7 CHAPTER ACIDS AND BASES
Nature of Acids and Bases
16. a. H2O(l) + H2O(l) H3O+(aq) + OH(aq) or H2O(l) H+(aq) + OH(aq) K = Kw = [H+][OH]
b. HF(aq) + H2O(l) F(aq) + H3O+(aq) or [H + ][F ] HF(aq) H+(aq) + F(aq) K = Ka = [ HF]
c. C5H5N(aq) + H2O(l) C5H5NH+(aq) + OH(aq)
K = Kb =
[C 5 H 5 NH + ][OH ] [C 5 H 5 N ]
17.
An acid is a proton (H+) donor, and a base is a proton acceptor. A conjugate acid-base pair differs by only a proton (H+) in the formulas. Acid a. b. c. H2CO3 C5H5NH+ C5H5NH+ Base H2O H2O HCO3 Conjugate Base of Acid HCO3 C5H5N C5H5N Conjugate Acid of Base H3O+ H3O+ H2CO3
18.
a. HClO4(aq) + H2O(l) H3O+(aq) + ClO4(aq). Only the forward reaction is indicated because HClO4 is a strong acid and is basically 100% dissociated in water. For acids, the dissociation reaction is commonly written without water as a reactant. The common abbreviation for this reaction is HClO4(aq) H+(aq) + ClO4(aq). This reaction is also called the Ka reaction because the equilibrium constant for this reaction is designated as Ka. b. Propanoic acid is a weak acid, so it is only partially dissociated in water. The dissociation reaction is CH3CH2CO2H(aq) + H2O(l) H3O+(aq) + CH3CH2CO2(aq) or CH3CH2CO2H(aq) H+(aq) + CH3CH2CO2(aq). c. NH4+ is a weak acid. Similar to propanoic acid, the dissociation reaction is: NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) or NH4+(aq) H+(aq) + NH3(aq)
192
CHAPTER 7
19.
ACIDS AND BASES
193
The dissociation reaction (the Ka reaction) of an acid in water commonly omits water as a reactant. We will follow this practice. All dissociation reactions produce H+ and the conjugate base of the acid that is dissociated. a. HC2H3O2(aq) H+(aq) + C2H3O2(aq) b. Co(H2O)63+(aq) H+(aq) + Co(H2O)5(OH)2+(aq) Ka = [H + ][C 2 H 3O 2 ] [HC 2 H 3O 2 ]
Ka =
[H + ][Co(H 2 O) 5 (OH) 2+ ] [Co(H 2 O) 3+ ] 6
[H + ][CH 3 NH 2 ] [CH 3 NH 3 ]
+
c. CH3NH3+(aq) H+(aq) + CH3NH2(aq)
20.
Ka =
Strong acids have a Ka >> 1 and weak acids have Ka < 1. Table 7.2 in the text lists some Ka values for weak acids. Ka values for strong acids are hard to determine so they are not listed in the text. However, there are only a few common strong acids so if you memorize the strong acids, then all other acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO3, HClO4, and H2SO4. a. b. c. d. HClO4 is a strong acid. HOCl is a weak acid (Ka = 3.5 108). H2SO4 is a strong acid. H2SO3 is a weak diprotic acid because the Ka1 and Ka2 values are less than 1.
21.
The beaker on the left represents a strong acid in solution; the acid HA is 100% dissociated into the H+ and A ions. The beaker on the right represents a weak acid in solution; only a little bit of the acid HB dissociates into ions, so the acid exists mostly as undissociated HB molecules in water. a. b. c. d. e. HNO2: weak acid beaker HNO3: strong acid beaker HCl: strong acid beaker HF: weak acid beaker HC2H3O2: weak acid beaker
22.
All Kb reactions refer to the base reacting with water to produce the conjugate acid of the base and OH. a. NH3(aq) + H2O(l) NH4 (aq) + OH (aq)
+
[OH ][ NH 4 ] Kb = [ NH 3 ] Kb = [OH ][HCN] [CN ] [OH ][C 5 H 5 NH + ] [C 5 H 5 N ]
+
b. CN(aq) + H2O(l) HCN(aq) + OH(aq) c. C5H5N(aq) + H2O(l) C5H5NH+(aq) + OH(aq)
Kb =
194
CHAPTER 7
d. C6H5NH2(aq) + H2O(l) C6H5NH3+(aq) + OH(aq) Kb =
ACIDS AND BASES
[OH ][C 6 H 5 NH 3 ] [C 6 H 5 NH 2 ]
+
23.
The Ka value is directly related to acid strength. As Ka increases, acid strength increases. For water, use Kw when comparing the acid strength of water to other species. The Ka values are: HClO4: strong acid (Ka >> 1); HClO2: Ka = 1.2 10 2 NH4+: Ka = 5.6 10 10 ; H2O: Ka = Kw = 1.0 10 14
From the Ka values, the ordering is: HClO4 > HClO2 > NH4+ > H2O 24. Except for water, these are the conjugate bases of the acids in the preceding exercise. In general, the weaker the acid, the stronger the conjugate base. ClO4 is the conjugate base of a strong acid; it is a terrible base (worse than water). The ordering is NH3 > ClO2 > H2O > ClO4 . a. H2SO4 is a strong acid and water is a very weak acid with Ka = Kw = 1.0 1014. H2SO4 is a much stronger acid than H2O. b. H2O, Ka = Kw = 1.0 1014; HOCl, Ka = 3.5 108; HOCl is a stronger acid than H2O because Ka for HOCl > Ka for H2O. c. NH4+ , Ka = 5.6 1010; HC2H2ClO2, Ka = 1.35 103; HC2H2ClO2 is a stronger acid than NH4+ because Ka for HC2H2ClO2 > Ka for NH4+. 26. a. H2O; the conjugate bases of strong acids are terrible bases (Kb < 1014). b. OCl ; the conjugate bases of weak acids are weak bases (1014 < Kb < 1); even though they are designated as weak bases, the conjugate bases of weak acids are all better bases than H2O. c. NH3; for a conjugate acid-base pair, Ka Kb = Kw. From this relationship, the stronger the acid, the weaker is the conjugate base (Kb decreases as Ka increases). Because HC2H2ClO2 is a stronger acid than NH4+ (Ka for HC2H2ClO2 > Ka for NH4+), NH3 will be a stronger base than C2H2ClO2 .
25.
27.
a. H2O and CH3CO2 b. An acid-base reaction can be thought of as a competition between two opposing bases. Because this equilibrium lies far to the left (Ka < 1), CH3CO2 is a stronger base than H2O. c. The acetate ion is a better base than water and produces basic solutions in water. When we put acetate ion into solution as the only major basic species, the reaction is: CH3CO2 + H2O CH3CO2H + OH
CHAPTER 7
ACIDS AND BASES
195
Now the competition is between CH3CO2 and OH for the proton. Hydroxide ion is the strongest base possible in water. The preceding equilibrium lies far to the left resulting in a Kb value of less than 1. Those species we specifically call weak bases (1014 < Kb < 1) lie between H2O and OH in base strength. Weak bases are stronger bases than water but are weaker bases than OH. 28. The NH4+ ion is a weak acid because it lies between H2O and H3O+ (H+) in terms of acid strength. Weak acids are better acids than water, thus their aqueous solutions are acidic. They are weak acids because they are not as strong as H3O+ (H+). Weak acids only partially dissociate in water and have Ka values of between 1014 and 1. In deciding whether a substance is an acid or a base, strong or weak, you should keep in mind a couple ideas: 1. There are only a few common strong acids and strong bases all of which should be memorized. Common strong acids = HCl, HBr, HI, HNO3, HClO4, and H2SO4. Common strong bases = LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. 2. All other acids and bases are weak and will have Ka and Kb values of less than 1 but greater than Kw (1014). Reference Table 7.2 for Ka values for some weak acids and Table 7.3 for Kb values for some weak bases. There are too many weak acids and weak bases to memorize them all. Therefore, use the tables of Ka and Kb values to help you identify weak acids and weak bases. Appendix 5 contains more complete tables of Ka and Kb values. a. c. e. g. i. weak acid (Ka = 4.0 104) weak base (Kb = 4.38 104) weak base (Kb = 1.8 105) weak acid (Ka = 1.8 104) strong acid b. d. f. h. strong acid strong base weak acid (Ka = 7.2 104) strong base
29.
Autoionization of Water and pH Scale
30. a. Because the value of the equilibrium constant increases as the temperature increases, the reaction is endothermic. In endothermic reactions, heat is a reactant, so an increase in temperature (heat) shifts the reaction to produce more products and increases K in the process. b. H2O(l) H+(aq) + OH(aq) Kw = 5.47 1014 = [H+][OH]
In pure water [H+] = [OH], so 5.47 1014 = [H+]2, [H+] = 2.34 107 M pH = log[H+] = log(2.34 107) = 6.631 A neutral solution of water at 50.C has: [H+] = [OH]; [H+] = 2.34 107 M; pH = 6.631 Obviously, the condition that [H+] = [OH] is the most general definition of a neutral solution.
196
c. Temp (C) 0 25 35 40. 50. Temp (K) 273 298 308 313 323 1/T (K1) 3.66 103 3.36 103 3.25 103 3.19 103 3.10 103
CHAPTER 7
Kw 1.14 1015 1.00 1014 2.09 1014 2.92 1014 5.47 1014
ACIDS AND BASES
ln Kw
34.408 32.236 31.499 31.165 30.537
From the graph: 37C = 310. K; 1/T = 3.23 103 K1 ln Kw = 31.38, Kw = e31.38 = 2.35 1014
d. At 37C, Kw = 2.35 1014 = [H+][OH] = [H+]2, [H+] = 1.53 107 M pH = log[H+] = log(1.53 107) = 6.815 31. a. H2O(l) H+(aq) + OH(aq) Kw = 2.92 10 14 = [H+][OH]
In pure water: [H+] = [OH], 2.92 10 14 = [H+]2, [H+] = 1.71 10 7 M = [OH] b. pH = !log[H+] = !log(1.71 10 7 ) = 6.767 c. [H+] = Kw/[OH] = (2.92 10 14 )/0.10 = 2.9 10 13 M; pH = !log(2.9 10 13 ) = 12.54 32. At 25C, the relationship [H+][OH] = Kw = 1.0 10 14 always holds for aqueous solutions. When [H+] is greater than 1.0 10 7 M, the solution is acidic; when [H+] is less than 1.0 10 7 M, the solution is basic; when [H+] = 1.0 10 7 M, the solution is neutral. In terms of [OH], an acidic solution has [OH] < 1.0 10 7 M, a basic solution has [OH] > 1.0 10 7 M, and a neutral solution has [OH] = 1.0 10 7 M. At 25C, pH + pOH = 14.00.
