 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy p3 p3 5.81 10 4 J/mol 1 1 ln e1.46 4.31 1.46 1 1 100 mmHg 8.314Jmol K 280.7 298.2 100 mmHg 1 atm 0.567atm K p p3 4.31 100mmHg 760 mmHg G o RT ln K p (8.314 10 3 kJmol1K 1 298.15K)ln(0.567) 1.42kJ/mol Conversion pathway approach: p R ln 2 H sub 1 1 p p1 ln 2 T T H sub 1 1 p1 R 1 2 T T 1 2 H sub 760mmHg 2.028 100mmHg Jmol1 5.81 10 4 Jmol1 1 1 3.49 10 5 7.5 273.15 32.4 273.15 8.314Jmol1K 1 ln
H sub 1 1 R T1 T2 p H sub 1 1 p3 p1e ln 3 p1 R T1 T2 p3 100mmHg e 5.8110 4 Jmol1 1 1 1 K 8.314 JK 1 mol1 280.7 298.2 431mmHg 1atm 0.567atm K p 760mmHg G o RT ln K p (8.314 10 3 kJmol1K 1 298.15K)ln(0.567) 1.42kJ/mol 11B (D) The standard entropy change for the reaction ( S o ) can be calculated from the known values of H o and G o . Stepwise approach: H o G o 454.8kJmol1 (323.1kJmol1 ) o o o o 441.7JK 1mol1 G H T S S T 298.15K Plausible chemical reaction for the production of ethylene glycol can also be written as: 2C(s)+3H 2 (g)+O 2 (g) CH 2OHCH 2 OH(l) o o Since S o {S products } {Sreac tan ts } it follows that:
o Srxn S o (CH 2OHCH 2 OH(l)) [2 S o (C(s)) 3 S o (H 2 (g)) S o (O 2 (g))] 441.7JK 1mol1 S o (CH 2 OHCH 2 OH(l)) [2 5.74JK 1mol1 3 130.7JK 1mol1 205.1JK 1mol1 ] S o (CH 2 OHCH 2 OH(l)) 441.7JK 1mol1 608.68JK 1mol1 167JK 1mol1 951 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Conversion pathway approach: H o G o 454.8kJmol1 (323.1kJmol1 ) 441.7JK 1mol1 G o H o T S o S o T 298.15K
441.7JK 1mol1 S o (CH 2OHCH 2 OH(l)) [2 5.74JK 1mol1 3 130.7JK 1mol1 205.1JK 1mol1 ] S o (CH 2 OHCH 2 OH(l)) 441.7JK 1mol1 608.68JK 1mol1 167JK 1mol1 EXERCISES
Spontaneous Change and Entropy
1. (E) (a) The freezing of ethanol involves a decrease in the entropy of the system. There is a reduction in mobility and in the number of forms in which their energy can be stored when they leave the solution and arrange themselves into a crystalline state. (b) The sublimation of dry ice involves converting a solid that has little mobility into a highly dispersed vapor which has a number of ways in which energy can be stored (rotational, translational). Thus, the entropy of the system increases substantially. The burning of rocket fuel involves converting a liquid fuel into the highly dispersed mixture of the gaseous combustion products. The entropy of the system increases substantially. (c) 2. 3. (E) Although there is a substantial change in entropy involved in (a) changing H 2 O (1iq., 1 atm) to H 2 O (g, 1 atm), it is not as large as (c) converting the liquid to a gas at 10 mmHg. The gas is more dispersed, (less ordered), at lower pressures. In (b), if we start with a solid and convert it to a gas at the lower pressure, the entropy change should be even larger, since a solid is more ordered (concentrated) than a liquid. Thus, in order of increasing S , the processes are: (a) (c) (b). (E) The first law of thermodynamics states that energy is neither created nor destroyed (thus, "The energy of the universe is constant"). A consequence of the second law of thermodynamics is that entropy of the universe increases for all spontaneous, that is, naturally occurring, processes (and therefore, "the entropy of the universe increases toward a maximum"). (E) When pollutants are produced they are usually dispersed throughout the environment. These pollutants thus start in a relatively compact form and end up dispersed throughout a large volume mixed with many other substances. The pollutants are highly dispersed, thus, they have a high entropy. Returning them to their original compact form requires reducing this entropy, which is a highly nonspontaneous process. If we have had enough foresight to retain these pollutants in a reasonably compact form, such as disposing of them in a secure landfill, rather than dispersing them in the atmosphere or in rivers and seas, the task of permanently removing them from the environment, and perhaps even converting them to useful forms, would be considerably easier. 4. 952 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 5. (E) (a) Increase in entropy because a gas has been created from a liquid, a condensed phase. (b) (c) Decrease in entropy as a condensed phase, a solid, is created from a solid and a gas. For this reaction we cannot be certain of the entropy change. Even though the number of moles of gas produced is the same as the number that reacted, we cannot conclude that the entropy change is zero because not all gases have the same molar entropy. 2H 2S g + 3O 2 g 2H 2 O g + 2SO 2 g Decrease in entropy since five moles of gas with high entropy become only four moles of gas, with about the same quantity of entropy per mole. (d) 6. (E) (a) At 75 C , 1 mol H 2 O (g, 1 atm) has a greater entropy than 1 mol H 2 O (1iq., 1 atm) since a gas is much more dispersed than a liquid. (b) 1 mol Fe = 0.896 mol Fe has a higher entropy than 0.80 mol Fe, both (s) 55.8 g Fe at 1 atm and 5 C , because entropy is an extensive property that depends on the amount of substance present. 50.0 g Fe 1 mol Br2 (1iq., 1 atm, 8 C ) has a higher entropy than 1 mol Br2 (s, 1atm, 8 C ) because solids are more ordered (concentrated) substances than are liquids, and furthermore, the liquid is at a higher temperature. 0.312 mol SO 2 (g, 0.110 atm, 32.5 C ) has a higher entropy than 0.284 mol O 2 (g, 15.0 atm, 22.3 C ) for at least three reasons. First, entropy is an extensive property that depends on the amount of substance present (more moles of SO2 than O2). Second, entropy increases with temperature (temperature of SO2 is greater than that for O2. Third, entropy is greater at lower pressures (the O2 has a much higher pressure). Furthermore, entropy generally is higher per mole for more complicated molecules. (c) (d) 7. (E) (a) Negative; A liquid (moderate entropy) combines with a solid to form another solid. (b) (c) (d) (e) Positive; One mole of high entropy gas forms where no gas was present before. Positive; One mole of high entropy gas forms where no gas was present before. Uncertain; The number of moles of gaseous products is the same as the number of moles of gaseous reactants. Negative; Two moles of gas (and a solid) combine to form just one mole of gas. 8. (M) The entropy of formation of a compound would be the difference between the absolute entropy of one mole of the compound and the sum of the absolute entropies of the appropriate amounts of the elements constituting the compound, with each species in its most stable form. 953 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Stepwise approach: It seems as though CS2 1 would have the highest molar entropy of formation of the compounds listed, since it is the only substance whose formation does not involve the consumption of high entropy gaseous reactants. This prediction can be checked by determining S o values from the data in Appendix D: f
(a)
C graphite + 2H 2 g CH 4 g Sfo CH 4 g = S o CH 4 g S o C graphite 2S o H 2 g 1 1 1 1 = 186.3 J mol K 5.74 J mol K 2 130.7 J mol1K 1 = 80.8 J mol1K 1 2C graphite + 3H 2 g + 1 O 2 g CH 3CH 2 OH l 2
1 Sfo CH3CH 2 OH l = S o CH3CH 2 OH l 2S o C graphite 3S o H 2 g S o O2 g 2 1 1 1 1 1 1 1 1 1 = 160.7 J mol K 2 5.74 J mol K 3 130.7 J mol K 2 205.1J mol K = 345.4 J mol1 K 1 (b) (c) C graphite + 2S rhombic CS2 l Sf CS2 l = S o CS2 l S o C graphite 2S o S rhombic = 151.3 J mol1 K 1 5.74 J mol1 K 1 2 31.80 J mol1 K 1 = 82.0 J mol1 K 1 Conversion pathway approach: CS2 would have the highest molar entropy of formation of the compounds listed, because it is the only substance whose formation does not involve the consumption of high entropy gaseous reactants. (a) C graphite + 2H 2 g CH 4 g Sfo CH 4 g = 186.3 J mol1K 1 5.74 J mol1K 1 2 130.7 J mol1K 1 80.8 J mol1K 1 2C graphite + 3H 2 g + 1 O 2 g CH 3CH 2 OH l 2
Sfo CH 3CH 2OH l = (160.7 2 5.74 3 130.7 1 205.1)J mol1 K 1 2 = 345.4 J mol1 K 1 (b) (c) C graphite + 2S rhombic CS2 l Sf CS2 l = 151.3 J mol1 K 1 5.74 J mol1 K 1 2 31.80 J mol1 K 1 = 82.0 J mol1 K 1 954 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Phase Transitions
9. (M) (a) H vap H f [H 2 O(g)] H f [H 2 O(l)] 241.8 kJ/mol (285.8 kJ/mol) 44.0 kJ/mol = S H 2 O g S o H 2 O l = 188.8 J mol1 K 1 69.91 J mol1K 1 1 1 = 118.9 J mol K There is an alternate, but incorrect, method of obtaining S vap .
S vap
o S vap = H vap T = 44.0 103 J/mol = 148 J mol1 K 1 298.15 K This method is invalid because the temperature in the denominator of the equation must be the temperature at which the liquidvapor transition is at equilibrium. Liquid water and water vapor at 1 atm pressure (standard state, indicated by ) are in equilibrium only at 100 C = 373 K.
(b) The reason why Hvap is different at 25 C from its value at 100 C has to do with the heat required to bring the reactants and products down to 298 K from 373 K. The specific heat of liquid water is higher than the heat capacity of steam. Thus, more heat is given off by lowering the temperature of the liquid water from 100 C to 25 C than is given off by lowering the temperature of the same amount of steam. Another way to think of this is that hydrogen bonding is more disrupted in water at 100 C than at 25 C (because the molecules are in rapidthermalmotion), and hence, there is not as much energy needed to convert liquid to vapor (thus H vap has a smaller value at 100 C . The reason why S vap has a larger value at 25 C than at 100 C has to do with dispersion. A vapor at 1 atm pressure (the case at both temperatures) has about the same entropy. On the other hand, liquid water is more disordered (better able to disperse energy) at higher temperatures since more of the hydrogen bonds are disrupted by thermal motion. (The hydrogen bonds are totally disrupted in the two vapors).
10. (M) In this problem we are given standard enthalpies of the formation ( H o ) of liquid f and gas pentane at 298.15 K and asked to estimate the normal boiling point of o o pentane, Gvap and furthermore comment on the significance of the sign of Gvap .
o The general strategy in solving this problem is to first determine H vap from the known enthalpies of formation. Trouton's rule can then be used to determine the o normal boiling point of pentane. Lastly, Gvap ,298 K can be calculated using Gvap = Hvap TS vap . 955 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Stepwise approach: o Calculate H vap from the known values of H o (part a): f C5H12(g) C5H12(l) o H f 173.5 kJmol1 146.9 kJmol1
o H vap 146.9 (173.5)kJmol1 26.6kJmol1 Determine normal boiling point using Trouton's rule (part a): o H vap o Svap 87Jmol1K 1 Tnbp
Tnbp o H vap o Svap 26.6kJmol 1 306K 87kJK 1mol 1 1000 Tnbp 32.9 o C Use Gvap = Hvap TS vap to calculate Gvap,298K (part b): Gvap = Hvap TS vap Gvap,298K 26.6kJmol1 298.15K Gvap,298K 0.66kJmol1 Comment on the value of Gvap,298K (part c): 87kJmol1K 1 1000 The positive value of Gvap indicates that normal boiling (having a vapor pressure of 1.00 atm) for pentane should be nonspontaneous (will not occur) at 298. The vapor pressure of pentane at 298 K should be less than 1.00 atm. Conversion pathway approach: C5H12(l) C5H12(g) H o 173.5 kJmol1 146.9 kJmol1 f
o H vap 146.9 (173.5)kJmol1 26.6kJmol1 o H vap S o vap Tnbp 87Jmol K Tnbp 1 1 Gvap = H vap T Svap 26.6kJmol1 306K o 87kJK 1mol1 Svap 1000 87kJmol1K 1 26.6kJmol1 298.15K 0.66kJmol1 1000 o H vap 11. (M) Trouton's rule is obeyed most closely by liquids that do not have a high degree of order within the liquid. In both HF and CH 3OH , hydrogen bonds create considerable order within the liquid. In C6 H 5CH 3 , the only attractive forces are nondirectional London forces, which have no preferred orientation as hydrogen bonds do. Thus, of the three choices, liquid C 6 H 5CH 3 would most closely follow Trouton's rule. 956 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 12. (E) H vap H f [Br2 (g)] H f [Br2 (l)] 30.91 kJ/mol 0.00 kJ/mol = 30.91 kJ/mol H vap H vap S vap S vap = Tvap 87 J mol K 1 1 or Tvap = 30.91103 J/mol = 3.5 102 K 1 1 87 J mol K The accepted value of the boiling point of bromine is 58.8 C = 332 K = 3.32 102 K . Thus, our estimate is in reasonable agreement with the measured value.
13. (M) The liquid watergaseous water equilibrium H 2 O (l, 0.50 atm) H 2 O (g, 0.50 atm) can only be established at one temperature, namely the boiling point for water under 0.50 atm external pressure. We can estimate the boiling point for water under 0.50 atm external pressure by using the ClausiusClapeyron equation: ln o P2 H vap 1 1 = R T1 T2 P 1 We know that at 373 K, the pressure of water vapor is 1.00 atm. Let's make P1 = 1.00 atm, P2 = 0.50 atm and T1 = 373 K. Thus, the boiling point under 0.50 atm pressure is T2. To find T2 we simply insert the appropriate information into the ClausiusClapeyron equation and solve for T2: ln 1 0.50 atm 40.7 kJ mol1 1 = 3 1 1 1.00 atm 8.3145 10 kJ K mol 373 K T2 1 1 1.416 104 K = 373 K T2 Solving for T2 we find a temperature of 354 K or 81C. Consequently, to achieve an equilibrium between gaseous and liquid water under 0.50 atm pressure, the temperature must be set at 354 K.
14. (M) Figure 1219 (phase diagram for carbon dioxide) shows that at 60C and under 1 atm of external pressure, carbon dioxide exists as a gas. In other words, neither solid nor liquid CO2 can exist at this temperature and pressure. Clearly, of the three phases, gaseous CO2 must be the most stable and, hence, have the lowest free energy when T = 60 C and Pext = 1.00 atm. Gibbs Energy and Spontaneous Change
15. (E) Answer (b) is correct. BrBr bonds are broken in this reaction, meaning that it is endothermic, with H 0 . Since the number of moles of gas increases during the reaction, S 0 . And, because G = H T S , this reaction is nonspontaneous ( G 0 ) at low temperatures where the H term predominates and spontaneous ( G 0 ) at high temperatures where the T S term predominates. 957 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 16. (E) Answer (d) is correct. A reaction that proceeds only through electrolysis is a reaction that is nonspontaneous. Such a reaction has G 0 . (E) (a) 17. H o 0 and S o 0 (since ngas 0 ) for this reaction. Thus, this reaction is case 2 of Table 191. It is spontaneous at low temperatures and nonspontaneous at high temperatures.
(b) We are unable to predict the sign of S o for this reaction, since ngas = 0 . Thus, no strong prediction as to the temperature behavior of this reaction can be made. Since Ho > 0, we can, however, conclude that the reaction will be nonspontaneous at low temperatures. (c) 18. H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 3 of Table 191. It is nonspontaneous at low temperatures, but spontaneous at high temperatures. (E) (a) H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 4 of Table 191. It is nonspontaneous at all temperatures.
(b) (c) H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 1 of Table 191.
It is spontaneous at all temperatures. H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 2 of Table 191.
It is spontaneous at low temperatures and nonspontaneous at high temperatures. 19. (E) First of all, the process is clearly spontaneous, and therefore G 0 . In addition, the gases are more dispersed when they are at a lower pressure and therefore S 0 . We also conclude that H = 0 because the gases are ideal and thus there are no forces of attraction or repulsion between them. (E) Because an ideal solution forms spontaneously, G 0 . Also, the molecules of solvent and solute that are mixed together in the solution are in a more dispersed state than the separated solvent and solute. Therefore, S 0 . However, in an ideal solution, the attractive forces between solvent and solute molecules equal those forces between solvent molecules and those between solute molecules. Thus, H = 0 . There is no net energy of interaction. (M) (a) An exothermic reaction (one that gives off heat) may not occur spontaneously if, at the same time, the system becomes more ordered (concentrated) that is, S o 0 . This is particularly true at a high temperature, where the TS term dominates the G expression. An example of such a process is freezing water (clearly exothermic because the reverse process, melting ice, is endothermic), which is not spontaneous at temperatures above 0 C . (b) 20. 21. A reaction in which S 0 need not be spontaneous if that process also is endothermic. This is particularly true at low temperatures, where the H term dominates the G expression. An example is the vaporization of water (clearly an endothermic process, one that requires heat, and one that produces a gas, so S 0 ), 958 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy which is not spontaneous at low temperatures, that is, below100 C (assuming Pext = 1.00 atm).
22. (M) In this problem we are asked to explain whether the reaction AB(g)A(g)+B(g) is always going to be spontaneous at high rather than low temperatures. In order to answer this question, we need to determine the signs of H , S and consequently G . Recall that G H T S . Stepwise approach: Determine the sign of S: We are generating two moles of gas from one mole. The randomness of the system increases and S must be greater than zero. Determine the sign of H: In this reaction, we are breaking AB bond. Bond breaking requires energy, so the reaction must be endothermic. Therefore, H is also greater than zero. Use G H T S to determine the sign of G : G H T S . Since H is positive and S is positive there will be a temperature at which T S will become greater than H . The reaction will be favored at high temperatures and disfavored at low temperatures. Conversion pathway approach: S for the reaction is greater than zero because we are generating two moles of gas from one mole. H for the reaction is also greater than zero because we are breaking AB (bond breaking requires energy). Because G H T S , there will be a temperature at which T S will become greater than H . The reaction will be favored at high temperatures and disfavored at low temperatures. Standard Gibbs Energy Change
23. (M) H o = H f NH 4Cl s H f NH 3 g H HCl g f 314.4 kJ/mol (46.11 kJ/mol 92.31 kJ/mol) 176.0 kJ/mol G o Gf NH 4Cl s Gf NH 3 g Gf HCl g 202.9 kJ/mol (16.48 kJ/mol 95.30 kJ/mol) 91.1 kJ/mol G o H o T S o H o G o 176.0 kJ/mol 91.1 kJ/mol 1000 J S 285 J mol1 1 kJ 298 K T
o 24. (M) (a) G o = Gf C2 H 6 g Gf C2 H 2 g 2Gf H 2 g G o = 2Gf SO 2 g + Gf O 2 g 2Gf SO 3 g = 32.82 kJ/mol 209.2 kJ/mol 2 0.00 kJ/mol = 242.0 kJ/mol (b) = 2 300.2 kJ/mol + 0.00 kJ/mol 2 371.1 kJ/mol = +141.8 kJ/mol 959 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (c) G o = 3Gf Fe s + 4Gf H 2 O g Gf Fe3 O 4 s 4Gf H 2 g = 3 0.00 kJ/mol + 4 228.6 kJ/mol 1015 kJ/mol 4 0.00 kJ/mol = 101 kJ/mol (d) G o = 2Gf Al3+ aq + 3Gf H 2 g 2Gf Al s 6Gf H + aq = 2 485 kJ/mol + 3 0.00 kJ/mol 2 0.00kJ/mol 6 0.00 kJ/mol = 970. kJ/mol = 2 222.4 J/K 2 311.7 J/K 205.1 J/K = 383.7 J/K 25. (M) (a) S o = 2 S o POCl3 1 2 S o PCl3 g S o O 2 g o G = H o T S o = 620.2 103 J 298 K 383.7 J/K = 506 103 J = 506 kJ (b) The reaction proceeds spontaneously in the forward direction when reactants and products are in their standard states, because the value of G o is less than zero. 26. o o o o + o o (M) (a) S = S Br2 l + 2 S HNO 2 aq 2 S H aq 2 S Br aq 2 S NO 2 g = 152.2 J/K + 2 135.6 J/K 2 0 J/K 2 82.4 J/K 2 240.1 J/K = 221.6 J/K G o = H o T S o = 61.6 103 J 298K 221.6 J/K = +4.4 103 J = +4.4 kJ
(b) 27. The reaction does not proceed spontaneously in the forward direction when reactants and products are in their standard states, because the value of G o is greater than zero. (M) We combine the reactions in the same way as for any Hess's law calculations. (a) N 2 O g N 2 g + 1 O 2 g G o = 1 +208.4 kJ = 104.2 kJ 2 2 N 2 g + 2 O 2 g 2 NO 2 g G o = +102.6 kJ 3 Net: N 2O g + 2 O 2 g 2NO 2 g G o = 104.2 +102.6 = 1.6 kJ This reaction reaches an equilibrium condition, with significant amounts of all species being present. This conclusion is based on the relatively small absolute value of G o . (b) 2N 2 g + 6H 2 g 4NH 3 g 4NO g 2N 2 g + 2O 2 g 4NH 3 g + 5O 2 g 4NO g + 6H 2 O l G o = 2 33.0 kJ = 66.0 kJ G o = 1010.5 kJ G o = 2 +173.1 kJ = 346.2 kJ Net: 6H 2 g + 3O 2 g 6H 2 O l G o = 66.0 kJ 1010.5 kJ 346.2 kJ = 1422.7 kJ This reaction is three times the desired reaction, which therefore has G o = 1422.7 kJ 3 = 474.3 kJ. The large negative G o value indicates that this reaction will go to completion at 25 C . 960 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (c) 2N 2 g + O 2 g 2N 2 O g 4NO g 2N 2 g + 2O 2 g 4NH 3 g + 5O 2 g 4NO g + 6H 2O l G o = 1010.5 kJ G o = 2 +173.1 kJ = 346.2 kJ
G o = +208.4 kJ = 1148.3kJ 4NH 3 g + 4O 2 g 2N 2 O g + 6H 2 O l G o = 1010.5 kJ 346.2 kJ + 208.4 kJ 28. This reaction is twice the desired reaction, which, therefore, has G o = 574.2 kJ. The very large negative value of the G o for this reaction indicates that it will go to completion. (M) We combine the reactions in the same way as for any Hess's law calculations.
(a) COS g + 2CO 2 g SO 2 g + 3CO g 2CO g + 2H 2O g 2CO 2 g + 2H 2 g G o = 246.4 kJ = +246.6 kJ G o = 2 28.6 kJ = 57.2 kJ COS g + 2H 2O g SO 2 g + CO g + 2H 2 g G o = +246.6 57.2 = +189.4 kJ This reaction is spontaneous in the reverse direction, because of the large positive value of G o
(b) COS g + 2CO 2 g SO 2 g + 3CO g 3CO g + 3H 2O g 3CO 2 g + 3H 2 g G o = 246.4 kJ = +246.6 kJ G o = 328.6 kJ = 85.8 kJ COS g + 3H 2 O g CO 2 g + SO 2 g + 3H 2 g G o = +246.6 85.8 = +160.8 kJ This reaction is spontaneous in the reverse direction, because of the large positive value of G o .
(c)
COS g + H 2 g CO g + H 2S g CO g + H 2 O g CO 2 g + H 2 g G o = +1.4 kJ G o = 28.6 kJ = 28.6 kJ COS g + H 2O g CO 2 g + H 2S g G o = 1.4 kJ 28.6 kJ = 30.0 kJ The negative value of the G o for this reaction indicates that it is spontaneous in the forward direction.
29. (D) (a) The combustion reaction is : C6 H 6 l + 15 O 2 g 6CO 2 g + 3H 2 O g or l 2 G o = 6Gf CO 2 g + 3Gf H 2 O l Gf C6 H 6 l 15 Gf O 2 g 2 = 6 394.4 kJ + 3 237.1 kJ +124.5 kJ 15 0.00 kJ = 3202 kJ 2 (b) G o = 6Gf CO 2 g + 3Gf H 2 O g Gf C6 H 6 l 15 Gf O 2 g 2 = 6 394.4 kJ + 3 228.6 kJ +124.5 kJ 15 0.00 kJ = 3177 kJ 2 961 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy We could determine the difference between the two values of G o by noting the difference between the two products: 3H 2 O l 3H 2O g and determining the value of G o for this difference: G o = 3Gf H 2 O g 3Gf H 2 O l = 3 228.6 237.1 kJ = 25.5 kJ 30. (M) We wish to find the value of the H o for the given reaction: F2 g 2 F(g)
H o = G o + T S o = 123.9 103 J + 298 K 114.8 J/K = 158.1 kJ/mole of bonds S o = 2S o F g S o F2 g = 2 158.8 J K 1 202.8 J K 1 = +114.8 J K 1 31. The value in Table 10.3 is 159 kJ/mol, which is in quite good agreement with the value found here. (M) (a) Srxn = Sproducts  Sreactants = [1 mol 301.2 J K1mol1 + 2 mol 188.8 J K1mol1] [2 mol 247.4 J K1mol1 + 1 mol 238.5 J K1mol1] = 54.5 J K1 Srxn = = 0.0545 kJ K1
(b) Hrxn= bonds broken in reactants (kJ/mol)) bonds broken in products(kJ/ mol))
= [4 mol (389 kJ mol1)NH + 4 mol (222 kJ mol1)OF] [4 mol (301 kJ mol1)NF + 4 mol (464 kJ mol1)OH] Hrxn= 616 kJ (c) (at 25 C ). Because both the entropy and enthalpy changes are negative, this reaction will be more highly favored at low temperatures (i.e., the reaction is enthalpy driven)
32. (D) In this problem we are asked to find G at 298 K for the decomposition of ammonium nitrate to yield dinitrogen oxide gas and liquid water. Furthermore, we are asked to determine whether the decomposition will be favored at temperatures above or below 298 K. In order to answer these questions, we first need the balanced chemical equation for the process. From the data in Appendix D, we can determine Hrxn and Srxn. Both quantities will be required to determine Grxn (Grxn=HrxnTSrxn). Finally the magnitude of Grxn as a function of temperature can be judged depending on the values of Hrxn and Srxn. Stepwise approach: First we need the balanced chemical equation for the process: NH4NO3(s) N2O(g) + 2H2O(l) Now we can determine Hrxn by utilizing H o values provided in Appendix D: f
H o f N2O(g) + 2H2O(l) NH4NO3(s) 1 365.6 kJmol 82.05 kJmol1 285.6 kJmol1 Grxn = Hrxn TSrxn = 616 kJ 298 K(0.0545 kJ K1) = 600 kJ Since the Grxn is negative, the reaction is spontaneous, and hence feasible Hrxn=Hfproducts Hfreactants Hrxn=[2 mol(285.8 kJ mol1) + 1 mol(82.05 kJ mol1)] [1 mol (365.6 kJ mol1)] 962 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Hrxn= 124.0 kJ Similarly, Srxn can be calculated utilizing S values provided in Appendix D NH4NO3(s) N2O(g) + 2H2O(l) 1 1 o S 15.1 Jmol K 219.9 Jmol1K1 69.91 Jmol1K1 Srxn=Sproducts Sreactants Srxn=[2 mol 69.91 J K1mol1 + 1 mol 219.9 J K1mol1 ] [1 mol 151.1 J K1mol1] Srxn=208.6 J K1 = 0.2086 kJ K1 To find Grxn we can either utilize Gf values provided in Appendix D or Grxn=HrxnTSrxn: Grxn=HrxnTSrxn=124.0 kJ 298.15 K0.2086 kJK1 Grxn=186.1 kJ Magnitude of Grxn as a function of temperature can be judged depending on the values of Hrxn and Srxn: Since Hrxn is negative and Srxn is positive, the decomposition of ammonium nitrate is spontaneous at all temperatures. However, as the temperature increases, the TS term gets larger and as a result, the decomposition reaction shift towards producing more products. Consequently, we can say that the reaction is more highly favored above 298 K (it will also be faster at higher temperatures) Conversion pathway approach: From the balanced chemical equation for the process NH4NO3(s) N2O(g) + 2H2O(l) we can determine Hrxn and Srxn by utilizing H o and S values provided in Appendix f D: Hrxn=[2 mol(285.8 kJ mol1) + 1 mol(82.05 kJ mol1)] [1 mol (365.6 kJ mol1)] Hrxn= 124.0 kJ Srxn=[2 mol 69.91 J K1mol1 + 1 mol 219.9 J K1mol1 ] [1 mol 151.1 J K1mol1] Srxn=208.6 J K1 = 0.2086 kJ K1 Grxn=HrxnTSrxn=124.0 kJ 298.15 K0.2086 kJK1 Grxn=186.1 kJ Since Hrxn is negative and Srxn is positive, the decomposition of ammonium nitrate is spontaneous at all temperatures. However, as the temperature increases, the TS term gets larger and as a result, the decomposition reaction shift towards producing more products. The reaction is highly favored above 298 K (it will also be faster). The Thermodynamic Equilibrium Constant
33. (E) In all three cases, Keq = Kp because only gases, pure solids, and pure liquids are present in the chemical equations. There are no factors for solids and liquids in Keq expressions, and gases appear as partial pressures in atmospheres. That makes Keq the same as Kp for these three reactions. 963 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy We now recall that K p = K c RT . Hence, in these three cases we have:
n (a) (b) (c) 2SO 2 g + O 2 g 2SO3 g ; ngas = 2 2 +1 = 1; K = K p = K c RT HI g 1 H 2 g + 1 I 2 g ; 2 2
ngas = 1 1 + 1 = 0; 2 2 1 K = Kp = Kc NH 4 HCO3 s NH 3 g + CO 2 g + H 2O l ; ngas = 2 0 = +2 K = K p = K c RT 2 34. (M) (a) (b) K= P{H 2 O g }4 P{H 2 g }4 Terms for both solids, Fe(s) and Fe 3O 4 s , are properly excluded from the thermodynamic equilibrium constant expression. (Actually, each solid has an activity of 1.00.) Thus, the equilibrium partial pressures of both H 2 g and H 2 O g do not depend on the amounts of the two solids present, as long as some of each solid is present. One way to understand this is that any chemical reaction occurs on the surface of the solids, and thus is unaffected by the amount present. We can produce H 2 g from H 2O g without regard to the proportions of Fe(s) and Fe 3O 4 s with the qualification, that there must always be some Fe(s) present for the production of H 2 g to continue. (c) 35. (M) In this problem we are asked to determine the equilibrium constant and the change in Gibbs free energy for the reaction between carbon monoxide and hydrogen to yield methanol. The equilibrium concentrations of each reagent at 483K were provided. We proceed by first determining the equilibrium constant. Gibbs free energy can be calculated using G o RT ln K . Stepwise approach: First determine the equilibrium constant for the reaction at 483K: CO(g)+2H 2 (g) CH3OH(g) [CH 3OH ( g )] 0.00892 K 14.5 [CO( g )][ H 2 ( g )] 0.0911 0.0822 2 Now use G o RT ln K to calculate the change in Gibbs free energy at 483 K: G o RT ln K G o 8.314 483 ln(14.5)Jmol1 1.1 10 4 Jmol1
G o 11kJmol1 Conversion pathway approach: [CH 3OH ( g )] 0.00892 14.5 K [CO( g )][ H 2 ( g )] 0.0911 0.0822 2 G o RT ln K 8.314 483 ln(14.5)Jmol1 1.1 10 4 Jmol1 G o 11kJmol1 964 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 36. (M) Gibbs free energy for the reaction ( G o H o T S o ) can be calculated using H o f and S o values for CO(g), H2(g) and CH3OH(g) from Appendix D. H o H o (CH 3OH(g)) [ H o (CO(g)) 2 H o (H 2 (g)] f f f H o 200.7kJmol1 (110.5kJmol1 0kJmol1 ) 90.2kJmol1 S o S o (CH 3OH(g)) [S o (CO(g)) 2S o (H 2 (g)]
S o 239.8JK 1mol1 (197.7JK 1mol1 2 130.7JK 1mol1 ) 219.3JK 1mol1 483K (219.3)kJK 1mol1 o 1 G 90.2kJmol 15.7kJmol1 1000 Equilibrium constant for the reaction can be calculated using G o RT ln K 15.7 1000Jmol 1 G o ln K 3.9 K e3.9 2.0 10 2 ln K 1 1 RT 8.314JK mol 483K The values are different because in this case, the calculated K is the thermodynamic equilibrium constant that represents the reactants and products in their standard states. In Exercise 35, the reactants and products were not in their standard states. Relationships Involving G, G o, Q and K
37. (M) G o = 2Gf NO g Gf N 2 O g 0.5 Gf O 2 g = 2 86.55 kJ/mol 104.2 kJ/mol 0.5 0.00 kJ/mol = 68.9 kJ/mol = RT ln K p = 8.3145 103 kJ mol1 K 1 298 K ln K p ln K p = 38. 68.9 kJ/mol = 27.8 8.3145 10 kJ mol1 K 1 298 K
3 K p = e 27.8 = 8 1013 o (M) (a) G = 2Gf N 2 O5 g 2Gf N 2 O 4 g Gf O 2 g = 2 115.1 kJ/mol 2 97.89 kJ/mol 0.00 kJ/mol = 34.4 kJ/mol
ln K p = G o 34.4 103 J/mol = = 13.9 8.3145 J mol1 K 1 298 K RT (b) G o = RT ln K p K p = e 13.9 = 9 107 39. (M) We first balance each chemical equation, then calculate the value of G o with data from Appendix D, and finally calculate the value of Keq with the use of G o = RT ln K . (a)
4HCl g + O 2 g 2H 2 O g + 2Cl2 s G o = 2Gf H 2 O g + 2Gf Cl2 g 4Gf HCl g Gf O 2 g kJ kJ kJ kJ kJ = 2 228.6 = 76.0 4 95.30 + 20 0 mol mol mol mol mol o 3 G +76.0 10 J/mol K = e +30.7 = 2 1013 ln K = = = +30.7 1 1 RT 8.3145 J mol K 298 K 965 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (b) 3Fe 2 O3 s + H 2 g 2Fe3O 4 s + H 2 O g G o = 2Gf Fe3O 4 s + Gf H 2 O g 3Gf Fe 2 O3 s Gf H 2 g = 32 kJ/mol ln K =
(c) = 2 1015 kJ/mol 228.6 kJ/mol 3 742.2 kJ/mol 0.00 kJ/mol G o 32 103 J/mol = = 13; RT 8.3145 J mol1 K 1 298 K K = e +13 = 4 105 2Ag + aq + SO 4 2 aq Ag 2SO 4 s G o = Gf Ag 2 SO 4 s 2Gf Ag + aq Gf SO 2 aq 4 = 618.4 kJ/mol 2 77.11kJ/mol 744.5 kJ/mol = 28.1kJ/mol G o 28.1 103 J/mol ln K = = = 11.3; K = e +11.3 = 8 104 1 1 RT 8.3145 J mol K 298 K
40. (E) S o = S o {CO g }+ S o {H g } S o {CO g } S o {H O g } 2 2 2
= 213.7 J mol1 K 1 +130.7 J mol1 K 1 197.7 J mol1 K 1 188.8 J mol1K 1 = 42.1 J mol1 K 1 41. (M) In this problem we need to determine in which direction the reaction 2SO 2 g + O 2 g 2SO3 g is spontaneous when the partial pressure of SO2, O2, and SO3 are 1.0104, 0.20 and 0.10 atm, respectively. We proceed by first determining the standard free energy change for the reaction ( G o ) using tabulated data in Appendix D. Change in Gibbs free energy for the reaction ( G ) is then calculated by employing the equation G G o RT ln Q p ,where Qp is the reaction quotient. Based on the sign of G , we can determine in which direction is the reaction spontaneous. Stepwise approach: First determine G o for the reaction using data in Appendix D: 2SO 2 g + O 2 g 2SO3 g G o = 2Gfo SO3 g 2Gfo SO 2 g Gfo O 2 g o G 2 (371.1 kJ/mol) 2 (300.2 kJ/mol) 0.0 kJ/mol G o 141.8kJ Calculate G by employing the equation G G o RT ln Q p , where Qp is the reaction quotient: G G o RT ln Q p
Qp = P{O 2 g }P{SO 2 g }2 P{SO3 g }2 966 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 0.10 atm Qp = 0.20 atm 1.0 104
2 atm 2 5.0 106 G = 141.8 kJ + (8.3145 103 kJ/Kmol )(298 K)ln(5.0 106) G = 141.8 kJ + 38.2 kJ = 104 kJ. Examine the sign of G to decide in which direction is the reaction spontaneous: Since G is negative, the reaction is spontaneous in the forward direction. Conversion pathway approach: G o = 2Gfo SO3 g 2Gfo SO 2 g Gfo O 2 g o G 2 (371.1 kJ/mol) 2 (300.2 kJ/mol) 0.0 kJ/mol=141.8kJ G G o RT ln Q p Qp 0.10 atm = P{O g }P{SO g } 0.20 atm 10 1.0
P{SO3 g }2
2 2 2 2 4 atm 2 5.0 106 G = 141.8 kJ + (8.3145 10 kJ/Kmol )(298 K)ln(5.0 106)=104 kJ. Since G is negative, the reaction is spontaneous in the forward direction. 3 42. (M) We begin by calculating the standard free energy change for the reaction: H 2 g + Cl2 g 2HCl g G = 2Gf HCl g Gf Cl2 g Gf H 2 g 2 (95.30 kJ/mol) 0.0 kJ/mol 0.0 kJ/mol 190.6 kJ Now we can calculate G by employing the equation G G o RT ln Q p , where 0.5 atm ; Qp = 1 Qp = P{H 2 g }P{Cl2 g } 0.5 atm 0.5 atm P{HCl g }2
2 G = 190.6 kJ + (8.3145 103 kJ/Kmol )(298 K)ln(1) G = 190.6 kJ + 0 kJ = 190.6 kJ. Since G is negative, the reaction is spontaneous in the forward direction.
43. (M) In order to determine the direction in which the reaction is spontaneous, we need to calculate the nonstandard free energy change for the reaction. To accomplish this, we will employ the equation G G o RT ln Qc , where 1.0 103 M 1.0 105 [H O + aq ] [CH 3CO 2 aq ] Qc = 3 ; Qc = (0.10 M) [CH 3CO 2 H aq ]
2 G = 27.07 kJ + (8.3145 103 kJ/Kmol )(298 K)ln(1.0 105) G = 27.07 kJ + (28.53 kJ) = 1.46 kJ. Since G is negative, the reaction is spontaneous in the forward direction. 967 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 44. (M) As was the case for exercise 39, we need to calculate the nonstandard free energy change for the reaction. Once again, we will employ the equation G G o RT ln Q , but this time
+ 1.0 103 M 1.0 105 [NH 4 aq ] [OH aq ] Qc = ; Qc = (0.10 M) [NH 3 aq ] 2 G = 29.05 kJ + (8.3145 103 kJ/Kmol)(298 K)ln(1.0 105) G = 29.05 kJ + (28.53 kJ) = 0.52 kJ. Since G is positive, the reaction is spontaneous in the reverse direction. 45. (E) The relationship So = (Go Ho)/T (Equation (b)) is incorrect. Rearranging this equation to put Go on one side by itself gives Go = Ho + TSo. This equation is not valid. The TSo term should be subtracted from the Ho term, not added to it. (E) The Go value is a powerful thermodynamic parameter because it can be used to determine the equilibrium constant for the reaction at each and every chemically reasonable temperature via the equation Go = RT ln K. (M) (a) To determine Kp we need the equilibrium partial pressures. In the ideal gas law, 46. 47. each partial pressure is defined by P = nRT / V . Because R, T, and V are the same for each gas, and because there are the same number of partial pressure factors in the numerator as in the denominator of the Kp expression, we can use the ratio of amounts to determine Kp .