CHAPTER 7
ACIDS AND BASES
Kw 1.0 10 14 = = 1.0 10-7 M; the solution is neutral. + 7 [H ] 1.0 10
197
a. [OH] =
pH = log[H+] = log(1.0 10 7 ) = 7.00; pOH = 14.00 7.00 = 7.00 b. [OH] = 1.0 10 14 = 12 M; the solution is basic. 8.3 10 16
pH = log(8.3 10 16 ) = 15.08; pOH = 14.00 15.08 = 1.08 c. [OH] =
1.0 10 14 = 8.3 10 16 M; the solution is acidic. 12
pH = log(12) = 1.08; pOH = 14.00 (1.08) = 15.08 d. [OH] = 1.0 10 14 = 1.9 10 10 M; the solution is acidic. 5 5.4 10
pH = log(5.4 10 5 ) = 4.27; pOH = 14.00 4.27 = 9.73 33. a. [H+] = 10pH, [H+] = 107.40 = 4.0 108 M pOH = 14.00 pH = 14.00 7.40 = 6.60; [OH] = 10pOH = 106.60 = 2.5 107 M or [OH] = Kw 1.0 10 14 = = 2.5 107 M; this solution is basic since pH > 7.00. + 8 [H ] 4.0 10
b. [H+] = 1015.3 = 5 1016 M; pOH = 14.00 15.3 = 1.3; [OH] = 10 (1.3) = 20 M; basic c. [H+] = 10 (1.0) = 10 M; pOH = 14.0 (1.0) = 15.0; [OH] = 10-15.0 = 1 1015 M; acidic d. [H+] = 103.20 = 6.3 104 M; pOH = 14.00 3.20 = 10.80; [OH] = 1010.80 = 1.6 1011 M; acidic e. [OH] = 105.0 = 1 105 M; pH = 14.0 pOH = 14.0 5.0 = 9.0; [H+] = 109.0 = 1 109 M; basic f. 34. [OH] = 109.60 = 2.5 1010 M; pH = 14.00 9.60 = 4.40; [H+] = 104.40 = 4.0 105 M; acidic
a. pOH = 14.00 9.63 = 4.37; [H+] = 10 9.63 = 2.3 10 10 M [OH] = 10 4.37 = 4.3 10 5 M; basic b. [H+] = 1.0 10 14 = 2.6 10 9 M; pH = log(2.6 10 9 ) = 8.59 3.9 10 6
pOH = 14.00 8.59 = 5.41; basic
198
CHAPTER 7
c. pH = log(0.027) = 1.57; pOH = 14.00 1.57 = 12.43 [OH] = 10 12.43 = 3.7 10 13 M; acidic d. pH = 14.0 12.2 = 1.8; [H+] = 10 1.8 = 1.6 10 2 M [OH] = 10 12.2 = 6 10 13 M; acidic
ACIDS AND BASES
Solutions of Acids
35. Strong acids are assumed to completely dissociate in water, for example, HCl(aq) + H2O(l) H3O+(aq) + Cl(aq) or HCl(aq) H+(aq) + Cl(aq). a. A 0.10 M HCl solution gives 0.10 M H+ and 0.10 M Cl because HCl completely dissociates. The amount of H+ from H2O will be insignificant. pH = log[H+] = log(0.10) = 1.00 b. 5.0 M H+ is produced when 5.0 M HClO4 completely dissociates. The amount of H+ from H2O will be insignificant. pH = log(5.0) = 0.70 (Negative pH values just indicate very concentrated acid solutions.) c. 1.0 1011 M H+ is produced when 1.0 1011 M HI completely dissociates. If you take the negative log of 1.0 1011, this gives pH = 11.00. This is impossible! We dissolved an acid in water and got a basic pH. What we must consider in this problem is that water by itself donates 1.0 107 M H+. We can normally ignore the small amount of H+ from H2O except when we have a very dilute solution of an acid (as in the case here). Therefore, the pH is that of neutral water (pH = 7.00) because the amount of HI present is insignificant. 36. 50.0 mL conc. HCl soln
1.19 g 38 g HCl 1 mol HCl = 0.62 mol HCl mL 100 g conc. HCl soln 36.5 g
70. g HNO 3 1 mol HNO 3 1.42 g = 0.32 mol HNO3 mL 100 g soln 63.0 g HNO 3
20.0 mL conc. HNO3 soln
HCl(aq) H+(aq) + Cl(aq) and HNO3(aq) H+(aq) + NO3(aq) (Both are strong acids.) So we will have 0.62 + 0.32 = 0.94 mol of H+ in the final solution. [H+] =
0.94 mol = 0.94 M; pH = log[H+] = log(0.94) = 0.027 = 0.03 1.00 L
[OH] =
Kw 1.0 10 14 = = 1.1 1014 M 0.94 [H + ]
CHAPTER 7
37.
ACIDS AND BASES
199
HCl is a strong acid. [H+] = 10 1.50 = 3.16 10 2 M (carrying one extra sig. fig.)
M1V1 = M2V2, V1 =
M 2 V2 3.16 10 2 mol/L 1.6 L = = 4.2 10 3 L M1 12 mol/L
4.2 mL of 12 M HCl with enough water added to make 1600 mL of solution will result in a solution having [H+] = 3.2 10 2 M and pH = 1.50. 38. [H+] = 10 5.10 = 7.9 10 6 M; HNO3(aq) H+(aq) + NO3(aq) Because HNO3 is a strong acid, we have a 7.9 10 6 M HNO3 solution. 0.2500 L 39. 7.9 10 6 mol HNO 3 63.02 g HNO 3 = 1.2 10 4 g HNO3 L mol HNO 3
a. Major species: H+(aq), Br(aq), and H2O(l). (HBr is a strong acid.) [H+] = 0.250 M pH = log[H+] = log(0.250) = 0.602 b. H+(aq), ClO4(aq), and H2O(l). (HClO4 is a strong acid.) pH = 0.602 c. H+(aq), NO3(aq), and H2O(l). (HNO3 is a strong acid.) pH = 0.602 d. HNO2 (Ka = 4.0 104) and H2O (Ka = Kw = 1.0 1014) are the major species. HNO2 is much stronger acid than H2O, so it is the major source of H+. However, HNO2 is a weak acid (Ka < 1), so it only partially dissociates in water. We must solve an equilibrium problem to determine [H+]. In this Solutions Guide, we will summarize the initial, change and equilibrium concentrations into one table called an ICE table. Solving the weak acid problem: HNO2(aq) Initial Change Equil. Ka =
H+(aq)
+
NO2(aq)
0.250 M ~0 0 x mol/L HNO2 dissociates to reach equilibrium x +x +x 0.250 x x x
x2 [H + ][ NO 2 ] , 4.0 10-4 = ; if we assume x << 0.250, then: [HNO 2 ] 0.250 x
4.0 10-4
x2 , 0.250
x=
4.0 10 4 (0.250) = 0.010 M
x 0.010 100 = 100 = 4.0% 0.250 0.250
We must check the assumption:
200
CHAPTER 7
ACIDS AND BASES
All the assumptions are good. The H+ contribution from water (107 M) is negligible and x is small compared to 0.250 (percent error = 4.0%). If the percent error is less than 5% for an assumption, we will consider it a valid assumption (called the 5% rule). Finishing the problem: x = 0.010 M = [H+]; pH = log[H+] = log(0.010) = 2.00 e. CH3CO2H (Ka = 1.8 105) and H2O (Ka = Kw = 1.0 1014) are the major species. CH3CO2H is the major source of H+. Solving the weak acid problem: CH3CO2H Initial Change Equil. Ka =
H+
+
CH3CO2
0.250 M ~0 0 x mol/L CH3CO2H dissociates to reach equilibrium x +x +x 0.250 x x x
[H + ][CH 3CO 2 ] x2 x2 (assuming x << 0.250) , 1.8 105 = [CH 3CO 2 H] 0.250 x 0.250 2.1 10 3 100 = 0.84%. Assumptions good. 0.250
x = 2.1 103 M; checking assumption:
[H+] = x = 2.1 103 M; pH = log(2.1 103) = 2.68 f. HCN (Ka = 6.2 1010) and H2O are the major species. HCN is the major source of H+. HCN Initial Change Equil.
H+
+
CN
0.250 M ~0 0 x mol/L HCN dissociates to reach equilibrium x +x +x 0.250 x x x [H + ][CN ] x2 x2 = (assuming x << 0.250) [HCN] 0.250 x 0.250
Ka = 6.2 1010 =
x = [H+] = 1.2 105 M; checking assumption: x is 0.0048% of 0.250.
Assumptions good. pH = log(1.2 105) = 4.92 40. At pH = 2.000, [H+] = 102.000 = 1.00 102 M; at pH = 4.000, [H+] = 104.000 = 1.00 104 M
Mol H+ present = 0.0100 L
0.0100 mol H + = 1.00 104 mol H+ L
Let V = total volume of solution at pH = 4.000: 1.00 104 mol/L = 0.0100 mol H + , V = 1.00 L V
Volume of water added = 1.00 L 0.0100 L = 0.99 L = 990 mL
CHAPTER 7
41.
ACIDS AND BASES
201
a. Major species: HOCl (Ka = 3.5 108) and water; major source of H+ = HOCl. Because Ka for HOCl is less than 1, HOCl is a weak acid and we must solve an equilibrium problem to determine [H+]. The setup is: HOCl (aq) Initial Change Equil.
H+(aq)
+
OCl(aq)
~0 0 0.20 M x mol/L HOCl dissociates to reach equilibrium x +x +x 0.20 x x x
x2 x2 [H + ][OCl ] = [HOCl] 0.20 x 0.20
Ka = 3.5 108 =
(assuming x << 0.20)
x = [H+] = 8.4 105 M
We have made two assumptions that we must check. 1. 0.20 x 0.20 (x/0.20) 100 = (8.4 105/0.20) 100 = 0.042%. Great assumption. If the percent error in the assumption is less than 5%, then the assumption is valid. 2. HOCl is the major source of H+, that is, we can ignore 1.0 107 M H+ already present in neutral H2O. [H+] from HOCl = 8.4 105 >> 1.0 107; this assumption is valid. In future problems we will always begin the problem solving process by making these assumptions, and we will always check them. However, we may not explicitly state that the assumptions are valid. We will always state when the assumptions are not valid and we have to use other techniques to solve the problem. Remember, anytime we make an assumption, we must check its validity before the solution to the problem is complete. Answering the question: [H+] = [OCl] = 8.4 105 M; [OH] = Kw/[H+] = 1.2 1010 M [HOCl] = 0.20 x = 0.20 M; pH = log[H+] = log(8.4 105) = 4.08 b. HOC6H5 (Ka = 1.6 1010) is the dominant producer of H+. Solving the weak acid problem: HOC6H5 Initial Change Equil.