Kp = P CO(g) P H 2 O(g) P CO2 (g) P H 2 (g) = n CO(g) n H 2 O(g) n CO2 (g) n H 2 (g) = 0.224 mol CO 0.224 mol H 2 O = 0.659 0.276 mol CO2 0.276 mol H 2 (b) (c) G o1000K = RT ln K p = 8.3145 J mol1 K 1 1000. K ln 0.659 = 3.467 103 J/mol = 3.467 kJ/mol 0.0340 mol CO 0.0650 mol H 2 O = 0.31 0.659 = K p 0.0750 mol CO 2 0.095 mol H 2 Since Qp is smaller than Kp , the reaction will proceed to the right, forming products, Qp = to attain equilibrium, i.e., G = 0. 48. (M) (a) We know that K p = K c RT . For the reaction 2SO 2 g + O 2 g 2SO3 g ,
n ngas = 2 2 +1 = 1 , and therefore a value of Kp can be obtained.
2.8 10 2 3.41 K 0.08206 L atm 1000 K mol K We recognize that K = K p since all of the substances involved in the reaction are K p = K c RT 1 gases. We can now evaluate G o . 968 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy G o = RT ln K eq = 8.3145 J 1000 K ln 3.41 = 1.02 104 J/mol = 10.2 kJ/mol mol K (b) We can evaluate Qc for this situation and compare the value with that of Kc = 2.8 102 to determine the direction of the reaction to reach equilibrium. 0.72 mol SO3 2 2.50 L [SO3 ] 45 2.8 102 K c Qc [SO 2 ]2 [O 2 ] 0.40 mol SO 2 0.18 mol O 2 2 2.50 L 2.50 L Since Qc is smaller than Kc the reaction will shift right, producing sulfur trioxide and consuming sulfur dioxide and molecular oxygen, until the two values are equal.
2 49. (M) (a) K = K c G o = RT ln K eq = 8.3145 103 kJ mol1 K 1 445 + 273 K ln 50.2 = 23.4 kJ G o = RT ln K p = 8.3145 103 kJ mol1 K 1 298 K ln 8.4 1013 G o = +68.9 kJ/mol K = K p = K c RT n g (b) = 1.7 1013 0.0821 298 1/2 = 8.4 1013 (c) K = K p = K c RT n = 4.61 103 0.08206 298 = 0.113
+1 G o = RT ln K p = 8.3145 10 3 kJ mol1 K 1 298K ln 0.113 = +5.40 kJ/mol
(d) K = K c = 9.14 10 6 G o = RT ln K c = 8.3145 103 kJ mol1 K 1 298 K ln 9.14 106 G o = +28.7 kJ/mol 50. (M) (a) The first equation involves the formation of one mole of Mg 2+ aq from only halfamole of Mg 2+ aq . We would expect a free energy change of half the size if only half as much product is formed.
(b) Mg OH 2 s and 2H + aq , while the second equation involves the formation of The value of K for the first reaction is the square of the value of K for the second reaction. The equilibrium constant expressions are related in the same fashion.
1/2 Mg 2 + Mg 2 + = = K 2 K1 = 2 2 H+ H+ 2 (c) The equilibrium solubilities will be the same regardless which expression is used. The equilibrium conditions (solubilities in this instance) are the same no matter how we choose to express them in an equilibrium constant expression. 969 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 51. (E) G o = RT ln K p = 8.3145 103 kJ mol1K 1 298 K ln 6.5 1011 = 67.4 kJ/mol
CO g + Cl2 g COCl2 g C graphite + 1 O 2 g CO g 2 C graphite + 1 O 2 g + Cl2 g COCl2 g 2 G o = 67.4 kJ/mol Gf = 137.2 kJ/mol Gf = 204.6 kJ/mol Gf of COCl2 g given in Appendix D is 204.6 kJ/mol, thus the agreement is excellent.
52. (M) In each case, we first determine the value of G o for the solubility reaction. From that, we calculate the value of the equilibrium constant, K sp , for the solubility reaction. (a)
AgBr s Ag + aq + Br aq G o = Gf Ag + aq + Gf Br aq Gf AgBr s 70.0 103 J/mol G o = = 28.2; RT 8.3145 J mol1 K 1 298.15 K = 77.11 kJ/mol 104.0 kJ/mol 96.90 kJ/mol = +70.0 kJ/mol K sp = e 28.2 = 6 1013 ln K =
(b) CaSO 4 s Ca 2+ aq + SO 4 2 aq o o 2 o G o = Gf Ca 2+ aq + Gf SO 4 aq Gf CaSO 4 s = 553.6 kJ/mol 744.5 kJ/mol 1332 kJ/mol = +34 kJ/mol 34 103 J/mol G o ln K = = = 14; RT 8.3145 J mol1 K 1 298.15 K
(c)
Fe OH 3 s Fe3+ aq + 3OH aq K sp = e 14 = 8 107 G o = Gf Fe 3+ aq + 3 Gf OH aq Gf Fe OH 3 s 220.2 103 J/mol G o = = 88.83 RT 8.3145 J mol1 K 1 298.15 K = 4.7 kJ/mol + 3 157.2 kJ/mol 696.5 kJ/mol = +220.2 kJ/mol K sp = e 88.83 = 2.6 1039 ln K =
53. (M)(a) We can determine the equilibrium partial pressure from the value of the equilibrium constant. G o 58.54 103 J/mol G o = RT ln K p ln K p = = = 23.63 RT 8.3145 J mol1K 1 298.15 K K p = P{O 2 g }1/2 = e 23.63 = 5.5 1011
(b) P{O 2 g } = 5.5 1011 = 3.0 1021 atm
2 Lavoisier did two things to increase the quantity of oxygen that he obtained. First, he ran the reaction at a high temperature, which favors the products (i.e., the side with molecular oxygen.) Second, the molecular oxygen was removed immediately after it was formed, which causes the equilibrium to shift to the right continuously (the shift towards products as result of the removal of the O2 is an example of Le Chtelier's principle). 970 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 54. (D) (a) We determine the values of H o and S o from the data in Appendix D, and then the value of G o at 25 C = 298 K.
H o = H f CH 3 OH g + H f H 2 O g H f CO 2 g 3 H f H 2 g = 200.7 kJ/mol + 241.8 kJ/mol 393.5 kJ/mol 3 0.00 kJ/mol = 49.0 kJ/mol S = S o CH 3OH g + S o H 2 O g S o CO 2 g 3 S o H 2 g o G o = H o T S o = 49.0 kJ/mol 298 K 0.1772 kJ mol1K 1 = +3.81 kJ/mol Because the value of G o is positive, this reaction does not proceed in the forward direction at 25 C .
(b) = (239.8 +188.8 213.7 3 130.7) J mol1K 1 = 177.2 J mol1K 1 Because the value of H o is negative and that of S o is negative, the reaction is nonspontaneous at high temperatures, if reactants and products are in their standard states. The reaction will proceed slightly in the forward direction, however, to produce an equilibrium mixture with small quantities of CH 3OH g and H 2 O g . Also, because the forward reaction is exothermic, this reaction is favored by lowering the temperature. That is, the value of K increases with decreasing temperature. (c) o G500K = H o T S o = 49.0 kJ/mol 500.K 0.1772 kJ mol1K 1 = 39.6 kJ/mol = 39.6 103 J/mol = RT ln K p ln K p (d) G 39.6 103 J/mol 9.53; RT 8.3145 J mol1 K 1 500. K 3H 2 g K p e 9.53 7.3 105 +H 2O g 0 atm +x atm x atm
3 Reaction: CO 2 g + Initial: 1.00 atm Changes: x atm Equil: (1.00 x ) atm K p = 7.3 105 = P CH 3OH P H 2 O P CO 2 P H 2 3 CH 3OH g 1.00 atm 0 atm 3x atm +x atm x atm 1.00 3x atm
= x x x 7.3 105 8.5 103 1.00 x 1.00 3x atm P CH 3OH Our assumption, that 3 x 1.00 x2 atm, is valid. Go and K as Function of Temperature
55 (M)(a) S o = S o Na 2 CO3 s + S o H 2 O l + S o CO 2 g 2 S o NaHCO3 s = 135.0
J K mol + 69.91 J K mol + 213.7 J J 2 101.7 = 215.2 K mol K mol K mol J 971 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (b) H o = H f Na 2 CO3 s + H f H 2 O l + H f CO 2 g 2H f NaHCO3 s = 1131
kJ mol 285.8 kJ mol 393.5 kJ kJ 2 950.8 = +91 mol mol mol kJ (c) G o = H o T S o = 91 kJ/mol 298 K 215.2 103 kJ mol1K 1 = 91 kJ/mol 64.13 kJ/mol = 27 kJ/mol 27 103 J/mol G o = = 10.9 RT 8.3145 J mol1 K 1 298 K (d) G o = RT ln K
K = e 10.9 = 2 105 ln K = 56. (M) (a) S o = S o CH3 CH 2 OH g + S o H 2 O g S o CO g 2S o H 2 g S o CH3 OH g S o = 282.7 J J J J J +188.8 197.7 2 130.7 239.8 K mol K mol K mol K mol K mol J K mol
2 S o = 227.4
3 H o = H f CH CH H o = 235.1 H o = 165.7 OH g + H f H 2 O g H f CO g 2H f H 2 g H f CH 3 OH g kJ kJ kJ kJ kJ 241.8 110.5 2 0.00 200.7 mol mol mol mol mol G o
(b) (c) kJ mol kJ = 165.7 mol kJ kJ kJ 298K 227.4 103 KkJ = 165.4 mol + 67.8 mol = 97.9 mol mol H o 0 for this reaction. Thus it is favored at low temperatures. Also, because ngas = +2 4 = 2 , which is less than zero, the reaction is favored at high pressures. First we assume that neither S o nor H o varies significantly with temperature. Then we compute a value for G o at 750 K. From this value of G o , we compute a value for Kp .
G o = H o T S o = 165.7 kJ/mol 750.K 227.4 103 kJ mol1 K 1 = 165.7 kJ/mol + 170.6 kJ/mol = +4.9kJ/mol = RT ln K p ln K p = 4.9 103 J/mol G o = = 0.79 RT 8.3145 J mol1 K 1 750.K K p = e 0.79 = 0.5 57. (E) In this problem we are asked to determine the temperature for the reaction between iron(III) oxide and carbon monoxide to yield iron and carbon dioxide given G o , H o , and S o . We proceed by rearranging G o =H o T S o in order to express the temperature as a function of G o , H o , and S o . 972 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Stepwise approach: Rearrange G o =H o T S o in order to express T as a function of G o , H o , and S o : G o =H o T S o T S o =H o G o H o G o S o Calculate T: 24.8 103 J 45.5 10 3 J = 1.36 103 K T= 15.2 J/K Conversion pathway approach: 3 3 H o G o 24.8 10 J 45.5 10 J o o o = 1.36 103 K G =H T S T 15.2 J/K S o T= 58. (E) We use the van't Hoff equation with H o = 1.8 105 J/mol, T1 = 800 . K, T2 = 100. C = 373 K, and K1 = 9.1 102 .
ln K 2 H o 1 1 1 1.8 10 5 J/mol 1 = = T T 8.3145 J mol1 K 1 800 K 373K = 31 K1 R 1 2 K2 K2 = e31 = 2.9 1013 = K1 9.1 102
59. K 2 = 2.9 1013 9.1102 = 3 1016 (M) We first determine the value of G o at 400 C , from the values of H o and S o , which are calculated from information listed in Appendix D. H o = 2H f NH 3 g H f N 2 g 3H f H 2 g = 2 46.11kJ/mol 0.00 kJ/mol 3 0.00 kJ/mol = 92.22 kJ/mol N 2 S o = 2 S o NH 3 g S o N 2 g 3S o H 2 g = 2 192.5 J mol1 K 1 191.6 J mol1 K 1 3 130.7 J mol1 K 1 = 198.7 J mol1 K 1 G o = H o T S o = 92.22 kJ/mol 673 K 0.1987 kJ mol1 K 1 = +41.51 kJ/mol = RT ln K p ln K p = G o 41.51 103 J/mol = = 7.42; 8.3145 J mol1K 1 673K RT K p = e 7.42 = 6.0 104 973 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 60. (M) (a) H o = H CO g + H H g H CO g H H O g f 2 f 2 f f 2 = 393.5 kJ/mol 0.00 kJ/mol 110.5 kJ/mol 241.8 kJ/mol = 41.2 kJ/mol S o = S o CO 2 g + S o H 2 g S o CO g S o H 2 O g = 213.7 J mol1 K 1 +130.7 J mol1 K 1 197.7 J mol1 K 1 188.8 J mol1 K 1 = 42.1 J mol1 K 1
G o H o T S o 41.2kJ/mol 298.15K (42.1 10 3 )kJ/molK G o 41.2kJ/mol 12.6kJ/mol 28.6kJ/mol (b) G o = H o T S o = 41.2 kJ/mol 875 K 42.1 10 3 kJ mol1 K 1 = 41.2 kJ/mol + 36.8 kJ/mol = 4.4 kJ/mol = RT ln K p ln K p = 61. G o 4.4 103 J/mol = = +0.60 8.3145 J mol1K 1 875K RT K p = e +0.60 = 1.8 (M) We assume that both H o and S o are constant with temperature. H o = 2H f SO 3 g 2H f SO 2 g H f O 2 g = 2 395.7 kJ/mol 2 296.8 kJ/mol 0.00 kJ/mol = 197.8 kJ/mol S o = 2S o SO3 g 2 S o SO 2 g S o O 2 g = 187.9 J mol1 K 1 G o = H o T S o = RT ln K T= = 2 256.8 J mol1 K 1 2 248.2 J mol1 K 1 205.1 J mol1 K 1 H o S o R ln K H o = T S o RT ln K T= 197.8 103 J/mol 650 K 187.9 J mol1 K 1 8.3145 J mol1 K 1 ln 1.0 106 This value compares very favorably with the value of T = 6.37 102 that was obtained in Example 1910.
62. (E) We use the van't Hoff equation to determine the value of H o (448 C = 721 K and 350 C = 623 K) .
K1 H o 1 1 H o 1 1 H o 50.0 4 ln = T T = ln 66.9 = 0.291 = R 623 721 = R 2.2 10 K2 R 2 1 H o 0.291 = = 1.3 103 K; R 2.2 104 K 1 H o = 1.3 103 K 8.3145 J K 1 mol1 = 11 103 J mol1 = 11 kJ mol1 974 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 63. (M) (a) ln K 2 H o 1 1 57.2 103 J/mol 1 1 = = = 2.11 1 1 K1 R T1 T2 8.3145 J mol K 298 K 273 K K2 = e 2.11 = 0.121 K1
(b) K 2 = 0.121 0.113 = 0.014 at 273 K K 2 H o 1 1 1 57.2 10 3 J/mol 1 0.113 ln = T T = 8.3145 J mol1 K 1 T 298 K = ln 1.00 = 2.180 K1 R 1 1 2 1 1 2.180 8.3145 1 K = 3.17 10 4 K 1 3 T 298K = 57.2 10 1 1 1 = 3.17 104 = 3.36 103 3.17 104 = 3.04 103 K 1 ; T1 298
64. T1 = 329 K (D) First we calculate G o at 298 K to obtain a value for Keq at that temperature. G o = 2Gf NO 2 g 2Gf NO g Gf O 2 g = 2 51.31 kJ/mol 2 86.55 kJ/mol 0.00 kJ/mol = 70.48 kJ/mol 70.48 103 J/mol K G 28.43 K e 28.43 2.2 1012 8.3145 J RT 298.15 K mol K Now we calculate H o for the reaction, which then will be inserted into the van't Hoff equation. H o = 2H f NO 2 g 2H f NO g H f O 2 g ln K = 2 33.18 kJ/mol 2 90.25 kJ/mol 0.00 kJ/mol = 114.14 kJ/mol K 2 H o 1 1 114.14 103 J/mol 1 1 ln = = = 9.26 1 1 K1 R T1 T2 8.3145 J mol K 298 K 373 K K2 = e 9.26 = 9.5 105 ; K 2 = 9.5 105 2.2 1012 = 2.1108 K1 Another way to find K at 100 C is to compute H o 114.14 kJ/mol from Hf values
o and S o 146.5 J mol1 K 1 from S o values. Then determine Go(59.5 kJ/mol), and find Kp with the expression G o = RT ln Kp . Not surprisingly, we obtain the same result, K p = 2.2 108 .
65. (M) First, the van't Hoff equation is used to obtain a value of H o . 200 C = 473K and 260 C = 533K . 1 H o 1 1 H o K 2.15 1011 1 = ln ln 2 = = 6.156 = 8 1 1 K1 R T1 T2 4.56 10 8.3145 J mol K 533K 473 K 6.156 6.156 = 2.9 105 H o H o = = 2.1 105 J / mol = 2.1 102 kJ / mol 2.9 105 975 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Another route to H o is the combination of standard enthalpies of formation. CO g + 3H 2 g CH 4 g + H 2O g H o = H f CH 4 g + H f H 2 O g H f CO g 3H f H 2 g = 74.81 kJ/mol 241.8 kJ/mol 110.5 3 0.00 kJ/mol = 206.1 kJ/mol Within the precision of the data supplied, the results are in good agreement.
66. (D) (a) t, C T,K 303 323 343 373 1/ T , K 1 Kp 1.66 105 3.90 104 6.27 103 2.31 101 lnKp 11.006 7.849 5.072 1.465 30. 50. 70. 100. 3.30 103 3.10 103 2.92 103 2.68 103 Plot of ln(Kp) versus 1/T
0.0026 0 2 ln Kp 4 6 8 10 y = 15402.12x + 39.83 0.0029 1/T(K1) 0.0032 The 12 of this graph is H o / R = 1.54 104 K slope o H = 8.3145 J mol1K 1 1.54 104 K = 128 103 J/mol = 128 kJ/mol
(b) When the total pressure is 2.00 atm, and both gases have been produced from NaHCO 3 s ,
P{H 2 O g } = P{CO 2 g } = 1.00 atm K p = P{H 2 O g }P{CO 2 g } = 1.00 1.00 = 1.00 Thus, ln K p = ln 1.00 = 0.000 . The corresponds to 1/T = 2.59 103 K 1 ;
T = 386 K . We can compute the same temperature from the van't Hoff equation. 976 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy ln K 2 H o 1 1 128 10 3 J/mol 1 1.66 10 5 1 = = = ln = 11.006 K1 R T1 T2 8.3145 J mol1 K 1 T1 303 K 1.00 1 1 11.006 8.3145 1 K = 7.15 10 4 K 1 3 T 303 K = 128 10 1 1 1 7.15 104 = 3.30 103 7.15 104 = 2.59 103 K 1 ; T1 = 386 K = T1 303 This temperature agrees well with the result obtained from the graph. Coupled Reactions
67. (E) (a) We compute G o for the given reaction in the following manner H o = H f TiCl4 l + H f O 2 g H f TiO 2 s 2H f Cl2 g = 804.2 kJ/mol + 0.00 kJ/mol 944.7 kJ/mol 2 0.00 kJ/mol = +140.5 kJ/mol S o = S o TiCl4 l + S o O 2 g S o TiO 2 s 2 S o Cl2 g = 39.1 J mol1 K 1 G o = H o T S o = +140.5 kJ/mol 298 K 39.1103 kJ mol1 K 1 = +140.5 kJ/mol +11.6 kJ/mol = +152.1 kJ/mol Thus the reaction is nonspontaneous at 25 C . (we also could have used values of Gf to calculate G o ).
(b)
o = 252.3 J mol 1 K 1 + 205.1 J mol1 K 1 50.33 J mol 1 K 1 2 223.1 J mol 1 K 1 For the cited reaction, G o = 2 Gf CO 2 g 2 Gf CO g Gf O 2 g G o = 2 394.4 kJ/mol 2 137.2 kJ/mol 0.00 kJ/mol = 514.4 kJ/mol Then we couple the two reactions.
TiO 2 s + 2Cl2 g TiCl4 l + O 2 g 2CO g + O 2 g 2CO 2 g _________________________________________________________________ G o = +152.1 kJ/mol G o = 514.4 kJ/mol TiO 2 s + 2Cl2 g + 2CO g TiCl4 l + 2CO 2 g ; G o = 362.3 kJ/mol The coupled reaction has G o 0 , and, therefore, is spontaneous. 977 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 68. (E) If G o 0 for the sum of coupled reactions, the reduction of the oxide with carbon is spontaneous. (a)
NiO s Ni s + 1 O 2 g 2 C s + 1 O 2 g CO g 2 Net : NiO s + C s Ni s + CO g G o = +115 kJ G o = 250 kJ G o = +115 kJ 250 kJ = 135 kJ G o = +280 kJ G o = 250 kJ G o = +280 kJ 250 kJ = +30 kJ G o = +630 kJ
G o = 2 250 kJ = 500 kJ Therefore the coupled reaction is spontaneous
(b)
MnO s Mn s + 1 O 2 g 2 C s + 1 O 2 g CO g 2 Net : MnO s +C s Mn s + CO g Therefore the coupled reaction is nonspontaneous
(c)
TiO 2 s Ti s + O 2 g 2C s + O 2 g 2CO g Net : TiO 2 s + 2C s Ti s + 2CO g G o = +630 kJ 500 kJ = +130 kJ Therefore the coupled reaction is nonspontaneous
69. (E) In this problem we need to determine if the phosphorylation of arginine with ATP is a spontaneous reaction. We proceed by coupling the two given reactions in order to calculate Gto for the overall reaction. The sign of Gto can then be used to determine whether the reaction is spontaneous or not. Stepwise approach: First determine Gto for the coupled reaction: ATP+H 2O ADP+P G to 31.5kJmol1 arginine+P phosphorarginine+H 2O G to 33.2kJmol1 ___________________________________________ ATP+arginine phosphorarginine+ADP G o (31.5 33.2)kJmol1 1.7kJmol1 Examine the sign of Gto : Gto 0 . Therefore, the reaction is not spontaneous. Conversion pathway approach: Gto for the coupled reaction is:
ATP+arginine phosphorarginine+ADP G o (31.5 33.2)kJmol1 1.7kJmol1 Since Gto 0 ,the reaction is not spontaneous. 978 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 70. (E) By coupling the two reactions, we obtain: Glu  +NH + Gln+H 2 O G to 14.8kJmol1 4 ATP+H 2O ADP+P G to 31.5kJmol1 _________________________________________ Glu  +NH + +ATP Gln+ADP+P G o (14.8 31.5)kJmol1 16.7kJmol1 4 Therefore, the reaction is spontaneous. INTEGRATIVE AND ADVANCED EXERCISES
71. (M) (a) The normal boiling point of mercury is that temperature at which the mercury vapor pressure is 1.00 atm, or where the equilibrium constant for the vaporization equilibrium reaction has a numerical value of 1.00. This is also the temperature where G 0 , since G RT ln K eq and ln (1.00) 0 . Hg(l) Hg(g) H H f [Hg(g)] H f [Hg(l)] 61.32 kJ/mol 0.00 kJ/mol 61.32 kJ/mol S S [Hg(g)] S [Hg(l)] 175.0 J mol 1 K 1 76.02 J mol 1 K 1 99.0 J mol 1 K 1 0 H T S 61.32 10 3 J/mol T 99.0 J mol 1 K 1 61.32 10 3 J/mol T 619 K 99.0 J mol 1 K 1
(b) The vapor pressure in atmospheres is the value of the equilibrium constant, which is related to the value of the free energy change for formation of Hg vapor.
Gf [Hg(g)] 31.82 kJ/mol RT ln K eq 31.82 103 J/mol K e 12.84 2.65 106 atm 12.84 8.3145 J mol1 K 1 298.15 K Therefore, the vapor pressure of Hg at 25C is 2.65106 atm. ln K 72. (M) (a) TRUE; It is the change in free energy for a process in which reactants and products are all in their standard states (regardless of whatever states might be mentioned in the statement of the problem). When liquid and gaseous water are each at 1 atm at 100 C (the normal boiling point), they are in equilibrium, so that G = G = 0 is only true when the difference of the standard free energies of products and reactants is zero. A reaction with G = 0 would be at equilibrium when products and reactants were all present under standard state conditions and the pressure of H2O(g) = 2.0 atm is not the standard pressure for H2O(g). (b) FALSE; G 0. The system is not at equilibrium. 979 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (c) FALSE; G can have only one value at any given temperature, and that is the
value corresponding to all reactants and products in their standard states, so at the normal boiling point G = 0 [as was also the case in answering part (a)]. Water will not vaporize spontaneously under standard conditions to produce water vapor with a pressure of 2 atmospheres. G 0. The process of transforming water to vapor at 2.0 atm pressure at 100C is not a spontaneous process; the condensation (reverse) process is spontaneous. (i.e. for the system to reach equilibrium, some H2O(l) must form) 73. (D) G 1 Gf [Br2 (g)] 1 Gf [Cl 2 (g)] Gf [BrCl(g)] 2 2 1 (3.11 kJ/mol) 1 (0.00 kJ/mol) (0.98 kJ/mol) 2.54 kJ/mol RT ln K p 2 2
(d) TRUE; ln K p 2.54 10 3 J/mol G 1.02 RT 8.3145 J mol 1 K 1 298.15 K K p e 1.02 0.361 For ease of solving the problem, we double the reaction, which squares the value of the equilibrium constant. K eq (0.357) 2 0.130 Reaction: 2 BrCl(g) Br2 (g) Cl2 (g) Initial: 1.00 mol 0 mol 0 mol x mol x mol Changes: 2 x mol Equil: (1.00 2 x)mol x mol x mol P{Br2 (g)} P{Cl 2 (g)} [n{Br2 (g)}RT / V ][n{Cl 2 (g)}RT / V ] n{Br2 (g)} n{Cl 2 (g)} Kp [n{BrCl(g)}RT / V ] 2 P{BrCl(g)}2 n{BrCl(g)}2 x2 x (0.361) 2 0.361 x 0.361 0.722 x 2 1.00 2 x (1.00 2 x) 0.361 x 0.210 mol Br2 0.210 mol Cl 2 1.00 2 x 0.580 mol BrCl 1.722 74. (M) First we determine the value of Kp for the dissociation reaction. If I2(g) is 50% dissociated, then for every mole of undissociated I2(g), one mole of I2(g) has dissociated, producing two moles of I(g). Thus, the partial pressure of I(g) is twice the partial pressure of I2(g) ( I 2 (g) 2 I(g) ). Ptotal 1.00 atm PI 2 (g) PI(g) PI 2 (g) 2 PI 2 (g) 3 PI 2 (g) PI 2 (g ) 0.333 atm Kp PI(g) 2 PI 2 (g ) (0.667) 2 1.34 0.333 ln K p 0.293 H 2 H f [I(g)] H f [I 2 (g)] 2 106.8 kJ/mol 62.44 kJ/mol 151.2 kJ/mol S 2 S [I(g)] S [I 2 (g)] 2 180.8 J mol 1 K 1 260.7 J mol 1 K 1 100.9 J mol 1 K 1 Now we equate two expressions for G and solve for T. 980 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy G H T S RT ln K p 151.2 10 3 100.9T 8.3145 T 0.293 151.2 10 3 1535 K 1.5 10 3 K 98.5 75. (M) (a) The oxide with the most positive (least negative) value of Gf is the one that most readily decomposes to the free metal and O2(g), since the decomposition is the reverse of the formation reaction. Thus the oxide that decomposes most readily is Ag2O(s). 151.2 10 3 100.9 T 2.44 T 98.4 T
T
(b) The decomposition reaction is 2 Ag 2 O(s) 4 Ag(s) O 2 (g) For this reaction K p PO 2 (g) . Thus, we need to find the temperature where K p 1.00 .Since
G RT ln K p and ln (1.00) 0 , we wish to know the temperature where G 0 . Note also that the decomposition is the reverse of the formation reaction. Thus, the following values are valid for the decomposition reaction at 298 K. H 31.05 kJ/mol G 11.20 kJ/mol We use these values to determine the value of S for the reaction. H G G H T S T S H G S T 3 3 31.05 10 J/mol 11.20 10 J/mol S 66.6 J mol 1 K 1 298 K Now we determine the value of T where G 0 . H G 31.05 10 3 J/mol 0.0 J/mol T 466 K 193 C S 66.6 J mol 1 K 1
76. (M) At 127 C = 400 K, the two phases are in equilibrium, meaning that G rxn 0 H rxn TS rxn [H f (yellow) H f (red)] T [ S (yellow) S (red)] [102.9 (105.4)] 10 3 J 400 K [ S ( yellow) 180 J mol 1 K 1 ] 2.5 10 3 J/mol 400 K S ( yellow) 7.20 10 4 J/mol (7.20 10 4 2.5 10 3 ) J/mol S ( yellow) 186 J mol 1 K 1 400 K Then we compute the value of the "entropy of formation" of the yellow form at 298 K. S f S [HgI 2 ] S [Hg(l)] S [I 2 (s)] [186 76.02 116.1] J mol1 K 1 6 J mol1 K 1 Now we can determine the value of the free energy of formation for the yellow form. kJ J 1kJ kJ Gf H f T S f 102.9 [298 K ( 6 ) ] 101.1 mol K mol 1000 J mol
77. (M) First we need a value for the equilibrium constant. 1% conversion means that 0.99 mol N2(g) are present at equilibrium for every 1.00 mole present initially. K Kp PNO(g) 2 PN2 (g) PO2 (g) [n{NO(g)}RT/V]2 n{NO(g)}2 [n{N 2 (g)}RT/V][n{O 2 (g)}RT/V] n{N 2 (g)} n{O 2 (g)} 981 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Reaction: Initial: Changes(1% rxn): Equil: N 2 (g) O 2 (g) 2 NO(g) 0 mol 0.020 mol 0.020 mol
K (0.020) 2 4.1 104 (0.99)(0.99) 1.00 mol 1.00 mol 0.010 mol 0.010 mol 0.99 mol 0.99 mol The cited reaction is twice the formation reaction of NO(g), and thus
H = 2Hf [NO(g)] = 2 90.25 kJ/mol = 180.50 kJ/mol S = 2S[NO(g)]  S[N 2 (g)]  S[O 2 (g)] =2 (210.7 J mol1 K)191.5 J mol1 K 1 205.0 J mol1 K 1 = 24.9 J mol1 K 1
G = RTlnK = 8.31447 JK 1mol1 (T)ln(4.1104 ) = 64.85(T) G = HTS = 64.85(T) = 180.5 kJ/mol  ( T)24.9 J mol1 K 1 180.5 10 J/mol = 64.85(T) + ( T)24.9 J mol1 K 1 = 89.75(T) 78. (E) Sr(IO3)2(s) Sr2+(aq) + 2 IO3(aq) T = 2.01 103 K G = (2 mol(128.0 kJ/mol) + (1 mol 500.5 kJ/mol)) (1 mol 855.1 kJ/mol) = 0.4 kJ G = RTlnK = 8.31447 JK1mol1(298.15 K)ln K = 0.4 kJ = 400 J ln K = 0.16 and K = 1.175 = [Sr2+][IO3]2 Let x = solubility of Sr(IO3)2 [Sr2+][IO3]2 = 1.175 = x(2x)2 = 4x3 x = 0.665 M for a saturated solution of Sr(IO3)2.
79. (M) PH2O = 75 torr or 0.0987 atm. K = (PH2O)2 = (0.0987)2 = 9.74 103 G = (2 mol(228.6 kJ/mol) + (1 mol)918.1 kJ/mol) (1 mol)1400.0 kJ/mol = 309.0 kJ H = (2 mol(241.8 kJ/mol) + (1 mol)1085.8.1 kJ/mol) (1 mol)1684.3 kJ/mol = 114.9 kJ S = (2 mol(188.J/K mol) + (1 mol146.J/K mol) + (1 mol)221.3.J/K mol = 302.3 J/K mol G = RT ln K eq = 8.3145 J mol1 K 1 T ln (9.74 103 ) = 38.5(T) G = 38.5(T) = HTS = 114,900 J mol1  (T) 302.3J K 1 mol1 114,900 = 340.8 K 1 (T) Hence: T = 337 K = 64 o C 131 103 J/mol G 80. (D) (a) G RT ln K 52.8 ln K RT 8.3145 J mol1 K 1 298.15 K
K e52.8 1.2 1023 atm 760 mmHg 8.9 1021 mmHg 1atm Since the system cannot produce a vacuum lower than 109 mmHg, this partial pressure of CO2(g) won't be detected in the system. 982 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (b) Since we have the value of G for the decomposition reaction at a specified temperature (298.15 K), and we need H and S for this same reaction to determine P{CO 2 (g)} as a function of temperature, obtaining either H or S will enable us to determine the other. (c) H H f [CaO(s)] H f [CO 2 (g)] H f [CaCO 3 (s)] 635.1 kJ/mol 393.5 kJ/mol (1207 kJ/mol) 178 kJ/mol G H T S S T S H G S H G T 178 kJ/mol 131 kJ/mol 1000 J 1.6 102 J mol1 K 1 298. K 1 kJ 1 atm 1.3 1012 760 mmHg K 1.0 109 mmHg G H T S RT ln K eq T H T S RT ln K 178 103 J/mol H 4.6 102 K S R ln K 1.6 102 J mol1 K 1 8.3145 J mol1 K 1 ln(1.3 1012 ) 81. (D) H H f [PCl 3 (g)] H f [Cl 2 (g)] H f [PCl 5 (g)] 287.0 kJ/mol 0.00 kJ/mol (374.9 kJ/mol) 87.9 kJ/mol S S [PCl3 (g)] S [Cl2 (g)] S [PCl5 (g)] 311.8 J mol1 K 1 + 223.1 J mol1 K 1 364.6 J mol1 K 1 = +170.3 J mol1 K 1 G H T S 87.9 10 3 J/mol 500 K 170.3 J mol 1 K 1 G 2.8 103 J/mol RT ln K p ln K p 2.8 10 J/mol G 0.67 RT 8.3145 J mol1 K 1 500 K K p e0.67 0.51 Pi [PCl 5 ] nRT 0.100 mol 0.08206 L atm mol 1 K 1 500 K 2.74 atm V 1.50 L Reaction: Initial: Changes: Equil: PCl5 (g) PCl3 (g) 0 atm x atm x atm Cl2 (g) 0 atm x atm x atm 2.74 atm x atm (2.74 x) atm 983 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Kp P[PCl3 ] P[Cl2 ] xx 0.51 P[PCl5 ] 2.74 x x 2 0.51(2.74 x) 1.4 0.51 x x 2 0.51 x 1.4 0
b b 2 4ac 0.51 0.26 5.6 0.96 atm, 1.47 atm 2a 2 Ptotal PPCl5 PPCl3 PCl 2 (2.74 x) x x 2.74 x 2.74 0.96 3.70 atm
x 82. (M) The value of H determined in Exercise 64 is H 128 kJ/mol . We use any one of the values of K p K eq to determine a value of G . At 30 C = 303 K,
G RT ln K eq (8.3145 J mol 1 K 1 )(303 K ) ln(1.66 10 5 ) 2.77 10 4 J/mol Now we determine S . G H T S H G 128 10 3 J/mol 2.77 10 4 J/mol S 331 J mol 1 K 1 T 303 K By using the appropriate S values in Appendix D, we calculate S 334 J mol 1 K 1 .
83. (M) In this problem we are asked to estimate the temperature at which the vapor pressure of cyclohexane is 100 mmHg. We begin by using Trouton's rule to determine the value of H vap for cyclohexane. The temperature at which the vapor pressure is 100.00 mmHg can then be determined using ClausiusClapeyron equation. Stepwise approach: Use Trouton's rule to find the value of H vap :
H vap Tnbp S vap 353.9 K 87 J mol 1 K 1 31 10 3 J/mol Next, use ClausiusClapeyron equation to find the required temperature: 100 mmHg P H vap 1 1 ln 2 T T ln 760 mmHg P1 R 1 2 31 10 3 J/mol 1 1 2.028 1 1 8.3145 J mol K 353.9 K T 1 2.028 8.3145 1 5.4 10 4 31 10 3 353.9 T 1 1 2.826 10 3 3.37 10 3 K 1 T T T 297 K 24 C 984 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Conversion pathway approach: P H vap 1 1 Tnbp Svap 1 1 ln 2 T T P1 R T1 T2 R 1 2
1 1 R P2 1 1 R P2 T T T S ln P T T T S ln P 1 2 nbp vap 1 2 1 nbp vap 1 1 1 8.314JK 1mol1 100mmHg ln 3.37 10 3 1 1 T2 353.9K 353.9K 87JK mol 760mmHg T2 297K=24 oC 84. (M)(a) 92. 1 2Ag(s) O 2 (g) Ag 2 O 2 1 G o (O 2 )} f 2 G o G o (Ag 2 O(s)){2G o ( Ag(s)) f f f 1 G o 11.2 kJ{2(0) (0)} 11.2 kJ Ag 2 O is thermodynamically stable at 25oC f 2 o o (b) Assuming H , S are constant from 25200oC (not so, but a reasonable assumption !)
So So (Ag 2 O){2 So (Ag(s)) 1 o 1 S (O 2 )} 121.3 (2(42.6) (205.1)) 66.5 J/K 2 2 (473 K)(66.5 J/K) G o 31.0 kJ H o T S o 0.45 kJ 1000 J/kJ thermodynamically unstable at 200oC 85. (M) G = 0 since the system is at equilibrium. As well, G = 0 because this process is under standard conditions. Since G = H  TS = 0. = H = TS =273.15 K21.99 J Since we are dealing with 2 moles of ice melting, the K1 mol1 = 6.007 kJ mol1. values of H and S are doubled. Hence, H = 12.01 kJ and S = 43.98 J K1. Note: The densities are not necessary for the calculations required for this question. 86. (D) First we determine the value of Kp that corresponds to 15% dissociation. We represent the initial pressure of phosgene as x atm. Reaction: COCl2 (g) CO(g) Initial : Changes: Equil: x atm 0.15 x atm 0.85 x atm Cl2 (g) 0 atm 0 atm 0.15 x atm 0.15 x atm 0.15 x atm 0.15 x atm 985 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Ptotal 0.85 x atm 0.15 x atm 0.15 x atm 1.15 x atm 1.00 atm
Kp PCO PCl2 PCOCl 2 (0.15 0.870) 2 0.0230 0.85 0.870 x 1.00 0.870 atm 1.15 Next we find the value of H for the decomposition reaction.
ln K 1 H 1 1 6.7 10 9 H 1 1 H ln 15.71 (1.18 10 3 ) 2 T T K2 R 2 R 668 373.0 R 4.44 10 1 H 15.71 1.33 10 4 , 3 R 1.18 10 H 1.33 10 4 8.3145 111 10 3 J/mol 111 kJ/mol And finally we find the temperature at which K = 0.0230. K1 H 1 1 0.0230 111 103 J/mol 1 1 ln 0.658 1 1 T T K2 R 2 0.0444 8.3145 J mol K 668 K T 1 1 1 0.658 8.3145 1 1 4.93 10 5 1.497 103 1.546 10 3 3 T T 668 T 111 10 T 647 K 374 C ln 87. (D) First we write the solubility reaction for AgBr. Then we calculate values of H and S for the reaction: AgBr(s) Ag (aq) Br (aq) K eq K sp [Ag ][Br ] s 2 H H f [Ag (aq)] H f [Br (aq)] H f [AgBr(s)] 105.6 kJ/mol 121.6 kJ/mol (100.4 kJ/mol) 84.4 kJ/mol S S [Ag (aq)] S [Br (aq)] S [AgBr(s)] 72.68 J mol 1 K 1 82.4 J mol 1 K 1 107.1 J mol 1 K 1 48.0 J mol 1 K 1 These values are then used to determine the value of G for the solubility reaction, and the standard free energy change, in turn, is used to obtain the value of K. G H T S 84.4 103 J mol1 (100 273) K 48.0 J mol1 K 1 66.5 103 J/mol 66.5 103 G ln K 21.4 K K sp e 21.4 5.0 1010 s 2 8.3145 J RT 373K mol K And now we compute the solubility of AgBr in mg/L. 187.77 g AgBr 1000 mg s 5.0 10 10 4.2 mg AgBr/L 1 mol AgBr 1g 986 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 88. (M) S298.15 = S274.68 + Sfusion + Sheating 298.15 12,660 J mol1 J J S298.15 = 67.15 J K mol1 + + 97.78 + 0.0586 T 274.68 274.68 274.68 K mol K mol K 2 S298.15 = 67.15 J K1 mol1 + 46.09 J K1 mol1 + 8.07 J K1 mol1 = 121.3 J K1 mol1 89. (M) S = Ssolid + Sfusion + Sheating + Svaporization + Spressure change J 134.0 dT 1 298.15 9866 J mol 33,850 J mol1 1 1 mol K + + S = 128.82 J K mol + 278.68 298.15 K T 278.68 K 95.13 torr + 8.3145 J K1 mol1 ln 760 torr S = 128.82 J K1 mol1 + 35.40 J K1 mol1 + 9.05 J K1 mol1 + 113.5 J K1 mol1 + (17.28 J K1 mol1 ) S = 269.53 J K1 mol1 90. (D) Start by labeling the particles A, B and C. Now arrange the particles among the states. One possibility includes A, B, and C occupying one energy state (=0,1,2 or 3). This counts as one microstate. Another possibility is two of the particles occupying one energy state with the remaining one being in a different state. This possibility includes a total of three microstates. The final set of combinations is one with each particle being in different energy state. This combination includes a total of six microstates. Therefore, the total number of microstates in the system is 10. See pictorial representation on the following page illustrating the three different cases. 987 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy All particles are in a single energy level ( i.e. =0) =3 =2 =1 =0
A B C 1 microstate Two particles are in one energy level (i.e. =0) and the remaining particle is in a different level (i.e. =2) =3 =2 =1 =0
A B C =3 =2 =1 =0
A B =3 =2 =1
C A 3 microstates B C =0 All particles are in different energy levels (i.e. =0, 1 and 2) 1 microstate 1 microstate 1 microstate =3 =2 =1 =0
C B =3 =2 =1 =0
C A =3 =2 =1 =0
B A A B C 1 microstate 1 microstate 1 microstate 6 microstates =3 =2 =1 =0
B C =3 =2 =1 =0
A C =3 =2 =1 =0
A B A B C 1 microstate 1 microstate 1 microstate 91. (M) (a) In the solid as the temperature increases, so do the translational, rotational, and vibrational degrees of freedom. In the liquid, most of the vibrational degrees of freedom are saturated and only translational and rotational degrees of freedom can increase. In the gas phase, all degrees of freedom are saturated. (b) The increase in translation and rotation on going from solid to liquid is much less than on going from liquid to gas. This is where most of the change in entropy is derived. 988 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 92. (D) Because KNO3 is a strong electrolyte, its solution reaction will be: KNO3 ( s ) H 2O K (aq ) NO3 (aq ) This reaction can be considered to be at equilibrium when the solid is in contact with a saturated solution, i.e. the conditions when crystallization begins. The solubility, s, of the salt, in moles per liter, can be calculated from the amount of salt weighted out and the volume of the solution. The equilibrium constant K for the reaction will be: K=[K + (aq)][NO3 (aq)]=(s)(s)=s 2 In the case of 25.0 mL solution at 340 K, the equilibrium constant K is: m 20.2g n 0.200mol n(KNO3 ) 0.200mol s= 8.0molL1 1 M 101.103gmol V 0.0250L
K s 2 8 2 64 The equilibrium constant K can be used to calculate G for the reaction using G=RTlnK: G 8.314JK 1mol1 340K ln 64 12kJmol1 The values for K and G at all other temperatures are summarized in the table below. Volume (mL) 25.0 29.2 33.4 37.6 41.8 46.0 51.0 T/(K) 340 329 320 313 310 306 303 1/T (K1) 0.00294 0.00304 0.00312 0.00320 0.00322 0.00327 0.00330 s (molL1) 8.0 6.9 6.0 5.3 4.8 4.3 3.9 K 64 48 36 28 23 18.5 15 lnK 4.2 3.9 3.6 3.3 3.1 2.9 2.7 G (kJmol1) 12 11 9.6 8.6 8.0 7.4 6.8 The plot of lnK v.s. 1/T provides H (slope=H/R) and S (yintercept=S/R) for the reaction:
4.5 y = 16.468  4145.7x R= 0.99156 4 ln K 3.5 3 2.5 0.0029 0.003 0.0031 0.0032
1 0.0033 0.0034 1/T (K ) H slope R 4145.7 8.314JK mol 34.5kJmol1
1 1 S y int ercept R 16.468 8.314JK 1mol1 136.9JK 1mol1 989 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy H for the crystallization process is 35.4 kJmol1. It is negative as expected because crystallization is an exothermic process. Furthermore, the positive value for S shows that crystallization is a process that decreases the entropy of a system. FEATURE PROBLEMS
93. (M) (a) The first method involves combining the values of Gfo . The second uses G o = H o T S o G o = Gf H 2 O g Gf H 2 O l = 228.572 kJ/mol 237.129 kJ/mol = +8.557 kJ/mol = 241.818 kJ/mol 285.830 kJ/mol = +44.012 kJ/mol H o = H f H 2 O g H f H 2 O l S = S o H 2 O g S o H 2 O l o = 188.825 J mol1 K 1 69.91 J mol1 K 1 = +118.92 J mol1 K 1
G o = H o T S o = 44.012 kJ/mol 298.15 K 118.92 103 kJ mol1 K 1 = +8.556 kJ/mol (b) We use the average value: G o = +8.558 103 J/mol = RT ln K 8558 J/mol ln K = = 3.452; K = e 3.452 = 0.0317 bar 8.3145 J mol1 K 1 298.15 K (c) P{H 2 O} = 0.0317 bar ln K = 1atm 760 mmHg = 23.8 mmHg 1.01325 bar 1atm (d) 8590 J/mol = 3.465 ; 8.3145 J mol1 K 1 298.15 K K = e 3.465 = 0.03127 atm ; 760 mmHg P{H 2 O} = 0.0313atm = 23.8 mmHg 1atm
94. (D) (a) When we combine two reactions and obtain the overall value of G o , we subtract the value on the plot of the reaction that becomes a reduction from the value on the plot of the reaction that is an oxidation. Thus, to reduce ZnO with elemental Mg, we subtract the values on the line labeled " 2 Zn + O 2 2 ZnO " from those on the line labeled " 2 Mg + O 2 2 MgO ". The result for the overall G o will always be negative because every point on the "zinc" line is above the corresponding point on the "magnesium" line (b) In contrast, the "carbon" line is only below the "zinc" line at temperatures above about 1000 C . Thus, only at these elevated temperatures can ZnO be reduced by carbon. 990 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy (c) The decomposition of zinc oxide to its elements is the reverse of the plotted reaction, the value of G o for the decomposition becomes negative, and the reaction becomes spontaneous, where the value of G o for the plotted reaction becomes positive. This occurs above about 1850 C . The "carbon" line has a negative slope, indicating that carbon monoxide becomes more stable as temperature rises. The point where CO(g) would become less stable than 2C(s) and O2(g) looks to be below 1000 C (by extrapolating the line to lower temperatures). Based on this plot, it is not possible to decompose CO(g) to C(s) and O 2 g in a spontaneous reaction.
0 500 1000 1500 2000 (d) Standard Free Energy Change (kJ) (e) 0 100 200 300 400 500 600 700 1 2 3
Temperature (C) Reaction 1 2 CO(g) + O2(g) 2 CO2(g) Reaction 2 C(s) + O2(g) CO2(g) Reaction 3 2 C(s) + O2(g) 2 CO(g) All three lines are straightline plots of G vs. T following the equation G = H TS. The general equation for a straight line is given below with the slightly modified Gibbs FreeEnergy equation as a reference: G = STH (here H assumed constant) y = mx + b (m = S = slope of the line) Thus, the slope of each line multiplied by minus one is equal to the S for the oxide formation reaction. It is hardly surprising, therefore, that the slopes for these lines differ so markedly because these three reactions have quite different S values (S for Reaction 1 = 173 J K1, S for Reaction 2 = 2.86 J K1, S for Reaction 3 = 178.8 J K1)
(f) Since other metal oxides apparently have positive slopes similar to Mg and Zn, we can conclude that in general, the stability of metal oxides decreases as the temperature increases. Put another way, the decomposition of metal oxides to their elements becomes more spontaneous as the temperature is increased. By contrast, the two reactions involving elemental carbon, namely Reaction 2 and Reaction 3, have negative slopes, indicating that the formation of CO2(g) and CO(g) from graphite becomes more favorable as the temperature rises. This means that the G for the reduction of metal oxides by carbon becomes more and more negative with 991 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy increasing temperature. Moreover, there must exist a threshold temperature for each metal oxide above which the reaction with carbon will occur spontaneously. Carbon would appear to be an excellent reducing agent, therefore, because it will reduce virtually any metal oxide to its corresponding metal as long as the temperature chosen for the reaction is higher than the threshold temperature (the threshold temperature is commonly referred to as the transition temperature). Consider for instance the reaction of MgO(s) with graphite to give CO2(g) and Mg metal: 2 MgO(s) + C(s) 2 Mg(s) + CO2(g) Srxn = 219.4 J/K and rxn = 809.9 kJ o H rxn 809.9 kJ Ttransition = = = 3691 K = Tthreshold o 0.2194 kJ K 1 Srxn Consequently, above 3691 K, carbon will spontaneously reduce MgO to Mg metal.
95. (D) (a) With a 36% efficiency and a condenser temperature (Tl) of 41 C = 314 K, Th 314 T T efficiency h 1 100% 36% 0.36 ; Th Th Th = (0.36 T h) + 314 K; 0.64 Th = 314 K; Th = 4.9 102 K The overall efficiency of the power plant is affected by factors other than the thermodynamic efficiency. For example, a portion of the heat of combustion of the fuel is lost to parts of the surroundings other than the steam boiler; there are frictional losses of energy in moving parts in the engine; and so on. To compensate for these losses, the thermodynamic efficiency must be greater than 36%. To obtain this higher thermodynamic efficiency, Th must be greater than 4.9 102 K. (b) (c) The steam pressure we are seeking is the vapor pressure of water at 4.9 102 K. We also know that the vapor pressure of water at 373 K (100 C) is 1 atm. The enthalpy of vaporization of water at 298 K is H = Hf[H2O(g) Hf[H2O(l)] = 241.8 kJ/mol (285.8 kJ/mol) = 44.0 kJ/mol. Although the enthalpy of vaporization is somewhat temperature dependent, we will assume that this value holds from 298 K to 4.9 102 K, and make appropriate substitutions into the ClausiusClapeyron equation. P2 1 44.0 kJ mol1 1 3 3 3 ln = = 5.29 10 2.68 10 2.04 10 3 1 1 atm 8.3145 10 kJ mol 373 K 490 K P P ln 2 = 3.39 ; 2 = 29.7 ; P2 30 atm 1 atm 1 atm The answer cannot be given with greater certainty because of the weakness of the assumption of a constant Hvapn. (d) It is not possible to devise a heat engine with 100% efficiency or greater. For 100% efficiency, Tl = 0 K, which is unattainable. To have an efficiency greater than 100 % would require a negative Tl, which is also unattainable. 992 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 96. (D) (a) Under biological standard conditions: ADP3 + HPO42 + H+ ATP4+ H2O Go' = 32.4 kJ/mol If all of the energy of combustion of 1 mole of glucose is employed in the conversion of ADP to ATP, then the maximum number of moles ATP produced is 2870 kJ mol1 88.6 moles ATP 32.4 kJ mol1 In an actual cell the number of ATP moles produced is 38, so that the efficiency is: number of ATP's actually produced 38 Efficiency = 0.43 Maximum number of ATP's that can be produced 88.6 Thus, the cell's efficiency is about 43%. Maximum number = The previously calculated efficiency is based upon the biological standard state. We now calculate the Gibbs energies involved under the actual conditions in the cell. To do this we require the relationship between G and G' for the two coupled reactions. For the combustion of glucose we have 6 6 aCO aH O o 2 2 G G RT ln 6 aglu aO 2 For the conversion of ADP to ATP we have aATP aH O o 2 RT ln G G aADP aP H + / 107 i Using the concentrations and pressures provided we can calculate the Gibbs energy for the combustion of glucose under biological conditions. First, we need to replace the activities by the appropriate effective concentrations. That is, 6 p / po a6 CO2 H 2 O o G G RT ln 6 o o glu / glu p / p O 2 using aH2O 1 for a dilute solution we obtain (b) (c) 6 6 RT ln 0.050 bar /1 bar 1 G G 6 glu /1 0.132 bar /1 bar The concentration of glucose is given in mg/mL and this has to be converted to molarity as follows: 1.0 mg g 1000 mL 1 [glu] = 0.00555 mol L1 , 1 mL 1000 mg L 180.16 g mol where the molar mass of glucose is 180.16 g mol1. Assuming a temperature of 37 oC for a biological system we have, for one mole of glucose: 993 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 6 0.050 16 G 2870 103 J 8.314 JK 1 310K ln 6 0.00555 /1 0.132 3 2.954 10 G 2870 103 J 8.314 JK 1 310K ln 0.00555 G 2870 103 J 8.314 JK 1 310 K ln 0.5323 G 2870 103 J 8.314 JK 1 310K 0.6305 G 2870 103 J 1.625 103 J G 2872 103 J In a similar manner we calculate the Gibbs free energy change for the conversion of ADP to ATP: ATP / 1 1 o G G RT ln 7 ADP / 1 Pi / 1 ( H / 10 ) 0.0001 G 32.4 103 J 8.314 JK 1 310 K ln 7 7 0.0001 0.0001 (10 /10 ) G 32.4 103 J 8.314 JK 1 310 K ln 104 G 32.4 103 J 8.314 JK 1 310 K 9.2103) 32.4 103 J 23.738 103 J 56.2 103 J (d) The efficiency under biological conditions is Efficiency = number of ATP's actually produced 38 0.744 Maximum number of ATP's that can be produced 2872/56.2 Thus, the cell's efficiency is about 74%. The theoretical efficiency of the diesel engine is: Th T1 1923 873 100% 100% 55% Th 1923 Thus, the diesel's actual efficiency is 0.78 55 % = 43 %. The cell's efficiency is 74% whereas that of the diesel engine is only 43 %. Why is there such a large discrepancy? The diesel engine supplies heat to the surroundings, which is at a lower temperature than the engine. This dissipation of energy raises the temperature of the surroundings and the entropy of the surroundings. A cell operates under isothermal conditions and the energy not utilized goes only towards changing the entropy of the cell's surroundings. The cell is more efficient since it does not heat the surroundings. Efficiency 994 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 97. (E) (a) In this case CO can exist in two states, therefore, W=2. There are N of these states in the crystal, and so we have S k ln 2 N 1.381 10 23 JK 1 6.022 10 23 mol1 ln 2 5.8JK 1mol1 (b) For water, W=3/2, which leads to 3 S k ln( )N 1.381 10 23 JK 1 6.022 10 23 mol1 ln1.5 3.4JK 1mol1 2 SELFASSESSMENT EXERCISES
98. (E) (a) Suniv or total entropy contains contributions from the entropy change of the system ( Ssys ) and the surroundings ( Ssurr ). According to the second law of thermodynamics, Suniv is always greater then zero. (b) G o or standard free energy of formation is the free energy change for a reaction in f which a substances in its standard state is formed from its elements in their reference forms in their standard states. (c) For a hypothetical chemical reaction aA+bB cC+dD , the equilibrium constant K is [C]c [D]d defined as K . [A]a [B]b
99. (E) (a) Absolute molar entropy is the entropy at zeropoint energy (lowest possible energy state) and it is equal to zero. (b) Coupled reactions are spontaneous reactions (G<0) that are obtained by pairing reactions with positive G with the reactions with negative G. (c) Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization has a value of approximately 87 Jmol1K1. (d) Equilibrium constant K for a certain chemical reaction can be evaluated using either G o or H o in conjunction with S o (which are often tabulated). The relationship used f f to calculate K is G o RT ln K .
100. (E) (a) A spontaneous process is a process that occurs in a system left to itself; once started, no external action form outsize the system is necessary to make the process continue. A nonspontaneous process is a process that will not occur unless some external action is continuously applied. (b) Second law of thermodynamics states that the entropy of universe is always greater than zero or in other words that all spontaneous processes produce an increase in the entropy of the universe. The third law of thermodynamics states that the entropy of perfect pure crystal at 0K is zero. (c) G is the Gibbs free energy defined as G=HTS. G0 is the standard free energy change. (E) Second law of thermodynamics states that all spontaneous processes produce an increase in the entropy of the universe. In other words, Suniv Ssys Ssurr 0 . 101. Therefore, the correct answer is (d). 995 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy 102. (E) The Gibbs free energy is a function of H, S and temperature T. It cannot be used to determine how much heat is absorbed from the surroundings or how much work the system does on the surroundings. Furthermore, it also cannot be used to determine the proportion of the heat evolved in an exothermic reaction that can be converted to various forms of work. Since Gibbs free energy is related to the equilibrium constant of a chemical reaction ( G RT ln K ) its value can be used to access the net direction in which the reaction occurs to reach equilibrium. Therefore, the correct answer is (c). (M) In order to answer this question, we must first determine whether the entropy change for the given reaction is positive or negative. The reaction produces three moles of gas from two moles, therefore there is an increase in randomness of the system, i.e. entropy change for the reaction is positive. Gibbs free energy is a function of enthalpy, entropy and temperature ( G H T S ). Since H 0 and S 0 , this reaction will be spontaneous at any temperature. The correct answer is (a). (M) Recall that G o RT ln K . If G o 0 , then it follows that G o RT ln K 0 . Solving for K yields: ln K 0 K e0 1 . Therefore, the correct answer is (b). (E) In this reaction, the number of moles of reactants equals the number of moles of products. Therefore, K is equal to Kp and Kc. The correct answers are (a) and (d). (M) (a) The two lines will intersect at the normal melting point of I2(s) which is 113.6 C. (b) G o for this process must be equal to zero because solid and liquid are at equilibrium and also in their standard states.
o 103. 104. 105. 106. 107. (M) (a) No reaction is expected because of the decrease in entropy and the expectation that the reaction is endothermic. As a check with data from Appendix D, G o =326.4 kJmol1 for the reaction as writtena very large value. (b) Based on the increase in entropy, the forward reaction should occur, at least to some extent. For this reaction G o =75.21 kJmol1. (c) S is probably small, and H is probably also small (one ClCl bond and one BrBr bonds are broken and two BrCl bonds are formed). G o should be small and the forward reaction should occur to a significant extent. For this reaction G o =5.07 kJmol1. (M) (a) Entropy change must be accessed for the system and its surroundings ( Suniv ), not just for the system alone. (b) Equilibrium constant can be calculated from G o ( G o RT ln K ), and K permits equilibrium calculations for nonstandard conditions. (D)
o f 108. 109.
H
o vap (a) First we need to determine
o f o H vap which is simply equal to: H [(C5 H 10 (g)] H [(C5 H 10 (l)] 77.2kJ/mol (105.9kJ/mol) 28.7kJ/mol . Now we use Trouton's rule to calculate the boiling point of cyclopentane: 996 Chapter 19: Spontaneous Change: Entropy and Gibbs Energy S o vap o H vap Tbp 87Jmol K Tbp 1 1 o H vap 87Jmol1K 1 28.7 1000Jmol1 330K 87Jmol1K 1 Tbp 330K 273.15K 57 o C
o o o (b) If we assume that H vap and Svap are independent of T we can calculate Gvap : 87 kJK 1mol1 2.8kJmol1 1000 o (c) Because Gvap,298 K >0, the vapor pressure is less than 1 atm at 298 K, consistent with
o o o Gvap,298 K H vap T Svap 28.7kJmol1 298.15K Tbp=57 oC.
110. (M) (a) We can use the data from Appendix D to determine the change in enthalpy and entropy for the reaction: H o H o (N 2O(g)) 2H o (H 2O(l)) H o (NH 4 NO3 (s)) f f f H o 82.05kJmol1 2 (285.8kJmol1 ) (365.6kJmol1 ) 124kJmol1 S o S o (N 2O(g)) 2S o (H 2O(l)) S o (NH 4 NO3 (s)) S o 219.9JK 1mol1 2 69.91JK 1mol1 151.1JK 1mol1 208.6JK 1mol1 (b) From the values of H o and S o determined in part (a) we can calculate G o at 298K: G o H o T S o G o 124kJmol1 298K Alternatively, G o 208.6kJmol1K 1 186.1kJmol1 1000 can also be calculated directly using G o values tabulated in f Appendix D. (c) The equilibrium constant for the reaction is calculated using G o RT ln K : G o RT ln K 186.1 1000Jmol1 8.314JK 1mol1 298K ln K
186100Jmol1 2477.6 ln K ln K 75.1 K e 75.1 4.1 10 32 (d) The reaction has H o 0 and S o 0 . Because G o H o T S o , the reaction will be spontaneous at all temperatures. 111. (M) Recall from exercise 104 that G o 0 when K=1. Therefore, we are looking for the diagram with smallest change in Gibbs free energy between the products and the reactants. The correct answer is diagram (a). Notice that diagrams (b) and (c) represent chemical reactions with small and large values of equilibrium constants, respectively. (M) Carbon dioxide is a gas at room temperature. The melting point of carbon dioxide is expected to be very low. At room temperature and normal atmospheric pressure this process is spontaneous. The entropy of the universe if positive. 112. 997 CHAPTER 20 ELECTROCHEMISTRY
PRACTICE EXAMPLES
1A (E) The conventions state that the anode material is written first, and the cathode material is written last. eAnode, oxidation: Sc s Sc 3+ aq + 3e Net reaction
1B Sc s + 3Ag aq Sc
+ ___________________________________ 3+ aq + 3Ag s
3+ anode salt bridge (E) Oxidation of Al(s) at the anode:
+ Reduction of Ag aq at the cathode: Ag aq + e
+ Al s Al aq + 3e Ag s ________________ Al NO3 Ag
 K + Al3+ Ag+ Overall reaction in cell: Al s + 3Ag + aq Al 3+ aq + 3Ag s 2A Diagram: Al(s)Al3+ (aq)Ag + (aq)Ag(s) 2B Cathode, reduction: {Cd 2 aq 2e Cd(s)} 3 ___________________________________________ Overall reaction in cell: 2In(s) 3Cd 2 (aq) 2In 3 (aq) 3Cd(s)
3A (M) We obtain the two balanced halfequations and the halfcell potentials from Table 201. Cathode, reduction: { Ag aq 1e Ag( s )} 2 ___________________________________________ Overall reaction in cell: Sn( s ) 2 Ag (aq ) Sn 2 (aq ) 2 Ag( s ) (E) Anode, oxidation: {In s In 3+ aq + 3e } 2 (E) Anode, oxidation: Sn s Sn 2 + aq + 2e Oxidation: {Fe 2+ aq Fe3+ aq + e } 2 Reduction: Cl2 g + 2e 2 Cl aq Net: 2Fe 2+ aq + Cl2 g 2Fe3+ aq + 2Cl aq ;
3B E o = 0.771V E o = +1.358V
o Ecell = +1.358V 0.771V = +0.587 V (M) Since we need to refer to Table 201, in any event, it is perhaps a bit easier to locate the two balanced halfequations in the table. There is only one halfequation involving both Fe 2+ aq and Fe 3+ aq ions. It is reversed and written as an oxidation below. The half equation involving MnO aq is also written below. [Actually, we need to know that in 4 acidic solution Mn 2+ aq is the principal reduction product of MnO aq .] 4 Oxidation: {Fe 2+ aq Fe3+ aq + e } 5 Reduction: MnO 4 aq + 8 H + aq + 5e Mn 2+ aq + 4 H 2 O(l) E o = 0.771V
E o = +1.51V Net: MnO 4 aq + 5 Fe 2+ aq + 8 H + aq Mn 2+ aq + 5 Fe3+ aq + 4H 2 O(l)
998 cathode Cathode, reduction: {Ag + aq + e Ag s } 3 Chapter 20: Electrochemistry o Ecell = +1.51 V 0.771 V = +0.74 V 4A (M) We write down the oxidation halfequation following the method of Chapter 5, and obtain the reduction halfequation from Table 201, along with the reduction halfcell potential. Oxidation: {H 2 C2 O 4 (aq) 2 CO 2 (aq) 2 H + (aq) 2 e } 3 Reduction: Cr2 O7
2 E o = +1.33V aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l) 2 Net: Cr2 O7 aq + 8 H + aq + 3H 2 C2 O 4 aq 2 Cr 3+ aq + 7 H 2 O(l) + 6 CO 2 g o Ecell = +1.81V = +1.33V E o {CO2 /H 2 C2 O4 }; E {CO 2 /H 2 C2 O 4 } E o{CO2 /H 2 C2 O4 } = 1.33V 1.81V = 0.48V 4B (M) The 2nd halfreaction must have O 2 g as reactant and H 2 O(l) as product. Oxidation: {Cr 2+ (aq) Cr 3+ (aq) e } 4 Reduction: O 2 g + 4H + aq + 4e 2H 2 O(l) Net: O 2 g + 4H aq 4 Cr (aq) 2H 2 O(l) + 4Cr
+ 2+ 3+ E {Cr 3 /Cr 2 } E o = +1.229V aq 3 ______________________________________ Ecell 1.653V 1.229V E {Cr / Cr } ;
o o 3 2 E {Cr /Cr 2 } 1.229V 1.653V 0.424V
o 5A (M) First we write down the two halfequations, obtain the halfcell potential for each, and o then calculate Ecell . From that value, we determine G o Oxidation: {Al s Al3+ aq +3e } 2 E o = +1.676V Reduction: {Br2 l +2e 2 Br aq } 3 E o = +1.065V
____________________________ o Net: 2 Al s + 3Br2 l 2 Al3+ aq + 6 Br aq Ecell = 1.676 V +1.065 V = 2.741V o G o = nFEcell = 6 mol e 96, 485 C 2.741V = 1.587 106 J = 1587 kJ 1mol e 5B (M) First we write down the two halfequations, one of which is the reduction equation from the previous example. The other is the oxidation that occurs in the standard hydrogen electrode. Oxidation: 2H 2 g 4H + aq + 4e Reduction: O 2 g + 4 H + aq + 4e 2 H 2 O(l) Net:
2 H 2 g + O2 g 2 H 2O l o n = 4 in this reaction. This net reaction is simply twice the formation reaction for H 2 O(l) and, therefore,
o G o = 2Gf H 2 O l = 2 237.1 kJ = 474.2 103 J = nFEcell 474.2 103 J G o o Ecell = = 1.229 V E o , as we might expect. 96,485C nF 4 mol e mol e 6A (M) Cu(s) will displace metal ions of a metal less active than copper. Silver ion is one example. 999 Chapter 20: Electrochemistry Oxidation: Cu s Cu 2+ aq + 2e Reduction: {Ag + aq + e Ag s } 2 Net: 2Ag + aq + Cu s Cu 2+ aq + 2Ag s 6B E o = 0.340V (from Table 20.1) E o = +0.800V
o Ecell = 0.340 V + 0.800 V = +0.460 V (M) We determine the value for the hypothetical reaction's cell potential. E o = +2.713V Oxidation: {Na s Na + aq + e } 2 Reduction: Mg 2+ aq +2e Mg s Net: 2 Na s +Mg 2+ aq 2 Na + aq + Mg s E o = 2.356 V o Ecell = 2.713V 2.356 V = +0.357 V The method is not feasible because another reaction occurs that has a more positive cell potential, i.e., Na(s) reacts with water to form H 2 g and NaOH(aq): Oxidation: {Na s Na + aq + e } 2
o Ecell = 2.713 V 0.828 V = +1.885 V . E o = +2.713V
E o =0.828V Reduction: 2H 2O+2e H 2 (g)+2OH  (aq) 7A (M) The oxidation is that of SO42 to S2O82, the reduction is that of O2 to H2O. Oxidation: {2 SO 4 Reduction: Net: aq S2O82 aq + 2e } 2 O 2 g + 4 H + aq + 4e 2H 2O(l)
2 E o = 2.01V
E o = +1.229 V
________________ O 2 g + 4 H aq + 2 SO 4
+ 2 aq S2O8 aq + 2H 2O(l)
2 Ecell = 0.78 V Because the standard cell potential is negative, we conclude that this cell reaction is nonspontaneous under standard conditions. This would not be a feasible method of producing peroxodisulfate ion.
7B (M) (1) The oxidation is that of Sn 2+ aq to Sn 4 + aq ; the reduction is that of O2 to H2O. E o = 0.154 V Oxidation: {Sn 2 + aq Sn 4 + aq + 2e } 2 Reduction: O 2 g + 4 H + aq + 4e 2H 2O(l) Net: E o = +1.229 V
___________ o O 2 g + 4H + aq + 2Sn 2+ aq 2Sn 4+ aq + 2H 2 O(l) Ecell = +1.075 V Since the standard cell potential is positive, this cell reaction is spontaneous under standard conditions.
(2) Oxidation: {Sn s Sn 2 + aq + 2e } The oxidation is that of Sn(s) to Sn 2+ aq ; the reduction is still that of O2 to H2O. 2 E o = +0.137 V
E o = +1.229 V
_______________ Reduction: O 2 g + 4 H + aq + 4e 2 H 2O(l) Net: O 2 g + 4 H + aq + 2Sn s 2Sn 2+ aq + 2 H 2O(l) o Ecell = 0.137 V +1.229 V = +1.366 V 1000 Chapter 20: Electrochemistry The standard cell potential for this reaction is more positive than that for situation (1). Thus, reaction (2) should occur preferentially. Also, if Sn 4 + aq is formed, it should react with Sn(s) to form Sn 2+ aq . Oxidation: Sn s Sn 2 + aq + 2e Reduction: Sn 4+ aq + 2 e Sn 2+ aq Net: Sn
8A
4+ E o = +0.137 V
E o = +0.154 V
___________________________________ aq + Sn s 2 Sn aq 2+ E o cell = +0.137 V + 0.154 V = +0.291V (M) For the reaction 2 Al s + 3Cu 2+ aq 2 Al3+ aq + 3Cu s we know n = 6 and
o Ecell = +2.013 V . We calculate the value of Keq .
o nEcell 0.0257 6 (+2.013) ln K eq ; ln K eq = = = 470; K eq = e 470 = 10204 n 0.0257 0.0257 The huge size of the equilibrium constant indicates that this reaction indeed will go to essentially 100% to completion. o Ecell = 8B o (M) We first determine the value of Ecell from the halfcell potentials. E o = +0.137 V Oxidation: Sn s Sn 2+ aq +2 e Reduction: Pb 2+ aq +2 e Pb s Net: Pb 2+ aq + Sn s Pb s + Sn 2+ aq o Ecell = E o = 0.125 V
_________________________________ o Ecell = +0.137 V 0.125 V = +0.012 V o nEcell 2 (+0.012) 0.0257 ln K eq ln K eq = = =0.93 K eq = e0.93 = 2.5 n 0.0257 0.0257 The equilibrium constant's small size (0.001 < K <1000) indicates that this reaction will not go to completion. 9A (M) We first need to determine the standard cell voltage and the cell reaction. E o = +1.676 V Oxidation: {Al s Al3+ aq + 3 e } 2 Reduction: {Sn 4+ aq + 2 e Sn 2+ aq } 3 Net: 2 Al s + 3 Sn
4+ E o = +0.154 V
_________________________ aq 2 Al aq + 3 Sn aq 3+ 2+ 2 3 o Ecell = +1.676 V + 0.154 V = +1.830 V Note that n = 6 . We now set up and substitute into the Nernst equation. Ecell Al3+ Sn 2+ 0.0592 = +1.830 0.0592 log 0.36 M 0.54 M =E log 3 n 6 Sn 4 + 0.086 M 3 = +1.830 V 0.0149 V = +1.815 V
2 o cell 3 9B (M) We first need to determine the standard cell voltage and the cell reaction. E o = 1.358 V Oxidation: 2 Cl 1.0 M Cl2 1 atm + 2e Reduction: PbO 2 s + 4 H + aq + 2 e Pb 2+ aq + 2 H 2 O(l) E o = +1.455 V
__________________________________________ 1001 Chapter 20: Electrochemistry Net: PbO 2 s + 4 H + 0.10 M + 2 Cl 1.0 M Cl 2 1 atm + Pb 2+ 0.050 M + 2 H 2O(l)
o Note that n = 2 . Substitute values into the Nernst Ecell = 1.358 V +1.455 V = +0.097 V equation. P{Cl2 } Pb 2+ 0.0592 = +0.097 0.0592 log 1.0 atm 0.050 M o Ecell =Ecell log 4 2 4 2 n 2 0.10 M 1.0 M H + Cl = +0.097 V 0.080 V = +0.017 V 10A (M) The cell reaction is 2 Fe3+ 0.35 M + Cu s 2 Fe 2+ 0.25 M + Cu 2+ 0.15 M with
o n = 2 and Ecell = 0.337 V + 0.771 V = 0.434 V concentrations into the Nernst equation. Next, substitute this voltage and the Ecell 2 Fe 2+ Cu 2+ 0.0592 = 0.434 0.0592 log 0.25 0.15 = 0.434 0.033 o = Ecell log 2 2 n 2 Fe3+ 0.35 2 Ecell = +0.467 V Thus the reaction is spontaneous under standard conditions as written.
10B (M) The reaction is not spontaneous under standard conditions in either direction when Ecell = 0.000 V . We use the standard cell potential from Example 2010. Ecell Ag + 0.0592 o = Ecell log 2+ ; n Hg 2 2 Ag + 0.0592 0.000 V = 0.054 V log 2+ 2 Hg Ag + = 101.82 = 0.0150 2+ Hg 2 2 Ag + 0.054 2 = 1.82; log 2+ = 0.0592 Hg o 11A (M) In this concentration cell Ecell = 0.000 V because the same reaction occurs at anode and cathode, only the concentrations of the ions differ. Ag + = 0.100 M in the cathode compartment. The anode compartment contains a saturated solution of AgCl(aq). Ksp = 1.8 1010 = Ag + Cl = s 2 ; Ag + 0.100 M Ag + 1.3 10 5 M Ecell = 0.000 s 1.8 1010 1.3 105 M Now we apply the Nernst equation. The cell reaction is 0.0592 1.3 105 M log = +0.23 V 1 0.100 M 11B (D) Because the electrodes in this cell are identical, the standard electrode potentials are o numerically equal and subtracting one from the other leads to the value Ecell = 0.000 V. However, because the ion concentrations differ, there is a potential difference between the 1002 Chapter 20: Electrochemistry two half cells (nonzero nonstandard voltage for the cell). Pb 2+ = 0.100 M in the cathode compartment, while the anode compartment contains a saturated solution of PbI 2 . We use the Nernst equation (with n = 2 ) to determine Pb 2+ in the saturated solution.
Ecell = +0.0567 V = 0.000 0.0592 xM log ; 2 0.100 M log xM 2 0.0567 = = 1.92 0.0592 0.100 M xM = 101.92 = 0.012; Pb 2+ anode = x M = 0.012 0.100 M = 0.0012 M; 0.100 M I = 2 0.0012 M 0.0024 M K sp = Pb 2 + I = 0.0012 0.0024 = 6.9 10 9 compared with 7.1 109 in Appendix D
2 2 12A (M) From Table 201 we choose one oxidations and one reductions reaction so as to get the least negative cell voltage. This will be the most likely pair of reactions to occur. E o = 0.535 V Oxidation: 2I aq I 2 s + 2 e 2 H 2 O(l) O 2 g + 4 H + aq + 4 e E o = 1.229 V E o = 2.924 V E o = 0.828 V Reduction: K + aq + e K s 2 H 2 O(l) + 2 e H 2 g + 2 OH aq The least negative standard cell potential 0.535 V 0.828 V = 1.363V occurs when I2 s is produced by oxidation at the anode, and H 2 g is produced by reduction at the cathode.
12B (M) We obtain from Table 201 all the possible oxidations and reductions and choose one of each to get the least negative cell voltage. That pair is the most likely pair of halfreactions to occur. Oxidation: 2 H 2 O(l) O2 g + 4 H + aq + 4e Ag s Ag + aq + e E o = 1.229V E o = 0.800V
E o = +0.800V E o = 0.828V Reduction: Ag + aq + e Ag s [We cannot further oxidize NO aq or Ag + aq .] 3
2 H 2 O(l) + 2e H 2 g + 2 OH aq Thus, we expect to form silver metal at the cathode and Ag + aq at the anode.
13A (M) The halfcell equation is Cu 2+ aq + 2e Cu s , indicating that two moles of electrons are required for each mole of copper deposited. Current is measured in amperes, or coulombs per second. We convert the mass of copper to coulombs of electrons needed for the reduction and the time in hours to seconds. 1003 Chapter 20: Electrochemistry 12.3 g Cu Current 1 molCu 2 mole 96,485 C 3.735 104 C 63.55g Cu 1 molCu 1mole 1.89 amperes 60 min 60 s 1.98 104 s 5.50 h 1h 1min 13B (D) We first determine the moles of O 2 g produced with the ideal gas equation. 1 atm 738 mm Hg 2.62 L 760 mm Hg moles O 2 (g) 0.104 mol O 2 0.08206 L atm 26.2 273.2 K mol K Then we determine the time needed to produce this amount of O2. 4 mol e 96,485 C 1s 1h elapsed time = 0.104 mol O 2 = 5.23 h 1 mol O 2 1mol e 2.13 C 3600 s INTEGRATIVE EXAMPLE
14A (D) In this problem we are asked to determine E o for the reduction of CO2(g) to C3H8(g) in an acidic solution. We proceed by first determining G o for the reaction using tabulated o values for G o in Appendix D. Next, Ecell for the reaction can be determined using f
o G o zFEcell . Given reaction can be separated into reduction and oxidation. Since we are in acidic medium, the reduction halfcell potential can be found in Table 20.1. Lastly, the oxidation halfcell potential can be calculated using o Ecell E o (reduction halfcell) E o (oxidation halfcell) . Stepwise approach First determine G o for the reaction using tabulated values for G o in Appendix D: f G o f C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) 23.3 kJ/mol 0 kJ/mol 394.4 kJ/mol 237.1 kJ/mol G o 3 G o (CO2 (g)) 4 G o (H 2O(l)) [G o (C3 H 8 (g)) 5 G o (O2 (g))] f f f f G o 3 (394.4) 4 (237.1) [23.3 5 0]kJ/mol G o 2108kJ/mol o o In order to calculate Ecell for the reaction using G o zFEcell , z must be first determined. We proceed by separating the given reaction into oxidation and reduction: Reduction: 5{O2(g)+4H+(aq)+4e 2H2O(l)} Eo=+1.229 V Oxidation: C3H8(g) + 6H2O(l) 3CO2(g) + 20H+ + 20e Eo=xV _____________________________________________________ Overall: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Eo=+1.229 VxV o o Since z=20, Ecell can now be calculated using G o zFEcell : 1004 Chapter 20: Electrochemistry o G o zFEcell 2108 1000J/mol 20mol e o Ecell 96485C o Ecell 1mol e 2108 1000 V 1.092V 20 96485 Finally, E o (reduction halfcell) can be calculated using
o Ecell E o (reduction halfcell) E o (oxidation halfcell) : 1.092 V = 1.229 V E o (oxidation halfcell) V E o (oxidation halfcell) =1.229 V 1.092 V = 0.137 V Therefore, Eo for the reduction of CO2(g) to C3H8(g) in an acidic medium is 0.137 V. Conversion pathway approach: Reduction: 5{O2(g)+4H+(aq)+4e 2H2O(l)} Eo=+1.229 V Oxidation: C3H8(g) + 6H2O(l) 3CO2(g) + 20H+ + 20e Eo=x V _____________________________________________________ Overall: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Eo=+1.229 Vx V 23.3 kJ/mol 0 kJ/mol 394.4 kJ/mol 237.1 kJ/mol G o f G o 3 G o (CO2 (g)) 4 G o (H 2O(l)) [G o (C3 H 8 (g)) 5 G o (O2 (g))] f f f f G o 3 (394.4) 4 (237.1) [23.3 5 0]kJ/mol G o 2108kJ/mol o G o zFEcell 2108 1000J/mol 20mol e o Ecell 96485C o Ecell 1mol e 2108 1000 V 1.092V 20 96485 1.092 V = 1.229 V E o (oxidation halfcell) V E o (oxidation halfcell) =1.229 V 1.092 V = 0.137 V
14B (D) This is a multi component problem dealing with a flow battery in which oxidation occurs at an aluminum anode and reduction at a carbonair cathode. Al3+ produced at the anode is complexed with OH anions from NaOH(aq) to form [Al(OH)4]. Stepwise approach: Part (a): The flow battery consists of aluminum anode where oxidation occurs and the formed Al3+ cations are complexes with OH anions to form [Al(OH)4]. The plausible halfreaction for the oxidation is: Oxidation: Al(s) + 4OH(aq) [Al(OH)4] + 3eThe cathode, on the other hand consists of carbon and air. The plausible halfreaction for the reduction involves the conversion of O2 and water to form OH anions (basic medium): Reduction: O2(g) + 2H2O(l) + 4e 4OH(aq) Combining the oxidation and reduction halfreactions we obtain overall reaction for the process: Oxidation: {Al(s) + 4OH(aq) [Al(OH)4] + 3e}4 1005 Chapter 20: Electrochemistry Reduction: {O2(g) + 2H2O(l) + 4e 4OH(aq)} 3 ____________________________________________ Overall: 4Al(s) + 4OH(aq) +3O2(g) + 6H2O(l) 4[Al(OH)4](aq) o Part(b): In order to find Eo for the reduction, use the known value for Ecell as well as Eo for the reduction halfreaction from Table 20.1: o Ecell E o (reduction halfcell) E o (oxidation halfcell)
o Ecell 0.401V E o (oxidation halfcell) 2.73V E o (oxidation halfcell) 0.401V 2.73V 2.329V o o Part (c): From the given value for Ecell (+2.73V) first calculate G o using G o zFEcell (notice that z=12 from part (a) above): 96485C o G o zFEcell 12mol e  2.73V 1mol e G o 3161kJ/mol Given the overall reaction (part (a)) and G o for OH(aq) anions and H2O(l), we can f
calculate the Gibbs energy of formation of the aluminate ion, [Al(OH)4]: Overall reaction: 4Al(s) + 4OH(aq) + 3O2(g) + 6H2O(l) 4[Al(OH)4](aq) o 157 kJ/mol 0 kJ/mol 237.2 kJ/mol x G f 0 kJ/mol G o 4 x [4 0 4 (157) 3 0 6 (237.2)kJ/mol 3161kJ/mol 4 x 3161 2051.2 5212.2kJ/mol x 1303kJ/mol Therefore, G o ([Al(OH )4 ] ) 1303kJ/mol f Part(d): First calculate the number of moles of electrons: 1mol enumber of mol e current(C / s) time(s) 96485C 60 min 60s C 1mol e number of mol e 4.00h 10.0 1h 1min s 96485C number of mol e 1.49mol e Now, use the oxidation halfreaction to determine the mass of Al(s) consumed: 1mol Al 26.98g Al mass(Al) 1.49mol e  13.4g 3mol e1mol Al Conversion pathway approach: Part (a): Oxidation: {Al(s) + 4OH(aq) [Al(OH)4] + 3e}4 Reduction: {O2(g) + 2H2O(l) + 4e 4OH(aq)} 3 ____________________________________________ Overall: 4Al(s) + 4OH(aq) +3O2(g) + 6H2O(l) 4[Al(OH)4](aq) Part (b): 1006 Chapter 20: Electrochemistry o Ecell E o (reduction halfcell) E o (oxidation halfcell) o Ecell 0.401V E o (oxidation halfcell) 2.73V E o (oxidation halfcell) 0.401V 2.73V 2.329V Part (c): 96485C o 2.73V G o zFEcell 12mol e  1mol e G o 3161kJ/mol Overall reaction: 4Al(s) + 4OH(aq) + 3O2(g) + 6H2O(l) 4[Al(OH)4](aq) G o 0 kJ/mol 157 kJ/mol 0 kJ/mol 237.2 kJ/mol x f G o 4 x [4 0 4 (157) 3 0 6 (237.2)kJ/mol 3161kJ/mol 4 x 3161 2051.2 5212.2kJ/mol x G o ([Al(OH )4 ] ) 1303kJ/mol f Part (d):
1mol enumber of mol e current(C / s) time(s) 96485C 60 min 60s C 1mol enumber of mol e 4.00h 10.0 1h 1min s 96485C number of mol e 1.49mol e 1mol Al 26.98g Al mass(Al) 1.49mol e 13.4g 3mol e1mol Al EXERCISES
Standard Electrode Potential
1. (E) (a) E o NO3 aq /NO g = 0.956 V . If it also does not dissolve in HCl, it has a reduction potential that is larger than E o H + aq /H 2 g = 0.000 V . If it displaces Ag + aq from solution, then it has a reduction potential that is smaller than E o Ag + aq /Ag s = 0.800 V . But if it does not displace Cu 2 + aq from solution, If the metal dissolves in HNO 3 , it has a reduction potential that is smaller than then its reduction potential is larger than E o Cu 2+ aq /Cu s = 0.340 V Thus, 0.340 V E o 0.800 V
(b) reduction potential is larger than E o Zn 2+ aq /Zn s = 0.763 V . If it also does not displace Fe 2+ aq from solution, its reduction potential is larger than
1007 If the metal dissolves in HCl, it has a reduction potential that is smaller than E o H + aq /H 2 g = 0.000 V . If it does not displace Zn 2+ aq from solution, its Chapter 20: Electrochemistry E o Fe2+ aq /Fe s = 0.440 V .