H+
+
OC6H5
Ka = 1.6 1010
~0 0 1.5 M x mol/L HOC6H5 dissociates to reach equilibrium x +x +x 1.5 x x x
202
Ka = 1.6 1010 =
CHAPTER 7
[H + ][OC 6 H 5 ] x2 x2 = [HOC6 H 5 ] 1 .5 x 1 .5
ACIDS AND BASES
(assuming x << 1.5)
x = [H+]= 1.5 105 M; assumptions good: 1.0 107 << 1.5 105 << 1.5
[H+] = [OC6H5 ] = 1.5 105 M; [OH] = 6.7 1010 M [HOC6H5] = 1.5 x = 1.5 M; pH = log(1.5 105) = 4.82 c. This is a weak acid in water. Solving the weak acid problem: HF Initial Change Equil.
H+
+
F
Ka = 7.2 104
~0 0 0.020 M x mol/L HF dissociates to reach equilibrium x +x +x 0.020 x x x (assuming x << 0.020)
x2 x2 [H + ][F ] = [HF] 0.020 x 0.020 x = [H+] = 3.8 103 M; Check assumptions:
Ka = 7.2 104 =
x 3.8 10 3 100 = H 100 = 19% 0.020 0.020
The assumption x << 0.020 is not good (x is more than 5% of 0.020). We must solve x2/(0.020 x) = 7.2 104 exactly by using either the quadratic formula or the method of successive approximations (see Appendix 1 of the text). Using successive approximations, we let 0.016 M be a new approximation for [HF]. That is, in the denominator try x = 0.0038 (the value of x we calculated making the normal assumption) so that 0.020 0.0038 = 0.016; then solve for a new value of x in the numerator.
x2 x2 = 7.2 104, x = 3.4 103 0.020 x 0.016
We use this new value of x to further refine our estimate of [HF], that is, 0.020 x = 0.020 0.0034 = 0.0166 (carrying an extra sig. fig.).
x2 x2 = 7.2 104, x = 3.5 103 0.020 x 0.0166
We repeat, until we get a self-consistent answer. This would be the same answer we would get solving exactly using the quadratic equation. In this case it is, x = 3.5 103. Thus: [H+] = [F] = x = 3.5 103 M; [OH] = Kw/[H+] = 2.9 1012 M
CHAPTER 7
ACIDS AND BASES
[HF] = 0.020 x = 0.020 0.0035 = 0.017 M; pH = 2.46
203
Note: When the 5% assumption fails, use whichever method you are most comfortable with to solve exactly. The method of successive approximations is probably fastest when the percent error is less than ~25% (unless you have a graphing calculator).
42.
Major species: HC2H2ClO2 (Ka = 1.35 103) and H2O; major source of H+: HC2H2ClO2 HC2H2ClO2 Initial Change Equil.
H+
+ C2H2ClO2
~0 0 0.10 M x mol/L HC2H2ClO2 dissociates to reach equilibrium x +x +x 0.10 x x x
x2 x2 , x = 1.2 102 M 0.10 x 0.10
Ka = 1.35 103 =
Checking the assumptions finds that x is 12% of 0.10 which fails the 5% rule. We must solve 1.35 103 = x2/(0.10 x) exactly using either the method of successive approximations or the quadratic equation. Using either method gives x = [H+] = 1.1 102 M. pH = log[H+] = log(1.1 102) = 1.96. 43. Major species: HIO3, H2O; major source of H+: HIO3 (a weak acid, Ka = 0.17) HIO3 Initial Change Equil. Ka = 0.17 =
H+
+
IO3
~0 0 0.010 M x mol/L HIO3 dissociates to reach equilibrium x +x +x 0.010 x
x
x
[H + ][IO3 ] x2 x2 = , x = 0.041; check assumption. [HIO3 ] 0.010 x 0.010
Assumption is horrible (x is more than 400% of 0.010). When the assumption is this poor, it is generally quickest to solve exactly using the quadratic formula (see Appendix 1 in text). Using the quadratic formula and carrying extra significant figures: 0.17 =
x2 , x2 = 0.17(0.010 x), x2 + (0.17)x 1.7 103 = 0 0.010 x
x=
0.17 [(0.17) 2 4(1)(1.7 10 3 )]1/ 2 0.17 0.189 = , x = 9.5 103 M 2(1) 2 (x must be positive)
x = 9.5 103 M = [H+]; pH = log(9.5 103) = 2.02
204
44.
CHAPTER 7
ACIDS AND BASES
HC3H5O2 (Ka = 1.3 105) and H2O (Ka = Kw = 1.0 1014) are the major species present. HC3H5O2 will be the dominant producer of H+ because HC3H5O2 is a stronger acid than H2O. Solving the weak acid problem: HC3H5O2 Initial Change Equil.
H+
+
C3H5O2
~0 0 0.100 M x mol/L HC3H5O2 dissociates to reach equilibrium x +x +x 0.100 x x x [H + ][C3 H 5 O 2 ] x2 x2 = [HC3 H 5 O 2 ] 0.100 x 0.100
Ka = 1.3 105 =
x = [H+] = 1.1 103 M; pH = log(1.1 103) = 2.96
Assumption follows the 5% rule (x is 1.1% of 0.100). [H+] = [C3H5O2] = 1.1 103 M; [OH] = Kw/[H+] = 9.1 1012 M [HC3H5O2] = 0.100 1.1 103 = 0.099 M Percent dissociation = 45. [H + ] 1.1 10 3 100 = 100 = 1.1% [HC3 H 5 O 2 ]0 0.100
This is a weak acid in water. We must solve a weak acid problem. Let HBz = C6H5CO2H. 0.56 g HBz 1 mol HBz = 4.6 103 mol; [HBz]0 = 4.6 103 M 122.1 g HBz Initial Change Equil.
H+
+
Bz-
4.6 103 M ~0 0 x mol/L HBz dissociates to reach equilibrium x +x +x 4.6 103 x x x
x2 x2 [H + ][Bz ] = [HBz] (4.6 10 3 x) 4.6 10 3
Ka = 6.4 105 =
x = [H+] = 5.4 104;
check assumptions:
x 5.4 10 4 100 = 100 = 12% 4.6 10 3 4.6 10 3
Assumption is not good (x is 12% of 4.6 103). When assumption(s) fail, we must solve exactly using the quadratic formula or the method of successive approximations (see Appendix 1 of text). Using successive approximations:
CHAPTER 7
ACIDS AND BASES
x2 = 6.4 105, x = 5.1 104 3 4 (4.6 10 ) (5.4 10 ) x2 = 6.4 105, x = 5.1 104 M (consistent answer) (4.6 10 3 ) (5.1 10 4 )
205
Thus: x = [H+] = [Bz] = [C6H5CO2] = 5.1 104 M [HBz] = [C6H5CO2H] = 4.6 103 x = 4.1 103 M pH = log(5.1 104) = 3.29; pOH = 14.00 pH = 10.71; [OH] = 1010.71 = 1.9 1011 M 46. Initial Change Equil. Ka = HBz C
H+ ~0
+
Bz0
HBz = C6H5CO2H
C = [HBz]0 = concentration of HBz that dissolves to give saturated solution. x mol/L HBz dissociates to reach equilibrium x +x +x x x Cx
x2 [H + ][Bz ] = 6.4 105 = , where x = [H+] [HBz] Cx
6.4 105 =
[ H + ]2 ; pH = 2.80; [H+] = 102.80 = 1.6 103 M + C [H ] (1.6 10 3 ) 2 = 4.0 102, C = (4.0 102) + (1.6 103) = 4.2 102 M 6.4 10 5
C 1.6 103 =
The molar solubility of C6H5CO2H is 4.2 102 mol/L. 4.2 10 2 mol C 6 H 5 CO 2 H 122.1 g C 6 H 5 CO 2 H 0.100 L = 0.51 g per 100. mL solution L mol C 6 H 5 CO 2 H
2 tablets 0.325 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 tablet 180.15 g = 0.0152 M 0.237 L H+ + C9H7O4
47.
[HC9H7O4] =
HC9H7O4 Initial Change Equil.
~0 0 0.0152 M x mol/L HC9H7O4 dissociates to reach equilibrium !x +x +x x x 0.0152 ! x
206
Ka = 3.3 10 4 =
CHAPTER 7
ACIDS AND BASES
[ H + ] [C 9 H 7 O 4 ] x2 x2 = , x = 2.2 10 3 M [HC9 H 7 O 4 ] 0.0152 x 0.0152 2.2 10 3 100 = 14% 0.0152
Assumption that 0.0152 x 0.0152 fails the 5% rule:
Using successive approximations or the quadratic equation gives an exact answer of x = 2.1 10 3 M. [H+] = x = 2.1 10 3 M; pH = !log(2.1 10 3 ) = 2.68 48. HF and HOC6H5 are both weak acids with Ka values of 7.2 104 and 1.6 1010, respectively. Because the Ka value for HF is much greater than the Ka value for HOC6H5, HF will be the dominant producer of H+ (we can ignore the amount of H+ produced from HOC6H5 because it will be insignificant). HF Initial Change Equil.
H+
+
F
1.0 M ~0 0 x mol/L HF dissociates to reach equilibrium x +x +x 1.0 x x x
Ka = 7.2 104 =
x2 x2 [H + ][F ] = 1 .0 x 1.0 [ HF]
x = [H+] = 2.7 102 M; pH = log(2.7 102) = 1.57; assumptions good.
Solving for [OC6H5] using HOC6H5 H+ + OC6H5 equilibrium: Ka = 1.6 1010 = [H + ][OC 6 H 5 ] (2.7 10 2 )[OC 6 H 5 ] = , [HOC 6 H 5 ] 1.0 [OC6H5] = 5.9 109 M
Note that this answer indicates that only 5.9 109 M HOC6H5 dissociates, which indicates that HF is truly the only significant producer of H+ in this solution. 49. a. HCl is a strong acid. It will produce 0.10 M H+. HOCl is a weak acid. Let's consider the equilibrium: HOCl Initial Change Equil.
H+
+
OCl
Ka = 3.5 108
0.10 M 0.10 M 0 x mol/L HOCl dissociates to reach equilibrium x +x +x 0.10 x 0.10 + x x
CHAPTER 7
ACIDS AND BASES
[H + ][OCl ] (0.10 + x)( x) = x, x = 3.5 108 M [HOCl] 0.10 x
207
Ka = 3.5 108 =
Assumptions are great (x is 0.000035% of 0.10). We are really assuming that HCl is the only important source of H+, which it is. The [H+] contribution from HOCl, x, is negligible. Therefore, [H+] = 0.10 M; pH = 1.00. b. HNO3 is a strong acid, giving an initial concentration of H+ equal to 0.050 M. Consider the equilibrium: HC2H3O2 Initial Change Equil.
H+
+
C2H3O2
Ka = 1.8 105
0.50 M 0.050 M 0 x mol/L HC2H3O2 dissociates to reach equilibrium x +x +x 0.50 x 0.050 + x x
5
[H + ][C 2 H 3O 2 ] (0.050 + x) x (0.050) x = Ka = 1.8 10 = [HC 2 H 3O 2 ] (0.50 x) 0.50
x = 1.8 104; assumptions are good (well within the 5% rule).
[H+] = 0.050 + x = 0.050 M and pH = 1.30 50. Let HA symbolize the weak acid. Set up the problem like a typical weak acid equilibrium problem. HA Initial Change Equil.
H+
+
A
~0 0 0.15 M x mol/L HA dissociates to reach equilibrium !x +x +x 0.15 ! x x x
If the acid is 3.0% dissociated, then x = [H+] is 3.0% of 0.15: x = 0.030 (0.15 M) = 4.5 10 3 M. Now that we know the value of x, we can solve for Ka. Ka = 51. Initial Change Equil. ( 4.5 10 3 ) 2 x2 [H + ][A ] = = = 1.4 10 4 3 0.15 x [HA] 0.15 ( 4.5 10 ) HX
H+
+
X
I ~0 0 where I = [HX]0 x mol/L HX dissociates to reach equilibrium x +x +x Ix x x
From the problem, x = 0.25(I) and I x = 0.30 M.
208
CHAPTER 7
I 0.25(I) = 0.30 M, I = 0.40 M and x = 0.25(0.40 M) = 0.10 M Ka = (0.10) 2 x2 [H + ][X ] = = = 0.033 I x 0.30 [HX]
ACIDS AND BASES
52.
In all parts of this problem, acetic acid (HC2H3O2) is the best weak acid present. We must solve a weak acid problem. a. Initial Change Equil. HC2H3O2
H+
+
C2H3O2
0.50 M ~0 0 x mol/L HC2H3O2 dissociates to reach equilibrium x +x +x 0.50 x x x [H + ][C 2 H 3O 2 ] x2 x2 = = [HC 2 H 3O 2 ] 0.50 x 0.50
Ka = 1.8 105 =
x = [H+] = [C2H3O2] = 3.0 103 M; assumptions good.
Percent dissociation = [H + ] 3.0 10 3 H 100 = H 100 = 0.60% [HC 2 H 3O 2 ]0 0.50
b. The setup for solutions b and c are similar to solution a except that the final equation is different because the new concentration of HC2H3O2 is different. Ka = 1.8 105 =
x2 x2 0.050 x 0.050
x = [H+] = [C2H3O2] = 9.5 104 M; assumptions good.
Percent dissociation = c. Ka = 1.8 105 = 9.5 10 4 100 = 1.9% 0.050
x2 x2 0.0050 x 0.0050
x = [H+] = [C2H3O2] = 3.0 104 M; check assumptions.
Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the method of successive approximations (see Appendix 1 of the text): x2 x2 = 1.8 105 = , x = 2.9 104 0.0047 0.0050 (3.0 10 4 ) Next trial also gives x = 2.9 104.
CHAPTER 7
ACIDS AND BASES
2.9 10 4 100 = 5.8% 5.0 10 3
209
Percent dissociation =
d. As we dilute a solution, all concentrations are decreased. Dilution will shift the equilibrium to the side with the greater number of particles. For example, suppose we double the volume of an equilibrium mixture of a weak acid by adding water; then:
[H + ]eq [X ]eq 2 2 Q= [HX]eq 2
=1K a 2
Q < Ka, so the equilibrium shifts to the right or toward a greater percent dissociation. e. [H+] depends on the initial concentration of weak acid and on how much weak acid dissociates. For solutions a-c, the initial concentration of acid decreases more rapidly than the percent dissociation increases. Thus [H+] decreases. 53. pH = 2.77, [H+] = 102.77 = 1.7 103 M HOCN Initial 0.0100 Equil. 0.0100 x
H+ ~0 x
+
OCN 0 x
x = [H+] = [OCN] = 1.7 103 M; [HOCN] = 0.0100 x = 0.0100 0.0017 = 0.0083 M
(1.7 10 3 ) 2 [H + ][OCN ] = = 3.5 104 3 [HOCN] 8.3 10
Ka = 54.
HClO4 is a strong acid with [H+] = 0.040 M. This equals the [H+] in the trichloroacetic acid solution. Set up the problem using the Ka equilibrium reaction for CCl3CO2H. CCl3CO2H Initial Equil. 0.050 M 0.050 x
H+ ~0 x
+
CCl3CO2 0 x
Ka =
[H + ][CCl 3CO 2 ] x2 = ; from the problem, x = [H+] = 4.0 102 M. [CCl 3CO 2 H ] 0.050 x (4.0 10 2 ) 2 = 0.16 0.050 ( 4.0 10 2 )
Ka = 55.
Major species: HCOOH and H2O; major source of H+: HCOOH
210
HCOOH Initial Change Equil.
CHAPTER 7
ACIDS AND BASES
H+
+ HCOO
C ~0 0 where C = [HCOOH]0 x mol/L HCOOH dissociates to reach equilibrium +x +x x Cx x x
Ka = 1.8 104 = 1.8 104 = 1.8 104 =
x2 [H + ][HCOO ] = , where x = [H+] Cx [HCOOH]
[ H + ]2 ; because pH = 2.70: [H+] = 102.70 = 2.0 103 M C [H + ] ( 2.0 10 3 ) 2 4.0 10 6 , C (2.0 103) = , C = 2.4 102 M C (2.0 10 3 ) 1.8 10 4
A 0.024 M formic acid solution will have pH = 2.70. 56. Let HSac = saccharin and I = [HSac]0. HSac Initial Equil. I I!x
H+ ~0 x
+
Sac 0 x
Ka = 1011.70 = 2.0 10 12
Ka = 2.0 10 12 = 2.0 10 12 =
x2 ; x = [H+] = 10 5.75 = 1.8 10 6 M Ix
(1.8 10 6 ) 2 , I = 1.6 M = [HSac]0. I (1.8 10 6 ) 1 mol 1L 1000 mL = 340 mL 183.19 g 1.6 mol L
100.0 g HC7H4NSO3
Solutions of Bases
57. NO3: Kb << Kw because HNO3 is a strong acid. All conjugate bases of strong acids have no base strength in water. H2O: Kb = Kw = 1.0 1014; NH3: Kb = 1.8 105; C5H5N: Kb =1.7 109 Base strength = NH3 > C5H5N > H2O > NO3 (As Kb increases, base strength increases.) 58. Excluding water, these are the conjugate acids of the bases in the preceding exercise. In general, the stronger the base, the weaker is the conjugate acid. Note: Even though NH4+ and C5H5NH+ are conjugate acids of weak bases, they are still weak acids with Ka values between Kw and 1. Prove this to yourself by calculating the Ka values for NH4+ and C5H5NH+ (Ka = Kw/Kb). Acid strength = HNO3 > C5H5NH+ > NH4+ > H2O
CHAPTER 7
59.
ACIDS AND BASES
b. C6H5NH2 c. OH d. CH3NH2
211
a. C6H5NH2
The base with the largest Kb value is the strongest base (K b , C6 H 5 NH 2 = 3.8 10 10 , K b , CH 3 NH 2 = 4.4 10 4 ). OH is the strongest base possible in water. 60. a. HClO4 (a strong acid) b. C6H5NH3+ c. C6H5NH3+
The acid with the largest Ka value is the strongest acid. To calculate Ka values for C6H5NH3+ and CH3NH3+, use Ka = Kw/Kb , where Kb refers to the bases C6H5NH2 or CH3NH2. 61. NaOH(aq) Na+(aq) + OH(aq); NaOH is a strong base that completely dissociates into Na+ and OH. The initial concentration of NaOH will equal the concentration of OH donated by NaOH. a. [OH] = 0.10 M; pOH = log[OH] = log(0.10) = 1.00 pH = 14.00 pOH = 14.00 1.00 = 13.00 Note that H2O is also present, but the amount of OH produced by H2O will be insignificant compared to the 0.10 M OH produced from the NaOH. b. The [OH] concentration donated by the NaOH is 1.0 1010 M. Water by itself donates 1.0 107 M. In this exercise, water is the major OH contributor, and [OH] = 1.0 107 M. pOH = log(1.0 107) = 7.00; pH = 14.00 7.00 = 7.00 c. [OH] = 2.0 M; pOH = log(2.0) = 0.30; pH = 14.00 (0.30) = 14.30 62. a. Ca(OH)2 Ca2+ + 2 OH; Ca(OH)2 is a strong base and dissociates completely. [OH] = 2(0.00040) = 8.0 104 M; pOH = log[OH] = 3.10; pH = 14.00 pOH = 10.90 b. 25 g KOH 1 mol KOH = 0.45 mol KOH/L L 56.1 g KOH KOH is a strong base, so [OH] = 0.45 M; pOH = log(0.45) = 0.35; pH = 13.65 c. 150.0 g NaOH 1 mol = 3.750 M; NaOH is a strong base, so [OH] = 3.750 M. L 40.00 g pOH = log(3.750) = 0.5740 and pH = 14.0000 (0.5740) = 14.5740 Although we are justified in calculating the answer to four decimal places, in reality the pH can only be measured to 0.01 pH units. 63. pH = 10.50; pOH = 14.00 10.50 = 3.50; [OH] = 103.50 = 3.2 104 M
212
CHAPTER 7
ACIDS AND BASES
Ba(OH)2(aq) Ba2+(aq) + 2 OH(aq); Ba(OH)2 donates 2 mol OH per mol Ba(OH)2. [Ba(OH)2] = 3.2 104 M OH
1 M Ba (OH) 2 = 1.6 104 M Ba(OH)2 2 M OH
A 1.6 104 M Ba(OH)2 solution will produce a pH = 10.50 solution. 64. pOH = 14.00 11.56 = 2.44; [OH] = [KOH] = 10 2.44 = 3.6 10 3 M 0.8000 L 65. 3.6 10 3 mol KOH 56.1 g KOH = 0.16 g KOH L mol KOH
Neutrally charged organic compounds containing at least one nitrogen atom generally behave as weak bases. The nitrogen atom has an unshared pair of electrons around it. This lone pair of electrons is used to form a bond to H+. These are solutions of weak bases in water. In each case, we must solve an equilibrium weak base problem. a. Initial Change Equil. (C2H5)3N + H2O
66.
(C2H5)3NH+ + OH
Kb = 4.0 10 4
0.20 M 0 ~0 x mol/L of (C2H5)3N reacts with H2O to reach equilibrium !x +x +x x x 0.20 ! x [(C 2 H 5 ) 3 NH + ][OH ] x2 x2 = , x = [OH] = 8.9 10 3 M [(C 2 H 5 ) 3 N ] 0.20 x 0.20
Kb = 4.0 10 4 =
Assumptions good (x is 4.5% of 0.20). [OH] = 8.9 10 3 M [H+] = b. Initial Equil.