2. 0.440 V E o 0.000 V (E) We would place a strip of solid indium metal into each of the metal ion solutions and see if the dissolved metal plates out on the indium strip. Similarly, strips of all the other metals would be immersed in a solution of In3+ to see if indium metal plates out. Eventually, we will find one metal whose ions are displaced by indium and another metal that displaces indium from solution, which are adjacent to each other in Table 201. The standard electrode potential for the In/In3+(aq) pair will lie between the standard reduction potentials for these two metals. This technique will work only if indium metal does not react with water, that is, if the standard reduction potential of In 3+ aq / In s is greater than about 1.8 V . The inaccuracy inherent in this technique is due to overpotentials, which can be as much as 0.200 V. Its imprecision is limited by the closeness of the reduction potentials for the two bracketing metals (M) We separate the given equation into its two halfequations. One of them is the reduction of nitrate ion in acidic solution, whose standard halfcell potential we retrieve from Table 201 and use to solve the problem. 3. Reduction: {NO3 aq +4 H + aq +3e NO g +2 H 2O(l)} 2 ; E o = +0.956 V Oxidation: {Pt(s) 4 Cl (aq) [PtCl4 ]2 (aq) 2 e } 3 ; E o {[PtCl4 ]2 (aq)/Pt(s)} Net: 3Pt s + 2 NO aq + 8 H + aq +12 Cl aq 3PtCl 4 3 2 aq + 2 NO g + 6H 2O l E o cell o 0.201V 0.956 V E {[PtCl 4 ] (aq)/Pt(s)}
o 2 E {[PtCl4 ]2 (aq)/Pt(s)} 0.956 V 0.201V 0.755 V
4. (M) In this problem, we are dealing with the electrochemical reaction involving the oxidation o of Na(in Hg) to Na+(aq) and the reduction of Cl2(s) to Cl(aq). Given that Ecell 3.20V , we are asked to find E o for the reduction of Na+ to Na(in Hg). We proceed by separating the given equation into its two halfequations. One of them is the reduction of Cl 2 (g) to Cl (aq) whose standard halfcell potential we obtain from Table 201 and use to solve the problem. Stepwise approach: Separate the given equation into two halfequations: E {Na (aq) / Na(in Hg)} Oxidation:{Na(in Hg) Na (aq) e } 2 Reduction: Cl2 (g) 2e 2Cl (aq) Net: 2Na(in Hg) Cl2 (g) 2 Na (aq) 2 Cl (aq) E 1.358 V Ecell 3.20 V o Use Ecell E o (reduction halfcell) E o (oxidation halfcell) to solve for E o (oxidation halfcell) : E o 3.20 V=+1.385 V E o{Na (aq) / Na(in Hg)} cell E {Na (aq) / Na(in Hg)} 1.358 V 3.20 V 1.84 V Conversion pathway approach: Oxidation:{Na(in Hg) Na (aq) e } 2 E {Na (aq) / Na(in Hg)} 1008 Chapter 20: Electrochemistry Reduction: Cl2 (g) 2e 2Cl (aq) Net: 2Na(in Hg) Cl2 (g) 2 Na (aq) 2 Cl (aq)
3.20V 1.358 E o {Na (aq) / Na(in Hg)} E o {Na (aq) / Na(in Hg)} 1.358V 3.20V 1.84V E 1.358 V Ecell 3.20 V 5. (M) We divide the net cell equation into two halfequations. Oxidation: {Al s + 4 OH aq [Al(OH) 4 ] aq + 3 e } 4 ; Reduction: {O 2 g + 2 H 2O(l) + 4 e 4 OH aq } 3 ;
o Ecell =2.71V=+0.401V E o {[Al(OH) 4 ] (aq)/Al(s)} E o {[Al(OH) 4 ] (aq)/Al(s)} E o = +0.401V
Ecell = 2.71V
o Net: 4 Al s + 3O 2 g + 6 H 2 O(l) + 4 OH aq 4[Al(OH) 4 ] aq E o {[Al(OH) 4 ] (aq)/Al(s)} = 0.401V 2.71V = 2.31V
6. (M) We divide the net cell equation into two halfequations. Oxidation: CH 4 g + 2H 2O(l) CO 2 g + 8 H + aq + 8e Reduction: {O 2 g + 4H + aq + 4 e 2H 2 O(l)} 2 Net: CH 4 g + 2 O 2 g CO 2 g + 2 H 2 O(l)
o Ecell 1.06 V 1.229 V E o CO2 g /CH 4 g E o CO 2 g /CH 4 g E o = +1.229 V
o Ecell =1.06 V E o CO 2 g /CH 4 g 1.229 V 1.06 V 0.17 V
7. (M) (a) We need standard reduction potentials for the given halfreactions from Table 10.1: Eo=+0.800 V Ag+(aq)+e Ag(s) Zn2+(aq)+2e Zn(s) Eo=0.763 V Cu2+(aq)+2e Cu(s) Eo=+0.340 V Al3+(aq)+3e Al(s) Eo=1.676 V Therefore, the largest positive cell potential will be obtained for the reaction involving the oxidation of Al(s) to Al3+(aq) and the reduction of Ag+(aq) to Ag(s): o Al(s) + 3Ag+(aq) 3Ag(s)+Al3+(aq) Ecell 1.676V 0.800V 2.476V Ag is the anode and Al is the cathode. (b) Reverse to the above, the cell with the smallest positive cell potential will be obtained for the reaction involving the oxidation of Zn(s) to Zn2+(aq) (anode) and the reduction of Cu+2(aq) to Cu(s) (cathode): o Zn(s) + Cu2+(aq) Cu(s)+Zn2+(aq) Ecell 0.763V 0.340V 1.103V (M) (a) The largest positive cell potential will be obtained for the reaction: Zn(s) + 4NH3(aq) + 2VO2+(aq) + 4H+(aq) [Zn(NH3)4]2+(aq)+2V3+(aq)+2H2O(l) o Ecell E o (reduction halfcell) E o (oxidation halfcell) 0.340V (1.015V ) 1.355V Zn is the anode and VO2+ is the cathode. 8. 1009 Chapter 20: Electrochemistry (b) Reverse to the above, the cell with the smallest positive cell potential will be obtained for the reaction involving the oxidation of Ti2+(aq) to Ti3+(aq) (anode) and the reduction of Sn+2(aq) to Sn(s) (cathode): 2Ti2+(aq) + Sn2+(aq) 2Ti3+(aq)+Sn(aq) o Ecell E o (reduction halfcell) E o (oxidation halfcell) 0.14V (0.37V ) 0.23V Predicting OxidationReduction Reactions
9. 10. 11. (E) (a) Ni2+, (b) Cd. (E) (a) potassium, (b) barium. (M) (a) Oxidation: Sn s Sn 2+ aq + 2 e 2+ E o = +0.137 V Reduction: Pb 2+ aq +2 e Pb s Net: Sn s + Pb
(b) E o = 0.125 V
________________________________ aq Sn aq + Pb s 2+ E o cell = +0.012 V Spontaneous Oxidation: 2I aq I 2 s + 2e Reduction: Cu 2+ aq +2e Cu s Net: 2I aq + Cu 2+ aq Cu s + I 2 s E o = 0.535 V E o = +0.340 V ________________________ o Ecell = 0.195 V Nonspontaneous E o = 1.229 V (c) Oxidation: {2H 2O(l) O 2 g + 4H + aq + 4 e } 3 Reduction: {NO3 (aq) + 4H + aq +3e NO g +2 H 2O(l)} 4 E o = +0.956 V
_ Net: 4 NO3 aq + 4 H + aq 3O 2 g + 4 NO g + 2 H 2 O(l) Nonspontaneous
(d) _______________________ o Ecell = 0.273 V Oxidation: Cl aq + 2 OH aq OCl aq + H 2 O(l) + 2e E o = 0.890 V Reduction: O3 g + H 2 O(l) + 2 e O 2 g + 2OH aq Net: Cl aq + O3 g OCl aq + O 2 g (basic solution) Spontaneous E o = +1.246 V
o Ecell = +0.356 V 12. (M) It is more difficult to oxidize Hg(l) to Hg 2 + 0.797 V than it is to reduce H+ to H 2 2 (0.000 V); Hg(l) will not dissolve in 1 M HCl. The standard reduction of nitrate ion to NO(g) in acidic solution is strongly spontaneous in acidic media ( +0.956 V ). This can help overcome the reluctance of Hg to be oxidized. Hg(l) will react with and dissolve in the HNO 3 aq . (M) (a) 13. Oxidation: Mg s Mg 2+ aq + 2 e E o = +2.356 V Reduction: Pb 2+ aq + 2 e Pb s Net: Mg s + Pb
2+ E o = 0.125 V
+ Pb s _____________________________________ aq Mg aq 2+ o Ecell = +2.231V 1010 Chapter 20: Electrochemistry This reaction occurs to a significant extent.
(b) Oxidation: Sn s Sn 2+ aq + 2 e Reduction: 2 H + aq H 2 g Net: Sn s + 2 H
+ E o = +0.137 V E o = 0.000 V
+ H2 g ________________________________________ aq Sn aq 2+ E o cell = +0.137 V This reaction will occur to a significant extent. Oxidation: Sn 2+ aq Sn 4+ aq + 2 e Reduction: SO 4
2 (c) E o = 0.154 V aq + 4H + aq +2 e SO2 g +2H 2O(l) E o = + 0.17 V
________________________ Net: Sn 2+ aq + SO4 2 aq + 4 H + aq Sn 4+ aq + SO2 g + 2H 2 O(l) This reaction will occur, but not to a large extent.
(d) o Ecell = + 0.02V Oxidation: {H 2O 2 aq O 2 g + 2 H + aq + 2e } 5 Reduction: {MnO 4 aq +8H + aq +5e  Mn 2+ aq +4H 2 O(l)} 2
 E o = 0.695 V E o = +1.51V o Net: 5H 2 O2 aq +2MnO4 aq +6H + aq 5O 2 g +2Mn 2+ aq +8H 2 O(l) Ecell = +0.82V This reaction will occur to a significant extent. (e) Oxidation: 2 Br aq Br2 aq + 2 e Reduction: I 2 s + 2e 2 I aq Net: 2 Br aq + I 2 s Br2 aq + 2 I aq E o = 1.065 V E o = +0.535 V
__________________________________ o Ecell = 0.530 V This reaction will not occur to a significant extent.
14. (M) In this problem we are asked to determine whether the electrochemical reaction between Co(s) and Ni2+(aq) to yield Co2+(aq) + Ni(s) will proceed to completion based on the known o Ecell value. This question can be answered by simply determining the equilibrium constant. Stepwise approach o First comment on the value of Ecell : o The relatively small positive value of Ecell for the reaction indicates that the reaction will proceed in the forward direction, but will stop short of completion. A much larger positive o value of Ecell would be necessary before we would conclude that the reaction goes to completion. 0.0257 o ln K eq : Calculate the equilibrium constant for the reaction using Ecell n o n Ecell 2 0.02 0.0257 o Ecell ln K eq ln K eq 2 0.0257 n 0.0257 K eq e 2 7 Comment on the value of Keq: 1011 Chapter 20: Electrochemistry Keq is small. A value of 1000 or more is needed before we can describe the reaction as one that goes to completion. Conversion pathway approach: 0.0257 o ln K eq Ecell n o n Ecell ln K eq 0.0257
K eq e 0.0257 e 0.0257 7
o n Ecell 20.02 Keq is too small. The reaction does not go to completion. o 15. (M) If Ecell is positive, the reaction will occur. For the reduction of Cr2O72 to Cr 3+ aq : Cr2 O7
2 aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l) E o = +1.33V If the oxidation has E o smaller (more negative) than 1.33V , the oxidation will not occur.
(a)
Sn 2+ aq Sn 4+ aq +2 e E o = 0.154 V Hence, Sn 2+ aq can be oxidized to Sn 4 + aq by Cr2O 2 aq . 7
(b) I 2 s + 6 H 2O(l) 2 IO3 aq +12 H + aq +10e E o = 1.20 V I2 s can be oxidized to IO aq by Cr2O 2 aq . 3 7
(c) Mn 2+ aq + 4 H 2 O(l) MnO 4 aq +8 H + aq +5e E o = 1.51V Mn 2+ aq cannot be oxidized to MnO aq by Cr2O 2 aq . 4 7
16. (M) In order to reduce Eu3+ to Eu2+, a stronger reducing agent than Eu2+ is required. From the list given, Al(s) and H2C2O4(aq) are stronger reducing agents. This is determined by looking at the reduction potentials (1.676 V for Al3+/Al(s) and 0.49 V for CO2, H+/H2C2O4(aq)), are more negative than 0.43 V). Co(s), H2O2 and Ag(s) are not strong enough reducing agents for this process. A quick look at their reduction potentials shows that they all have more positive reduction potentials than that for Eu3+ to Eu2+ (0.277 V for Co2+/Co(s), +0.695 V for O2, H+/H2O2(aq) and +0.800 V for Ag+/Ag(s). (M) (a) 17. Oxidation: {Ag s Ag + aq + e } 3 ______________ E o = 0.800 V Reduction: NO3 aq + 4 H + aq + 3 e NO g + 2 H 2 O(l) Ag(s) reacts with HNO 3 aq to form a solution of AgNO 3 aq .
(b) E o = +0.956 V
_____________ o Net: 3Ag s +NO3 aq +4 H + aq 3Ag + aq +NO g +2 H 2 O(l) Ecell = +0.156 V Oxidation: Zn s Zn 2+ aq + 2 e E o = +0.763V Reduction: 2 H + aq +2 e H 2 g Net: Zn s + 2 H + aq Zn 2+ aq + H 2 g E o = 0.000 V
o Ecell = +0.763V Zn(s) reacts with HI(aq) to form a solution of ZnI 2 aq . 1012 Chapter 20: Electrochemistry (c) Oxidation: Au s Au 3+ aq + 3e Reduction: NO3 aq + 4 H + aq + 3e NO g + 2H 2 O(l) E o = 1.52 V E o = +0.956 V Au(s) does not react with 1.00 M HNO 3 aq .
18. o Net: Au s + NO3 aq + 4 H + aq Au 3+ aq + NO g + 2 H 2 O(l) ; Ecell = 0.56 V (M) In each case, we determine whether Ecell is greater than zero; if so, the reaction will occur. (a) Oxidation: Fe s Fe 2+ aq + 2 e E o 0.440 V Reduction: Zn 2+ aq + 2e Zn s Net: Fe s + Zn 2+ aq Fe 2+ aq + Zn s Oxidation: {2Cl aq Cl2 g + 2 e } 5 E o = 0.763V o Ecell 0.323V The reaction is not spontaneous under standard conditions as written (b) E o = 1.358 V Reduction: {MnO 4 aq +8 H + aq +5e Mn 2+ aq +4 H 2 O(l)}2 ; E o = 1.51V Net: 10Cl aq + 2 MnO 4 aq +16 H + aq 5Cl2 g + 2 Mn 2+ aq + 8 H 2O(l)
o Ecell = +0.15 V . The reaction is spontaneous under standard conditions as written. (c) Oxidation: {Ag s Ag + aq + e } 2 Reduction: 2 H + aq + 2 e H 2 g Net: 2 Ag s + 2 H + aq 2 Ag + aq + H 2 g Oxidation: {2 Cl aq Cl2 g + 2 e } 2 Reduction: O 2 g + 4 H + aq + 4 e 2 H 2 O(l) E o = 0.800 V E o = +0.000 V o Ecell = 0.800 V The reaction is not spontaneous under standard conditions as written. (d) E o = 1.358 V E o = +1.229 V
_______________________________________ Net: 4 Cl aq + 4 H + aq + O 2 g 2 Cl2 g + 2 H 2 O(l) o Ecell = 0.129 V The reaction is not spontaneous under standard conditions as written. Galvanic Cells
19. (M) (a) Oxidation: {Al s Al3+ aq + 3e } 2
2 Al s + 3 Sn 2+ aq 2 Al3+ aq + 3 Sn s E o = +1.676 V Reduction: {Sn 2+ aq + 2e Sn s } 3 Net:
(b) E o = 0.137 V
____________________________ o Ecell = +1.539 V Oxidation: Fe 2+ aq Fe3+ aq + e Reduction: Ag + aq + e Ag s Net:
Fe
2+ E o = 0.771 V E o = +0.800 V
____________________ aq + Ag aq Fe aq +Ag s + 3+ E o cell = +0.029 V (c) Oxidation: {Cr(s) Cr 2+ (aq)+2e } 3 1013 E o = 0.90V Chapter 20: Electrochemistry Reduction: {Au 3+ (aq)+3e Au(s)} 2 Net:
(d) E o = 1.52V
____________________ 3Cr(s)+2Au 3+ (aq) 3Cr 2+ (aq)+2Au(s) o Ecell = 2.42V Oxidation: 2H 2 O(l) O 2 (g)+4H + (aq)+4eReduction: O 2 (g)+2H 2O(l)+4e 4OH  (aq) Net: H 2 O(l) H (aq)+OH (aq)
+  E o = 1.229V E o = 0.401V
____________________ E o cell = 0.828V 20. (M) In this problem we are asked to write the halfreactions, balanced chemical equation o and determine Ecell for a series of electrochemical cells. (a) Stepwise approach First write the oxidation and reduction halfreactions and find Eo values from Appendix D: Oxidation: Cu(s) Cu 2+ (aq)+2e E o = 0.340 V E o = +0.520 V Reduction: Cu + (aq)+e Cu(s) In order to obtain balanced net equation, the reduction halfreaction needs to be multiplied by 2: 2Cu + (aq)+2e 2Cu(s) Add the two halfreactions to obtain the net reaction: Cu(s) Cu 2+ (aq)+2e2Cu + (aq)+2e 2Cu(s)
____________________ Net: 2Cu (aq) Cu (aq)+Cu(s) o Determine Ecell :
+ 2+ o Ecell =0.340V+0.520V= +0.18 V Conversion pathway approach: Oxidation: Cu(s) Cu 2+ (aq)+2eReduction: {Cu + (aq)+e Cu(s)} 2 Net: 2Cu (aq) Cu (aq)+Cu(s) Follow the same methodology for parts (b), (c), and (d). Oxidation: Ag(s)+I  (aq) AgI(s)+e+ 2+ E o = 0.340 V E o = +0.520 V
___________ E o cell 0.18V (b) E o = 0.152 V Reduction: AgCl(s)+e Ag(s)+Cl (aq) Net:
(c) E o = 0.2223 V
____________________ AgCl(s)+I (aq) AgI(s)+Cl (aq)
 E o cell = +0.3743 V E o = 1.76 V Oxidation: {Ce3+ (aq) Ce 4+ (aq)+e } 2 Reduction: I 2 (s)+2e 2I  (aq) Net:
2Ce (aq)+I 2 (s) 2Ce (aq)+2I (aq)
3+ 4+  E o = +0.535 V
____________________ o Ecell = 1.225 V 1014 Chapter 20: Electrochemistry (d) Oxidation: {U(s) U 3+ (aq)+3e } 2 Reduction: {V 2+ (aq)+2e V(s)} 3 Net: 2U(s)+3V (aq) 2U (aq)+3V(s)
2+ 3+ E o = 1.66V E o = 1.13V
____________________ o Ecell = +0.53 V 21. (M) In each case, we determine whether Ecell is greater than zero; if so, the reaction will occur. (a) Oxidation: H 2 (g) 2H + 2 e aq E o 0.000 V Reduction: F2 (g) + 2 e 2F aq E o = 2.866 V o Net: H 2 (g) + F2 (g) 2 H + (aq) + 2 F (aq) Ecell 2.866 V The reaction is spontaneous under standard conditions as written (b) Oxidation: Cu(s) Cu 2+ (aq) + 2 e Reduction: Ba 2+ (aq) + 2 e Ba(s) ; E o = 0.340 V E o = 2.92 V o Net: Cu(s) + Ba 2+ (aq) Cu 2+ (aq) + Ba(s) Ecell = 3.26 V . The reaction is not spontaneous under standard conditions as written. (c) Oxidation: Fe2+ (aq) Fe3+ (aq) + e 2 Reduction: Fe 2+ (aq) + 2e Fe(s) Net: 3 Fe2+ (aq) Fe(s) + 2 Fe3+ (aq) The reaction is not spontaneous as written. E o = 0.771 V E o = 0.440 V
o Ecell = 1.211 V (d) Oxidation: 2 Hg(l) + 2 Cl Hg 2 Cl2 (s) + 2 e E o = (0.2676) V Reduction: 2 HgCl2 (aq) + 2 e Hg 2 Cl2 (s) + 2 Cl (aq) E o = +0.63V
o Net: 2 Hg(l) + 2 HgCl2 (aq) 2 Hg 2 Cl2 (s) Ecell = 0.36 V (divide by 2 to get Hg(l) + HgCl 2 (aq) Hg 2 Cl2 (s) ) The reaction is spontaneous under standard conditions as written. 1015 Chapter 20: Electrochemistry 22. (M) (a)
e salt bridge
a n o d e Cu Cu2+ Fe3+ Fe2+ Anode, oxidation: Cu s Cu 2+ aq + 2 e Net: Cu s + 2 Fe3+ aq Cu 2+ aq + 2 Fe 2+ aq (b)
e c a t h o d e Pt E o = 0.340 V Cathode, reduction: {Fe3+ aq + e Fe 2+ aq } 2 ; E o = +0.771V
o Ecell = +0.431V salt bridge
a n o d e Al Al3+ Pb2+ Anode, oxidation: {Al s Al3+ aq + 3e } 2 Cathode, reduction: {Pb 2+ aq + 2 e Pb s } 3 ; Net: 2Al s + 3Pb 2+ aq 2Al3+ aq + 3Pb s (c)
e c a t h o d e Pb E o = +1.676 V E o = 0.125 V
o Ecell = +1.551 V salt bridge
a n o d e Pt H+ H2O O2(g) Cl c a t h o d e Pt Cl2(g) Anode, oxidation: 2 H 2 O(l) O 2 g + 4 H + aq + 4 e E o = 1.229 V Cathode, reduction: {Cl2 g + 2 e 2 Cl aq } 2 E o = +1.358 V 1016 Chapter 20: Electrochemistry
o Net: 2 H 2 O(l) + 2 Cl2 g O 2 g + 4 H + aq + 4 Cl aq Ecell = +0.129 V (d)
e salt bridge
a n o d e Zn Zn+2 HNO3 c a t h o d e Pt NO(g) Anode, oxidation: {Zn s Zn 2+ aq + 2 e } 3 E o = +0.763V Cathode, reduction: {NO3 aq +4 H + aq +3e NO g +2 H 2 O(l)} 2 ; E o = +0.956 V
o Net: 3Zn s + 2NO3 aq + 8H + aq 3Zn 2+ aq + 2NO g + 4H 2 O(l) Ecell = +1.719 V 23. (M) In each case, we determine whether Ecell is greater than zero; if so, the reaction will occur. (a) Oxidation: Ag(s) Ag + (aq)+e E o 0.800 V Reduction: Fe3+ (aq)+e  Fe 2+ (aq) Net: Ag(s)+Fe3+ (aq) Ag + (aq)+Fe 2+ (aq) E o = 0.771 V o Ecell 0.029 V The reaction is not spontaneous under standard conditions as written. (b) Oxidation: Sn(s) Sn 2+ (aq)+2eReduction: Sn 4+ (aq)+2e Sn +2 (aq) Net: Sn(s)+Sn 4+ (aq) 2Sn 2+ (aq) E o 0.137 V E o = 0.154 V (c) o Ecell 0.291V The reaction is spontaneous under standard conditions as written. Oxidation: 2Br  (aq) Br2 (l)+2e E o 1.065V Reduction: 2Hg 2+ (aq)+2e Hg + (aq) 2 Net: 2Br  (aq)+2Hg 2+ (aq) Br2 (l)+Hg + 2 E o = 0.630 V (d) o Ecell 0.435V The reaction is not spontaneous under standard conditions as written. Oxidation: {NO3 (aq)+2H + (aq)+e  NO 2 (g)+H 2O(l)} 2 E o 2.326V Reduction: Zn(s) Zn 2+ (aq)+2e E o = 1.563V o Net: 2NO3 (aq)+4H + (aq)+Zn(s) 2NO 2 (g)+2H 2O(l)+Zn 2 (aq) Ecell 0.435V The reaction is not spontaneous under standard conditions as written. 1017 Chapter 20: Electrochemistry 24. (M) (a) Oxidation: Fe s Fe 2+ aq + 2 e Fe s 2 + aq aq Cl2 g s Fe Cl Pt E o = +0.440 V E o = +1.358 V
o Ecell = +1.798 V Reduction: Cl2 g + 2 e 2Cl aq Net: Fe s + Cl2 g Fe 2+ aq + 2Cl aq (b) Zn s 2+ aq aq s Zn Ag Ag Oxidation: Zn s Zn 2+ aq + 2 e Reduction: {Ag + aq + e Ag s } 2 Net: Zn s + 2 Ag + aq Zn 2+ aq + 2 Ag s E o = +0.763V E o = +0.800 V
o Ecell = +1.563V (c) Pt s + aq ,Cu 2 + aq + aq s Cu Cu Cu Oxidation: Cu + aq Cu 2+ aq + e Reduction: Cu + aq + e Cu s Net: 2 Cu + aq Cu 2+ aq + Cu s E o = 0.159 V E o = +0.520 V
o Ecell = +0.361V (d) Oxidation: Mg s Mg 2+ aq + 2 e Mg s 2 + aq aq Br2 l s Mg Br Pt E o = +2.356 V E o = +1.065 V
o Ecell = +3.421V Reduction: Br2 l + 2 e 2 Br aq Net: Mg s + Br2 l Mg 2+ aq + 2 Br aq 34(a) 24(a) anode eCl2(g) salt bridge cathode 34(b) 24(b) anode esalt bridge cathode Ag Zn2+ Ag+
Pt Mg2+ Br2 Brcathode Fe Fe2+
34(c) 24(c) cathode esalt bridge Pt Cl Zn 34(d) 24(d) esalt bridge anode Pt Cu+ Cu2+ Cu+ Cu Mg 1018 anode Chapter 20: Electrochemistry o G o , Ecell , and K 25. (M) (a) Oxidation: {Al s Al3+ aq +3e } 2 E o = +1.676 V E o = +0.337 V
______________________ Reduction: {Cu 2+ aq +2 e Cu s } 3 Net: 2 Al s + 3Cu 2+ aq 2 Al3+ aq + 3Cu s o G o = nFEcell = 6 mol e 96, 485 C/mol e 2.013V o Ecell = +2.013V G o = 1.165 106 J = 1.165 103 kJ
(b) Oxidation: {2 I aq I 2 s +2 e } 2 Reduction: O 2 g + 4 H + aq + 4 e 2 H 2 O(l) E o = 0.535 V E o = +1.229 V
_________________________________________ o Net: 4 I aq + O 2 g + 4 H + aq 2 I 2 s + 2 H 2 O(l) Ecell = +0.694 V o G o = nFEcell = 4 mol e 96, 485 C/mol e 0.694 V = 2.68 105 J = 268 kJ (c) Oxidation: {Ag s Ag + aq + e } 6 Reduction: Cr2 O7
2 E o = 0.800 V E o = +1.33V aq +14 H + aq +6 e 2 Cr 3+ aq +7 H 2O(l)
2 Net: 6 Ag s + Cr2 O7 aq +14 H + aq 6 Ag + aq + 2 Cr 3+ aq + 7 H 2O(l) o G o = nFEcell = 6mol e 96,485C/mol e 0.53V = 3.1105 J = 3.1102 kJ o Ecell = 0.800 V +1.33V = +0.53V 26. (M) In this problem we need to write the equilibrium constant expression for a set of redox o reactions and determine the value of K at 25 oC. We proceed by calculating Ecell from standard electrode reduction potentials (Table 20.1). Then we use the expression o G o = nFEcell = RTln K to calculate K. (a) Stepwise approach: o Determine Ecell from standard electrode reduction potentials (Table 20.1): Oxidation: Ni(s) Ni2+ (aq)+2eReduction: {V 3+ (aq)+e V 2+ (aq)} 2 Net: Ni(s)+2V (aq) Ni (aq)+2V (aq)
3+ 2+ 2+ E o 0.257 V E o 0.255 V
_________________________ o Ecell =0.003V o Use the expression G o = nFEcell = RTln K to calculate K: 1019 Chapter 20: Electrochemistry o G o = nFEcell = RTln K; G o 2mole 96485 C 0.003V 578.9J mol G o RT ln K 578.9J 8.314JK 1mol 1 298.15K ln K
2 ln K 0.233 K e0.233 1.26 Ni2+ V 2+ K 1.26 2 3+ V Conversion pathway approach: o Determine Ecell from standard electrode reduction potentials (Table 20.1): Oxidation: Ni(s) Ni2+ (aq)+2eReduction: {V 3+ (aq)+e V 2+ (aq)} 2 Net: Ni(s)+2V 3+ (aq) Ni2+ (aq)+2V 2+ (aq)
o o cell E o 0.257 V E o 0.255 V
_________________________ o Ecell =0.003V o nFEcell n 96485Cmol 1 Eo G = nFE = RTln K ln K eq RT 8.314JK 1mol 1 298.15K cell n 2 mol e 0.003V o ln K eq Ecell 0.233 0.0257 0.0257 Ni2+ V 2+ 0.233 K eq e 1.26 2 3+ V 2 Similar methodology can be used for parts (b) and (c) (b) Oxidation: 2Cl aq Cl2 g + 2 e E o = 1.358 V E o = +1.23V Reduction: MnO 2 s + 4H + aq + 2e Mn 2+ aq + 2H 2O(l) o Net: 2 Cl aq + MnO 2 s + 4 H + aq Mn 2+ aq + Cl 2 g + 2 H 2O(l) Ecell = 0.13V (c) Mn 2+ P{Cl2 g } 2 mol e 0.13 V 10.1 5 ln K eq = = 10.1 ; K eq = e = 4 10 = 4 2 0.0257 Cl H + Oxidation: 4 OH aq O 2 g + 2H 2 O(l) + 4 e E o = 0.401 V Reduction: {OCl (aq)+H 2 O(l)+2e Cl (aq)+2 OH } 2 Net: 2OCI (aq) 2CI (aq) O 2 (g) 4 mol e (0.489 V) 76.1 ln K eq 0.0257
2 E o 0.890 V
o Ecell 0.489 V K eq e 76.1 Cl P O 2 (g) 1 10 2 OCl 33 27. o (M) First calculate Ecell from standard electrode reduction potentials (Table 20.1). Then use o G o = nFEcell = RTln K to determine G o and K. 1020 Chapter 20: Electrochemistry (a) Oxidation: {Ce 3 (aq) Ce 4 (aq) e } 5 Reduction: MnO4 (aq)+8H + (aq)+5e Mn 2+ (aq)+4H 2O(l)
(b)
o G o = nFEcell = RTln K; E o = 1.61 V
E o = +1.51V o Net: MnO4 (aq)+8H + (aq)+5Ce 3+ (aq) Mn 2+ (aq)+4H 2 O(l)+5Ce 4+ (aq) Ecell 0.100V G o 5 96485
(c) C (0.100V ) 48.24kJmol1 mol G o RT ln K 48.24 1000Jmol1 8.314JK 1mol1 298.15K ln K ln K 19.46 K e19.46 3.5 109
(d) Since K is very small the reaction will not go to completion. 28.
o (M) First calculate Ecell from standard electrode reduction potentials (Table 20.1). Then use o G o = nFEcell = RTln K to determine G o and K. (a) Oxidation: Pb 2+ (aq) Pb 4+ (aq)+2e Reduction: Sn 4+ (aq)+2e  Sn 2+ (aq) Net: Pb 2+ (aq)+Sn 4+ (aq) Pb 4+ (aq)+Sn 2+ (aq) Eo 0.180V Eo 0.154V
o Ecell 0.026V (b) o G o = nFEcell = RTln K G o 2 96485 C (0.026) 5017Jmol1 mol (c) (d) 29. G o RT ln K 5017Jmol1 8.314JK 1mol1 298.15K ln K ln K 2.02 K e2.02 0.132 The value of K is small and the reaction does not go to completion.
o o A negative value of Ecell ( 0.0050 V ) indicates that G o = nFEcell is positive (M) (a) which in turn indicates that Keq is less than one K eq 1.00 ; G o = RTlnKeq . Keq = Cu 2+ Cu + 2 2 Sn 2+ Sn 4+ Thus, when all concentrations are the same, the ion product, Q, equals 1.00. From the negative standard cell potential, it is clear that Keq must be (slightly) less than one. Therefore, all the concentrations cannot be 0.500 M at the same time.
(b) In order to establish equilibrium, that is, to have the ion product become less than 1.00, and equal the equilibrium constant, the concentrations of the products must decrease and those of the reactants must increase. A net reaction to the left (towards the reactants) will occur.
1021 Chapter 20: Electrochemistry 30. (D) (a) First we must calculate the value of the equilibrium constant from the standard cell potential. o nEcell 0.0257 2 mol e ( 0.017) V o Ecell = ln K eq ; ln K eq = = = 1.32 n 0.0257 0.0257 K eq = e 1.32 = 0.266 To determine if the described solution is possible, we compare . Thus, when 2 2 H + BrO4 Ce3+ 4+ 3+ BrO 4 = Ce 0.675 M , BrO 3 = Ce 0.600 M and pH=1 ( H + =0.1M 0.600 0.675 2 = 112.5 0.266 = K eq . Therefore, the the ion product, Q = 0.12 0.675 0.600 2 described situation can occur
(b) Keq with Q. Now K eq = BrO3 Ce 4+ 2 In order to establish equilibrium, that is, to have the ion product (112.5) become equal to 0.266, the equilibrium constant, the concentrations of the reactants must increase and those of the products must decrease. Thus, a net reaction to the left (formation of reactants) will occur. 31. (M) Cell reaction: Zn s + Ag 2O s ZnO s + 2Ag s . We assume that the cell operates at 298 K. G o = Gfo ZnO s + 2Gfo Ag s Gfo Zn s Gfo Ag 2O s = 318.3 kJ/mol + 2 0.00 kJ/mol 0.00 kJ/mol 11.20 kJ/mol o = 307.1 kJ/mol = nFEcell o Ecell = G o 307.1 103 J/mol = = 1.591V 2 mol e /mol rxn 96, 485C/mol e nF 32. (M) From equation (20.15) we know n = 12 and the overall cell reaction. First we must compute value of G o . 96485 C o G o = n FEcell = 12 mol e 2.71 V = 3.14 106 J = 3.14 103 kJ 1 mol e Then we will use this value, the balanced equation and values of Gf to calculate
o Gfo Al OH 4 . 4 Al s + 3O 2 g + 6 H 2O(l) + 4 OH aq 4 Al OH 4 aq G o = 4Gfo Al OH 4 4Gfo [Al s ] 3Gfo [O 2 g ] 6Gfo [H 2 O l ] 4Gfo [OH aq ] 1022 Chapter 20: Electrochemistry 3.14 103 kJ = 4Gfo Al OH 4 4 0.00 kJ 3 0.00 kJ 6 237.1 kJ 4 157.2 = 4Gfo Al OH 4 + 2051.4 kJ Gfo Al OH 4 = 3.14 103 kJ 2051.4 kJ 4 = 1.30 103 kJ/mol 33. (D) From the data provided we can construct the following Latimer diagram.
IrO2 (IV) 0.223 V Ir3+ 1.156 V (III) Ir (Acidic conditions) (0) Latimer diagrams are used to calculate the standard potentials of nonadjacent halfcell couples. Our objective in this question is to calculate the voltage differential between IrO2 and iridium metal (Ir), which are separated in the diagram by Ir3+. The process basically involves adding two halfreactions to obtain a third halfreaction. The potentials for the two halfreactions cannot, however, simply be added to get the target halfcell voltage because the electrons are not cancelled in the process of adding the two halfreactions. Instead, to find E1/2 cell for the target halfreaction, we must use free energy changes, which are additive. To begin, we will balance the relevant halfreactions in acidic solution: 4 H+(aq) + IrO2(s) + e Ir3+(aq) + 2 H2O(l) Ir3+(aq)+ 3 e Ir(s) E1/2red(a) = 0.223 V E1/2red(b) = 1.156V 4 H+(aq) + IrO2(s) + 4e 2 H2O(l) + Ir(s) E1/2red(c) = ? E1/2red(c) E1/2red(a) + E1/2red(b) but G(a) + G(b) = G(c) and G = nFE 4F(E1/2red(c)) = 1F(E1/2red(a)) + 3F(E1/2red(b) 4F(E1/2red(c)) = 1F(0.223) + 3F(1.156 1F (0.223) 3F (1.156) 1(0.223) 3(1.156) E1/2red(c) = = = 0.923 V 4 F 4 In other words, E(c) is the weighted average of E(a) and E(b)
34. (D) This question will be answered in a manner similar to that used to solve 31. Let's get underway by writing down the appropriate Latimer diagram:
H2MoO4 0.646 V (VI) MoO2 (IV) 0.428 V ?V Mo3+ (III) (Acidic conditions) This time we want to calculate the standard voltage change for the 1 e reduction of MoO2 to Mo3+. Once again, we must balance the halfcell reactions in acidic solution: H2MoO4(aq) + 2 e + 2 H+(aq) MoO2(s) + 2 H2O(l) MoO2(s) + 4 H+(aq) + 1 e Mo3+(aq) + 2 H2O(l) H2MoO4(aq) + 3 e + 6 H+(aq) Mo3+(aq) + 4 H2O(l) So, 3F(E1/2red(c)) = 2F(E1/2red(a)) + 1F(E1/2red(b) 3F(0.428 V) = 2F(0.646) + 1F(E1/2red(b) 1FE1/2red(b) = 3F(0.428 V) + 2F(0.646) E1/2red(a) = 0.646 V E1/2red(b) = ? V E1/2red(c) = 0.428 V 1023 Chapter 20: Electrochemistry E1/2red(c) = 3F (0.428 V) 2F (0.646) = 1.284 V 1.292 V = 0.008 V 1F Concentration Dependence of Ecell the Nernst Equation
35. (M) In this problem we are asked to determine the concentration of [Ag+] ions in electrochemical cell that is not under standard conditions. We proceed by first determining o Ecell . Using the Nerst equation and the known value of E, we can then calculate the concentration of [Ag+]. Stepwise approach: o First, determine Ecell : Oxidation: Zn s Zn 2+ aq + 2 e E o = +0.763V Reduction: {Ag + aq + e Ag s } 2 Net: Zn s + 2 Ag + aq Zn 2+ aq + 2Ag s E o = +0.800 V
o Ecell = +1.563V Use the Nerst equation and the known value of E to solve for [Ag]+: Zn 2+ 0.0592 = +1.563V 0.0592 log 1.00 = +1.250 V o log E =Ecell 2 2 n x2 Ag + 1.00 M 2 (1.250 1.563) log 10.6 ; x 2.5 1011 5 106 M 2 x 0.0592 Therefore, [Ag+] = 5 106 M Conversion pathway approach: E o = +0.763V Oxidation: Zn s Zn 2+ aq + 2 e Reduction: {Ag + aq + e Ag s } 2 Net: Zn s + 2 Ag + aq Zn 2+ aq + 2Ag s E=E
o cell E o = +0.800 V
o Ecell = +1.563V Zn 2+ 2+ 0.0592 log Zn n (E E o ) log cell 2 2 n 0.0592 Ag + Ag + Zn 2+ 10 n o ( E Ecell ) 0.0592 Ag + 2 Ag + Zn 2+ 10 n o ( E Ecell ) 0.0592 Ag + 1.00
2 (1.2501.563) 0.0592 5 106 36. 10 (M) In each case, we employ the equation Ecell = 0.0592 pH . (a) (b) Ecell = 0.0592 pH = 0.0592 5.25 = 0.311 V pH = log 0.0103 = 1.987 Ecell = 0.0592 pH = 0.0592 1.987 = 0.118 V 1024 Chapter 20: Electrochemistry (c) H + C2 H 3O x2 x2 2 5 Ka = = 1.8 10 = 0.158 x 0.158 HC 2 H 3O 2 x 0.158 1.8 105 1.7 103 M pH log (1.7 103 ) 2.77 Ecell = 0.0592 pH = 0.0592 2.77 = 0.164 V 37. o (M) We first calculate Ecell for each reaction and then use the Nernst equation to calculate Ecell . (a) Oxidation: {Al s Al3+ 0.18 M + 3 e } 2 E o = +1.676 V Reduction: {Fe 2+ 0.85 M + 2 e Fe s } 3 Net: 2 Al s + 3Fe 2+ 0.85 M 2 Al3+ 0.18 M + 3Fe s 2 o Ecell = Ecell E o = 0.440 V
o Ecell = +1.236 V 2 Al3+ 0.0592 = 1.236 V 0.0592 log 0.18 1.249 V log 3 3 6 n 0.85 Fe2+ (b) Oxidation: {Ag s Ag + 0.34 M + e } 2 Reduction: Cl2 0.55 atm + 2 e 2 Cl 0.098 M 2 2 E o = 0.800 V E o = +1.358 V o Net: Cl2 0.55atm + 2 Ag s 2 Cl 0.098 M + 2 Ag + 0.34 M ; Ecell = +0.558 V Ecell = E o cell = 0.0592 n 2 2 Cl Ag + = +0.558 0.0592 log 0.34 0.098 = +0.638 V log 38. (M) (a) 0.40 M + 2 e Reduction: {Cr 3+ 0.35 M +1e Cr 2+ 0.25 M } 2 Net: 2 Cr 3+ 0.35 M + Mn s 2 Cr 2+ 0.25 M + Mn 2+ 0.40 M 2+ Oxidation: Mn s Mn P{Cl 2 g } 2 0.55 E o = +1.18 V E o = 0.424 V
o Ecell = +0.76 V o Ecell = Ecell 2 Cr 2+ Mn 2+ 0.0592 = +0.76 V 0.0592 log 0.25 0.40 = +0.78 V log 2 2 n 2 Cr 3+ 0.35 2 (b) Oxidation: {Mg s Mg 2+ 0.016M + 2e } 3 E o = +2.356 V Reduction:{ Al OH 0.25M +3 e 4OH 0.042 M +Al s } 2 ; E o = 2.310V 4 Net: 3 Mg s +2[Al OH ] 0.25M 3Mg 4 2 0.016 M +8OH 0.042M +2Al s ; ___________ o Ecell = +0.046 V 1025 Chapter 20: Electrochemistry Ecell 3 8 Mg 2 + OH 0.0592 = +0.046 0.0592 log 0.016 0.042 o = Ecell log 2 6 6 0.25 2 Al OH 4 = 0.046 V + 0.150 V = 0.196 V 3 8 39. (M) All these observations can be understood in terms of the procedure we use to balance halfequations: the ionelectron method. (a) The reactions for which E depends on pH are those that contain either H + aq or OH aq in the balanced halfequation. These reactions involve oxoacids and oxoanions whose central atom changes oxidation state. (b) H + aq will inevitably be on the left side of the reduction of an oxoanion because reduction is accompanied by not only a decrease in oxidation state, but also by the loss 2 of oxygen atoms, as in ClO3 CIO 2 , SO 4 SO 2 , and NO3 NO . These oxygen atoms appear on the righthand side as H 2 O molecules. The hydrogens that are added to the righthand side with the water molecules are then balanced with H + aq on the lefthand side. If a halfreaction with H + aq ions present is transferred to basic solution, it may be (c) originally present. This results in H 2 O(l) on the side that had H + aq ions (the left side in this case) and OH aq ions on the other side (the right side.) rebalanced by adding to each side OH aq ions equal in number to the H + aq 40. (M) Oxidation: 2 Cl aq Cl2 g + 2 e E o = 1.358 V E o = +1.455 V Reduction: PbO 2 s + 4 H + aq + 2 e Pb 2+ aq + 2 H 2 O(l) o Net: PbO 2 s + 4 H + aq + 2 Cl aq Pb 2+ aq + 2 H 2 O(l) + Cl2 g ; Ecell = +0.097 V We derive an expression for Ecell that depends on just the changing H + . Ecell = E
o cell 0.0592 P{Cl 2 }[Pb 2 ] (1.00 atm)(1.00 M) log = +0.097 0.0296 log 4 2 2 [H ] [Cl ] [H ]4 (1.00) 2 0.097 4 0.0296 log[H ] 0.097 0.118log[H + ] 0.097 0.118 pH
(a) Ecell = +0.097 + 0.118 log 6.0 = +0.189 V Forward reaction is spontaneous under standard conditions Ecell = +0.097 + 0.118 log 1.2 = +0.106 V Forward reaction is spontaneous under standard conditions (b) 1026 Chapter 20: Electrochemistry (c) Ecell = +0.097 0.118 4.25 = 0.405 V Forward reaction is nonspontaneous under standard conditions The reaction is spontaneous in strongly acidic solutions (very low pH), but is nonspontaneous under standard conditions in basic, neutral, and weakly acidic solutions.
41. (M) Oxidation: Zn s Zn 2+ aq 2 e E o = +0.763V E o = +0.337 V
o Ecell = +1.100 V Reduction: Cu 2+ aq + 2 e Cu s Net: Zn s + Cu 2+ aq Cu s + Zn 2+ aq (a) We set E = 0.000V , Zn 2+ = 1.00M , and solve for Cu 2+ in the Nernst equation. Ecell = E
o cell log 1.0 M 0.000 1.100 = = 37.2; 2+ 0.0296 Cu Zn 2+ 0.0592 log 2+ ; 2 Cu 0.000 = 1.100 0.0296 log 1.0 M Cu 2+ Cu 2+ = 1037.2 = 6 1038 M (b) If we work the problem the other way, by assuming initial concentrations of Cu 2+ = 6 1038 M and = 1.0 M and Zn 2+ = 0.0 M , we obtain Cu 2+ initial initial final Zn 2+ final = 1.0 M . Thus, we would conclude that this reaction goes to completion.
42. (M) Oxidation:
Sn s Sn 2+ aq + 2 e E o = +0.137 V Reduction: Pb 2+ aq + 2 e Pb s Net: Sn s + Pb 2 + aq Sn 2+ aq + Pb s E o = 0.125 V
o Ecell = +0.012 V Now we wish to find out if Pb 2+ aq will be completely displaced, that is, will Pb 2+ reach 0.0010 M, if Sn 2+ is fixed at 1.00 M? We use the Nernst equation to determine if the cell voltage still is positive under these conditions. Sn 2+ 0.0592 0.0592 1.00 o Ecell = Ecell log = +0.012 log = +0.012 0.089 = 0.077 V 2+ 2 2 0.0010 Pb The negative cell potential tells us that this reaction will not go to completion under the conditions stated. The reaction stops being spontaneous when Ecell = 0 . We can work this the another way as well: assume that Pb 2 + = 1.0 x M and calculate Sn 2+ = x M at equilibrium, that is, Sn 2+ 0.0592 0.0592 x o where Ecell = 0 . Ecell = 0.00 = Ecell log = +0.012 log 2+ 2 2 1.0 x Pb 1027 Chapter 20: Electrochemistry 2 0.012 2.6 x = = 0.41 x = 10 0.41 1.0 x = 2.6 2.6 x x = = 0.72 M 0.0592 3.6 1.0 x We would expect the final Sn 2+ to equal 1.0 M (or at least 0.999 M) if the reaction went to log completion. Instead it equals 0.72 M and consequently, the reaction fails to go to completion.
43.
o (M) (a) The two halfequations and the cell equation are given below. Ecell = 0.000 V Oxidation: H 2 g 2 H + 0.65 M KOH + 2 e Reduction: Net: 2 H + 1.0 M + 2 e H 2 g 2 H + 1.0 M 2 H + 0.65 M KOH Kw 1.00 1014 M 2 H+ = = = 1.5 1014 M base OH 0.65 M H + base 1.5 10 14 0.0592 0.0592 Ecell = E log = 0.000 log 2 2 2 1.0 2 H + acid For the reduction of H 2 O(l) to H 2 g in basic solution,
o cell 2 2 = +0.818 V (b) 2 H 2 O(l) + 2 e 2 H 2 g + 2 OH aq , E o = 0.828 V . This reduction is the 44. reverse of the reaction that occurs in the anode of the cell described, with one small difference: in the standard halfcell, [OH] = 1.00 M, while in the anode halfcell in the case at hand, [OH] = 0.65 M. Or, viewed in another way, in 1.00 M KOH, [H+] is smaller still than in 0.65 M KOH. The forward reaction (dilution of H + ) should be even more spontaneous, (i.e. a more positive voltage will be created), with 1.00 M o KOH than with 0.65 M KOH. We expect that Ecell (1.000 M NaOH) should be a little larger than Ecell (0.65 M NaOH), which, is in fact, the case. (M) (a) Because NH 3 aq is a weaker base than KOH(aq), [OH] will be smaller than in the previous problem. Therefore the [H+] will be higher. Its logarithm will be less negative, and the cell voltage will be less positive. Or, viewed as in Exercise 41(b), the difference in [H+] between 1.0 M H+ and 0.65 M KOH is greater than the difference in [H+] between 1.0 M H+ and 0.65 M NH3. The forward reaction is "less spontaneous" and Ecell is less positive.
(b) Reaction: Initial: Changes: Equil: NH 4 + OH xx x2 = 1.8 105 = Kb = 0.65 x 0.65 NH 3 x = OH = 0.65 1.8 105 3.4 103 M; NH 3 aq + H 2 O(l) 0.65M x M 0.65 x M NH + aq + 4 0M +x M xM OH aq 0 M +x M xM [H 3O + ] 1.00 1014 2.9 1012 M 3.4 103 1028 Chapter 20: Electrochemistry Ecell = E o cell [ H ]2 (2.9 1012 ) 2 0.0592 0.0592 base log 2 0.000 log 0.683 V [ H ]acid (10) 2 . 2 2 45. (M) First we need to find Ag + in a saturated solution of Ag 2 CrO 4 . 1.1 10 12 6.5 10 5 M 4 o The cell diagrammed is a concentration cell, for which Ecell = 0.000 V , n = 1, K sp = Ag + CrO 2 = 2 s s = 4 s 3 = 1.1 10 12 s 4 2 2
4 Ag + anode = 2 s = 1.3 10 M 3 Cell reaction: Ag s + Ag + 0.125 M Ag s + Ag + 1.3 10 4 M
o Ecell = Ecell 0.0592 1.3 10 M log = 0.000 + 0.177 V = 0.177 V 1 0.125 M 4 46. (M) First we need to determine Ag + in the saturated solution of Ag 3 PO 4 .
o The cell diagrammed is a concentration cell, for which Ecell = 0.000V, n = 1 . Cell reaction: Ag s + Ag + 0.140 M Ag s + Ag + x M 0.0592 xM xM 0.180 o log ; log = = 3.04 Ecell = 0.180V = Ecell 1 0.140 M 0.140 M 0.0592
x M = 0.140 M 103.04 = 0.140 M 9.110 4 = 1.3 104 M = Ag + K sp = Ag + PO3 = 3s s = 1.3 10 4 3 3 4 3 1.3 10 4 3 = 9.5 1017 E o = +0.137 V anode 47. (D) (a) Reduction: Pb 2 + 0.600 M + 2 e Pb s Oxidation: Sn s Sn 2+ 0.075 M + 2 e E o = 0.125 V
_____ o Net: Sn s + Pb 2+ 0.600 M Pb s + Sn 2+ 0.075 M ; Ecell = +0.012 V Ecell = E o cell Sn 2+ 0.0592 0.075 log 2+ = 0.012 0.0296 log = 0.012 + 0.027 = 0.039 V 2 0.600 Pb (b) As the reaction proceeds, [Sn2+] increases while [Pb2+] decreases. These changes cause the driving force behind the reaction to steadily decrease with the passage of time. This decline in driving force is manifested as a decrease in Ecell with time. When Pb 2+ = 0.500 M = 0.600 M 0.100 M , Sn 2+ = 0.075 M + 0.100 M , because the stoichiometry of the reaction is 1:1 for Sn 2+ and Pb 2+ . Sn 2+ 0.0592 0.175 o Ecell = Ecell log 2+ = 0.012 0.0296 log = 0.012 + 0.013 = 0.025 V 2 0.500 Pb (c) 1029 Chapter 20: Electrochemistry (d) Reaction: Initial: Changes: Final: Sn(s) + Pb 2+ aq 0.600 M x M Pb s + Sn 2 + aq 0.075 M +x M 0.600 x M 0.075 + x M Ecell = E
log o cell Sn 2+ 0.0592 0.075 + x log = 0.020 = 0.012 0.0296 log 2+ 2 0.600 x Pb 0.075 + x Ecell 0.012 0.020 0.012 0.075 + x = = = 0.27; = 100.27 = 0.54 0.0296 0.0296 0.600 x 0.600 x 0.324 0.075 0.075 + x = 0.54 0.600 x = 0.324 0.54 x; x = = 0.162 M 1.54 Sn 2+ = 0.075 + 0.162 = 0.237 M (e) Here we use the expression developed in part (d). 0.075 + x Ecell 0.012 0.000 0.012 log = = = +0.41 0.600 x 0.0296 0.0296 0.075 + x = 10+0.41 = 2.6; 0.075 + x = 2.6 0.600 x = 1.6 2.6 x 0.600 x 1.6 0.075 x= = 0.42 M 3.6 2+ 2+ Sn = 0.075 + 0.42 = 0.50 M; Pb = 0.600 0.42 = 0.18 M Oxidation: Ag s Ag + 0.015 M + e E o = 0.800 V 48. (D) (a) Reduction: Fe3+ 0.055 M + e Fe 2+ 0.045 M E o = +0.771V o Net: Ag s + Fe3+ 0.055 M Ag + 0.015 M + Fe 2+ 0.045 M Ecell = 0.029 V
o cell Ecell = E Ag + Fe 2+ 0.0592 = 0.029 0.0592 log 0.015 0.045 log 3+ 1 0.055 Fe = 0.029 V + 0.113 V = +0.084 V
(b) As the reaction proceeds, Ag + and Fe 2+ will increase, while Fe 3+ decrease. These changes cause the driving force behind the reaction to steadily decrease with the passage of time. This decline in driving force is manifested as a decrease in Ecell with time. (c) When Ag + = 0.020 M = 0.015 M + 0.005 M , Fe 2+ = 0.045 M + 0.005 M = 0.050 M and Fe 3+ = 0.055 M 0.005 M = 0.500 M , because, by the stoichiometry of the reaction, a mole of Fe 2+ is produced and a mole of Fe 3+ is consumed for every mole of Ag + produced. 1030 Chapter 20: Electrochemistry Ecell = E o cell Ag + Fe 2+ 0.0592 0.020 0.050 log = 0.029 0.0592log 3+ 1 0.050 Fe = 0.029 V + 0.101 V = +0.072 V
(d) Reaction: Initial: Changes: Final:
o Ecell = Ecell Ag s + Fe3+ 0.055 M 0.055 M x M 0.055 x M Ag + 0.015 M + 0.015 M +x M 0.015 + x M Fe 2 + 0.045 M 0.045 M +x M 0.045 + x M Ag + Fe 2 + 0.0592 = 0.029 0.0592 log 0.015 + x 0.045 + x log 3+ 1 Fe 0.055 x 0.015 + x 0.045 + x = Ecell + 0.029 = 0.010 + 0.029 = 0.66 log 0.0592 0.0592 0.055 x 0.015 + x 0.045 + x = 10 0.66 = 0.22 0.055 x 0.00068 + 0.060 x + x 2 = 0.22 0.055 x = 0.012 0.22 x x 2 + 0.28 x 0.011 = 0 b b 2 4ac 0.28 (0.28) 2 + 4 0.011 x= = = 0.035 M 2a 2 Ag + = 0.015 M + 0.035 M = 0.050 M Fe 2+ = 0.045 M + 0.035 M = 0.080 M Fe 3+ = 0.055 M 0.035 M = 0.020 M
(e) We use the expression that was developed in part (d).
log 0.015 + x 0.045 + x = Ecell + 0.029 = 0.000 + 0.029 = 0.49 0.0592 0.0592 0.055 x 0.015 + x 0.045 + x = 10 0.49 = 0.32 0.055 x 0.00068 + 0.060 x + x 2 = 0.32 0.055 x = 0.018 0.32 x x 2 + 0.38 x 0.017 = 0 b b 2 4ac 0.38 (0.38) 2 + 4 0.017 x= = = 0.040 M 2a 2 Ag + = 0.015 M + 0.040 M = 0.055 M Fe 2+ = 0.045 M + 0.040 M = 0.085 M Fe 3+ = 0.055 M 0.040 M = 0.015 M 1031 Chapter 20: Electrochemistry 49. (M) First we will need to come up with a balanced equation for the overall redox reaction. Clearly, the reaction must involve the oxidation of Cl(aq) and the reduction of Cr2O72(aq): 14 H+(aq) + Cr2O72(aq) + 6 e 2 Cr3+(aq) + 7 H2O(l) {Cl(aq) 1/2 Cl2(g) + 1 e} 6 E1/2red = 1.33 V E1/2ox = 1.358 V
_____________________________________________ 14 H+(aq) + Cr2O72(aq) + 6 Cl(aq) 2 Cr3+(aq) + 7 H2O(l) + 3 Cl2(g) Ecell = 0.03 V A negative cell potential means, the oxidation of Cl(aq) to Cl2(g) by Cr2O72(aq) at standard conditions will not occur spontaneously. We could obtain some Cl2(g) from this reaction by driving it to the product side with an external voltage. In other words, the reverse reaction is the spontaneous reaction at standard conditions and if we want to produce some Cl2(g) from the system, we must push the nonspontaneous reaction in its forward direction with an external voltage, (i.e., a DC power source). Since Ecell is only slightly negative, we could also drive the reaction by removing products as they are formed and replenishing reactants as they are consumed.
50. 50. (D) We proceed by first deriving a balanced equation for the reaction occurring in the cell: Oxidation: Fe(s) Fe 2+ (aq)+2e Reduction: {Fe 3+ (aq)+e Fe 2+ (aq)} 2 Net: Fe(s)+2Fe 3+ (aq) 3Fe 2+ (aq) (a) G o and the equilibrium constant Keq can be calculated using o G o = nFEcell RT ln K eq :
o G o = nFEcell = 2 96485Cmol1 1.21V 233.5kJmol1 G o RT ln K 8.314JK 1mol1 298.15K ln K 233.5 1000Jmol1 ln K 94.2 K e94.2 8.1 1040 (b) Before calculating voltage using the Nernst equation, we need to rewrite the net reaction to take into account concentration gradient for Fe2+(aq): Oxidation: Fe(s) Fe 2+ (aq,1.0 10 3M)+2e Reduction: {Fe 3+ (aq,1.0 10 3M)+e  Fe 2+ (aq,0.10M)} 2 Net: Fe(s)+2Fe 3+ (aq,1.0 103 ) Fe 2+ (aq,1.0 10 3M)+2Fe 2+ (aq,0.10M) Therefore, 1.0 10 3 (0.10)2 Q 10 (1.0 10 3 )2 Now, we can apply the Nerst equation to calculate the voltage: 0.0592 0.0592 o Ecell Ecell log Q =1.21 V log10 1.18 V 2 n (c) From parts (a) and (b) we can conclude that the reaction between Fe(s) and Fe3+(aq) is spontaneous. The reverse reaction (i.e. disproportionation of Fe2+(aq)) must therefore be nonspontaneous. 1032 Chapter 20: Electrochemistry Batteries and Fuel Cells
51. (M) Stepwise approach: (a) The cell diagram begins with the anode and ends with the cathode. Cell diagram: Cr s Cr 2 + aq ,Cr 3+ aq Fe 2 + aq , Fe 3+ aq Fe s (b) Oxidation: Cr 2+ aq Cr 3+ aq + e Reduction: Fe3+ aq + e Fe 2+ aq Net: Cr 2+ aq + Fe3+ aq Cr 3+ aq + Fe 2+ aq E o = +0.424 V E o = +0.771V
o Ecell = +1.195 V Conversion pathway approach: Cr s Cr 2 + aq ,Cr 3+ aq Fe 2 + aq , Fe 3+ aq Fe s E o = +0.424 V
52. (M) (a) E o = +0.771V E o = +0.763V o Ecell = 0.424V 0.771V 1.195V Oxidation: Zn s Zn 2+ aq + 2 e + Reduction: 2MnO 2 s + H 2 O(l) + 2 e Mn 2 O3 s + 2OH aq Acidbase: {NH 4 aq + OH aq NH 3 g + H 2 O l } 2
___________________________________________________ Complex: Zn 2+ aq + 2NH 3 aq + 2Cl aq Zn NH 3 2 Cl2 s Net: Zn s + 2 MnO 2 s + 2 NH 4 + aq + 2 Cl aq Mn 2 O 3 s + H 2 O l + [Zn NH 3 2 ]Cl 2 s (b)
o G o = nFEcell = 2 mol e 96485 C/mol e 1.55 V = 2.99 105 J/mol This is the standard free energy change for the entire reaction, which is composed of the four reactions in part (a). We can determine the values of G o for the acidbase and complex formation reactions by employing the appropriate data from Appendix D and pK f = 4.81 (the negative log of the Kf for [Zn(NH3)2]2+).
Gao b = RTlnK b = 8.3145 J mol1 K 1 298.15 K ln 1.8 105 = 5.417 104 J/mol
2 2 o Gcmplx = RTlnK f = 8.3145 J mol 1 K 298.15 K ln 104.81 = 2.746 104 J/mol o o o Then Gtotal = Gredox + Gao b + Gcmplx o o o Gredox = Gtotal Gao b Gcmplx = 2.99 105 J/mol 5.417 104 + 2.746 104 J/mol = 3.26 105 J/mol Thus, the voltage of the redox reactions alone is 3.26 105 J o E = = 1.69 V 1.69V = +0.763V + E o MnO 2 /Mn 2 O3 2 mol e 96485 C / mol e
E o MnO 2 /Mn 2O3 = 1.69V 0.763V = +0.93V The electrode potentials were calculated by using equilibrium constants from Appendix D. These calculations do not take into account the cell's own internal resistance to the flow of electrons, which makes the actual voltage developed by the electrodes less than the theoretical values derived from equilibrium constants. Also because the solid species 1033 Chapter 20: Electrochemistry (other than Zn) do not appear as compact rods, but rather are dispersed in a paste, and since very little water is present in the cell, the activities for the various species involved in the electrochemical reactions will deviate markedly from unity. As a result, the equilibrium constants for the reactions taking place in the cell will be substantially different from those provided in Appendix D, which apply only to dilute solutions and reactions involving solid reactants and products that posses small surface areas. The actual electrode voltages, therefore, will end up being different from those calculated here.
53. (M) (a) Cell reaction: 2H 2 g + O 2 g 2H 2 O l
o o Grxn = 2Gf H 2 O l = 2 237.1 kJ/mol = 474.2 kJ/mol o G 474.2 103 J/mol o = = 1.229 V Ecell = 4 mol e 96485 C/mol e nF
(b) Anode, oxidation: {Zn s Zn 2+ aq + 2e } 2 Cathode, reduction: O 2 g + 4 H + aq + 4e 2H 2 O(l) E o = +0.763V E o = +1.229 V o Net: 2 Zn s + O 2 g + 4 H + aq 2 Zn 2+ aq + 2 H 2 O(l) Ecell = +1.992 V (c) Anode, oxidation: Mg s Mg 2+ aq + 2e Cathode, reduction: I 2 s + 2e 2 I aq Net: Mg s + I 2 s Mg 2+ aq + 2 I aq E o = +2.356 V E o = +0.535 V
o Ecell = +2.891V 54. (M) (a) Zn(s) Zn2+(aq) + 2 e Precipitation: Zn2+(aq) + 2 OH(aq) Zn(OH)2(s) Reduction: 2 MnO2(s) + H2)(l) + 2 e Mn2O3(s) + 2 OH(aq) Oxidation: __ Net : Zn(s) + 2 MnO2(s) + H2O(l) + 2 OH(aq) Mn2O3(s) + Zn(OH)2(s) (b) In 50, we determined that the standard voltage for the reduction reaction is +0.93 V (n = 2 e). To convert this voltage to an equilibrium constant (at 25 C) use: nE o 2(0.93) K red 1031.42 3 1031 31.4 ; log K red 0.0592 0.0592 and for Zn(s) Zn2+(aq) + 2 e (E = 0.763 V and n = 2 e) nE o 2(0.763) 25.8 ; log K ox K ox 1025.78 6 1025 0.0592 0.0592 Gtotal = Gprecipitation + Goxidation + Greduction 1 Gtotal = RT ln + (RT lnKox) + (RT lnKred) K sp, Zn(OH)2 1034 Chapter 20: Electrochemistry Gtotal = RT (ln 1 K sp, Zn(OH)2 + lnKox + lnKred) 1 kJ (298 K) (ln + ln(6.0 1025) + ln(2.6 1031)) 17 1.2 10 K mol Gtotal = 423 kJ = nFEtotal 423103 J Hence, Etotal = Ecell = = 2.19 V (2 mol)(96485 C mol1 ) Gtotal = 0.0083145 55. (M) AluminumAir Battery: 2 Al(s) + 3/2 O2(g) Al2O3(s) ZincAir Battery: Zn(s) + O2(g) ZnO(s) IronAir Battery Fe(s) + O2(g) FeO(s) Calculate the quantity of charge transferred when 1.00 g of metal is consumed in each cell. AluminumAir Cell: 1 mol Al(s) 3 mol e 96, 485C 1.00 g Al(s) 1.07 104 C 26.98 g Al(s) 1 mol Al(s) 1 mol e ZincAir Cell:
1.00 g Zn(s) 1 mol Zn(s) 2 mol e 96, 485 C 2.95 103 C 65.39 g Zn(s) 1 mol Zn(s) 1 mol e 1 mol Fe(s) 2 mol e 96, 485 C 3.46 103 C 55.847 g Fe(s) 1 mol Fe(s) 1 mol e IronAir Cell:
1.00 g Fe(s) As expected, aluminum has the greatest quantity of charge transferred per unit mass (1.00 g) of metal oxidized. This is because aluminum has the smallest molar mass and forms the most highly charged cation (3+ for aluminum vs 2+ for Zn and Fe).
56. (M) (a) A voltaic cell with a voltage of 0.1000 V would be possible by using two halfcells whose standard reduction potentials differ by approximately 0.10 V, such as the following pair. Oxidation: 2 Cr 3+ aq + 7 H 2O(l) Cr2O7 2 aq +14 H + aq + 6 e E o = 1.33 V Reduction: {PbO2 s + 4 H + aq + 2e Pb2+ aq + 2 H 2O(l)} 3
3+ 22+ + E o = +1.455 V ______________________________
o Net: 2Cr aq +3PbO 2 s +H 2 O(l) Cr2 O 7 aq +3Pb aq +2H aq Ecell =0.125 V The voltage can be adjusted to 0.1000 V by a suitable alteration of the concentrations. Pb 2+ or H + could be increased or Cr 3+ could be decreased, or any combination of the three of these.
(b) To produce a cell with a voltage of 2.500 V requires that one start with two halfcells whose reduction potentials differ by about that much. An interesting pair follows. Oxidation: Al s Al3+ aq + 3e E o = +1.676 V 1035 Chapter 20: Electrochemistry Reduction: {Ag + aq + e Ag s } 3 Net:
Al s + 3Ag aq Al
+ 3+ E o = +0.800 V
___________________________ aq + 3Ag s E o cell = +2.476 V Again, the desired voltage can be obtained by adjusting the concentrations. In this case increasing Ag + and/or decreasing Al 3+ would do the trick. Since no known pair of halfcells has a potential difference larger than about 6 volts, we conclude that producing a single cell with a potential of 10.00 V is currently impossible. It is possible, however, to join several cells together into a battery that delivers a voltage of 10.00 V. For instance, four of the cells from part (b) would deliver ~10.0 V at the instant of hookup. (M) Oxidation (anode): Li(s) Li+ (aq)+e E o 3.040V Reduction (cathode): MnO 2 (s)+2H 2 O(l)+e Mn(OH)3 (s)+OH  (aq) E o 0.20V
(c)
________________________________________________________________________________________________________ 57. Net: MnO 2 (s)+2H 2 O(l)+Li(s) Mn(OH)3 (s)+OH  (aq)+Li (aq ) Cell diagram: Li( s ), Li (aq ) KOH ( satd ) MnO2 ( s ), Mn(OH )3 ( s )
58. (M) (a) Oxidation (anode): Zn(s) Zn +2 (aq)+2eReduction (cathode): Br2 (l)+2e 2Br  (aq) o Ecell 2.84V E o 0.763V E o 1.065V
o Ecell 1.828V E o 3.040V E o 2.866V o Ecell 4.868V ________________________________________________________________________________________________________ Net: Zn(s)+Br2 (l) Zn 2+ (aq)+2Br  (aq) (b) Oxidation (anode): {Li(s) Li+ (aq)+e } 2 Reduction (cathode): F2 (g)+2e 2F  (aq) Net: 2Li(s)+F2 (g) 2Li+ (aq)+2F  (aq) ________________________________________________________________________________________________________ Electrochemical Mechanism of Corrosion
59. (M) (a) Because copper is a less active metal than is iron (i.e. a weaker reducing agent), this situation would be similar to that of an iron or steel can plated with a tin coating that has been scratched. Oxidation of iron metal to Fe2+(aq) should be enhanced in the body of the nail (blue precipitate), and hydroxide ion should be produced in the vicinity of the copper wire (pink color), which serves as the cathode. (b) (c) Because a scratch tears the iron and exposes "fresh" metal, it is more susceptible to corrosion. We expect blue precipitate in the vicinity of the scratch. Zinc should protect the iron nail from corrosion. There should be almost no blue precipitate; the zinc corrodes instead. The pink color of OH should continue to form. 60. (M) The oxidation process involved at the anode reaction, is the formation of Fe2+(aq). This occurs far below the water line. The reduction process involved at the cathode, is the formation of OH aq from O2(g). It is logical that this reaction would occur at or near the water line close to the atmosphere (which contains O2). This reduction reaction requires O2(g) from the atmosphere and H2O(l) from the water. The oxidation reaction, on the other hand simply 1036 Chapter 20: Electrochemistry requires iron from the pipe together with an aqueous solution into which the Fe 2 + aq can disperse and not build up to such a high concentration that corrosion is inhibited. Anode, oxidation: Cathode, reduction:
61. Fe s Fe 2 + aq + 2e O 2 g + 2H 2 O(l) + 4e 4OH aq (M)During the process of corrosion, the metal that corrodes loses electrons. Thus, the metal in these instances behaves as an anode and, hence, can be viewed as bearing a latent negative polarity. One way in which we could retard oxidation of the metal would be to convert it into a cathode. Once transformed into a cathode, the metal would develop a positive charge and no longer release electrons (or oxidize). This change in polarity can be accomplished by hooking up the metal to an inert electrode in the ground and then applying a voltage across the two metals in such a way that the inert electrode becomes the anode and the metal that needs protecting becomes the cathode. This way, any oxidation that occurs will take place at the negatively charged inert electrode rather than the positively charged metal electrode. (M) As soon as the iron and copper came into direct contact, an electrochemical cell was created, in which the more powerfully reducing metal (Fe) was oxidized. In this way, the iron behaved as a sacrificial anode, protecting the copper from corrosion. The two halfreactions and the net cell reaction are shown below: 62. Anode (oxidation) Cathode (reduction) Net: Fe(s) Fe2+(aq) + 2 e Cu2+(aq) + 2 e Cu(s) Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E = 0.440 V E = 0.337 V Ecell = 0.777 V Note that because of the presence of iron and its electrical contact with the copper, any copper that does corrode is reduced back to the metal. Electrolysis Reactions
63. (M) Here we start by calculating the total amount of charge passed and the number of moles of electrons transferred. 60 s 2.15 C 1 mol e mol e = 75 min = 0.10 mol e 1 min 1s 96485 C (a) mass Zn = 0.10 mol e 1 mol Zn 2+ 1 mol Zn 65.39 g Zn = 3.3 g Zn 2 mol e 1 mol Zn 2+ 1 mol Zn (b) 1 mol Al3+ 1 mol Al 26.98 g Al mass Al = 0.10 mol e = 0.90 g Al 3 mol e 1 mol Al3+ 1 mol Al mass Ag = 0.10 mol e 1 mol Ag + 1 mol Ag 107.9 g Ag = 11 g Ag 1 mol e 1 mol Ag + 1 mol Ag (c) 1037 Chapter 20: Electrochemistry (d) 64. mass Ni = 0.10 mol e 1 mol Ni 2+ 1 mol Ni 58.69 g Ni = 2.9 g Ni 2 mol e 1 mol Ni 2+ 1 mol Ni 1 mol Cu 2 mol e 1mol H 2 g = 0.0516 mol H 2 g 63.55g Cu 1 mol Cu 2 mol e Then we use the ideal gas equation to find the volume of H 2 g . 0.08206 L atm 0.0516 mol H 2 (273.2 28.2) K mol K Volume of H 2 (g) 1.27 L 1 atm 763 mmHg 760 mmHg This answer assumes the H 2 g is not collected over water, and that the H2(g) formed is the only gas present in the container (i.e. no water vapor present) Conversion pathway approach: Cu 2 + aq + 2 e Cu s and 2 H + aq + 2 e H 2 g mol H 2 g = 3.28 g Cu 1 mol Cu 2 mol e 1mol H 2 g = 0.0516 mol H 2 g 63.55g Cu 1 mol Cu 2 mol e 0.08206 L atm (273.2 28.2)K 0.0516 molH 2 molK V (H 2 (g)) 1.27 L 1 atm 763 mmHg 760 mmHg mol H 2 g = 3.28 g Cu (M) We proceed by first writing down the net electrochemical reaction. The number of moles of hydrogen produced in the reaction can be calculated from the reaction stoichiometry. Lastly, the volume of hydrogen can be determined using ideal gas law. Stepwise approach: The two half reactions follow: Cu 2 + aq + 2 e Cu s and 2 H + aq + 2 e H 2 g Thus, two moles of electrons are needed to produce each mole of Cu(s) and two moles of electrons are needed to produce each mole of H 2 g . With this information, we can compute the moles of H 2 g that will be produced. 65. (M) Here we must determine the standard cell voltage of each chemical reaction. Those chemical reactions that have a negative voltage are the ones that require electrolysis. (a) Oxidation: 2 H 2 O(l) 4H + aq + O 2 g + 4 e Reduction: {2 H + aq + 2 e H 2 g } 2 Net: 2 H 2 O(l) 2 H 2 g + O 2 g Oxidation: Zn s Zn 2+ aq + 2 e Reduction: Fe 2+ aq + 2 e Fe s E o = 1.229 V E o = 0.000 V
o Ecell = 1.229 V This reaction requires electrolysis, with an applied voltage greater than +1.229V .
(b) E o = +0.763 V E o = 0.440 V 1038 Chapter 20: Electrochemistry Net: Zn s + Fe2+ aq Fe s + Zn 2+ aq (c) o Ecell = +0.323V This is a spontaneous reaction under standard conditions. Oxidation: {Fe 2 + aq Fe 3+ aq + e } 2 Reduction: I 2 s + 2 e 2I aq E o = 0.771 V
E o = +0.535 V o Net: 2 Fe 2 + aq + I2 s 2 Fe 3+ aq + 2 I aq Ecell = 0.236 V This reaction requires electrolysis, with an applied voltage greater than +0.236V . (d) Oxidation: Cu s Cu 2+ aq + 2 e Reduction: {Sn 4+ aq + 2 e Sn 2+ aq } 2 E o = 0.337 V E o = +0.154 V o Net: Cu s + Sn 4 + aq Cu 2 + aq + Sn 2+ aq Ecell = 0.183V This reaction requires electrolysis, with an applied voltage greater than +0.183V . 66. (M) (a) Because oxidation occurs at the anode, we know that the product cannot be H 2 (H2 is produced from the reduction of H 2 O ), SO 2 , (which is a reduction product of SO 4 ), or SO 3 (which is produced from SO 4 without a change of oxidation state; it is the dehydration product of H 2SO 4 ). It is, in fact O2 that forms at the anode. The oxidation of water at the anode produces the O2(g).