Kw 1.0 10 14 = = 1.1 10 12 M; pH = 11.96 3 [OH ] 8.9 10
HONH2 + H2O 0.20 M 0.20 ! x
HONH3+ 0 x
+
OH ~0 x
Kb = 1.1 10 8
Kb = 1.1 10 8 =
x2 x2 , x = [OH] = 4.7 10 5 M; assumptions good. 0.20 x 0.20
[H+] = 2.1 10 10 M; pH = 9.68 67. This is a solution of a weak base in water. We must solve a weak base equilibrium problem.
CHAPTER 7
ACIDS AND BASES
C2H5NH2 + H2O
213
C2H5NH3+ +
OH
Kb = 5.6 104
Initial Change Equil. Kb =
0.20 M 0 ~0 x mol/L C2H5NH2 reacts with H2O to reach equilibrium !x +x +x 0.20 ! x x x
+
[C 2 H 5 NH 3 ][OH ] x2 x2 = (assuming x << 0.20) [C 2 H 5 NH 2 ] 0.20 x 0.20
1.1 10 2 100 = 5.5% 0.20 The assumption fails the 5% rule. We must solve exactly using either the quadratic equation or the method of successive approximations (see Appendix 1 of the text). Using successive approximations and carrying extra significant figures:
x = 1.1 10 2 ; checking assumption:
x2 x2 = = 5.6 104, x = 1.0 10 2 M (consistent answer) 0.20 0.011 0.189 x = [OH] = 1.0 10 2 M; [H+] =
68. Initial Change Equil. (C2H5)2NH + H2O Kw 1.0 10 14 = 1.0 10 12 M; pH = 12.00 = 2 [OH ] 1.0 10
(C2H5)2NH2+
+
OH
Kb = 1.3 10 3
0.050 M 0 ~0 x mol/L (C2H5)2NH reacts with H2O to reach equilibrium !x +x +x x x 0.050 ! x [(C 2 H 5 ) 2 NH 2 ][OH ] x2 x2 = [(C 2 H 5 ) 2 NH] 0.050 x 0.050
+
Kb = 1.3 10 3 =
x = 8.1 10 3 ; assumption is bad (x is 16% of 0.20).
Using successive approximations: x2 1.3 10 3 = , x = 7.4 10 3 0.050 0.081 1.3 10 3 =
x2 , x = 7.4 10 3 (consistent answer) 0.050 0.074
[OH] = x = 7.4 10 3 M; [H+] = Kw/[OH] = 1.4 10 12 M; pH = 11.85 69. 5.0 10 3 g 1 mol = 1.7 103 M = [codeine]0; let cod = codeine (C18H21NO3). 0.0100 L 299.4 g
214
Solving the weak base equilibrium problem: cod + H2O Initial Change Equil.
CHAPTER 7
ACIDS AND BASES
codH+
+
OH
Kb = 106.05 = 8.9 107
0 ~0 1.7 103 M x mol/L codeine reacts with H2O to reach equilibrium x +x +x x x 1.7 103 x
Kb = 8.9 107 =
x2 x2 , x = 3.9 105 ; assumptions good. (1.7 10 3 x) 1.7 10 3
[OH] = 3.9 105 M; [H+] = Kw/[OH] = 2.6 1010 M; pH = log[H+] = 9.59 70. Codeine = C18H21NO3; codeine sulfate = C36H44N2O10S The formula for codeine sulfate works out to (codeine H+)2SO42, where codeine H+ = HC18H21NO3+. Two codeine molecules are protonated by H2SO4, forming the conjugate acid of codeine. The SO42 then acts as the counter ion to give a neutral compound. Codeine sulfate is an ionic compound that is more soluble in water than codeine, allowing more of the drug into the bloodstream. 71. To solve for percent ionization, just solve the weak base equilibrium problem. a. NH3 + H2O Initial 0.10 M Equil. 0.10 x Kb = 1.8 105 =
NH4+ 0 x
+
OH ~0 x
Kb = 1.8 105
x2 x2 , x = [OH] = 1.3 103 M; assumptions good. 0.10 x 0.10
[OH ] 1.3 10 3 M 100 = 100 = 1.3% 0.10 M [ NH 3 ]0
Percent ionization = b. Initial Equil. 1.8 105 =
NH3 + H2O 0.010 M 0.010 x
NH4+ 0 x
+
OH ~0 x
x2 x2 , x = [OH] = 4.2 104 M; assumptions good. 0.010 x 0.010
4.2 10 4 100 = 4.2% 0.010
Percent ionization =
Note: For the same base, the percent ionization increases as the initial concentration of base decreases.
CHAPTER 7
c.
ACIDS AND BASES
CH3NH2 + H2O
215
CH3NH3+ 0 x
+
OH ~0 x
Kb = 4.38 104
Initial Equil.
0.10 M 0.10 x
x2 x2 , x = 6.6 103; assumption fails the 5% rule (x is 0.10 x 0.10 6.6% of 0.10). Using successive approximations and carrying extra significant figures:
4.38 104 =
x2 x2 = = 4.38 104, x = 6.4 103 0.10 0.0066 0.093
Percent ionization = 6.4 10 3 100 = 6.4% 0.10
(consistent answer)
72.
1.0 g quinine 1 mol quinine = 1.6 103 M quinine; let Q = quinine = C20H24N2O2. 1.9000 L 324.4 g quinine Q Initial Change Equil. + H2O
QH+ +
OH
Kb = 105.1 = 8 106
1.6 103 M 0 ~0 x mol/L quinine reacts with H2O to reach equilibrium +x +x x 3 1.6 10 x x x [QH + ][OH ] x2 x2 = [Q ] (1.6 10 3 x) 1.6 10 3
Kb = 8 106 =
x = 1 104; assumption fails 5% rule (x is 6% of 0.0016). Using successive approximations: x2 = 8 106, x = 1 104 M (consistent answer) 3 4 (1.6 10 1 10 ) x = [OH] = 1 104 M; pOH = 4.0; pH = 10.0
73. Using the Kb reaction to solve where PT = p-toluidine (CH3C6H4NH2): PT Initial Change Equil. + H2O
PTH+
+
OH
0.016 M 0 ~0 x mol/L of PT reacts with H2O to reach equilibrium x +x +x 0.016 x x x
216
Kb =
CHAPTER 7
x2 [PTH + ][OH ] = 0.016 x [PT ]
ACIDS AND BASES
Since pH = 8.60: pOH = 14.00 8.60 = 5.40 and [OH] = x = 105.40 = 4.0 106 M Kb = 74. Initial Equil. (4.0 10 6 ) 2 = 1.0 109 0.016 ( 4.0 10 6 ) HONH2 + H2O I Ix
HONH3+ + OH 0 x ~0 x
Kb = 1.1 10 8 I = [HONH2]0
Kb = 1.1 10 8 =
x2 Ix
From problem, pH = 10.00, so pOH = 4.00 and x = [OH] = 1.0 10 4 M. 1.1 10 8 = (1.0 10 4 ) 2 , I = 0.91 M I (1.0 10 4 ) 0.91 mol HONH 2 33.03 g HONH 2 = 7.5 g HONH2 L mol HONH 2
Mass HONH2 = 0.2500 L
Polyprotic Acids
75. H3C6H5O7(aq) H2C6H5O7 (aq) + H (aq)
+
[H 2 C 6 H 5 O 7 ][H + ] K a1 = [ H 3C 6 H 5 O 7 ]
Ka2 = K a3 = [HC 6 H 5 O 7 ][H + ] [H 2 C 6 H 5O 7 ] [C 6 H 5 O 7 ][H + ] [HC 6 H 5 O 7 ]
2 3 2
H2C6H5O7(aq) HC6H5O72(aq) + H+(aq) HC6H5O72(aq) C6H5O73(aq) + H+(aq) 76.
H2CO3 is a weak acid with K a1 = 4.3 10 7 and K a 2 = 4.8 10 11 . The [H+] concentration in solution will be determined from the K a1 reaction because K a1 >> K a 2 . Because K a1 << 1, the [H+] < 0.10 M; only a small percentage of the 0.10 M H2CO3 will dissociate into HCO3 and H+. So statement a best describes the 0.10 M H2CO3 solution. H2SO4 is a strong acid as well as a very good weak acid ( K a1 >> 1, K a 2 = 1.2 10 2 ). All of the 0.10 M H2SO4 solution will dissociate into 0.10 M H+ and 0.10 M HSO4. However, because HSO4 is a good weak acid due to the relatively large Ka value, some of the 0.10 M HSO4 will dissociate into some more H+ and SO42. Therefore, the [H+] will be greater than 0.10 M but will not reach 0.20 M because only some of 0.10 M HSO4 will dissociate. Statement c is best for a 0.10 M H2SO4 solution.
CHAPTER 7
77.
ACIDS AND BASES
217
The reactions are: H3AsO4 H+ + H2AsO4 K a1 = 5 103
H2AsO4 H+ + HAsO42 K a 2 = 8 108 HAsO42 H+ + AsO43 K a 3 = 6 1010
We will deal with the reactions in order of importance, beginning with the largest Ka, K a1 . H3AsO4 Initial Equil. 5 103 = 0.20 M 0.20 - x
H+ ~0 x
+
H2AsO4 0 x
K a1 = 5 103 =
[H + ][H 2 AsO 4 ] [H 3 AsO 4 ]
x2 x2 , x = 3 102 M; assumption fails the 5% rule. x 0.20 0.20
Solving by the method of successive approximations: 5 103 = x2/(0.20 0.03), x = 3 102 (consistent answer) [H+] = [H2AsO4] = 3 102 M; [H3AsO4] = 0.20 - 0.03 = 0.17 M Because K a 2 =
[H + ][HAsO 4 ] [H 2 AsO 4 ]
2
= 8 108 is much smaller than the K a1 value, very little of
H2AsO4 (and HAsO42) dissociates compared to H3AsO4. Therefore, [H+] and [H2AsO4] will not change significantly by the K a 2 reaction. Using the previously calculated concentrations of H+ and H2AsO4 to calculate the concentration of HAsO42: (3 10 2 )[HAsO 4 ] , [HAsO42] = 8 108 M 8 10 = 2 3 10
8
2
The assumption that the K a 2 reaction does not change [H+] and [H2AsO4] is good. We repeat the process using K a 3 to get [AsO43]. K a 3 = 6 1010 = [H + ][AsO 4 ] [HAsO 4 ]
2 3
=
(3 10 2 )[AsO 4 ] 8 10 8
3
[AsO43] = 1.6 1015 2 1015 M; assumption good.