(b) 2 2 Reduction should occur at the cathode. The possible species that can be reduced are H 2 O to H 2 g , K + aq to K s , and SO 2 aq to perhaps SO 2 g . Because 4 potassium is a highly active metal, it will not be produced in aqueous solution. In order for SO 2 aq to be reduced, it would have to migrate to the negatively charged 4 cathode, which is not very probable since like charges repel each other. Thus, H 2 g is produced at the cathode. At the anode: At the cathode:
2 H 2 O(l) 4H + aq + O 2 g + 4 e {2H + aq + 2 e H 2 g } 2 (c) E o = 1.229 V E o = 0.000 V
o Ecell = 1.229 V Net cell reaction: 2 H 2 O(l) 2 H 2 g + O 2 g A voltage greater than 1.229 V is required. Because of the high overpotential required for the formation of gases, we expect that a higher voltage will be necessary.
67. (M) (a) (b) At the anode: 2 H 2 O(l) 4 H + aq + O 2 g + 4 e At the cathode: {2H + aq + 2 e H 2 g } 2 Net cell reaction: 2 H 2 O(l) 2 H 2 g + O 2 g The two gases that are produced are H 2 g and O 2 g . E o = 1.229 V E o = 0.000 V
o Ecell = 1.229 V 68. (M) The electrolysis of Na 2 SO 4 aq produces molecular oxygen at the anode. E o O 2 g /H 2 O = 1.229 V . The other possible product is S2O 2 aq . It is however, 8 1039 Chapter 20: Electrochemistry unlikely to form because it has a considerably less favorable halfcell potential. 2 2 E o S2 O8 aq /SO 4 aq = 2.01 V . H 2 g is formed at the cathode. 3600 s 2.83 C 1 mol e 1 mol O 2 mol O 2 = 3.75 h = 0.0990 mol O 2 1h 1s 96485 C 4 mol e The vapor pressure of water at 25 C , from Table 122, is 23.8 mmHg. nRT 0.0990 mol 0.08206 L atm mol1 K 1 298 K 2.56 L O 2 (g) V 1 atm P (742 23.8) mmHg 760 mmHg
69. (M) (a) Zn 2+ aq + 2e Zn s 60 s 1.87 C 1 mol e 1 mol Zn 65.39 g Zn mass of Zn = 42.5 min = 1.62 g Zn 1 min 1s 96, 485 C 2 mol e 1 mol Zn 2I aq I2 s + 2e time needed = 2.79 g I2 1mol I2 2 mol e 96, 485 C 1 s 1 min = 20.2 min 253.8g I2 1mol I2 1 mol e 1.75C 60 s
2.68 C 1 mol e 1 mol Cu 2+ 1000 mmol 1s 96, 485 C 2 mol e 1mol (b) 70. (M) (a) Cu 2+ aq + 2e Cu s mmol Cu 2+ consumed = 282 s = 3.92 mmol Cu 2+ decrease in Cu 2+ = 3.92 mmol Cu 2+ = 0.00922 M 425 mL final Cu 2+ = 0.366 M 0.00922 M = 0.357 M
(b)
mmol Ag + consumed = 255 mL 0.196 M 0.175 M = 5.36 mmol Ag +
time needed = 5.36 mmol Ag + 1 mol Ag + 1 mol e 96485C 1s 1000 mmol Ag + 1 mol Ag + 1 mol e 1.84 C = 281 s 2.8 102 s 71. (M) (a) charge = 1.206 g Ag 1 mol Ag 1 mol e 96,485 C = 1079 C 107.87 g Ag 1 mol Ag 1 mol e (b) current = 1079 C = 0.7642 A 1412 s 1040 Chapter 20: Electrochemistry 72. (D) (a) Anode, oxidation: 2H 2 O(l) 4H + aq + 4e + O 2 g E o = 1.229V Cathode, reduction: {Ag + aq + e Ag s } 4 Net: 2H 2 O(l) + 4Ag + aq 4H + aq + O 2 g + 4Ag s (b) E o = +0.800 V
o Ecell = 0.429 V charge = 25.8639 25.0782 g Ag current = 1 mol Ag 1 mol e 96, 485 C = 702.8C 107.87 g Ag 1 mol Ag 1 mol e 702.8 C 1h = 0.0976 A 2.00 h 3600 s (c) The gas is molecular oxygen. 1 mol e 1mol O 2 L atm 702.8 C 0.08206 (23+273) K 96485 C 4 mol e mol K nRT V 1 atm P 755 mmHg 760 mmHg 1000 mL 0.0445 L O 2 44.5 mL of O 2 1L 73. (D) (a) The electrochemical reaction is: Anode (oxidation): {Ag(s) Ag + (aq)+e (aq)} 2 Cathode (reduction): Cu 2+ (aq)+2e Cu(s) E o 0.800V Eo 0.340V
o Ecell 0.46V ___________________________________________________________________________________________________ Net: 2Ag(s)+Cu 2+ (aq) Cu(s)+2Ag + (aq) Therefore, copper should plate out first. charge 1h (b) current = = 0.75A charge 6750C 2.50hmin 3600 s mass=6750 C 1 mole 1mol Cu 63.546 g Cu =2.22 g Cu 96485 C 2 mol e1 mol Cu (c) The total mass of the metal is 3.50 g out of which 2.22g is copper. Therefore, the mass of silver in the sample is 3.50g2.22g=1.28g or (1.28/3.50)x100=37%.
74. First, calculate the number of moles of electrons involved in the electrolysis: 60 sec 1mole e1.20C / s 32.0 min 0.0239mol e1min 96485C From the known mass of platinum, determine the number of moles: 1mol Pt 0.0109molPt 2.12gPt 195.078gPt Determine the number of electrons transferred: 0.0239 mol e 2.19 0.0109 mol Pt
(D) 1041 Chapter 20: Electrochemistry Therefore, 2.19 is shared between Pt2+ and Ptx+. Since we know the mole ratio between Pt2+ and Ptx+, we can calculate x: 2.19=0.90(+2) + 0.10 (x). 2.19 1.80 0.10 x x 4 (a) The oxidation state of the contaminant is +4. INTEGRATIVE AND ADVANCED EXERCISES
75. (M) Oxidation:
Reduction: Net: V 3 H 2 O VO 2 2 H e V 3 H 2 O Ag VO 2 2 H Ag(s) E a
E o 0.800 V Eo 0.439 V cell Ag e Ag(s) 0.439 V E o a 0.800 V E a 0.800 V 0.439 V 0.361 V Oxidation: Reduction: Net: V 2 V 3 e VO 2 2 H e V 3 H 2 O V 2 VO 2 2 H 2 V3 H 2 O V 3 e V 2 E ob E o 0.361 V Eo 0.616 V cell E o = 0.255 V 0.616 V E o b 0.361 V E o b 0.361 V 0.616 V 0.255 V Thus, for the cited reaction: 76. (M)The cell reaction for discharging a lead storage battery is equation (20.24). Pb(s) PbO 2 (s) 2 H (aq) 2 HSO 4 (aq) PbSO 4 (s) 2 H 2 O(l) The halfreactions with which this equation was derived indicates that two moles of electrons are transferred for every two moles of sulfate ion consumed. We first compute the amount of H2SO4 initially present and then the amount of H2SO4 consumed. 5.00 mol H 2 SO 4 initial amount H 2 SO 4 1.50 L 7.50 mol H 2 SO 4 1 L soln 2 1 mol H 2 SO 4 3600 s 2.50 C 1 mol e 2 mol SO 4 amount H 2 SO 4 consumed 6.0 h 2 96485 C 1h 1s 2 mol e 1 mol SO 4 0.56 mol H 2 SO 4 final [H 2 SO 4 ] 7.50 mol 0.56 mol 4.63 M 1.50 L 77. (M) The cell reaction is 2 Cl (aq) 2 H 2 O(l) 2 OH (aq) H 2 (g) Cl 2 (g) We first determine the charge transferred per 1000 kg Cl2. 1 mol Cl2 1000 g 2 mol e 96, 485 C charge 1000 kg Cl2 2.72 109 C 1 kg 70.90 g Cl2 1 mol Cl2 1 mol e 1042 Chapter 20: Electrochemistry 1J 1 kJ 9.38 10 6 kJ 1 V C 1000 J 1W s 1h 1 kWh (b) energy 9.38 10 9 J 2.61 10 3 kWh 1J 3600 s 1000 W h
(a) energy 3.45 V 2.72 10 9 C 78. (D) We determine the equilibrium constant for the reaction. Oxidation : Fe(s) Fe (aq) 2e _ E 0.440 V Reduction : {Cr 3 (aq) e _ Cr 2 (aq)} 2 E 0.424 V
2+ Net:
G nFE
o o cell Fe(s) + 2 Cr (aq) 3+ Fe (aq) + 2 Cr (aq) 2+ E = +0.016 V cell RT 2 mol e 96485 Coul/mol e _ 0.016 V ln K 1.2 8.3145 J mol 1 K 1 298 K
_ RT ln K ln K o nFEcell K e1..2 3.3 Reaction: Fe(s) 2 Cr 3 (aq) Fe 2 (aq) Initial : 1.00 M 0M Changes : 2x M x M x M Equil: (1.00 2x)M K [Fe 2 ][Cr 2 ]2 x (2 x) 2 3.3 [Cr 3 ]2 (1.00 2 x) 2 2Cr 2 (aq) 0 M 2x M 2x M Let us simply substitute values of x into this cubic equation to find a solution. Notice that x cannot be larger than 0.50, (values > 0.5 will result in a negative value for the denominator. x 0.40 x 0.37 0.40 (0.80) 2 K 6.4 3.3 (1.00 0.80) 2 K 0.37 (0.74) 2 3.0 3.3 (1.00 0.74) 2 x 0.35 x 0.38 0.35 (0.70) 2 K 1.9 3.3 (1.00 0.70) 2 K 0.38 (0.76) 2 3.8 3.3 (1.00 0.76) 2 Thus, we conclude that x = 0.37 M = [Fe2+].
79. (D) First we calculate the standard cell potential. Oxidation: Fe 2 (aq) Fe3 (aq) e Reduction: Ag (aq) e Ag(s) E o 0.771 V E o 0.800 V o Net: Fe 2 (aq) Ag (aq) Fe3 (aq) Ag(s) Ecell 0.029 V Next, we use the given concentrations to calculate the cell voltage with the Nernst equation. Ecell Eocell 0.0592 1 log [Fe3 ] [Fe ][Ag ]
2 0.029 0.0592 log 0.0050 0.0050 2.0 0.029 0.018 0.047 V The reaction will continue to move in the forward direction until concentrations of reactants decrease and those of products increase a bit more. At equilibrium, Ecell = 0, and we have the following. 1043 Chapter 20: Electrochemistry E o cell 0.0592 [Fe3 ] 0.029 log 1 [Fe 2 ][Ag ] [Fe3 ] 0.029 0.49 log 2 [Fe ][Ag ] 0.0592 Fe3 (aq) 0.0050 M x M (0.0050 x ) M Ag(s) Reaction: Fe2 (aq) Ag (aq) Initial: 0.0050 M 2.0 M x M x M Changes: Equil: (0.0050 x) M (2.0  x) M 0.0050 x [Fe3 ] 0.0050 x 100.49 3.1 2 [Fe ][Ag ] (0.0050 x)(2.0 x) 2.0(0.0050 x) 0.026 6.2(0.0050 x) 0.0050 x 0.031 6.2 x 7.2 x 0.026 x 0.0036 M 7.2 Note that the assumption that x 2.0 is valid. [Fe2+] = 0.0050 M 0.0036 M = 0.0014 M
80. (D) We first note that we are dealing with a concentration cell, one in which the standard oxidation reaction is the reverse of the standard reduction reaction, and consequently its standard cell potential is precisely zero volts. For this cell, the Nernst equation permits us to determine the ratio of the two silver ion concentrations. 0.0592 [Ag (satd AgI)] 0.0860 V log Ecell 0.000 V 1 [Ag (satd AgCl, x M Cl )] log 0.0860 [Ag (satd AgI)] 1.45 [Ag (satd AgCl, x M Cl )] 0.0592 [Ag (satd AgI)] 101.45 0.035 [Ag (satd AgCl, x M Cl )] We can determine the numerator concentration from the solubility product expression for AgI(s)
K sp [Ag ][I ] 8.5 1017 s 2 s 8.5 1017 9.2 109 M This permits the determination of the concentration in the denominator. 9.2 10 9 [Ag (satd AgCl, x M Cl )] 2.6 10 7 M 0.035 We now can determine the value of x. Note: Cl arises from two sources, one being the dissolved AgCl. K sp [Ag ][Cl ] 1.8 10 10 (2.6 10 7 )(2.6 10 7 x) 6.8 10 14 2.6 10 7 x 1.8 10 10 6.8 10 14 x 6.9 10 4 M [Cl ] 7 2.6 10
81. (M) The Faraday constant can be evaluated by measuring the electric charge needed to produce a certain quantity of chemical change. For instance, let's imagine that an electric circuit contains the halfreaction Cu 2 (aq) 2 e Cu(s) . The electrode on which the solid copper plates out is weighed before and after the passage of electric current. The mass gain is the mass of copper reduced, which then is converted into the moles of copper reduced. The number of moles of electrons involved in the reduction then is computed from the stoichiometry for the reduction halfreaction. In addition, the amperage is measured during the reduction, and the time is recorded. For simplicity, we assume the amperage is constant. Then the total charge (in coulombs) equals the current (in amperes, that is, 1044 Chapter 20: Electrochemistry coulombs per second) multiplied by the time (in seconds). The ratio of the total charge (in coulombs) required by the reduction divided by the amount (in moles) of electrons is the Faraday constant. To determine the Avogadro constant, one uses the charge of the electron, 1.602 1019 C and the Faraday constant in the following calculation.
NA 96,485 C 1 electron electrons 6.023 10 23 19 1 mol electrons 1.602 10 C mole 82. (M) In this problem we are asked to determine G o for N2H4(aq) using the electrochemical f data for hydrazine fuel cell. We first determine the value of G o for the cell reaction, a reaction in which n = 4. G o can then be determined using data in Appendix D. f Stewise approach: Calculate G o for the cell reaction (n=4): 96,485 C o 1.559 V 6.017 10 5 J 601.7 kJ G o n FEcell 4 mol e 1 mol e Using the data in Appendix D, determine G o for hydrazine (N2H4): f 601.7 kJ Gf o[N 2 (g)] 2 Gf o[H 2 O(l)] Gf o[N 2 H 4 (aq)] Gf o[O 2 (g)] 0.00 kJ 2 ( 237.2 kJ) Gf o[N 2 H 4 (aq)] 0.00 kJ Gf o[N 2 H 4 (aq)] 2 ( 237.2 kJ) 601.7 kJ 127.3 kJ Conversion pathway approach: 96,485 C o 1.559 V 6.017 10 5 J 601.7 kJ G o n FEcell 4 mol e 1 mol e o 601.7 kJ Gf [N 2 (g)] 2 Gf o[H 2 O(l)] Gf o[N 2 H 4 (aq)] Gf o[O 2 (g)] 0.00 kJ 2 ( 237.2 kJ) Gf o[N 2 H 4 (aq)] 0.00 kJ Gf o[N 2 H 4 (aq)] 2 ( 237.2 kJ) 601.7 kJ 127.3 kJ
83. (M) In general, we shall assume that both ions are present initially at concentrations of 1.00 M. Then we shall compute the concentration of the more easily reduced ion when its reduction potential has reached the point at which the other cation starts being reduced by electrolysis. In performing this calculation we use the Nernst equation, but modified for use with a halfreaction. We find that, in general, the greater the difference in E values for two reduction halfreactions, the more effective the separation. (a) In this case, no calculation is necessary. If the electrolysis takes place in aqueous solution, H2(g) rather than K(s) will be produced at the cathode. Cu2+ can be separated from K+ by electrolysis. (b) Cu 2 (aq) 2 e Cu(s) E o 0.340 V Ag (aq) e Ag(s) E o 0.800 V Ag+ will be reduced first. Now we ask what [Ag+] will be when E = +0.337 V. 0.0592 1 1 0.800 0.340 0.337 V 0.800 V log log 7.77 1 0.0592 [Ag ] [Ag ] 1045 Chapter 20: Electrochemistry [Ag+] = 1.7 108 M. Separation of the two cations is essentially complete.
(c) Pb 2 (aq) 2 e Pb(s) E o 0.125 V Sn 2 (aq) 2 e Sn(s) E o 0.137 V Pb2+ will be reduced first. We now ask what [Pb2+] will be when E = 0.137 V. 0.0592 1 1 2(0.137 0.125) 0.137 V 0.125 V log log 0.41 2 2 2 0.0592 [Pb ] [Pb ] [Pb2+] = 100.41 = 0.39 M Separation of Pb2+ from Sn2+ is not complete. 84. (D) The efficiency value for a fuel cell will be greater than 1.00 for any exothermic reaction ( H o 0 ) that has G that is more negative than its H o value. Since G o H o T S o , this means that the value of S o must be positive. Moreover, for this to be the case, ngas is usually greater than zero. Let us consider the situation that might lead to this type of reaction. The combustion of carbonhydrogenoxygen compounds leads to the formation of H2O(l) and CO2(g). Since most of the oxygen in these compounds comes from O2(g) (some is present in the CHO compound), there is a balance in the number of moles of gas producedCO2(g) and those consumedO2(g)which is offset in a negative direction by the H2O(l) produced. Thus, the combustion of these types of compounds will only have a positive value of ngas if the number of oxygens in the formula of the compound is more than twice the number of hydrogens. By comparison, the decomposition of NOCl(g), an oxychloride of nitrogen, does produce more moles of gas than it consumes. Let us investigate this decomposition reaction. NOCl(g) 1 N 2 (g) 1 O 2 (g) 1 Cl 2 (g) 2 2 2 H o 1 H of [ N 2 (g)] 1 H o [O 2 (g)] 1 H o [Cl 2 (g)] H o [ NOCl(g)] f f f 2 2 2 0.500 (0.00 kJ/mol 0.00 kJ/mol 0.00 kJ/mol) 51.71 kJ/mol 51.71 kJ/mol G o 1 G o [ N 2 (g)] 1 G o [O 2 (g)] 1 G of [Cl 2 (g)] G o [ NOCl(g)] f f f 2 2 2 0.500 (0.00 kJ/mol 0.00 kJ/mol 0.00 kJ/mol) 66.08 kJ/mol 66.08 kJ/mol G o 66.08 kJ/mol 1.278 H o 51.71 kJ/mol Yet another simple reaction that meets the requirement that G o H o is the combustion of graphite: C(graphite) O 2 (g) CO 2 (g) We see from Appendix D that o G f [CO 2 (g)] 394.4 kJ/mol is more negative than H fo [CO 2 (g)] 393.5 kJ/mol . (This reaction is accompanied by an increase in entropy; S =213.75.74205.1 =2.86 J/K, = 1.002.) G o H o is true of the reaction in which CO(g) is formed from the elements. From Appendix D, H fo {CO(g)} 110.5 kJ/mol , and Gfo {CO(g)} 137.2 kJ/mol , producing = (137.2/110.5) = 1.242. Note that any reaction that has > 1.00 will be spontaneous under standard conditions at all temperatures. (There, of course, is another category, namely, an endothermic reaction that has S o 0 . This type of reaction is nonspontaneous under standard conditions at all temperatures. As such it consumes energy while it is running, which is clearly not a desirable result for a fuel cell.) 1046 Chapter 20: Electrochemistry 85. (M) We first write the two halfequations and then calculate a value of G o from thermochemical data. This value then is used to determine the standard cell potential. Oxidation: CH 3 CH 2 OH(g) 3 H 2 O(l) 2 CO 2 ( g ) 12 H (aq) 12 e _ Reduction: {O 2 (g ) 4 H (aq) 4e _ 2 H 2 O(l)} 3 Overall: CH 3 CH 2 OH(g) 3 O 2 (g ) 3 H 2 O(l) 2 CO 2 ( g ) Thus, n 12 (a) G o 2Gfo [CO 2 (g)] 3G o [ H 2 O(l)] Gfo [CH 3 CH 2 OH(g)] 3Gfo [O 2 (g)] f 2( 394.4 kJ/mol) 3( 237.1 kJ/mol) (168.5 kJ/mol) 3(0.00 kJ/mol) 1331.6 kJ/mol 1331.6 103 J/mol G Ecell 1.1501 V n F 12 mol e 96, 485 C/mol e (b) E cell E [O 2 (g)/H 2 O] E [CO 2 (g ) /CH 3 CH 2 OH(g )] 1.1501 V 1.229 V E [CO 2 (g ) / CH 3 CH 2 OH(g)] E [CO 2 (g ) / CH 3 CH 2 OH(g)] 1.229 1.1501 0.079 V 86. (M) First we determine the change in the amount of H+ in each compartment.
Oxidation: 2 H 2 O(l) O 2 ( g ) 4 H (aq) 4 e amount H 212 min Reduction: 2 H (aq) 2 e H 2 (g) 60 s 1.25 C 1 mol e 1 mol H 0.165 mol H 1 min 1s 96, 485 C 1 mol e Before electrolysis, there is 0.500 mol H2PO4 and 0.500 mol HPO42 in each compartment. The electrolysis adds 0.165 mol H+ to the anode compartment, which has the effect of transforming 0.165 mol HPO42 into 0.165 mol H2PO4, giving a total of 0.335 mol HPO42 (0.500 mol 0.165 mol) and 0.665 mol H2PO4 (0.500 mol + 0.165 mol). We can use the HendersonHasselbalch equation to determine the pH of the solution in the anode compartment.
pH pK a2 log [HPO 4 ] 0.335 mol HPO 4 / 0.500 L 7.20 log 6.90 [H 2 PO 4 ] 0.665 mol H 2 PO 4 / 0.500 L
2 2 Again we use the HendersonHasselbalch equation in the cathode compartment. After electrolysis there is 0.665 mol HPO42 and 0.335 mol H2PO4. Again we use the HendersonHasselbalch equation.
pH pK a2 log [HPO 4 ] [H 2 PO 4 ] 2 7.20 log 0.665 mol HPO 4 2 / 0.500 L 7.50 0.335 mol H 2 PO 4 / 0.500 L 87. (M) We first determine the change in the amount of M2+ ion in each compartment. 3600 s 0.500 C 1mol e 1mol M 2+ =0.0933 mol M 2+ M 2+ =10.00h 1h 1s 96485 C 2mol e This change in amount is the increase in the amount of Cu2+ and the decrease in the amount of Zn2+. We can also calculate the change in each of the concentrations. 1047 Chapter 20: Electrochemistry 0.0933 mol Cu 2 0.0933 mol Zn 2 2 [Cu ] 0.933 M [Zn ] 0.933 M 0.1000 L 0.1000 L Then the concentrations of the two ions are determined. [Cu 2 ] 1.000 M 0.933 M 1.933 M [Zn 2 ] 1.000 M 0.933 M 0.067 M Now we run the cell as a voltaic cell, first determining the value of Ecell. Oxidation : Zn(s) Zn 2 (aq) 2 e E 0.763 V 2 Reduction : Cu 2 (aq) 2 e Cu(s) E 0.340 V 2 Net: Zn(s) Cu (aq) Cu(s) Ecell 1.103 V Then we use the Nernst equation to determine the voltage of this cell. 0.067 M 0.0592 [Zn 2 ] 0.0592 1.103 1.103 0.043 1.146 V Ecell E log log cell 2 2 2 1.933 M [Cu ] 88. (M)(a) Anode: Zn(s) Zn 2 (aq) 2e E 0.763 V Cathode: {AgCl(s) + e (aq) Ag(s) + Cl (1 M)} 2 E=+0.2223 V 2+ Net: Zn(s) + 2AgCl(s) Zn (aq) + 2Ag(s) + 2Cl (1 M) E = +0.985 V cell
(b) The major reason why this electrode is easier to use than the standard hydrogen electrode is that it does not involve a gas. Thus there are not the practical difficulties involved in handling gases. Another reason is that it yields a higher value of E , cell thus, this is a more spontaneous system. Oxidation: Reduction: Ag(s) Ag (aq) e AgCl(s) e Ag(s) Cl (aq) E o 0.800 V E 0.2223 V (c) o Net: AgCl(s) Ag (aq) Cl (aq) Ecell 0.578 V The net reaction is the solubility reaction, for which the equilibrium constant is Ksp. G nFE RT ln K sp ln K sp nFE RT 1 mol e 96485 C 1 J (0.578 V) 1 V.C 1 mol e 22.5 1 1 8.3145 J mol K 298.15 K K sp e 22.5 1.7 1010 This value is in good agreement with the value of 1.8 1010 given in Table 181. 89. (D) (a) Ag(s) Ag+(aq) + eE = 0.800 V AgCl(s) + e Ag(s) + Cl (aq) E = 0.2223 V AgCl(s) Ag+(aq) + Cl(aq) Ecell = 0.5777 V [1.00][1.00 103 ] 0.0592 [Ag + ][Cl ] 0.0592 = 0.5777 V log log Ecell = Ecell 1 1 1 1 (b) = 0.400 V 10.00 mL of 0.0100 M CrO42 + 100.0 mL of 1.00 103 M Ag+ (Vtotal = 110.0 mL) 1048 Chapter 20: Electrochemistry Concentration of CrO42after dilution: 0.0100 M10.00 mL /110.00 mL = 0.000909 M Concentration of Ag+ after dilution: 0.00100 M100.0 mL /110.00 mL = 0.000909 M K sp 1.11012 2Ag+(aq) + CrO42(aq) Ag2CrO4(s) Initial 0.000909 M 0.000909 M Change(100% rxn) 0.000909 M 0.000455 M New initial 0M 0.000455 M Change +2x +x Equilibrium 2x 0.000455 M+x 0.000455 M 12 2 1.1 10 = (2x) (0.000454) x = 0.0000246 M Note: 5.4% of 0.000455 M (assumption may be considered valid) (Answer would be x = 0.0000253 using method of successive approx.) [Ag+] = 2x = 0.0000492 M (0.0000506 M using method of successive approx.)
Ecell E cell 0.0592 0.0592 log [ Ag ][Cl ] 0.5777 log [1.00M ][4.92 104 M ] 1 1 0.323 V (0.306 V for method of successive approximations) (c) 10.00 mL 0.0100 M NH3 + 10.00 mL 0.0100 M CrO42 + 100.0 mL 1.00103 M Ag+ (Vtotal = 120.0 mL) Concentration of NH3 after dilution: 10.0 M10.00 mL /120.00 mL = 0.833 M Concentration of CrO42after dilution: 0.0100 M10.00 mL /110.00 mL = 0.000833 M Concentration of Ag+ after dilution: 0.00100 M100.0 mL /110.00 mL = 0.000833 M In order to determine the equilibrium concentration of free Ag+(aq), we first consider complexation of Ag+(aq) by NH3(aq) and then check to see if precipitation occurs. Ag+(aq) + 0.000833 M 0.000833 M 0M +x x x
K f 1.610 2NH3(aq) 0.833 M 0.00167 M 0.831 M +2x (0.831+2x) M 0.831 M
7 Initial Change(100% rxn) New initial Change Equilibrium Equilibrium (x 0) Ag(NH3)2+(aq) 0M +0.000833 M 0.000833 M x (0.000833 x) M 0.000833 M x = 7.541011 M = [Ag+] 1.6 107 = 0.000833 /x(0.831)2 Note: The assumption is valid Now we look to see if a precipitate forms: Qsp = (7.541011)2(0.000833) = 4.7 1024 Since Qsp < Ksp (1.1 1012), no precipitate forms and [Ag+] = 7.54 1011 M E cell = E cell E cell
90. 0.0592 0.0592 log [Ag + ][Cl ]= 0.5777 V log [1.00M][7.54 1011 M] 1 1 = 0.0215 V (M) We assume that the Pb2+(aq) is "ignored" by the silver electrode, that is, the silver electrode detects only silver ion in solution. 1049 Chapter 20: Electrochemistry Oxidation : H 2 (g) 2 H (aq) 2 e Reduction : {Ag (aq) e Ag(s)} 2 Net : H 2 (g) 2 Ag E 0.000 V E 0.800 V 0.0592 2 1.002 [Ag ]2 2 H (aq) 2 Ag(s) E cell 0.800 V log [H ]2 [Ag ]2 0.503 V 0.800 V log Ecell Ecell log 0.0592 2 2(0.800 0.503) 1.00 2 10.0 2 0.0592 [Ag ] 1.00 2 10 10.0 1.0 1010 [Ag ] 2 [Ag + ]2 =1.0 1010 M 2 [Ag + ]=1.0 105 mass Ag 0.500 L % Ag 1.0 10 5 mol Ag 1 mol Ag 107.87 g Ag 5.4 10 4 g Ag 1 L soln 1 mol Ag 1 mol Ag 5.4 10 4 g Ag 100 % 0.051% Ag (by mass) 1.050 g sample 91. (M) 250.0 mL of 0.1000 M CuSO4 = 0.02500 moles Cu2+ initially. moles of Cu2+ plated out =
1 mol e 1 mol Cu 2+ 3.512C 1368 s 0.02490 mol Cu 2+ 96,485 C s 2 mole moles of Cu2+ in solution = 0.02500 mol 0.02490 mol = 0.00010 mol Cu2+ [Cu2+] = 0.00010 mol Cu2+/0.250 L = 4.0 104 M Initial Change(100% rxn) New initial Change Equilibrium Cu2+(aq) + 0.00040 M 0.00040M 0M +x x
K f 1.110 4NH3(aq) 0.10 M maintained 0.10 M maintained 0.10 M
13 Cu(NH3)42+(aq) 0M +0.00040 M 0.00040 M x (0.00040 x) M 0.00040 M [Cu(NH 3 ) 4 2+ ] 0.00040 1.1 1013 Kf = = [Cu 2+ ] = 3.6 1013 M 2+ 4 2+ 4 [Cu ][NH3 ] [Cu ](0.10) Hence, the assumption is valid. The concentration of Cu(NH3)42+(aq) = 0.00040 M which is 40 times greater than the 1 105 detection limit. Thus, the blue color should appear.
92. (M) First we determine the molar solubility of AgBr in 1 M NH3 .
Sum Initial Equil. K AgBr(s) 2NH 3 (aq) Ag(NH 3 ) 2 (aq) Br (aq) 1.00 M 1.002s 0 M s 0M s {s AgBr molar solubility} s 2.81 10 3 M (also [Br ]) K K sp K f 8.0 106 [Ag(NH 3 ) 2 ][Br ] s2 8.0 106 [NH 3 (aq)] (1 2 s ) 2 K sp [Br ] 5.0 1013 1.8 1010 M 3 2.81 10 So [Ag ] 1050 Chapter 20: Electrochemistry AgBr(s) Ag Br Ag 2NH 3 Ag(NH 3 ) 2 K sp 5.0 1013 K f 1.6 107 (sum) AgBr(s) 2NH 3 (aq) Ag(NH 3 ) 2 Br K K sp K f 8.0 106 Now, let's construct the cell, guessing that the standard hydrogen electrode is the anode. Oxidation: H 2 (g) 2 H 2 e E o 0.000 V Reduction: {Ag (aq) e Ag(s) } 2 Net: 2 Ag (aq) H 2 (g) 2 Ag(s) 2 H (aq) From the Nernst equation:
o E o 0.800 V
o Ecell 0.800V 0.0592 [ H ]2 0.0592 12 o 0.0592 0.800 V E E log 10 Q E log log10 n n 2 [ Ag ]2 (1.78 1010 ) 2 and E 0.223 V. Since the voltage is positive, our guess is correct and the standard hydrogen electrode is the anode (oxidizing electrode). 93. (M) (a) Anode: 2H2O(l) 4 e + 4 H+(aq) + O2(g) Cathode: 2 H2O(l) + 2 e 2 OH(aq) + H2(g) Overall: 2 H2O(l) + 4 H2O(l) 4 H+(aq) + 4 OH(aq) + 2 H2(g) + O2(g) 2 H2O(l) 2 H2(g) + O2(g)
(b) 21.5 mA = 0.0215 A or 0.0215 C s1 for 683 s
mol H 2SO 4 = 0.0215 C s 683s 1 mol e96485C 1 mol H + 1 mol e
 1 mol H 2SO 4 2 mol H + = 7.61 105 mol H 2SO 4 7.61 105 mol H 2SO 4 in 10.00 mL. Hence [H 2SO 4 ] = 7.61 105 mol / 0.01000 L = 7.61 103 M 94. (D) First we need to find the total surface area
Outer circumference = 2r = 2(2.50 cm) = 15.7 cm Surface area = circumference thickness = 15.7 cm 0.50 cm = 7.85 cm2 Inner circumference = 2r = 2(1.00 cm) = 6.28 cm Surface area = circumference thickness = 6.28 cm 0.50 cm = 3.14 cm2 Area of small circle = r2 = (1.00 cm)2 = 3.14 cm2 Area of large circle = r2 = (2.50 cm)2 = 19.6 cm2 Total area = 7.85 cm2 + 3.14 cm2 + 2(19.6 cm2) 2(3.14 cm2) = 43.91 cm2 Volume of metal needed = surface area thickness of plating = 43.91 cm2 0.0050 cm = 0.22 cm3 8.90 g 1 mol Ni 2 mol e 96485 C Charge required = 0.22 cm 3 6437.5 C 58.693 g Ni 1 mol Ni 1 mol ecm 3 Time = charge/time = 6437.5 C / 1.50 C/s = 4291.7 s or 71.5 min 1051 Chapter 20: Electrochemistry 95. (M) The overall reaction for the electrolytic cell is: o Ecell 1.103V Cu(s)+Zn 2+ (aq) Cu 2+ (aq)+Zn(s) Next, we calculate the number of moles of Zn2+(aq) plated out and number of moles of Cu2+(aq) formed: 0.500C 60 min 60s 1 mol e 1 mol Cu 2+ 10 h 0.0935mol Cu 2+ n(Cu 2+ ) = 96,485 C s 1h 1min 2 mol e 2+ 2+ n(Zn ) n(Cu )=0.0935mol Initially, solution contained 1.00molL1 0.100L=0.100 mol Zn2+(aq). Therefore, at the end of electrolysis we are left with: 6.5 103mol n(Zn 2+ ) LEFT =(0.1000.0935)mol=6.5 103mol [Zn 2+ ]= =6.5 102 M 0.1L 0.0935mol n(Cu 2+ ) FORMED =0.0935mol [Cu 2+ ]= =0.936M 0.1L The new potential after the cell was switched to a voltaic one can be calculated using Nernst equation: 0.0592 [Zn 2 ] 0.0592 6.5 103 M log 1.103 log 1.103 0.064 1.167 V Ecell E cell 2 2 0.935 M [Cu 2 ]
96. (M) (a) The metal has to have a reduction potential more negative than 0.691 V, so that its oxidation can reverse the tarnishing reaction's 0.691 V reduction potential. Aluminum is a good candidate, because it is inexpensive, readily available, will not react with water and has an Eo of 1.676 V. Zinc is also a possibility with an Eo of 0.763 V, but we don't choose it because there may be an overpotential associated with the tarnishing reaction. (b) Oxidation: Reduction: Net: {Al(s) Al3 (aq) 3 e } 2 {Ag 2 S(s) 2 e Ag(s) S2 (aq) } 3 2 Al(s) 3 Ag 2 S(s) 6Ag(s) 2 Al3 (aq) 3 S2 (aq) (c) (d) The dissolved NaHCO3(s) serves as an electrolyte. It would also enhance the electrical contact between the two objects. There are several chemicals involved: Al, H2O, and NaHCO3. Although the aluminum plate will be consumed very slowly because silver tarnish is very thin, it will, nonetheless, eventually erode away. We should be able to detect loss of mass after many uses. Overall: C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Anode: 6 H2O(l) + C3H8(g) 3 CO2(g) + 20 H+(g) + 20 eCathode: 20 e + 20 H+(g) + 5 O2(g) 10 H2O(l) 97. (M) (a) Grxn = 3(394.4 kJ/mol) + 4(237.1 kJ/mol) 1(23.3 kJ/mol) = 2108.3 kJ/mol 1052 Chapter 20: Electrochemistry Grxn = 2108.3 kJ/mol = 2,108,300 J/mol Grxn = nFEcell = 20 mol e (96485 C/mol e) Ecell Ecell = +1.0926 V 20 e + 20 H+(g) + 5 O2(g) Hence 10 H2O(l) Ecathode = +1.229 V +1.0926 V = +1.229 V Eanode Ecell = Ecathode Eanode Eanode = +0.136 V (reduction potential for 3 CO2(g) + 20 H+(g) + 20 e 6 H2O(l) + C3H8(g)) (b) Use thermodynamic tables for 3 CO2(g) + 20 H+(g) + 20 e 6 H2O(l) + C3H8(g) Grxn = 6(237.1 kJ/mol) + 1(23.3 kJ/mol) [(394.4 kJ/mol) + 20( 0kJ/mol)]= 262.7 kJ/mol Gred = 262.7 kJ/mol = 262,700 J/mol = nFEred = 20 mol e(96,485 C/mol e)Ered Ered = 0.136 V (Same value, as found in (a))
o 98. (D) (a) Equation 20.15 ( G o zFEcell ) gives the relationship between the standard Gibbs energy of a reaction and the standard cell potential. Gibbs free energy also varies with temperature ( G o H o T S o ). If we assume that H o and S o do not vary significantly over a small temperature range, we can derive an equation for the temperature o variation of Ecell : H o T S o G H T S zFE E zF Considering two different temperatures one can write: H o T1S o H o T2 S o o o Ecell (T1 ) and Ecell (T2 ) zF zF o o o H T1S H T2 S o o o Ecell (T1 ) Ecell (T2 ) zF zF o o T S T2 S S o o o Ecell (T1 ) Ecell (T2 ) 1 T2 T1 zF zF (b) Using this equation, we can now calculate the cell potential of a Daniel cell at 50 oC: 10.4JK 1mol 1 o o Ecell (25 o C) Ecell (50 o C) 50 25 K 0.00135 2 96485Cmol 1 o Ecell (50 o C) 1.103V 0.00135 1.104V
o o o o cell o cell 99. (D) Recall that under nonstandard conditions G G o RT ln K . G o H o T S o and G zFEcell one obtains: zFEcell H o T S o RT lnQ For two different temperatures (T1 and T2) we can write: Substituting 1053 Chapter 20: Electrochemistry zFEcell (T1 ) H o T1S o RT1 lnQ zFEcell (T2 ) H o T2 S o RT2 lnQ H o T1S o RT1 lnQ H o T2 S o RT2 lnQ Ecell (T1 ) Ecell (T2 ) zF o T S RT1 lnQ T2 S o RT2 lnQ Ecell (T1 ) Ecell (T2 ) 1 zF o T S RT1 lnQ T2 S o RT2 lnQ Ecell (T1 ) Ecell (T2 ) 1 zF zF o S R lnQ S o R lnQ Ecell (T1 ) Ecell (T2 ) T1 T2 zF zF S o R lnQ Ecell (T1 ) Ecell (T2 ) (T1 T2 ) zF o o The value of Q at 25 oC can be calculated from Ecell and Ecell . First calculate Ecell : Oxidation: Cu(s) Cu 2+ (aq)+2eEo 0.340V 3+ 2+ Reduction: 2Fe (aq)+2e 2Fe (aq) Eo 0.771V _____________________________________________________________________________________________________________
o Cu(s)+2Fe 3 Cu 2+ (aq)+2Fe 2 Ecell 0.431V 0.0592 0.370 0.431 logQ 0.296 logQ 0.431 0.370 0.061 2 logQ 0.206 Q 10 0.206 1.61 Now, use the above derived equation and solve for S o : S o 8.314 ln1.61 0.394 0.370 (50 25) 2 96485 Overall: S o 3.96 ) S o 3.96 185.3 S o 189.2JK 1 192970 o o Since G H o T S o zFEcell we can calculate G o , K (at 25 oC) and H o : o G o zFEcell 2 96485 0.431 83.2kJ 0.024 25 ( G o RT ln K 83.2 1000 8.314 298.15 ln K ln K 33.56 K e33.56 3.77 1014 189.2 83.2kJ=H o 298.15 kJ H o 83.2 56.4 26.8kJ 1000 o Since we have H and S o we can calculate the value of the equilibrium constant at 50 o C: 189.2 G o H o T S o 26.8kJ (273.15 50)K kJK 1 87.9kJ 1000 1 1 87.9 1000J 8.314JK mol (273.15 50)K ln K ln K 32.7 K e 32.7 1.59 1014 1054 Chapter 20: Electrochemistry Choose the values for the concentrations of Fe2+, Cu2+ and Fe3+ that will give the value of the above calculated Q. For example: Fe 2+ Cu 2+ =1.61 Q= 3+ 2 Fe 0.12 1.61 1.61 0.12 Determine the equilibrium concentrations at 50 oC. Notice that since Q<K, a net change occurs from left to right (the direction of the forward reaction): Cu(s)+2Fe 3 Cu 2+ (aq)+2Fe 2 Initial: 0.1 1.61 0.1 Change: 0.1x 1.61+x 0.1+x
2 Fe 3+ eq Obviously, the reaction is almost completely shifted towards products. First assume that the reaction goes to completion, and then let the equilibrium be shifted towards reactants: Cu(s)+2Fe 3 Cu 2+ (aq)+2Fe 2 Initial: 0.1 1.61 0.1 Final 0 1.71 0.2 Equilibrium 0+x 1.71x 0.2x K= Fe 2+ eqCu 2+ eq Fe 3+ eq 2 2 K= Fe 2+ eqCu 2+ eq 2 2 =1.59 1014 (0.1 x)2 (1.61 x) (0.1 x)2 =1.59 1014 (0.2 x)2 (1.71 x) 0.2 2 1.71 (x)2 x2 0.2 2 1.71 4.3 10 16 14 1.59 10 x 2.1 10 8 M Therefore, [Cu2+]1.7M, [Fe2+]0.2M and [Fe3+]2.1x108M x2 100.
o (D) This problem can be solved by utilizing the relationship between G o and Ecell ( G o =o zF Ecell ): Consider a hypothetical set of the following reactions: o o A ne A n E1 and G1 B me B m E o and G o 2 2 Overall: A B ne me A n B m o E o ? and G o G1 G o rxn rxn 2 o o G o G1 G o nFE1o mFE2 rxn 2 o o (n m)FErxn nFE1o mFE2 o o o (n m)FErxn nFE1o mFE2 Erxn o nFE1o mFE2 nm Therefore, for nsets of halfreactions: 1055 Chapter 20: Electrochemistry Eo n E
i o i n i The Eo for the given halfreaction can be determined by combining four halfreactions: H 6 IO 6 +H + +2e IO3 +3H 2 O E o 1.60V 1 E o 1.19V IO3 +6H + +5e I2 +3H 2O 2 I2 +2H 2O 2HIO+2H + +2e E o 1.45V 2I I 2 2e E o 0.535V Overall : 1 H 6 IO 6 +H + +2e IO3 +6H + +5e I2 +2H 2 O+2I IO3 +3H 2 O+ I2 +3H 2O+2HIO+2H + +2e I 2 2e 2 1 H 6 IO 6 +5H + +2I +3e I2 +4H 2 O+2HIO 2 1.60 3 1.19 5 1.45 2 0.535 2 2.26V Eo 2522 FEATURE PROBLEMS
101. (D) (a) Anode: H 2 g,1 atm 2H + 1 M + 2e E o = 0.0000 V E o = 0.800 V
_________________________________________ Cathode: {Ag + x M + e Ag s } 2 Net: H 2 g,1atm + 2Ag + aq 2H + 1 M + 2Ag s (b) o Ecell = 0.800 (c) Since the voltage in the anode halfcell remains constant, we use the Nernst equation to calculate the halfcell voltage in the cathode halfcell, with two moles of electrons. This is then added to E for the anode halfcell. Because E o = 0.000 for the anode half cell, Ecell = E cathode 0.0592 1 log E = Eo 0.800 + 0.0592 log Ag + 0.800 0.0592 log x + 2 2 Ag (i) Initially Ag + = 0.0100 ; E = 0.800 + 0.0592 log 0.0100 = 0.682 V = Ecell Note that 50.0 mL of titrant is required for the titration, since both AgNO 3 and KI have the same concentrations and they react in equimolar ratios.
(ii) After 20.0 mL of titrant is added, the total volume of solution is 70.0 mL and the unreacted Ag + is that in the untitrated 30.0 mL of 0.0100 M AgNO 3 aq . 1056 Chapter 20: Electrochemistry 30.0 mL 0.0100 M Ag + Ag = = 0.00429M 70.0 mL
+ E = 0.800 + 0.0592 log 0.00429 = 0.660 V = Ecell (iii) After 49.0 mL of titrant is added, the total volume of solution is 99.0 mL and the unreacted Ag + is that in the untitrated 1.0 mL of 0.0100 M AgNO 3 aq . Ag + = 1.0 mL 0.0100 M Ag + = 0.00010 M 99.0 mL E = 0.800 + 0.0592 log 0.00010 = 0.563 V = Ecell (iv) At the equivalence point, we have a saturated solution of AgI, for which Ag + = K sp (AgI) 8.5 1017 9.2 109 E = 0.800 + 0.0592 log 9.2 109 = 0.324 V = Ecell . After the equivalence point, the Ag + is determined by the I resulting from the excess KI(aq).
(v) When 51.0 mL of titrant is added, the total volume of solution is 101.0 mL and the excess I is that in 1.0 mL of 0.0100 M KI(aq). I = 1.0 mL 0.0100 M I =9.9 105 M 101.0 mL K 8.5 1017 13 Ag + = sp = I  0.000099 =8.6 10 M E = 0.800 + 0.0592 log 8.6 1013 = 0.086 V = Ecell (vi) When 60.0 mL of titrant is added, the total volume of solution is 110.0 mL and the excess I is that in 10.0 mL of 0.0100 M KI(aq). 10.0 mL0.0100 M I I = =0.00091 M 110.0 mL K 8.51017 Ag + = sp = = 9.31014 M 0.00091 I E = 0.800 + 0.0592 log 9.3 10 14 = 0.029 V = Ecell 1057 Chapter 20: Electrochemistry (d) The titration curve is presented below.
0.8 Plot of Voltage versus volum e of I(aq) 0.6 Voltage (V) 0.4 0.2 0 0 20 Volum e of I (m L) 40 60 102. (D) (a) (1) anode: Na s Na + in ethylamine + e cathode: Na (in ethylamine) Na(amalgam, 0.206 %) Net: Na s Na amalgam, 0.206% (2) anode: 2Na amalgam, 0.206% 2Na + 1 M + 2e cathode: 2H + (aq,1M) + 2e H 2 g,1 atm Net: 2Na (amalgam, 0.206 %) + 2H + (aq,1M) 2 Na + (1 M) + H 2 (g,1 atm) (b) (1) (2) 96, 485 C 0.8453V = 8.156 104 J or 81.56 kJ 1mol e 96, 485C G = 2 mol e 1.8673V = 36.033 104 J or 360.33 kJ 1 mol e G = 1 mol e 2Na(s) 2Na(amalgam, 0.206%) G1 2 8.156 104 J 2Na (amalg, 0.206%) + 2H + (aq,1M) 2 Na (1M) + H 2 (g, 1 atm) G2 36.033 104 J Overall: 2 Na(s) + 2H + (aq) 2 Na (1M) + H 2 (g, 1 atm) G = G1 + G2 = 16.312 104 J 36.033 104 J = 52.345 104 J or 523.45 kJ Since standard conditions are implied in the overall reaction, G G o .
o (c) (1) (2) (d) Ecell E o Na + 1 M /Na s = 2.713V . This is precisely the same as the value in
Appendix D. 52.345 104 J = = E o H + 1 M /H 2 1 atm E o Na + 1 M /Na s 96, 485 C 1J 2 mol e 1 VC 1 mol e 1058 Chapter 20: Electrochemistry 103. (D) The question marks in the original Latimer diagram have been replaced with letters in the diagram below to make the solution easier to follow: (c) (a) (b) BrO4 1.025 V BrO3Br2 BrOBr(d) (e) 0.584 V By referring to Appendix D and by employing the correct procedure for adding together halfreactions of the same type we obtain: o Ec 1.065 V (Appendix D) (c) Br2 (l) 2 e 2 Br (aq) (d) (b) (a) (b) (c) (e) BrO (aq) H 2 O(l) 2 e Br (aq) 2 OH (aq) 2 BrO (aq) 2 H 2O(l) 2 e Br2 (l) 4 OH (aq) Ed 0.766 V (Appendix D)
o o Eb 0.455 V (Appendix D) Ea ?
o {BrO3 (aq) 2 H 2 O(l) 4 e BrO (aq) 4 OH (aq)} 2 2 BrO (aq) 2 H 2O(l) 2 e Br2 (l) 4 OH (aq) Br2 (l) 2 e 2 Br (aq)
2 BrO3 (aq) 6 H 2 O(l) 12 e 12 OH (aq) 2 Br (aq)
o o o o GTotal G(a) G(b) G(c) o 12 F Eeo 8F Eao 2 F Eb 2 F Eco Eb 0.455 V
o Ec 1.065 V
o
__________________________________ Ee 0.584 V
o 12 F 0.584 V 8F Eao 2 F 0.455 V 2 F 1.065 V Eao 104. (D) (a) 12 F 0.584 V 2 F 0.455 V 2 F 1.065 V 8 F = 0.496 V The capacitance of the cell membrane is given by the following equation, C 0 A
l where 0 = 3 8.854 1012 C2 N1 m2; A = 1 106 cm2; and l = 1 106 cm. Together with the factors necessary to convert from cm to m and from cm2 to m2, these data yield
2 C C2 1m 2 3 8.854 1012 1 2 (1106 cm 2 ) N m 100 cm = 2.66 1013 C Nm 1m (1106 cm) 100 cm C 2.66 1013 1F C = 2.66 1013 F N m C2 1 N m 2 1059 Chapter 20: Electrochemistry (b) Since the capacitance C is the charge in coulombs per volt, the charge on the membrane, Q, is given by the product of the capacitance and the potential across the cell membrane. C Q = 2.66 1013 0.085 V = 2.26 1014 C V The number of K+ ions required to produce this charge is Q 2.26 1014 C = 1.41105 K + ions e 1.602 1019 C/ion The number of K+ ions in a typical cell is 1L 23 ions 3 mol (1108 cm3 )= 9.3 1011 ions 6.022 10 155 10 3 mol L 1000 cm The fraction of the ions involved in establishing the charge on the cell membrane is 1.4 105 ions =1.5 107 (~ 0.000015 %) 9.3 1011 ions Thus, the concentration of K+ ions in the cell remains constant at 155 mM. (c) (d) (e) 105. (M) Reactions with a positive cell potential are reactions for which G o 0 , or reactions for which K>1. S o , H o and U o cannot be used alone to determine whether a particular electrochemical reaction will have a positive or negative value. 106. (M) The halfreactions for the first cell are: Anode (oxidation): X(s) X + (aq)+eo E X / X Cathode (reduction): 2H + (aq)+2e  H 2 (g) E o 0V Since the electrons are flowing from metal X to the standard hydrogen electrode, o E X / X 0V . The halfreactions for the second cell are: Anode (oxidation): X(s) X + (aq)+eCathode (reduction): Y2+ +2e  Y(s)
o EY 2 /Y o E X / X o o Since the electrons are flowing from metal X to metal Y, E X / X + EY 2 /Y >0. o o o From the first cell we know that E X / X 0V . Therefore, E X / X > EY 2 /Y . 107. (M) The standard reduction potential of the Fe2+(aq)/Fe(s) couple can be determined from: Fe 2+ (aq) Fe 3+ (aq)+e Eo 0.771V
Fe 3+ (aq)+3e Fe(s) E o 0.04V Overall: Fe 2+ (aq)+2e Fe(s) We proceed similarly to the solution for 100: 1060 Chapter 20: Electrochemistry Eo n E
i o i n 0.771 (1) 0.04 3 0.445V 31 i SELFASSESSMENT EXERCISES
108. (E) (a) A standard electrode potential E o measures the tendency for a reduction process to occur at an electrode. (b) F is the Faraday constant, or the electric charge per mole of electrons (96485 C/mol). (c) The anode is the electrode at which oxidation occurs. (d) The cathode is the electrode at which reduction occurs. 109. (E) (a) A salt bridge is a devise used to connect the oxidation and reduction halfcells of a galvanic (voltaic) cell. (b) The standard hydrogen electrode (abbreviated SHE), also called normal hydrogen electrode (NHE), is a redox electrode which forms the basis of the thermodynamic scale of oxidationreduction potentials. By definition electrode potential for SHE is 0. (c) Cathodic protection is a technique commonly used to control the corrosion of a metal surface by making it work as a cathode of an electrochemical cell. This is achieved by placing in contact with the metal to be protected another more easily corroded metal to act as the anode of the electrochemical cell. (d) A fuel cell is an electrochemical cell that produces electricity from fuels. The essential process in a fuel cell is fuel+oxygen oxidation products. 110. (E) (a) An overall cell reaction is a combination of oxidation and reduction halfreactions. (b) In a galvanic (voltaic) cell, chemical change is used to produce electricity. In an electrolytic cell, electricity is used to produce a nonspontaneous rection. (c) In a primary cell, the cell reaction is not reversible. In a secondary cell, the cell reaction can be reversed by passing electricity through the cell (charging). o (d) Ecell refers to the standard cell potential (the ionic species are present in aqueous solution at unit activity (approximately 1M), and gases are at 1 bar pressure (approximately 1 atm). 111. (M) (a) False. The cathode is the positive electrode in a voltaic cell and negative in electrolytic cell. (b) False. The function of the salt bridge is to permit the migration of the ions not electrons. (c) True. The anode is the negative electrode in a voltaic cell. (d) True. (e) True. Reduction always occurs at the cathode of an electrochemical cell. Because of the removal of electrons by the reduction halfreaction, the cathode of a voltaic cell is positive. Because of the electrons forced onto it, the cathode of an electrolytic cell is negative. For both types, the cathode is the electrode at which electrons enter the cell. 1061 Chapter 20: Electrochemistry (f) False. Reversing the direction of the electron flow changes the voltaic cell into an electrolytic cell. (g) True. The cell reaction is an oxidationreduction reaction. 112. (M) The correct answer is (b), Hg 2 (aq ) is more readily reduced than H (aq ) . 113. (M) Under nonstandard conditions, apply the Nernst equation to calculate Ecell : 0.0592 o Ecell Ecell log Q z 0.0592 0.10 Ecell 0.66 log 0.63V 2 0.01 The correct answer is (d). 114. (E) (c) The displacement of Ni(s) from the solution will proceed to a considerable extent, but the reaction will not go to completion. 115. (E) The gas evolved at the anode when K2SO4(aq) is electrolyzed between Pt electrodes is most likely oxygen. 116. (M) The electrochemical reaction in the cell is: Anode (oxidation): {Al(s) Al3+ (aq)+3e } 2 Cathode (reduction): {H 2 (g)+2e 2H + (aq)} 3 Overall: 2Al(s)+3H 2 (g) 2Al 3+ (aq)+6H + (aq) 1mol Al 3 mol H 2 4.5g Al 0.250mol H 2 26.98g Al 2 mol Al 22.4L H 2 0.250 mol H 2 =5.6L H 2 1mol H 2 117. (E)
The correct answer is (a) G . E o 0.763V E o 0.956V 118. (M) Anode (oxidation): {Zn(s) Zn 2+ (aq)+2e } 3 Cathode (reduction): {NO3 (aq)+4H + (aq)+3e  NO(g)+2H 2 O(l)} 2
Overall: _________________________________________________________________________________________________________
o 3Zn(s)+2NO3 (aq)+8H + (aq) 3Zn 2+ (aq)+2NO(g)+4H 2O(l) Ecell 1.719V Cell diagram: Zn(s) Zn 2+ (1M) H + (1M),NO3 (1M) NO(g,1atm) Pt(s) 119. (M) Apply the Nernst equation: 1062 Chapter 20: Electrochemistry 0.0592 log Q z 0.0592 0.108 0 log x 2 log x 2 3.65 2 2 3.65 x 10 x 0.0150 M pH log(0.0150) 1.82
o Ecell Ecell o 120. (M) (a) Since we are given Ecell , we can calculate K for the given reaction: RT o Ecell ln K nF 8.314 JK 1mol 1 298 K 0.0050V ln K ln K 0.389 2 96485Cmol 1 K e0.389 0.68 Since for the given conditions Q=1, the system is not at equilibrium. (b) Because Q>K, a net reaction occurs to the left. o 121. (M) (a) Fe(s)+Cu 2+ (1M) Fe 2+ (1M)+Cu(s) , Ecell 0.780V , electron flow from B to A o (b) Sn 2+ (1M)+2Ag + (1M) Sn 4+ (aq)+2Ag(s) , Ecell 0.646V , electron flow from A to B. o (c) Zn(s)+Fe 2+ (0.0010M) Zn 2+ (0.10M)+Fe(s) , Ecell 0.264V ,electron flow from A to B. 122. (M) (a) (b) (c) (d) Cl2(g) at anode and Cu(s) at cathode. O2(g) at anode and H2(g) and OH(aq) at cathode. Cl2(g) at anode and Ba(l) at cathode. O2(g) at anode and H2(g) and OH(aq) at cathode. 1063 CHAPTER 24 COMPLEX IONS AND COORDINATION COMPOUNDS
PRACTICE EXAMPLES
1A (E) There are two different kinds of ligands in this complex ion, I and CN. Both are monodentate ligands, that is, they each form only one bond to the central atom. Since there are five ligands in total for the complex ion, the coordination number is 5: C.N. = 5 . Each CN ligand has a charge of 1, as does the I ligand. Thus, the O.S. must be such that: O.S. + 4 +1 1 = 3 = O.S. 5 . Therefore, O.S. = +2 . (E) The ligands are all CN . Fe3+ is the central metal ion. The complex ion is Fe CN 6 . 3 1B 2A (E) There are six "Cl" ligands (chloride), each with a charge of 1. The platinum metal 2 center has an oxidation state of +4 . Thus, the complex ion is PtCl6 , and we need two K + to balance charge: K 2 PtCl6 . (E) The "SCN" ligand is the thiocyanato group, with a charge of 1bonding to the central metal ion through the sulfur atom. The "NH3" ligand is ammonia, a neutral ligand. There are five (penta) ammine ligands bonded to the metal. The oxidation state of the cobalt atom is +3 . The complex ion is not negatively charged, so its name does not end with "ate". The name of the compound is pentaamminethiocyanatoScobalt(III) chloride. (M) The oxalato ligand must occupy two cis positions. Either the two NH3 or the two Cl ligands can be coplanar with the oxalate ligand, leaving the other two ligands axial. The other isomer has one NH3 and one Cl ligand coplanar with the oxalate ligand. The structures are sketched below. 2B 3A ox Cl
3B Cl Co NH3 NH3 ox H3N NH3 Co Cl Cl ox H3N Cl Co NH3 Cl (M) We start out with the two pyridines, C5 H 5 N , located cis to each other. With this assignment imposed, we can have the two Clligands trans and the two CO ligands cis, the two CO ligands trans and the two Cl ligands cis, or both the Clligands and the two CO ligands cis. If we now place the two pyridines trans, we can have either both other pairs trans, or both other pairs cis. There are five geometric isomers. They follow, in the order described.
1163 Chapter 24: Complex Ions and Coordination Compounds 4A (M) The F ligand is a weak field ligand. MnF6 2 is an octahedral complex. Mn 4+ has
2 three 3d electrons. The ligand field splitting diagram for MnF6 is sketched below. There are three unpaired electrons.
eg t2g 4B (M) Co2+ has seven 3d electrons. Cl is a weak field ligand. H2O is a moderate field ligand. There are three unpaired electrons in each case. The number of unpaired electrons has no dependence on geometry for either metal ion complex.
E CoCl 4 2 E Co H 2O 6 2+ E 5A (M) CN is a strong field ligand. Co 2+ has seven 3d electrons. In the absence of a crystal field, all five d orbitals have the same energy. Three of the seven d electrons in this case will be unpaired. We need an orbital splitting diagram in which there are three orbitals of the same energy at higher energy. This is the case with a tetrahedral orbital diagram. t2 g eg eg t2 g In absence of a crystal field Tetrahedral geometry
2 Octahedralgeometry Thus, Co CN 4 must be tetrahedral (3 unpaired electrons) and not octahedral (1 unpaired electron) because the magnetic behavior of a tetrahedral arrangement would agree with the experimental observations (3 unpaired electrons).
5B (M) NH3 is a strong field ligand. Cu 2+ has nine 3d electrons. There is only one way to arrange nine electrons in five dorbitals and that is to have four fully occupied orbitals (two electrons in each orbital), and one halffilled orbital. Thus, the complex ion must be paramagnetic to the extent of one unpaired electron, regardless of the geometry the ligands adopt around the central metal ion. (M) We are certain that Co H 2O 6 2 6A 2 is octahedral with a moderate field ligand. Tetrahedral CoCl 4 has a weak field ligand. The relative values of ligand field splitting for the same ligand are t = 0.44 o . Thus, Co H 2O 6 2+ absorbs light of higher energy, 1164 Chapter 24: Complex Ions and Coordination Compounds blue or green light, leaving a light pink as the complementary color we observe. CoCl 4 absorbs lower energy red light, leaving blue light to pass through and be seen.
CO NC5H5 Cl Mo Cl OC NC5H5 Cl NC5H5 OC Mo CO Cl NC5H5 Cl NC5H5 Cl Mo CO OC NC5H5
2+ 2 NC5H5 CO Cl Mo Cl OC NC5H5
4 NC5H5 CO Cl Mo CO Cl NC5H5 6B (M) In order, the two complex ions are Fe H 2O 6 and Fe CN 6 . We know that CN is a strong field ligand; it should give rise to a large value of o and absorb light of Fe H 2O 6 NO 3 2 , contains the weak field ligand H2O and thus should be green. (The weak field would result in the absorption of light of long wavelength (namely, red light), which would leave green as the color we observe.) the shorter wavelength. We would expect the cyano complex to absorb blue or violet light and thus K 4 Fe CN 6 3 H 2 O should appear yellow. The compound INTEGRATIVE EXAMPLE
A (M) (a) There is no reaction with AgNO3 or en, so the compound must be transchlorobis(ethylenediamine)nitritoNcobalt(III) nitrite
+
H2 N H2N NO2 Cl Co N H2 H2 N NO2 (b) If the compound reacts with AgNO3, but not with en, it must be transbis(ethylenediamine)dinitritoNcobalt(III) chloride.
+
H N HN NO2 NO2 Co N H H N Cl (c) If it reacts with AgNO3 and en and is optically active, it must be cisbis(ethylenediamine)dinitritoNcobalt(III) chloride.
+
H2 N H2N NO2 NO2 Co N H2 H2 N Cl 1165 Chapter 24: Complex Ions and Coordination Compounds B (D) We first need to compute the empirical formula of the complex compound: 0.236 1 mol Pt =0.236 mol Pt =1 mol Pt 46.2 g Pt 0.236 195.1 g Pt 0.946 1 mol Cl =0.946 mol Cl =4 mol Pt 33.6 g Cl 0.236 35.5 g Cl 1.18 1 mol N =1.18 mol N =5 mol N 16.6 g N 0.236 14.01 g N 3.6 1 mol H =3.6 mol H =15 mol H 3.6 g Pt 0.236 1gH The nitrogen ligand is NH3, apparently, so the empirical formula is Pt(NH3)5Cl4. T 0.74 o C 0.4m . The effective The effective molality of the solution is m K fp 1.86 o C / m molality is 4 times the stated molality, so we have 4 particles produced per mole of Pt complex, and therefore 3 ionizable chloride ions. We can write this in the following way: [Pt(NH3)5Cl][Cl]3. Only one form of the cation (with charge 3+) shown below will exist.
+3 NH3 H3N Pt H3N Cl NH3 NH3 EXERCISES
Nomenclature
1. (E) (a) (b) (c) CrCl4 NH 3 2 3 diamminetetrachlorochromate(III) ion hexacyanoferrate(III) ion Fe CN 6 Cr en 3 Ni CN 4 . tris(ethylenediamine)chromium(III) tetracyanonickelate(II) 3 2 ion 2. (E) (a) Co NH 3 6 The coordination number of Co is 6; there are six monodentate NH 3 ligands attached to Co. Since the NH 3 ligand is neutral, the oxidation state of cobalt is +2 , the same as the charge for the complex ion; hexaamminecobalt(II) ion.
2+ 1166 Chapter 24: Complex Ions and Coordination Compounds (b) AlF6 The coordination number of Al is 6; F is monodentate. Each F has a 1 charge; thus the oxidation state of Al is +3 ; hexafluoroaluminate(III) ion.
3 (c) Cu CN 4 The coordination number of Cu is 4; CN is monodentate. CN has a 1 charge; thus the oxidation state of Cu is +2 ; tetracyanocuprate(II) ion CrBr2 NH 3 4 The coordination number of Cr is 6; NH 3 and Br are monodentate. NH 3 has no charge; Br has a 1 charge. The oxidation state of chromium is +3 ; tetraamminedibromochromium(III) ion
+ 2 (d) (e) Co ox 3 The coordination number of Co is 6; oxalate is bidentate. C2O 2 ox 4 has a 2 charge; thus the oxidation state of cobalt is +2 ; trioxalatocobaltate(II) ion. 4 (f) Ag S2 O 3 2 The coordination number of Ag is 2; S2 O 32 is monodentate. S2 O 32 has a 2 charge; thus the oxidation state of silver is +1 ; dithiosulfatoargentate(I) ion. ( Although +1 is by far the most common oxidation state for silver in its compounds, stable silver(III) complexes are known. Thus, strictly speaking, silver is not a nonvariable metal, and hence when naming silver compounds, the oxidation state(s) for the silver atom(s) should be specified).
3 2+ 3. (M) (a) Co OH H 2 O 4 NH 3 (b) (c) (d) (e) Co ONO 3 NH 3 3 Pt H 2O 4 PtCl6 amminetetraaquahydroxocobalt(III) ion triamminetrinitritoOcobalt(III) tetraaquaplatinum(II) hexachloroplatinate(IV) diaquadioxalatoferrate(III) ion silver(I) tetraiodomercurate(II) potassium hexacyanoferrate(III) bis(ethylenediamine)copper(II) ion pentaaquahydroxoaluminum(III) chloride amminechlorobis(ethylenediammine)chromium(III) sulfate
tris(ethylenediamine)iron(III) hexacyanoferrate(II) Fe ox 2 H 2O 2 Ag 2 HgI 4 4. (M) (a) K 3 Fe CN 6 (b) (c) (d) (e) Cu en 2 2+ Al OH H 2O 5 Cl2 CrCl en 2 NH 3 SO 4 Fe en 3 4 Fe CN 6 3 1167 Chapter 24: Complex Ions and Coordination Compounds Bonding and Structure in Complex Ions
5. (E) The Lewis structures are grouped together at the end. (a) (b) (c) (d) H2O has 12 + 6 = 8 valence electrons, or 4 pairs.
CH 3 NH 2 has 4 + 3 1+ 5 + 2 1 = 14 valence electrons, or 7 pairs. ONO has 2 6 + 5 +1 = 18 valence electrons, or 9 pairs. The structure has a 1 formal charge on the oxygen that is singly bonded to N. SCN has 6 + 4 + 5 +1 = 16 valence electrons, or 8 pairs. This structure, appropriately, gives a 1 formal charge to N. H H H N C H O N O S C N O H H H
(a) (b) (c) (d) 6. (M) The Lewis structures are grouped together at the end. (a) (b) (c) (d) OH has 6 +1 + 1 = 8 valence electrons, or 4 pairs. SO42 has 6 + 6 4 + 2 = 32 valence electrons, or 16 pairs. C2O42 has 2 4 + 6 4 + 2 = 34 valence electrons, or 17 pairs. SCN has 6 + 4 + 5 +1 = 16 valence electrons, or 8 pairs. This structure, appropriately, gives a 1 formal charge to N.
O O H O
2+ O O O O O
(c) O N C O
monodentate or bidentate S O S (a) 7. (M) We assume that PtCl 4
2 (b) (d) is square planar by analogy with PtCl2 NH 3 2 in Figure 245. The other two complex ions are octahedral.
2Cl Cl Pt Cl Cl (b) H2O H2O 2+ NH3 NH3 Co NH3 OH2 (c) H2O H2O Cl OH2 Cr OH2 OH2 2+ (a) 1168 Chapter 24: Complex Ions and Coordination Compounds 8. (M) The structures for Pt ox 2 2 2 and Cr ox 3 3 are drawn below. The structure of
2 Fe EDTA is the same as the generic structure for M EDTA drawn in Figure 2423, n+ 2+ with M = Fe . O (a) (b) 3O O O O
C C O C Pt O C C O 2 C C O O C O O (c) See Figure 2423.
+ O Cr O C O C O O C O O 9. (M) (a)
H3N H3N NH3 NH3 Cr NH3 OSO 3 (b) O C C O O C O O 3 O Co O C O C O O C O O pentaamminesulfateochromium(III) (c)
factriamminedichloronitroOcobalt(III) NH3 NH3 Cl Co NH3 ONO Cl trioxalatocobaltate(III) mertriamminecisdichloronitroOcobalt(III) NH3 NH3 Cl Co ONO Cl NH3 mertriamminetransdichloronitroOcobalt(III) NH3 NH3 Cl Co Cl ONO NH3 10. (M) (a) pentaamminenitrotoNcobalt(III) ion copper(II)
NH3 NH3 Co NH3 NH3 2+ (b) ethylenediaminedithiocyanatoSCH2 CH2 NH2 H2N NCS Cu O2N H3N SCN 1169 Chapter 24: Complex Ions and Coordination Compounds (c) hexaaquanickel(II) ion 2+ H2O OH2 H2O Ni OH2 H2O H2O Isomerism
11. (E) (a) cistrans isomerism cannot occur with tetrahedral structures because all of the ligands are separated by the same angular distance from each other. One ligand cannot be on the other side of the central atom from another. (b)
Square planar structures can show cistrans isomerism. Examples are drawn following, with the cisisomer drawn on the left, and the transisomer drawn on the right. A A B B
M A B cisisomer M B A transisomer (c) 12. Linear structures do not display cistrans isomerism; there is only one way to bond the two ligands to the central atom.
2+ (M) (a) CrOH NH 3 5 2+ NH3 NH3 HO Cr NH3 H3N NH3 (b)
+ has one isomer. CrCl2 H 2 O NH 3 3 has three isomers. + + + Cl Cl Cl NH3 Cl NH3 H2O Cr NH3 H2O Cr NH3 H2O Cr Cl H3N H3N H3N Cl NH3 NH3 CrCl2 en 2 has two geometric isomers, cis and trans. + + Cl C N Cl C Cl N Cr N N Cr N C N C N Cl C C N C C
+ (c) 1170 Chapter 24: Complex Ions and Coordination Compounds (d) CrCl 4 en has only one isomer since the ethylenediamine (en) ligand cannot bond trans to the central metal atom. Cl Cl Cl Cr N Cl C N C (e) Cr en 3 3+ N N Cr C N N C has only one geometric isomer; it has two optical isomers. 3+ C C
N N C C 13. (M) (a) There are three different square planar isomers, with D, C, and B, respectively, trans to the A ligand. They are drawn below.
A Pt C D D B A Pt C D
2+ B A Pt C B A Zn C B D B Zn C A D (b) Tetrahedral ZnABCD are drawn above. does display optical isomerism. The two optical isomers 14. (M) There are a total of four coordination isomers. They are listed below. We assume that the oxidation state of each central metal ion is +3 . Co en 3 Cr ox 3 tris(ethylenediamine)cobalt(III) trioxalatochromate(III) Co ox en 2 Cr ox 2 en bis(ethylenediamine)oxalatocobalt(III) (ethylenediamine)dioxalatochromate(III) Cr ox en 2 Co ox 2 en bis(ethylenediamine)oxalatochromium(III) (ethylenediamine)dioxalatocobaltate(III) Cr en 3 Co ox 3 tris(ethylenediamine)chromium(III) trioxalatocobaltate(III) 1171 Chapter 24: Complex Ions and Coordination Compounds 15. (M) The cisdichlorobis(ethylenediamine)cobalt(III) ion is optically active. The two optical isomers are drawn below. The transisomer is not optically active: the ion and its mirror image are superimposable.
Cl Cl Cl Co N C N C N N Co N N C N C C C C N C cisisomers Cl C N Cl Co C N Cl N C N C transisomer 16. (M) Complex ions (a) and (b) are identical; complex ions (a) and (d) are geometric isomers; complex ions (b) and (d) are geometric isomers; complex ion (c) is distinctly different from the other three complex ions (it has a different chemical formula). Crystal Field Theory
17. (E) In crystal field theory, the five d orbitals of a central transition metal ion are split into two (or more) groups of different energies. The energy spacing between these groups often corresponds to the energy of a photon of visible light. Thus, the transitionmetal complex will absorb light with energy corresponding to this spacing. If white light is incident on the complex ion, the light remaining after absorption will be missing some of its components. Thus, light of certain wavelengths (corresponding to the energies absorbed) will no longer be present in the formerly white light. The resulting light is colored. For example, if blue light is absorbed from white light, the remaining light will be yellow in color. (E) The difference in color is due to the difference in the value of , which is the ligand field splitting energy. When the value of is large, short wavelength light, which has a blue color, is absorbed, and the substance or its solution appears yellow. On the other hand, when the value of is small, light of long wavelength, which has a red or yellow color, is absorbed, and the substance or its solution appears blue. The cyano ligand is a strong field ligand, producing a large value of , and thus yellow complexes. On the other hand, the aqua ligands are weak field ligands, which produce a small value of , and hence blue complexes. 18. 19. (M) We begin with the 7 electron dorbital diagram for Co2+ [Ar]
The strong field and weak field diagrams for octahedral complexes follow, along with the number of unpaired electrons in each case. Strong Field eg Weak Field eg
(1 unpaired electron) (3 unpaired electrons) t2g t2g 1172 Chapter 24: Complex Ions and Coordination Compounds 20. (M) We begin with the 8 electron dorbital diagram for Ni 2+ Ar The strong field and weak field diagrams for octahedral complexes follow, along with the number of unpaired electrons in each case. Strong Field Weak Field eg eg (2 unpaired electrons) (2 unpaired electrons) t2g t2g The number of unpaired electrons is the same in both cases. 21. (M) (a) Both of the central atoms have the same oxidation state, +3. We give the electron configuration of the central atom to the left, then the completed crystal field diagram in the center, and finally the number of unpaired electrons. The chloro ligand is a weak field ligand in the spectrochemical series.
Mo 3+
Kr 4d 3 weak field eg Kr 4d t2g 3 unpaired electrons; paramagnetic eg t2g no unpaired electrons; diamagnetic The ethylenediamine ligand is a strong field ligand in the spectrochemical series. Co 3+
Ar Ar 3d 6 strong field (b) In CoCl 4 the oxidation state of cobalt is 2+. Chloro is a weak field ligand. The electron configuration of Co 2+ is Ar 3d 7 or Ar The tetrahedral ligand field diagram is shown on the right. weak field t2g eg 3 unpaired electrons 2 22. (M) (a) In Cu py 4 the oxidation state of copper is +2. Pyridine is a strong field ligand. The electron configuration of Cu 2+ is Ar 3d 9 or Ar 2+ There is no possible way that an odd number of electrons can be paired up, without at least one electron being unpaired. Cu py 4 2+ is paramagnetic. 1173 Chapter 24: Complex Ions and Coordination Compounds (b) In Mn CN 6 the oxidation state of manganese is +3. Cyano is a strong field ligand. The electron configuration of Mn 3+ is Ar 3d 4 or Ar The ligand field diagram follows, on the lefthand side. In FeCl4 the oxidation 3 state of iron is +3. Chloro is a weak field ligand. The electron configuration of Fe 3+ is Ar 3d 5 or [Ar] The ligand field diagram follows, below. t2g eg strong field weak field eg t2g 2 unpaired electrons 5 unpaired electrons There are more unpaired electrons in FeCl 4
23. (M) The electron configuration of Ni 2+ than in Mn CN 6 . 8 is Ar 3d or Ar 3 Ammonia is a strong field ligand. The ligand field diagrams follow, octahedral at left, tetrahedral in the center and square planar at right. Octahedral eg t2g Tetrahedral t2g eg Square Planar dx2y2 dxy dz2 dxz, dyz Since the octahedral and tetrahedral configurations have the same number of unpaired electrons (that is, 2 unpaired electrons), we cannot use magnetic behavior to determine whether the ammine complex of nickel(II) is octahedral or tetrahedral. But we can determine if the complex is square planar, since the square planar complex is diamagnetic with zero unpaired electrons.
24. (M) The difference is due to the fact that Fe CN 6 2+ 4 is a strong field complex ion, while Fe H 2O 6 is a weak field complex ion. The electron configurations for an iron atom and an iron(II) ion, and the ligand field diagrams for the two complex ions follow. Fe CN 6 4 [Fe H 2 O 6 ] 2+ iron atom iron(II) ion Ar 3d 4s eg t2g [Ar] 3d eg t2g 4s 1174 Chapter 24: Complex Ions and Coordination Compounds ComplexIon Equilibria
25. (E) (a) (b) Zn OH 2 s + 4 NH3 aq Zn NH3 4 2+ aq + 2 OH aq Cu 2+ aq + 2 OH aq Cu OH 2 s The blue color is most likely due to the presence of some unreacted [Cu(H2O)4]2+ (pale blue) Cu OH 2 s + 4 NH3 aq Cu NH3 4 2+ aq, dark blue + 2 OH aq 2+ Cu NH 3 4 26. 2+ aq + 4 H3O+ aq Cu H 2O 4 aq + 4 NH 4 + aq 2 (M) (a) CuCl2 s + 2 Cl aq CuCl4 aq, yellow 2 CuCl4 2 aq + 4H 2O(l)
2+ CuCl4 aq, yellow Cu H 2O 4 aq, pale blue + 4 Cl aq 2 or 2 Cu H 2 O 2 Cl2 aq, green + 4 Cl aq Either Or
(b) CuCl4 aq + 4 H 2O(l) Cu H 2O 4 aq + 4 Cl aq 2 2+ Cu H 2 O 2 Cl2 aq + 2H 2 O(l) Cu H 2O 4 2+ aq + 2 Cl aq The blue solution is that of [Cr(H2O)6]2+. This is quickly oxidized to Cr 3+ by O 2 g from the atmosphere. The green color is due to CrCl2 H 2 O 4 . 3+ 4 [Cr(H 2 O)6 ]2 (aq, blue)+ 4 H + (aq) +8 Cl (aq) + O 2 (g) 4 [CrCl2 (H 2O) 4 ] (aq, green)+10 H 2 O(l) Over a period of time, we might expect volatile HCl(g) to escape, leading to the formation of complex ions with more H 2 O and less Cl . H + aq + Cl aq HCl g CrCl 2 H 2O 4 aq, green + H 2 O(l) CrCl H 2O 5 + 2+ aq, bluegreen + Cl aq 1175 Chapter 24: Complex Ions and Coordination Compounds CrCl H 2O 5 2+ aq, bluegreen + H 2O(l) Cr H 2O 6 aq, blue + Cl aq 3+ Actually, to ensure that these final two reactions actually do occur in a timely fashion, it would be helpful to dilute the solution with water after the chromium metal has dissolved.
27. (M) Co en 3 should have the largest overall Kf value. We expect a complex ion with polydentate ligands to have a larger value for its formation constant than complexes that contain only monodentate ligands. This phenomenon is known as the chelate effect. Once one end of a polydentate ligand becomes attached to the central metal, the attachment of the remaining electron pairs is relatively easy because they already are close to the central metal (and do not have to migrate in from a distant point in the solution).
3+ 28. (M) (a) Zn NH 3 4 2+ 4 = K1 K2 K3 K4 = 3.9 102 2.1 102 1.0 102 50.= 4.1 108 .