218
So in 0.20 M analytical concentration of H3AsO4:
CHAPTER 7
ACIDS AND BASES
[H3AsO4] = 0.17 M; [H+] = [H2AsO4] = 3 102 M; [HAsO42] = 8 108 M; [AsO43] = 2 1015 M; [OH] = Kw/[H+] = 3 1013 M 78. The relevant reactions are: H2CO3 H+ + HCO3
K a1 = 4.3 107; HCO3 H+ + CO32
K a 2 = 4.8 1011
Initially, we deal only with the first reaction (since K a1 >> K a 2 ), and then let those results control values of concentrations in the second reaction. H2CO3 Initial Equil. 0.010 M 0.010 x
H+ ~0 x
+ HCO3 0 x
K a1 = 4.3 107 =
[H + ][HCO 3 ] x2 x2 = [H 2 CO 3 ] 0.010 x 0.010
x = 6.6 105 M = [H+] = [HCO3]; assumptions good.
HCO3 Initial Equil. 6.6 105 M 6.6 105 y
H+ 6.6 105 M 6.6 105 + y
+
CO32 0 y
[H + ][CO 3 ] [HCO 3 ]
2
If y is small, then [H+] = [HCO3] and K a 2 = 4.8 1011 =
y.
y = [CO32] = 4.8 1011 M; assumptions good.
The amount of H+ from the second dissociation is 4.8 1011 M or: 4.8 10 11 100 = 7.3 105 % 5 6.6 10 This result justifies our treating the equilibria separately. If the second dissociation contributed a significant amount of H+, we would have to treat both equilibria simultaneously. The reaction that occurs when acid is added to a solution of HCO3 is: HCO3(aq) + H+(aq) H2CO3(aq) H2O(l) + CO2(g) The bubbles are CO2(g) and are formed by the breakdown of unstable H2CO3 molecules. We should write H2O(l) + CO2(aq) or CO2(aq) for what we call carbonic acid. It is for convenience, however, that we write H2CO3(aq).
CHAPTER 7
79.
ACIDS AND BASES
219
For H2C6H6O6. K a1 = 7.9 10 5 and K a 2 = 1.6 10 12 . Because K a1 >> K a 2 , the amount of H+ produced by the K a 2 reaction will be negligible. 0.500 g [H2C6H6O6]0 = 1 mol H 2 C 6 H 6 O 6 176.12 g = 0.0142 M 0.2000 L
H2C6H6O6(aq) Initial Equil. 0.0142 M 0.0142 ! x
HC6H6O6(aq) 0 x
+ H+(aq) ~0 x
K a1 = 7.9 10 5
K a1 = 7.9 10 5 =
x2 x2 , x = 1.1 10 3 ; assumption fails the 5% rule. 0.0142 0.0142 x
Solving by the method of successive approximations: 7.9 10 5 =
x2 , x = 1.0 10 3 M (consistent answer) 0.0142 1.1 10 3
Because H+ produced by the K a 2 reaction will be negligible, [H+] = 1.0 10 3 and pH = 3.00. 80. Because K a 2 for H2S is so small, we can ignore the H+ contribution from the K a 2 reaction. H2S Initial Equil. 0.10 M 0.10 ! x
H+ ~0 x
HS 0 x
K a1 = 1.0 10 7
K a1 = 1.0 10 7 =
x2 x2 , x = [H+] = 1.0 10 4 ; assumptions good. 0.10 x 0.10
pH = !log(1.0 10 4 ) = 4.00 Use the K a 2 reaction to determine [S2]. Initial Equil. HS 1.0 10 4 M 1.0 10 4 ! x
H+ + 4 1.0 10 M 1.0 10 4 + x
S2 0 x
K a 2 = 1.0 10 19 =
(1.0 10 4 + x) x (1.0 10 4 )x (1.0 10 4 x) 1.0 10 4
x = [S2] = 1.0 10 19 M; assumptions good.
220
81.
CHAPTER 7
ACIDS AND BASES
The dominant H+ producer is the strong acid H2SO4. A 2.0 M H2SO4 solution produces 2.0 M HSO4- and 2.0 M H+. However, HSO4 is a weak acid that could also add H+ to the solution. HSO4 Initial Change Equil.
H+
+
SO42
2.0 M 2.0 M 0 x mol/L HSO4 dissociates to reach equilibrium x +x +x 2.0 x 2.0 + x x
[H + ][SO 4 ] [HSO 4 ]
2
K a 2 = 1.2 102 =
=
(2.0 + x) x 2.0( x) , x = 1.2 102 M 2.0 x 2 .0
Because x is 0.60% of 2.0, the assumption is valid by the 5% rule. The amount of additional H+ from HSO4 is 1.2 102 M. The total amount of H+ present is: [H+] = 2.0 + (1.2 102) = 2.0 M; pH = log(2.0) = -0.30
Note: In this problem H+ from HSO4 could have been ignored. However, this is not usually the case in more dilute solutions of H2SO4.
82.
For H2SO4, the first dissociation occurs to completion. The hydrogen sulfate ion, HSO4, is a weak acid with K a 2 = 1.2 102. We will consider this equilibrium for additional H+ production: HSO4 Initial Change Equil.
H+
+
SO42
0.0050 M 0.0050 M 0 x mol/L HSO4 dissociates to reach equilibrium x +x +x 0.0050 x 0.0050 + x x (0.0050 + x) x x, x = 0.012; assumption is horrible (240% error). 0.0050 x
K a 2 = 0.012 =
Using the quadratic formula: 6.0 105 (0.012)x = x2 + (0.0050)x, x2 + (0.017)x 6.0 105 = 0
x=
0.017 (2.9 10 4 + 2.4 10 4 )1/ 2 0.017 0.023 = , x = 3.0 103 M 2 2
[H+] = 0.0050 + x = 0.0050 + 0.0030 = 0.0080 M; pH = 2.10
Note: We had to consider both H2SO4 and HSO4 for H+ production in this problem.
CHAPTER 7
ACIDS AND BASES
221
Acid-Base Properties of Salts
83. a. These are strong acids like HCl, HBr, HI, HNO3, H2SO4, or HClO4. b. These are salts of the conjugate acids of the bases in Table 7.3. These conjugate acids are all weak acids. NH4Cl, CH3NH3NO3, and C2H5NH3Br are three examples. Note that the anions used to form these salts are conjugate bases of strong acids; this is so because they have no acidic or basic properties in water (with the exception of HSO4, which has weak acid properties). c. These are strong ases like LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. d. These are salts of the conjugate bases of the neutrally charged weak acids in Table 7.2. The conjugate bases of weak acids are weak bases themselves. Three examples are NaClO2, KC2H3O2, and CaF2. The cations used to form these salts are Li+, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, or Ba2+ because these cations have no acidic or basic properties in water. Notice that these are the cations of the strong bases you should memorize. e. There are two ways to make a neutral salt. The easiest way is to combine a conjugate base of a strong acid (except for HSO4) with one of the cations from a strong base. These ions have no acidic/basic properties in water, so salts of these ions are neutral. Three examples are NaCl, KNO3, and SrI2. Another type of strong electrolyte that can produce neutral solutions are salts that contain an ion with weak acid properties combined with an ion of opposite charge having weak base properties. If the Ka for the weak acid ion is equal to the Kb for the weak base ion, then the salt will produce a neutral solution. The most common example of this type of salt is ammonium acetate (NH4C2H3O2). For this salt, Ka for NH4+ = Kb for C2H3O2 = 5.6 10 10 . This salt at any concentration produces a neutral solution. 84. Ka Kb = Kw, log(Ka Kb) = log Kw
log Ka log Kb = log Kw, pKa + pKb = pKw = 14.00 (Kw = 1.0 1014 at 25C)
85. One difficult aspect of acid-base chemistry is recognizing what types of species are present in solution, that is, whether a species is a strong acid, strong base, weak acid, weak base, or a neutral species. Below are some ideas and generalizations to keep in mind that will help in recognizing types of species present. a. b. c. Memorize the following strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4 Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2 Weak acids have a Ka value of less than 1 but greater than Kw. Some weak acids are listed in Table 7.2 of the text. Weak bases have a Kb value of less than 1 but greater than Kw. Some weak bases are listed in Table 7.3 of the text.
222
d.
CHAPTER 7
ACIDS AND BASES
Conjugate bases of weak acids are weak bases, that is, all have a Kb value of less than 1 but greater than Kw. Some examples of these are the conjugate bases of the weak acids listed in Table 7.2 of the text. Conjugate acids of weak bases are weak acids, that is, all have a Ka value of less than 1 but greater than Kw. Some examples of these are the conjugate acids of the weak bases listed in Table 7.3 of the text. Alkali metal ions (Li+, K+, Na+, Rb+, Cs+) and some alkaline earth metal ions (Ca2+, Sr2+, Ba2+) have no acidic or basic properties in water. Conjugate bases of strong acids (Cl, Br-, I, NO3, ClO4, HSO4) have no basic properties in water (Kb << Kw), and only HSO4- has any acidic properties in water. Strong acid; HF: weak acid (Ka = 7.2 104) F is the conjugate base of the weak acid HF, so F is a weak base. The Kb value for F = Kw/Ka, HF = 1.4 1011. Na+ has no acidic or basic properties. Neutral (pH = 7.0); Na+ and I have no acidic/basic properties. In order of increasing pH, we place the compounds from most acidic (lowest pH) to most basic (highest pH). Increasing pH: HI < HF < NaI < NaF.
e.
f. g.
Lets apply these ideas to this problem to see what types of species are present. a. HI: NaF: NaI:
b. NH4Br: HBr: KBr: NH3:
NH4+ is a weak acid (Ka = 5.6 1010), and Br- is a neutral species. Strong acid Neutral; K+ and Br- have no acidic/basic properties. Weak base, Kb = 1.8 105
Increasing pH: HBr < NH4Br < KBr < NH3 Most Most acidic basic c. C6H5NH3NO3: NaNO3: NaOH: HOC6H5: KOC6H5: C6H5NH3+ is a weak acid (K a = K w /K b, C6 H 5 NH 2 = 1.0 1014/3.8 1010 = 2.6 105), and NO3 is a neutral species. Neutral; Na+ and NO3 have no acidic/basic properties. Strong base Weak acid (Ka = 1.6 1010) OC6H5 is a weak base (K b = K w /K a, HOC6 H 5 = 6.3 105), and K+ is a neutral species. C6H5NH2: HNO3: Weak base (Kb = 3.8 1010) Strong acid
CHAPTER 7
ACIDS AND BASES
223
This is a little more difficult than the previous parts of this problem because two weak acids and two weak bases are present. Between the weak acids, C6H5NH3+ is a stronger weak acid than HOC6H5 since the Ka value for C6H5NH3+ is larger than the Ka value for HOC6H5. Between the two weak bases, because the Kb value for OC6H5- is larger than the Kb value for C6H5NH2, OC6H5 is a stronger weak base than C6H5NH2. Increasing pH: HNO3 < C6H5NH3NO3 < HOC6H5 < NaNO3 < C6H5NH2 < KOC6H5 < NaOH Most acidic Most basic 86. Reference Table 7.6 of the text and the solution to Exercise 85 for some generalizations on acid-base properties of salts. The letters in parenthesis is(are) the generalization(s) listed in Exercise 85 that identifies that species. CaBr2: KNO2: HClO4: HNO2: HONH3ClO4: Neutral; Ca2+ and Br- have no acidic/basic properties (f and g). NO2 is a weak base, Kb = (1.0 1014)/(4.0 104) = 2.5 1011 (c and d). Ignore K+ (f). Strong acid (a) Weak acid, Ka = 4.0 104 (c) HONH3+ is a weak acid, Ka = K w /K b, HONH 2 = (1.0 1014)/(1.1 108 ) = 9.1 107 (c and e). Ignore ClO4 (g). NH4NO2: NH4+ is a weak acid, Ka = 5.6 1010 (c and e). NO2 is a weak base, Kb = 2.5 1011 (c and d). Because the Ka value for NH4+ is a slightly larger than Kb for NO2, the solution will be slightly acidic with a pH a little lower than 7.0.
Using the information above (identity and the Ka or Kb values), the ordering is: Most acidic most basic: HClO4 > HNO2 > HONH3ClO4 > NH4NO2 > CaBr2 > KNO2 87. Reference Table 7.6 of the text and the solution to Exercise 85 for some generalizations on acid-base properties of salts. a. KCl K+ + Cl neutral; K+ and Cl have no effect on pH. b. C2H5NH3CN C2H5NH3+ + CN basic; C2H5NH3+ is a weak acid (K a = K w /K b,C 2 H 5 NH 2 = 1.0 1014/5.6 104 = 1.8 1011), and CN is a weak base (Kb = Kw/Ka, HCN = 1.0 1014/6.2 1010 = 1.6 105). Because K b , CN > K a , C 2 H 5 NH3+ ,
the solution of C2H5NH3CN will be basic. c. C5H5NHF C5H5NH+ + F acidic; C5H5NH+ is a weak acid (K a = K w /K b,C5 H 5 N = 5.9 10 6 ) , and F is a weak base (Kb = Kw/Ka, HF = 1.4 1011). Because K a , C5 H5 NH + > K b, F , the solution of C5H5NHF will be acidic.
224
CHAPTER 7
ACIDS AND BASES
d. NH4C2H3O2 NH4+ + C2H3O2 neutral; NH4+ is a weak acid (Ka = 5.6 1010), and C2H3O2 is a weak base (K b = K w /K a, HC 2 H 3O 2 = 5.6 10 10 ). Because K a , NH + =
4
K b, C
2 H 3O 2
, the solution of NH4C2H3O2 will have pH = 7.00.
e. NaHSO3 Na+ + HSO3 acidic; Na+ has no acidic/basic properties. HSO3 is a weak acid (K a 2 = 1.0 107), and HSO3 is also a weak base (K b = K w /K a1 , H 2SO 3 = 1.0 1014/1.5 102 = 6.7 1013). HSO3 is a stronger acid than a base because Ka > Kb. Therefore, the solution is acidic. f. NaHCO3 Na+ + HCO3 basic; ignore Na+; HCO3 is a weak acid (K a 2 = 4.8 1011), and HCO3 is a weak base (Kb = K w /K a1 , H 2CO3 = 2.3 108) . HCO3 is a stronger base than an acid because Kb > Ka. Therefore, the solution is basic. 88. a. CH3NH3Cl CH3NH3+ + Cl: CH3NH3+ is a weak acid. Cl is the conjugate base of a strong acid. Cl has no basic (or acidic) properties. CH3NH3+ CH3NH2 + H+ Ka = [CH 3 NH 2 ][H + ] [CH 3 NH 3 ] + H+
+
=
Kw 1.00 10 14 = Kb 4.38 10 4
= 2.28 10 11 CH3NH3+ Initial
CH3NH2
0 ~0 0.10 M x mol/L CH3NH3+ dissociates to reach equilibrium Change !x +x +x Equil. 0.10 ! x x x 2 2 x x (assuming x << 0.10) Ka = 2.28 10 11 = 0.10 x 0.10
x = [H+] = 1.5 10 6 M; pH = 5.82; assumptions good.
b. NaCN Na+ + CN: CN is a weak base. Na+ has no acidic (or basic) properties. CN + H2O Initial Change Equil.
HCN
+
OH
Kb =
Kw 1.0 10 14 = Ka 6.2 10 10
0.050 M 0 ~0 Kb = 1.6 10 5 x mol/L CN reacts with H2O to reach equilibrium !x +x +x 0.050 ! x x x
Kb = 1.6 10 5 =
x2 [HCN][OH ] x2 = 0.050 x 0.050 [CN ]
x = [OH] = 8.9 10 4 M; pOH = 3.05; pH = 10.95; assumptions good.
CHAPTER 7
89.
ACIDS AND BASES
225
a. KNO2 K+ + NO2: NO2 is a weak base. Ignore K+. NO2 + H2O Initial 0.12 M Equil. 0.12 ! x Kb = 2.5 10 11 =
HNO2 + OH 0 x ~0 x =
Kb =
Kw 1.0 10 14 = 2.5 10 11 = 4 Ka 4.0 10
[OH ][HNO 2 ] [ NO 2 ]
x2 x2 0.12 x 0.12
x = [OH] = 1.7 10 6 M; pOH = 5.77; pH = 8.23; assumptions good.
b. NaOCl Na+ + OCl: OCl is a weak base. Ignore Na+. OCl + H2O Initial 0.45 M Equil. 0.45 ! x Kb = 2.9 10 7 =
HOCl + OH 0 x ~0 x
Kb =
Kw 1.0 10 14 = = 2.9 10 7 Ka 3.5 10 8
x2 x2 [HOCl][OH ] = 0.45 x 0.45 [OCl ]
x = [OH] = 3.6 10 4 M; pOH = 3.44; pH = 10.56; assumptions good.
c. NH4ClO4 NH4+ + ClO4: NH4+ is a weak acid. ClO4 is the conjugate base of a strong acid. ClO4 has no basic (or acidic) properties. 14 K NH4+ NH3 + H+ Ka = w = 1.0 10 5 = 5.6 10 10 Kb 1.8 10 Initial 0.40 M Equil. 0.40 ! x Ka = 5.6 10 10 = 0 x [ NH 3 ][H + ] [ NH 4 ]
+
~0 x =
x2 x2 0.40 x 0.40
x = [H+] = 1.5 10 5 M; pH = 4.82; assumptions good.
90. Solution is acidic from HSO4 H+ + SO42. Solving the weak acid problem: HSO4 Initial Equil. 0.10 M 0.10 - x
H+ ~0 x
+
SO42 0 x
Ka = 1.2 102
226
Ka = 1.2 102 = [H + ][SO 4 ] [HSO 4 ]
2
CHAPTER 7
=
ACIDS AND BASES
x2 x2 , x = 0.035 0.10 x 0.10
Assumption is not good (35% error). Using successive approximations:
x2 x2 = 1.2 102, x = 0.028 0.10 x 0.10 0.035 x2 = 1.2 102, x = 0.029 M (consistent answer) 0.10 0.028 x = [H+] = 0.029 M; pH = 1.54
If we add Na2CO3 to a solution of NaHSO4, the base CO32 will react with the acid HSO4. Depending on relative amounts, two reactions are possible. CO32(aq) + HSO4(aq) or
HCO3(aq) + SO42(aq)
CO32 (aq) + 2 HSO4(aq) 2 SO42(aq) + H2O(l) + CO2(g) 91. NaN3 Na+ + N3; Azide (N3) is a weak base since it is the conjugate base of a weak acid. All conjugate bases of weak acids are weak bases (Kw < Kb < 1). Ignore Na+. N3 + H2O Initial Change Equil. Kb =
HN3 +
OH
Kb =
Kw 1.0 10 14 = = 5.3 1010 5 Ka 1.9 10
0.010 M 0 ~0 x mol/L of N3 reacts with H2O to reach equilibrium x +x +x 0.010 x x x , 5.3 1010 =
[HN 3 ][OH ] [N3 ]
x2 x2 (assuming x << 0.010) 0.010 x 0.010
x = [OH] = 2.3 106 M; [H+] =
1.0 10 14 = 4.3 109 M; assumptions good. 6 2.3 10
[HN3] = [OH] = 2.3 106 M; [Na+] = 0.010 M; [N3] = 0.010 2.3 106 = 0.010 M 92. C2H5NH3Cl C2H5NH3+ + Cl; C2H5NH3+ is the conjugate acid of the weak base C2H5NH2 (Kb = 5.6 104). As is true for all conjugate acids of weak bases, C2H5NH3+ is a weak acid. Cl has no basic (or acidic) properties. Ignore Cl. Solving the weak acid problem:
CHAPTER 7
ACIDS AND BASES
C2H5NH3+
227
+ H+ Ka = Kw/(5.6 104) = 1.8 1011
C2H5NH2
Initial Change Equil.
0 ~0 0.25 M x mol/L C2H5NH3+ dissociates to reach equilibrium x +x +x 0.25 x x x
[C 2 H 5 NH 2 ][H + ] [C 2 H 5 NH 3 ]
+
Ka = 1.8 1011 =
=
x2 x2 (assuming x << 0.25) 0.25 x 0.25
x = [H+] = 2.1 106 M; pH = 5.68; assumptions good.
[C2H5NH2] = [H+] = 2.1 106 M; [C2H5NH3+] = 0.25 M; [Cl] = 0.25 M [OH] = Kw/[H+] = 4.8 109 M 93. All these salts contain Na+, which has no acidic/basic properties and a conjugate base of a weak acid (except for NaCl, where Cl is a neutral species). All conjugate bases of weak acids are weak bases since Kb values for these species are between Kw and 1. To identify the species, we will use the data given to determine the Kb value for the weak conjugate base. From the Kb value and data in Table 7.2 of the text, we can identify the conjugate base present by calculating the Ka value for the weak acid. We will use A- as an abbreviation for the weak conjugate base. A + H2O Initial Change Equil. Kb =
HA
+
OH
0.100 mol/1.00 L 0 ~0 x mol/L A reacts with H2O to reach equilibrium x +x +x x x 0.100 x
x2 [HA ][OH ] = ; from the problem, pH = 8.07: 0.100 x [A ]
pOH = 14.00 8.07 = 5.93; [OH] = x = 105.93 = 1.2 106 M Kb = (1.2 10 6 ) 2 = 1.4 1011 = Kb value for the conjugate base of a weak acid. 6 0.100 (1.2 10 ) 1.0 10 14 = 7.1 104 1.4 10 11
The Ka value for the weak acid equals Kw/Kb: Ka =
From Table 7.2 of the text, this Ka value is closest to HF. Therefore, the unknown salt is NaF.
228
94.