(b) Ni H 2O 2 NH 3 4 2+ 4 = K1 K2 K3 K4 = 6.3 102 1.7 102 54 15 = 8.7 107
29. (M) First: Fe H 2 O 6 3+ aq + en aq Fe H 2O 4 en aq + 2H 2O(l) 3+ 3+ K1 = 104.34
Second: Fe H 2 O 4 en Third: aq + en aq Fe H 2O 2 en 2 en aq Fe en 3 3+ 3+ 3+ aq + 2H 2O(l) K2 = 103.31 Fe H 2 O 2 en 2 3+ 3+ aq + aq + 2H 2O(l) K3 = 102.05 Fe H 2 O 6 aq + 3 en aq Fe en 3 aq + 6H 2 O(l) Net: K f = K1 K2 K3 log Kf = 4.34 + 3.31+ 2.05 = 9.70 Kf = 109.70 = 5.0 109 = 3 30. (E) Since the overall formation constant is the product of the individual stepwise formation constants, the logarithm of the overall formation constant is the sum of the logarithms of the stepwise formation constants.
log Kf = log K1 + log K2 + log K3 + log K4 = 2.80 +1.60 + 0.49 + 0.73 = 5.62 Kf = 105.62 = 4.2 105 1176 Chapter 24: Complex Ions and Coordination Compounds 31. (M) (a) Aluminum(III) forms a stable (and soluble) hydroxo complex but not a stable ammine complex. Al H 2 O 3 OH 3 s + OH aq Al H 2 O 2 OH 4 aq + H 2 O(l) (b) Although zinc(II) forms a soluble stable ammine complex ion, its formation constant is not sufficiently large to dissolve highly insoluble ZnS. However, it is sufficiently large to dissolve the moderately insoluble ZnCO 3 . Said another way, ZnS does not produce sufficient Zn 2+ to permit the complex ion to form. ZnCO3 s + 4 NH 3 aq Zn NH 3 4 2+ aq + CO32 aq (c) Chloride ion forms a stable complex ion with silver(I) ion, that dissolves the AgCl(s) that formed when Cl is low. AgCl s Ag + aq + Cl aq and Ag + aq + 2 Cl aq AgCl2 aq Overall: AgCl s + Cl aq AgCl2 aq 32. (M) (a) Because of the large value of the formation constant for the complex ion, Co NH 3 6 aq , the concentration of free Co 3+ aq is too small to enable it to oxidize water to O2 g . Since there is not a complex ion present, except, of course,
3+ Co H 2 O 6 aq , when CoCl 3 is dissolved in water, the Co 3+ is sufficiently high for the oxidationreduction reaction to be spontaneous.
3+ 4 Co3+ aq + 2 H 2 O(l) 4 Co 2+ aq + 4 H + aq + O 2 g (b) Although AgI(s) is often described as insoluble, there is actually a small concentration of Ag + aq present because of the solubility equilibrium: 2 AgI(s) Ag(aq) +I(aq) These silver ions react with thiosulfate ion to form the stable dithiosulfatoargentate(I) complex ion: Ag + aq + 2S2 O32 aq Ag S2 O3 2 aq 1177 Chapter 24: Complex Ions and Coordination Compounds AcidBase Properties
33. (E) Al H 2 O 6 Al H 2O 6 3+ 3+ aq is capable of releasing H+:
2+ aq + H 2O(l) AlOH H 2O 5 aq + H 3O + aq The value of its ionization constant ( pKa = 5.01) approximates that of acetic acid.
34. (M) (a) CrOH H 2 O 5 (b) CrOH H 2 O 5 2+ 2+ aq + OH aq Cr OH 2 H 2O 4 aq + + H 2 O(l) aq + H 3O + aq Cr H 2 O 6 3+ aq + H 2 O(l) Applications
35. (M) (a) Solubility: AgBr s Ag + aq + Br aq K f =1.710 Ag + aq + 2S2 O32 aq Ag S2 O3 2 13 Ksp =5.01013 Cplx. Ion Formation: 3 aq Net: AgBr s + 2S2 O32 aq Ag S2 O3 2 K overall = 5.0 1013 1.7 1013 = 8.5 3 aq + Br  (aq) K overall = K sp K f With a reasonably high S2 O 3
(b) 2 , this reaction will go essentially to completion.
+ NH 3 aq cannot be used in the fixing of photographic film because of the relatively small value of Kf for Ag NH 3 2 aq , Kf = 1.6 107 . This would produce a 6 value of K = 8.0 10 in the expression above, which is nowhere large enough to drive the reaction to completion. 36. aq Co NH aq e Reduction: O aq 2e 2OH aq H
(M) Oxidation: Co NH 3 6 2 2 2+ 3+ 3 6 3+ 2 E 0.10V. E 0.88V Net: H 2 O 2 aq + 2 Co NH 3 6 2+ aq 2 Co NH3 6 aq + 2OH (aq) E cell = +0.88 V+ 0.10 V = +0.78 V. The positive value of the standard cell potential indicates that this is a spontaneous reaction.
37. (M) To make the cis isomer, we must use ligands that show a strong tendency for directing incoming ligands to positions that are trans to themselves. I has a stronger tendency than does Cl or NH3 for directing incoming ligands to the trans positions, and so it is beneficial to convert K2[PtCl4] to K2[PtI4] before replacing ligands around Pt with NH3 molecules. 1178 Chapter 24: Complex Ions and Coordination Compounds 38. (M) Transplatin is more reactive and is involved in more side reactions before it reaches its target inside of cancer cells. Thus, although it is more reactive, it is less effective at killing cancer cells. cisPt(NH3)2Cl2 transPt(NH3)2Cl2
Cl Pt H3N NH3 Cl Cl Pt Cl NH3 NH3 INTEGRATIVE AND ADVANCED EXERCISES
39.
(a) (b) (c) (d) (e) (M)
cupric tetraammine ion dichlorotetraamminecobaltic chloride platinic(IV) hexachloride ion disodium copper tetrachloride dipotassium antimony(III) pentachloride tetraamminecopper(II) ion tetraamminedichlorocobalt(III) chloride hexachloroplatinate(IV) ion sodium tetrachlorocuprate(II) potassium pentachloroantimonate(III) [Cu(NH3)4]2+ [CoCl2(NH3)4]Cl [PtCl6]2 Na2[CuCl4] K2[SbCl5] 40. (E)[Pt(NH3)4][PtCl4] tetraammineplatinum(II) tetrachloroplatinate(II) 41. (M) The four possible isomers for [CoCl2(en)(NH3)2]+ are sketched below:
NH3 Cl Co Cl H 2N NH3 NH2 H3N Co H 3N Cl H 2N Cl NH2 Cl Co H3N H2N NH3 Cl Cl NH2 H3N Co H 2N NH3 Cl NH2 42. (M)The color of the green solid, as well as that of the green solution, is produced by the complex ion [CrCl2(H2O)4]+. Over time in solution, the chloro ligands are replaced by aqua ligands, producing violet [Cr(H2O)6]3+(aq). When the water is evaporated, the original complex is reformed as the concentration of chloro ligand, [Cl], gets higher and the chloro ligands replace the aqua ligands.
3 Cl H2 O H2O Cl Cr OH2 OH2 +2H20 H2O H2 O OH2 OH2 Cr OH2 OH2 2Cl +2H20 H2O H2O Cl Cl Cr OH2 OH2 43. (M)The chloro ligand, being lower in the spectrochemical series than the ethylenediamine ligand, is less strongly bonded to the central atom than is the ethylenediamine ligand. Therefore, of the two types of ligands, we expect the chloro ligand to be replaced more readily. In the cis isomer, the two chloro ligands are 90 from each other. This is the angular spacing that can be readily spanned by the oxalato ligand, thus we expect reaction with the cis isomer to occur rapidly. On the other hand, in the trans isomer, the two chloro ligands are located 180 from each other. After 1179 Chapter 24: Complex Ions and Coordination Compounds the chloro ligands are removed, at least one end of one ethylenediamine ligand would have to be relocated to allow the oxalato ligand to bond as a bidentate ligand. Consequently, replacement of the two chloro ligands by the oxalato ligand should be much slower for the trans isomer than for the cis isomer.
44. (M) Cl2 + 2 e 2 Cl[Pt(NH3)4]2+ [Pt(NH3)4]4+ + 2 eCl2 + [Pt(NH3)4]2+ [Pt(Cl2NH3)4]2+
Cl H 3N H3N Pt NH3 H 3N Pt H3N Cl NH3 NH3 + ClCl
NH3 45. (M) The successive acid ionizations of a complex ion such as [Fe(OH)6]3+ are more nearly equal in magnitude than those of an uncharged polyprotic acid such as H3PO4 principally because the complex ion has a positive charge. The second proton is leaving a species which has one fewer positive charge but which is nonetheless positively charged. Since positive charges repel each other, successive ionizations should not show a great decrease in the magnitude of their ionization constants. In the case of polyprotic acids, on the other hand, successive protons are leaving a species whose negative charge is increasingly greater with each step. Since unlike charges attract each other, it becomes increasingly difficult to remove successive protons. 46. (M)
1
Cl Pt Cl Cl NH3 Cl +1
NH3 Pt Cl NH3 H3N Cl Pt NH3 NH3 K+ ClQuickTimeTM and a TIFF (LZW) decompressor are needed to see this picture. (a) +2
H3N Pt H3N NH3 NH3 Cl (b) 2
Cl Pt Cl Cl (c) 2Cl 2K+ (d) (e) 47. (M) (a) Fe(H 2 O)63+ (aq) Initial 0.100 M Equil. (0.100 x) M K 9.0 10 H 2 O(l) 4 Fe(H 2 O)5 OH 2 (aq) 0M x C o n d u c t i v i t y Number of Cl ligands H3 O 0M x {where x is the molar quantity of Fe(H 2O)63+ (aq) hydrolyzed}
[[Fe(H 2 O)5 OH]2+ ] [H 3 O + ] x2 9.0 104 K = 3+ 0.100x [[Fe(H 2 O)6 ] ] Solving, we find x 9.5 103 M, which is the [H 3 O + ] so: pH log(9.5 103 ) 2.02 1180 Chapter 24: Complex Ions and Coordination Compounds (b) [Fe(H 2 O)6 ]3+ (aq) H 2 O(l) initial 0.100 M equil. (0.100 x) M K 9.0 10 4 [Fe(H 2 O)5OH]2 (aq) 0M H 3O 0.100 M xM (0.100 x) M {where x is [[Fe(H 2 O)6 ]3+ ] reacting} K [[Fe(H 2 O)5OH]2 ] [H 3O + ] x (0.100 x) 9.0 104 3+ [[Fe(H 2 O)6 ] ] (0.100x ) Solving, we find x 9.0 104 M, which is the [[Fe(H 2 O)5 OH]2 ] (c) We simply substitute [[Fe(H 2 O)5 OH]2 ] = 1.0 106 M into the K a expression with [Fe(H 2 O)6 ]3+ 0.100 M and determine the concentration of H 3 O + [H 3 O + ] K a [[Fe(H 2 O)6 ]3+ ] 9.0 104 (0.100 M) 90. M [[Fe(H 2 O)5 OH]2 ] [1.0 106 M] To maintain the concentration at this level requires an impossibly high concentration of H 3 O + 48. (D) Let us first determine the concentration of the uncomplexed (free) Pb2+(aq). Because of the large value of the formation constant, we assume that most of the lead(II) ion is present as the EDTA complex ion ([Pb(EDTA)]2). Once equilibrium is established, we can see if there is sufficient lead(II) ion present in solution to precipitate with the sulfide ion. Note: the concentration of EDTA remains constant at 0.20 M)
Reaction: Initial: Change: Equilibrium:
Kf = Pb2+(aq) + EDTA4(aq) 0M +x M x M 0.20 M 0.20 M constant 210 18 [Pb(EDTA)]2(aq) 0.010 M x M (0.010x) M 0.010 M
x = 2.5 1020 M = [Pb2+ ] [Pb(EDTA)2 ] 0.010 = 2 1018 = 2+ 40.20(x) [Pb ][EDTA ] Now determine the [S2 ] required to precipitate PbS from this solution. K sp = [Pb2 ][S 2 ] 8 1028 [S2 ]= 8 1028 = 2.5 108 M Recall (i) that a saturated H 2S solution is ~0.10 M H 2S 20 (5 10 ) and (ii) That the [S2 ] = K a2 = 1 1014 Qsp = [Pb2 ][S 2 ] 2.5 1020 1 1014 2.5 1034 Qsp << K sp K sp 8 1028 Hence, PbS(s) will not precipitate from this solution. 49. (M) The formation of [Cu(NH3)4]2+(aq) from [Cu(H2O)4]2+(aq) has K1 = 1.9 104, K2 = 3.9 103, K3 = 1.0 103, and K4 = 1.5 102. Since these equilibrium constants are all considerably larger than 1.00, one expects that the reactions they represent, yielding ultimately the ion [Cu(NH3)4]2+(aq), will go essentially to completion. However, if the concentration of NH3 were limited to less than the stoichiometric amount, that is, to less than 4 mol NH3 per mol of 1181 Chapter 24: Complex Ions and Coordination Compounds Cu2+(aq), one would expect that the ammineaqua complex ions would be present in significantly higher concentrations than the [Cu(NH3)4]2+(aq) ions.
50. (D) For [Ca(EDTA)]2, Kf = 4 1010 and for [Mg(EDTA)]2, Kf = 4 108. In Table 181, the least soluble calcium compound is CaCO3, Ksp = 2.8 109, and the least soluble magnesium compound is Mg3(PO4)2, Ksp = 1 1025. We can determine the equilibrium constants for adding carbonate ion to [Ca(EDTA)]2(aq) and for adding phosphate ion to [Mg(EDTA)]2(aq) as follows: [Ca(EDTA)]2 (aq) Ca 2 (aq) EDTA 4 (aq) Precipitation: Ca 2 (aq) CO32 (aq) CaCO3 (s)
Instability: Net:
K K i 1/4 1010 K ppt 1/2.8 109 [Ca(EDTA)]2 + CO32 (aq) CaCO3 (s) + EDTA 4 (aq) K K i K ppt
1 (4 10 )( 2.8 109 )
10 9 103 The small value of the equilibrium constant indicates that this reaction does not proceed very far toward products. {[Mg(EDTA)]2 (aq) Mg2 (aq) EDTA4 (aq)} 3 Precipitation: 3 Mg2 (aq) 2 PO43 (aq) Mg3 (PO4 )2 (s)
Instability: Net :
K Ki3 1/(4 108 )3 Kppt 1/1 1025 K Ki3 Kppt 3 [Mg(EDTA)] 2+2 PO43 (aq) Mg3 (PO4 )2 (s) +3 EDTA4 (aq)
8 3 1 0.16 (4 10 ) (1 1025 ) This is not such a small value, but again we do not expect the formation of much product, particularly if the [EDTA4 ] is kept high. We can approximate what the concentration of precipitating anion must be in each case, assuming that the concentration of complex ion is 0.10 M and that of [EDTA4 ] also is 0.10 M.
Reaction: K [Ca(EDTA)]2 CO32 (aq) CaCO3 (s) EDTA 4 (aq) [CO32 ] 1 102 M [EDTA 4 ] 0.10 M 9 103 2 2 [[Ca(EDTA)] ][CO3 ] 0.10 M [CO32 ] This is an impossibly high [CO32]. Reaction: 3[Mg(EDTA)]2 2 PO 43 (aq) Mg 3 (PO 4 ) 2 (s) 3 EDTA 4 (aq) [PO 43 ] 2.5 M [EDTA 4 ]3 (0.10 M)3 K 0.16 [[Mg(EDTA)]2 ]3[PO 43 ]2 (0.10 M)3 [PO 43 ]2 Although this [PO43] is not impossibly high, it is unlikely that it will occur without the deliberate addition of phosphate ion to the water. Alternatively, we can substitute [[M(EDTA)]2 ] = 0.10 M and [EDTA4 ] = 0.10 M into the formation constant [[M(EDTA)]2 ] expression: K f [M 2 ][EDTA 4 ] 1182 Chapter 24: Complex Ions and Coordination Compounds This substitution gives [M 2 ] 1 , hence, [Ca2+] = 2.5 1011 M and [Mg2+] = 2.5 109 M, Kf which are concentrations that do not normally lead to the formation of precipitates unless the concentrations of anions are substantial. Specifically, the required anion concentrations are [CO32] = 1.1 102 M for CaCO3, and [PO43] = 2.5 M, just as computed above.
51. (M) If a 99% conversion to the chloro complex is achieved, which is the percent conversion necessary to produce a yellow color, [[CuCl4]2] = 0.99 0.10 M = 0.099 M, and [[Cu(H2O)4]2+] = 0.01 0.10 M = 0.0010 M. We substitute these values into the formation constant expression and solve for [Cl].
[[CuCl 4 ] 2 ] 0.099 M 4.2 10 5 Kf 4 2 [[Cu(H 2 O) 4 ] ] [Cl ] 0.0010 M [Cl ] 4
4 [Cl ] 0.099 0.0010 4.2 10 5 0.12 M This is the final concentration of free chloride ion in the solution. If we wish to account for all chloride ion that has been added, we must include the chloride ion present in the complex ion. total [Cl] = 0.12 M free Cl + (4 0.099 M) bound Cl = 0.52 M
52. (M)(a) Oxidation: 2H 2 O(l) O 2 (g) 4 H (aq) 4 e
3 2 E 1.229 V
E 1.82 V Reduction: {Co ((aq) e Co (aq)} Net: 4 Co3 (aq) 2 H2 O(l) 4 Co2 (aq) 4 H (aq) O2 (g) E 0.59 V cell (b) Reaction: Co3 (aq) 6 NH 3 (aq) [Co(NH 3 )6 ]3 (aq) Initial: 1.0 M Changes: xM Equil: (1.0 x)M 0.10 M constant 0.10 M 0M xM xM [[Co(NH 3 )6 ]3 ] x Kf 4.5 1033 3 6 (1.0 x)(0.10)6 [Co ][NH 3 ] Thus [[Co(NH3)6]3+] = 1.0 M because K is so large, and 1 [Co 3 ] 2.2 10 28 M 4.5 10 27 x 4.5 1027 1.0 x 1183 Chapter 24: Complex Ions and Coordination Compounds (c) The equilibrium and equilibrium constant for the reaction of NH3 with water follows. NH 3 (aq) H 2 O(l) NH 4 (aq) OH (aq) [NH 4 ][OH  ] [OH ]2 K b 1.8 10 [NH 3 ] 0.10
5 [OH ] 0.10(1.8 105 ) 0.0013 M In determining the [OH ], we have noted that [NH4+] = [OH ] by stoichiometry, and also that Kw 1.0 10 14 7.7 10 12 M [NH3] = 0.10 M, as we assumed above. [H 3 O ] 0.0013 [OH ] We use the Nernst equation to determine the potential of reaction (24.12) at the conditions described. E E cell [Co 2 ]4 [H ]4 P[O 2 ] 0.0592 log 4 [Co3 ]4 0.0592 (1 104 ) 4 (7.7 1012 ) 4 0.2 E 0.59 log 0.59 V 0.732 V 0.142 V 4 (2.2 1028 ) 4 The negative cell potential indicates that the reaction indeed does not occur.
53. (M) We use the Nernst equation to determine the value of [Cu2+]. The cell reaction follows.
Cu(s) 2 H (aq) Cu 2 (aq) H 2 (g) E E 2 E 0.337 V cell [Cu 2 ] [Cu ] 0.0592 0.0592 log 0.08 V 0.337 log 2 2 [H ] 2 (1.00) 2 2 (0.337 0.08) 14.1 0.0592 [Cu 2 ] 8 10 15 M log [Cu 2 ] Now we determine the value of Kf. Kf [[Cu(NH 3 ) 4 ] 2 ] [Cu ] [NH 3 ]
2 4 1.00 1 1014 4 15 8 10 (1.00) This compares favorably with the value of Kf = 1.1 1013 given in Table 182, especially considering the imprecision with which the data are known. (Ecell = 0.08 V is known to but one significant figure.) 1184 Chapter 24: Complex Ions and Coordination Compounds 54. (M) We first determine [Ag+] in the cyanide solution.
[[Ag(CN) 2 ] ] 0.10 5.6 1018 2 [Ag ][CN ] [Ag ] (0.10.) 2 The cell reaction is as follows. It has 0.10 [Ag ] 1.8 10 18 5.6 1018 (0.10) 2 E 0.000 V ; the same reaction occurs at both anode and cathode and thus the Nernstian cell voltage is influenced only by the Ag+ concentration. Kf Ag (0.10 M) Ag (0.10 M [Ag(CN) 2 ] , 0.10 M KCN) E E [Ag ]CN 0.0592 1.8 10 18 log 0.000 0.0592 log 0.99 V 1 0.10 [Ag ]
+ OH2 H3N H3N Cl Co NH3 NH3 ClH3N H3N OH2 OH2 Co NH3 NH3 2Cl+2 55. (M) A 0.10 mol/L solution of these compounds would result in ion concentration of 0.20 mol/L for [Co Cl (H2O(NH3)4)]ClH2O or 0.30 mol/L for [Co(H2O)2 (NH3)4]Cl2. Observed freezing point depression = 0.56 C. T = Kfmi = 0.56 deg = 1.86 mol kg1 deg(0.10 mol/L)i Hence, i = 3.0 mol/kg or 3.0 mol/L. This suggests that the compound is [Co(H2O)2 (NH3)4]Cl2. 56. (M) Sc(H2O)63+ and Zn(H2O)42+ have outer energy shells with 18 electrons, and thus they do not have any electronic transitions in the energy range corresponding to visible light. Fe(H2O)63+ has 17 electrons (not 18) in its outer shell; thus, it does have electronic transitions in the energy range corresponding to visible light. 57. (M) The Cr3+ ion would have 3 unpaired electrons, each residing in a 3d orbital and would be sp3d2 hybridized. The hybrid orbitals would be hybrids of 4s, 4p, and 3d (or 4d) orbitals. Each CrNH3 coordinate covalent bond is a bond formed when a lone pair in an sp3 orbital on N is directed toward an empty sp3d2 orbital on Cr3+. The number of unpaired electrons predicted by valence bond theory would be the same as the number of unpaired electrons predicted by crystal field theory. 58. (M) Since these reactions involve Ni2+ binding to six identical nitrogen donor atoms, we can assume the H for each reaction is approximately the same. A large formation constant indicates a more negative free energy. This indicates that the entropy term for these reactions is different. The large increase in entropy for each step, can be explained by the chelate effect. The same number of water molecules are displaced by fewer ligands, which is directly related to entropy change. 1185 Chapter 24: Complex Ions and Coordination Compounds 59. (M) Co(acac)3
CH3 CH3 O Co O O H3C CH3 CH3 H3C H3C O H3C H3C O Co O O CH3 O H3C O O O O CH3 This compound has a nonsuperimposable mirror image (therefore it is optically active). These are enantiomers. One enantiomer rotates plane polarized light clockwise, while the other rotates plane polarized light counterclockwise. Polarimetry can be used to determine which isomer rotates plane polarized light in a particular direction. This compound has a superimposable mirror image (therefore it is optically inactive). These are the same compound. This compound will not rotate plane polarized light (net rotation = 0). trans[Co(acac)2(H2O)2]Cl2 H3C OH2 O Co O OH2 O O CH3 H3C O OH2 O Co O OH2 O CH3 H3C CH3 H3C CH3 cis[Co(acac)2(H2O)2]Cl2
CH3 H3C H3C O O Co O O OH2 OH H2O HO O O Co O O CH3 H3 C OH2 H2O CH3 This compound has a nonsuperimposable mirror image (therefore it is optically active). These are enantiomers. One enantiomer rotates plane polarized light clockwise, while the other rotates plane polarized light counterclockwise. By using a polarimeter, we can determine which isomer rotates plane polarized light in a particular direction. 3+ 2+ 60. (D)[Co(H2O)6] (aq) + e [Co(H2O)6] (aq) log K = nE/0.0592 = (1)(1.82)/0.0592 = 30.74 [Co(en)3]3+(aq) + 6 H2O(l) [Co(H2O)6]2+(aq) + 3 en [Co(H2O)6]3+(aq) + e[Co(en)3]3+(aq) + e K = 1030.74 K = 1/1047.30 K = 1012.18 K = 1030.74 Koverall = ? [Co(H2O)6]3+(aq) + 3 en [Co(en)3]2+(aq) + 6 H2O(l) [Co(H2O)6]2+(aq) [Co(en)3]2+(aq) 1186 Chapter 24: Complex Ions and Coordination Compounds Koverall = 1/1047.301012.181030.74 = 104.38 = 0.0000417 E = (0.0592/n)log K = (0.0592/1)log(0.0000417) = 0.26 V 4 Co3+(aq) + 4 e 4 Co2+(aq) Ecathode = 1.82 V + 2 H2O(l) 4 H (aq) + O2(g) + 4 e Eanode = 1.23 V 4 Co3+(aq) +2 H2O(l) 4 H+(aq) + O2(g) + 4 Co2+(aq) Ecell = Ecathode Eanode = 1.82V 1.23 V = +0.59 V (Spontaneous) 4 [Co(en)3]3+(aq) + 4 e 4 [Co(en)3]2+(aq) Ecathode = 0.26 V Eanode = 1.23 V 2 H2O(l) 4 H+(aq) + O2(g) + 4 e3+ + 4 [Co(en)3] (aq) +2 H2O(l) 4 H (aq) + O2(g) + 4 [Co(en)3]2+(aq) Ecell = Ecathode E = 0.26 V 1.23 V = 1.49 V (nonspontaneous) 61. (M) There are two sets of isomers for Co(gly)3. Each set comprises two enantiomers (nonsuperimposable mirror images). One rotates plane polarized light clockwise (+), the other counterclockwise (). Experiments are needed to determine which enantiomer is (+) and which is ().
O O H2 N O O Co
H2N O O NH2 O Co H2N O O
NH2 O
OO O NH2 NH2 H2N H2N O O Co
NH2 O NH2 O O Co
H2N O O O O O O (+)factris(glycinato)cobalt(III) ()factris(glycinato)cobalt(III) (+)mertris(glycinato)cobalt(III) ()mertris(glycinato)cobalt(III)
O O [Co(gly)2Cl(NH3)] has 10 possible isomer, four pairs of enantiomers and two achiral isomers.
NH3 O O Co
H2N Cl NH3 O NH2 O
Cl O Co
H3N O NH2 NH2 H2N H2N O
Cl Co
NH3 O O O Co
H2N Cl H2 N O O
O O 1187 Chapter 24: Complex Ions and Coordination Compounds H2N
Cl O O NH2 O Co O NH2
Cl NH3 H2N
Cl O O NH2 O Co H2N O
NH3 Cl O Co O H2N
O O O Co NH2 O H3N H3N O O H2 N H3N Cl O O Co NH2 O O O NH2 Co H2 N O NH3 Cl O O 62. (M) The coordination compound is facecentered cubic, K+ occupies tetrahedral holes, while PtCl62 occupies octahedral holes. 63.
# NH3 Formula: Total # ions (per formula unit) # NH3 Formula: Total # ions (per formula unit) 4 [PtCl2(NH3)4]Cl2 3 5 [PtCl(NH3)5]Cl3 4 6 [Pt(NH3)6]Cl4 5 (M)
0 K2[PtCl6] 3 1 K[PtCl5(NH3)] 2 2 PtCl4(NH3)2 0 3 [PtCl3(NH3)3]Cl 2 1188 Chapter 24: Complex Ions and Coordination Compounds FEATURE PROBLEMS
A trigonal prismatic structure predicts three geometric isomers for [CoCl2(NH3)4]+, 64. (M)(a) which is one more than the actual number of geometric isomers found for this complex ion. All three geometric isomers arising from a trigonal prism are shown below.
Cl Cl Cl Cl (i) (ii) (iii) Cl Cl The fact that the trigonal prismatic structure does not afford the correct number of isomers is a clear indication that the ion actually adopts some other structural form (i.e., the theoretical model is contradicted by the experimental result). We know now of course, that this ion has an octahedral structure and as a result, it can exist only in cis and trans configurations. (b) All attempts to produce optical isomers of [Co(en)3]3+ based upon a trigonal prismatic structure are shown below. The ethylenediamine ligand appears as an arc in diagrams below: (i) (ii) (iii) Only structure (iii), which has an ethylenediamine ligand connecting the diagonal corners of a face can give rise to optical isomers. Structure (iii) is highly unlikely, however, because the ethylenediamine ligand is simply too short to effectively span the diagonal distance across the face of the prism. Thus, barring any unusual stretching of the ethylenediamine ligand, a trigonal prismatic structure cannot account for the optical isomerism that was observed for [Co(en)3]3+. 65. (D) Assuming that each hydroxide ligand bears its normal 1 charge, and that each ammonia ligand is neutral, the total contribution of negative charge from the ligands is 6 . Since the net charge on the complex ions is 6+, the average oxidation state for each Co atom must be +3 (i.e., each Co in the complex can be viewed as a Co3+ 3d 6 ion surrounded by six ligands.) The five 3d orbitals on each Co are split by the octahedrally arranged ligands into three lower energy orbitals, called t2g orbitals, and two higher energy orbitals, called eg orbitals. We are told in the question that the complex is low spin. This is simply another way of saying that all six 3d electrons on each Co are paired up in the t2g set as a result of the eg and t2g orbitals being separated by a relatively large energy gap (see below). Hence, there should be no unpaired electrons in the hexacation (i.e., the cation is expected to be diamagnetic). 1189 Chapter 24: Complex Ions and Coordination Compounds Low Spin Co3+ (Octahedral Environment): eg Relatively Large Crystal Field Splitting
t2g The Lewis structures for the two optical isomers (enantiomers) are depicted below:
H3N H3N H3N H3N Co H3N H H H O O HO Co O Co H3N NH3 NH3 NH3 NH3 Co O H O H NH3 NH3 H3N NH3 H3N Co NH3 OH H H3N NH3 H O Co O Co NH3 O H O H NH3 H3N O H Co H3N NH3 NH3 Nonsuperimposable mirror images of one another 66. (D) The data used to construct a plot of hydration energy as a function of metal ion atomic number is collected in the table below. The graph of hydration energy (kJ/mol) versus metal ion atomic number is located beneath the table.
Metal ion Atomic number Ca2+ 20 Sc2+ 21 Ti2+ 22 2+ V 23 2+ Cr 24 2+ Mn 25 Fe2+ 26 Co2+ 27 2+ Ni 28 2+ Cu 29 2+ Zn 30 Hydration energy 2468 kJ/mol 2673 kJ/mol 2750 kJ/mol 2814 kJ/mol 2799 kJ/mol 2743 kJ/mol 2843 kJ/mol 2904 kJ/mol 2986 kJ/mol 2989 kJ/mol 2936 kJ/mol
2+ (a)
20 Hydration Energy (kJ/mol) 2600 Hydration Enthalpy of M ions (Z = 2030)
22 24 26 28 30 2800 3000 Atomic Number (Z) 1190 Chapter 24: Complex Ions and Coordination Compounds (b) When a metal ion is placed in an octahedral field of ligands, the five dorbitals are split into eg and t2g subsets, as shown in the diagram below: dz 2 dx2  y 2
3 eg /5o dorbitals (symmerical field) 2 /5o t2g dxy dyz dxz (octahedral field) Since water is a weak field ligand, the magnitude of the splitting is relatively small. As a consequence, highspin configurations result for all of the hexaaqua complexes. The electron configurations for the metal ions in the highspin hexaaqua complexes and their associated crystal field stabilization energies (CFSE) are provided in the table below Metal Ion Configuration t2g eg number of unpaired e CFSE(o) Ca2+ 3d 0 0 0 0 0 2+ 1 Sc 3d 1 0 1 2/ 5 Ti2+ 3d 2 2 0 2 4/ 5 V2+ 3d 3 3 0 3 6/ 5 Cr2+ 3d 4 3 1 4 3/ 5 2+ 5 Mn 3d 3 2 5 0 Fe2+ 3d 6 4 2 4 2/ 5 Co2+ 3d 7 5 2 3 4/ 5 Ni2+ 3d 8 6 2 2 6/ 5 2+ 9 Cu 3d 6 3 1 3/ 5 2+ 10 Zn 3d 6 4 0 0 2+ 2+ Thus, the crystal field stabilization energy is zero for Ca . Mn and Zn2+. (c) The lines drawn between those ions that have a CFSE = 0 show the trend for the enthalpy of hydration after the contribution from the crystal field stabilization energy has been subtracted from the experimental values. The Ca to Mn and Mn to Zn lines are quite similar. Both line have slopes that are negative and are of comparable magnitude. This trend shows that as one proceeds from left to right across the periodic table, the energy of hydration for dications becomes increasingly more negative. The hexaaqua complexes become progressively more stable because the Zeff experienced by the bonding electrons in the valence shell of the metal ion steadily increases as we move further and further to the right. Put another way, the Zeff climbs steadily as we move from left to right and this leads to the positive charge density on the metal becoming larger and larger, which results in the water ligands steadily being pulled closer and closer to the nucleus. Of course, the closer the approach of the water ligands to the metal, the greater is the energy released upon successful coordination of the ligand. 1191 Chapter 24: Complex Ions and Coordination Compounds (d) Those ions that exhibit crystal field stabilization energies greater than zero have heats of hydration that are more negative (i.e. more energy released) than the hypothetical heat of hydration for the ion with CFSE subtracted out. The heat of hydration without CFSE for a given ion falls on the line drawn between the two flanking ions with CFSE = 0 at a position directly above the point for the experimental hydration energy. The energy difference between the observed heat of hydration for the ion and the heat of hydration without CFSE is, of course, approximately equal to the CFSE for the ion. (e) As was mentioned in the answer to part (c), the straight line drawn between manganese and zinc (both ions with CFSE = 0) on the previous plot, describes the enthalpy trend after the ligand field stabilization energy has been subtracted from the experimental values for the hydration enthalpy. Thus, o for Fe2+ in [Fe(H2O)6]2+ is approximately equal to 5/2 of the energy difference (in kJ/mol) between the observed hydration energy for Fe2+(g) and the point for Fe2+ on the line connecting Mn2+ and Zn2+, which is the expected enthalpy of hydration after the CFSE has been subtracted out. Remember that the crystal field stabilization energy for Fe2+ that is obtained from the graph is not o, but rather 2/5o, since the CFSE for a 3d6 ion in an octahedral field is just 2/5 of o. Consequently, to obtain o , we must multiply the enthalpy difference by 5/2. According to the graph, the highspin CFSE for Fe2+ is 2843 kJ/mol (2782 kJ/mol) or 61 kJ/mol. Consequently, o = 5/2(61 kJ/mol) = 153 kJ/mol, or 1.5 102 kJ/mol is the energy difference between the eg and t2g orbital sets. (f) The color of an octahedral complex is the result of the promotion of an electron on the metal from a t2g orbital eg orbital. The energy difference between the eg and t2g orbital sets is o. As the metalligand bonding becomes stronger, the separation between the t2g and eg orbitals becomes larger. If the eg set is not full, then the metal complex will exhibit an absorption band corresponding to a t2g eg transition. Thus, the [Fe(H2O)6]2+complex ion should absorb electromagnetic radiation that has Ephoton = o. Since o 150 kJ/mol (calculated in part (e) of this question) for a mole of [Fe(H2O)6]2+(aq),
1mol [Fe(H 2 O)6 ]2+ 1.5 102 kJ 1000 J 2+ 23 2+ 1mol[Fe(H 2 O)6 ] 6.022 10 [Fe(H 2 O)6 ] 1kJ 19 = 2.5 10 J per ion E 2.5 1019 J = = = 3.8 1014 s1 h 6.626 1034 J s Ephoton = = So, the [Fe(H2O)6]2+ ion will absorb radiation with a wavelength of 780 nm, which is red light in the visible part of the electromagnetic spectrum. c 2.998 108 m s 1 = = 7.8 107 m (780 nm) h 3.8 1014 s 1 1192 Chapter 24: Complex Ions and Coordination Compounds SELFASSESSMENT EXERCISES
67. (E) (a) Coordination number is the number of ligands coordinated to a transition metal complex. (b) 0 is the crystal field splitting parameter which depends on the coordination geometry of a transition metal complex. (c) Ammine complex is the complex that contains NH3 ligands. (d) An enantiomer is one of two stereoisomers that are mirror images of each other that are "nonsuperposable" (not identical). (E) (a) A spectrochemical series is a list of ligands ordered on ligand strength and a list of metal ions based on oxidation number, group and identity. (b) Crystal field theory (CFT) is a model that describes the electronic structure of transition metal compounds, all of which can be considered coordination complexes. (c) Optical isomers are two compounds which contain the same number and kinds of atoms, and bonds (i.e., the connectivity between atoms is the same), and different spatial arrangements of the atoms, but which have nonsuperimposable mirror images. (d) Structural isomerism in accordance with IUPAC, is a form of isomerism in which molecules with the same molecular formula have atoms bonded together in different orders. (E) (a) Coordination number is the number of ligands coordinated to a transition metal complex. Oxidation number of a central atom in a coordination compound is the charge that it would have if all the ligands were removed along with the electron pairs that were shared with the central atom. (b) A monodentate ligand has one point at which it can attach to the central atom. Polydentate ligand has many points at which it can coordinate to a transition metal. (c) Cis isomer has identical groups on the same side. In trans isomer, on the other hand, identical groups are on the opposite side. (d) Dextrorotation and levorotation refer, respectively, to the properties of rotating plane polarized light clockwise (for dextrorotation) or counterclockwise (for levorotation). A compound with dextrorotation is called dextrorotary, while a compound with levorotation is called levorotary. (e) When metal is coordinated to ligands to form a complex, its "d" orbital splits into high and low energy groups of suborbitals. Depending on the nature of the ligands, the energy difference separating these groups can be large or small. In the first case, electrons of the d orbital tend to pair in the low energy suborbitals, a configuration known as "low spin". If the energy difference is low, electrons tend to distribute unpaired, giving rise to a "high spin" configuration. High spin is associated with paramagnetism (the property of being attracted to magnetic fields), while low spin is associated to diamagnetism (inert or repelled by magnets). (E) (d) (E) (e) 68. 69. 70. 71. 1193 Chapter 24: Complex Ions and Coordination Compounds 72. 73. 74. 75. 76. 77. (E) (b) (E) (a) (E) (d) (E) (c) (E) (b) (M) (a) pentaamminebromocobalt(III)sulfate, no isomerism. (b) hexaamminechromium(III)hexacyanocobaltate(III), no isomerism. (c) sodiumhexanitritoNcobaltate(III), no isomerism (d) tris(ethylenediamine)cobalt(III)chloride, two optical isomers. (M) (a) [Ag(CN)2]; (b) [Pt(NO2)(NH3)3]+; (c) [CoCl(en)2(H2O)]2+; (d) K4[Cr(CN)6] (M)
Cl O Cl Fe Cl H2N Cl Cl O O H2N O Fe NH2 Cl H3N Cr H3N OH NH3 78. 79. +
Cl NH3 Cl Pt Cl Cl 2 H2 N Cl (a) (b) (c) (d) 80. (M) (a) one structure only (all positions are equivalent for the H2O ligand; NH3 ligands attach at the remaining five sites).
+3 NH3 H2O Co H3N NH3 NH3 NH3 (b) two structures, cis and trans, based on the placement of H2O.
+3 NH3 H2O Co H3N NH3 OH2 NH3 H2O Co H2O NH3 NH3 NH3 NH3 +3 trans cis 1194 Chapter 24: Complex Ions and Coordination Compounds (c) two structures, fac and mer.
+3 NH3 H3N Co H2O OH2 OH2 NH3 H2O Co H2O NH3 NH3 NH3 OH2 +3 fac
+3 NH3 H2O Co H2O NH3 OH2 OH2 H3N mer
+3 OH2 OH2 Co H 3N OH2 OH2 (d) two structures, cis and trans, based on the placement of NH3. trans cis 81. (M) (a) coordination isomerism, based on the interchange of ligands between the complex cation and complex anion. (b) linkage isomerism, based on the mode of attachment of the SCN ligand (either SCN or NCS). (c) no isomerism (d) geometric isomerism, based on whether Cl ligands are cis or trans. (e) geometric isomerism, based on whether the NH3 or OH ligands are fac or mer. (M) (a) geometric isomerism (cis and trans) and optical isomerism in the cis isomer. (b) geometric isomerism (cis and trans), optical isomerism in the cis isomer, and linkage isomerism in the thiocyanate ligand. (c) no isomers. (d) no isomers for this squareplanar complex. (e) two geometric isomers, one with the tridentate ligand occupying meridional positions and the other with the tridentate ligand occupying facial positions. (M) Because ethylenediamine (en) is a stronger field ligand than H2O, more energy must be absorbed by [Co(en)3]3+ than by [Co(H2O)6]3+ to simulate an electronic transition. This means that [Co(en)3]3+(aq) absorbs shorter wavelength light and transmits longer wavelength light than does [Co(H2O)6]3+(aq). Thus [Co(en)3]3+(aq) is yellow and [Co(H2O)6]3+(aq) is blue. 82. 83. 1195 ...
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