CHAPTER 7
ACIDS AND BASES
BHCl BH+ + Cl; Cl is the conjugate base of the strong acid HCl, so Cl has no acidic/ basic properties. BH+ is a weak acid because it is the conjugate acid of a weak base B. Determining the Ka value for BH+: BH+ Initial Change Equil. Ka =
B
+
H+
0 ~0 0.10 M x mol/L BH+ dissociates to reach equilibrium x +x +x x x 0.10 x
x2 [B][H + ] = ; from the problem, pH = 5.82: 0.10 x [BH + ]
(1.5 10 6 ) 2 = 2.3 1011 6 0.10 (1.5 10 )
[H+] = x = 105.82 = 1.5 106 M; Ka =
Kb for the base B = Kw/Ka = (1.0 1014)/(2.3 1011) = 4.3 104. From Table 7.3 of the text, this Kb value is closest to CH3NH2, so the unknown salt is CH3NH3Cl. 95. Major species: Co(H2O)63+ (Ka = 1.0 105), Cl (neutral), and H2O (Kw = 1.0 1014); Co(H2O)63+ will determine the pH since it is a stronger acid than water. Solving the weak acid problem in the usual manner: Co(H2O)63+ Initial Equil. 0.10 M 0.10 - x
Co(H2O)5(OH)2+ 0 x
+
H+ ~0 x
Ka = 1.0 105
Ka = 1.0 105 =
x2 x2 , x = [H+] = 1.0 103 M 0.10 x 0.10
pH = log(1.0 103) = 3.00; assumptions good. 96. Major species present are H2O, C5H5NH+ [Ka = K w /K b, C5H 5 N = (1.0 1014)/(1.7 109) = 5.9 106], and F [Kb = K w /K a, HF = (1.0 1014)/(7.2 104) = 1.4 1011]. The reaction to consider is the best acid present (C5H5NH+) reacting with the best base present (F). Lets solve by first setting up an ICE table. C5H5NH+(aq) Initial Change Equil. 0.200 M x 0.200 x + F(aq)
C5H5N(aq) + HF(aq) 0 +x x 0 +x x
0.200 M x 0.200 x
CHAPTER 7
ACIDS AND BASES
1 K a , HF = 5.9 106 1 = 8.2 103 7.2 10 4
229
K = K a, C H
5
5 NH
+
K=
[C 5 H 5 N][HF] x2 , 8.2 103 = ; taking the square root of both sides: [C 5 H 5 NH + ][F ] (0.200 x) 2
0.091 =
x , x = 0.018 (0.091)x, x = 0.016 M 0.200 x
From the setup to the problem, x = [C5H5N] = [HF] = 0.016 M, and 0.200 x = 0.200 0.016 = 0.184 M = [C5H5NH+] = [F]. To solve for the [H+], we can use either the Ka equilibrium for C5H5NH+ or the Ka equilibrium for HF. Using C5H5NH+ data:
K a, C H
5
5
= 5.9 106 = NH +
[C 5 H 5 N ][H + ] (0.016)[H + ] = , [H+] = 6.8 105 M + 0.184 [C 5 H 5 NH ]
pH = log(6.8 105) = 4.17 As one would expect, because the Ka for the weak acid is larger than the Kb for the weak base, a solution of this salt should be acidic. 97. Major species: NH4+, OCl, and H2O; Ka for NH4+ = (1.0 1014)/(1.8 105) = 5.6 10 10 and Kb for OCl = (1.0 1014)/(3.5 108) = 2.9 10 7 . Because OCl is a better base than NH4+ is an acid, the solution will be basic. The dominant equilibrium is the best acid (NH4+) reacting with the best base (OCl) present. NH4+ Initial Change Equil. 0.50 M x 0.50 x 1 K a , HOCl + OCl
NH3 0 +x x
+ HOCl 0 +x x
0.50 M x 0.50 x
K = K a , NH +
4
= (5.6 1010)/(3.5 108) = 0.016
x(x) (0.50 x)(0.50 x)
K = 0.016 =
[ NH 3 ][HOCl] [ NH 4 ][OCl ]
+
=
x2 x = 0.016, = (0.016)1/2 = 0.13, x = 0.058 M 2 0.50 x (0.50 x)
To solve for the H+, use any pertinent Ka or Kb value. Using Ka for NH4+:
K a , NH
4
+
= 5.6 10 10 =
[ NH 3 ][H + ] [ NH 4 ]
+
=
(0.058)[H + ] , [H+] = 4.3 10 9 M, pH = 8.37 0.50 0.058
230
98.
CHAPTER 7
ACIDS AND BASES
Major species: Na+, PO43 (a weak base), and H2O; the Kb value for PO43 is much larger than the Kb values for HPO42 and H2PO4. We can ignore the contribution of OH from the Kb reactions for HPO42 and H2PO4 . K b for PO 4
3
=
Kw 1.0 10 14 = = 0.021 K a3 4.8 10 13
Note: Kb for HPO4 = K w /K a 2 = 1.6 10 7 and Kb for H2PO4 = K w /K a1 = 1.3 10 12.
Indeed, Kb for PO43 >> Kb values for HPO4 and and H2PO4. PO43 + H2O Initial Equil. 0.10 M 0.10 ! x
HPO42 + OH 0 x ~0 x
Kb = 0.021
Kb = 0.021 =
x2 ; because Kb is so large, the 5% assumption will not hold. Solving 0.10 x
using the quadratic equation:
x2 + (0.021)x 0.0021 = 0, x = [OH] = 3.7 10 2 , pOH = 1.43, pH = 12.57
Solutions of Dilute Acids and Bases
99. Initial Change Equil. HBrO
H+
+
BrO
Ka = 2 109
~0 0 1.0 106 M x mol/L HBrO dissociates to reach equilibrium x +x +x 6 x x 1.0 10 x
Ka = 2 109 =
x2 x2 , x = [H+] = 4 108 M; pH = 7.4 6 6 (1.0 10 x) 1.0 10
Lets check the assumptions. This answer is impossible! We can't add a small amount of an acid to a neutral solution and get a basic solution. The highest pH possible for an acid in water is 7.0. In the correct solution we would have to take into account the autoionization of water. 100. C6H5OH Initial 4.0 105 M Equil. 4.0 105 x
C6H5O 0 x
+
H+ ~0 x
C6H5OH = phenol
CHAPTER 7
Ka =
ACIDS AND BASES
231
x2 x2 , 1.6 1010 , x = [H+] = 8.0 108 M 5 5 (4.0 10 x) 4.0 10
Check assumptions. The assumption that the H+ contribution from water is negligible is poor. Whenever the calculated pH is greater than 6.0 ([H+] < 1 106 M) for an acid solution, the H+ contribution from water should be considered. From Section 7.9 of the text, try [H+] = (Ka[HA]0 + Kw)1/2 . [H+] = [(1.6 1010)(4.0 105) + (1.0 1014)]1/2 = 1.3 107 M This equation will work if [HA]0 = 4.0 105 >> good. [H+] = 1.3 107 M; pH = 6.89 [ H + ]2 K w = 5.3 108. Assumption [H + ]
Note: If the assumption that ([H+]2 Kw)/[H+] << Ka is bad, then the full equation derived in Section 7.9 of the text should be used.
101. HCN Initial 5.0 104 M Equil. 5.0 104 x Ka =
H+ ~0 x
+
CN 0 x
Ka = 6.2 1010
x2 x2 = 6.2 1010, x = 5.6 107; check assumptions. 4 4 (5.0 10 x) 5.0 10
The assumption that the H+ contribution from water is negligible is poor. Whenever the calculated pH is greater than 6.0 for a weak acid, the water contribution to [H+] must be considered. From Section 7.9 in text: if [ H + ]2 K w << [HCN]0 = 5.0 104, then we can use [H+] = (Ka[HCN]o + Kw)1/2. + [H ]
Using this formula: [H+] = [(6.2 1010)(5.0 104) + (1.0 1014)]1/2, [H+] = 5.7 107 M Checking assumptions: [ H + ]2 K w = 5.5 107 << 5.0 104 + [H ]
Assumptions good. pH = log(5.7 107) = 6.24 102. We can't neglect the [H+] contribution from H2O since this is a very dilute solution of the strong acid. Following the strategy developed in Section 7.10 of the text, we first determine the charge balance equation and then manipulate this equation to get into one unknown. Charge balance: [H+] = [NO3] + [OH], [H+] = [NO3] + Kw/[H+] [H+]2 1.0 1014 = [H+](5.0 108), [H+]2 (5.0 108)[H+] 1.0 1014 = 0
232
CHAPTER 7
Using the quadratic formula: [H+] = 1.3 107 M; pH = 6.89
ACIDS AND BASES
103.
We can't neglect the [H+] contribution from H2O since this is a very dilute solution of the strong acid. Following the strategy developed in Section 7.10 of the text, we first determine the charge balance equation and then manipulate this equation to get into one unknown. [Positive charge] = [negative charge] [H+] = [Cl] + [OH] = 7.0 107 + Kw Kw (because [Cl] = 7.0 107 and [OH] = ) + [H ] [H + ]
[ H + ]2 K w = 7.0 107, [H+]2 (7.0 107)[H+] 1.0 1014 = 0 + [H ] Using the quadratic formula to solve: [H+] = (7.0 10 7 ) [(7.0 10 7 ) 2 4(1)(1.0 10 14 )]1/ 2 2(1)
[H+] = 7.1 107 M; pH = log(7.1 107) = 6.15 104. Because this is a very dilute solution of NaOH, we must worry about the amount of OH donated from the autoionization of water. NaOH Na+ + OH H2O H+ + OH Kw = [H+][OH] = 1.0 1014
This solution, like all solutions, must be charged balanced; that is, [positive charge] = [negative charge]. For this problem, the charge balance equation is: [Na+] + [H+] = [OH], where [Na+] = 1.0 107 M and [H+] = Substituting into the charge balance equation: 1.0 107 + 1.0 10 14 = [OH], [OH]2 (1.0 107)[OH] 1.0 1014 = 0 [OH ]
Kw [OH ]
Using the quadratic formula to solve: [OH] = (1.0 10 7 ) [(1.0 10 7 ) 2 4(1)(1.0 10 14 )]1/ 2 2(1)
[OH] = 1.6 107 M; pOH = log(1.6 107) = 6.80; pH = 7.20
CHAPTER 7
ACIDS AND BASES
233
Additional Exercises
105. a. NH3 + H3O+ NH4+ + H2O Keq =
+ K 1.8 10 5 [ NH 4 ] 1 = = b= = 1.8 109 + + 14 Kw 1.0 10 [ NH 3 ][H ] K a for NH 4
b. NO2 + H3O+
H2O + HNO2
Keq =
[HNO 2 ] [ NO 2 ][ H ] Keq =
+
=
1 1 = = 2.5 103 Ka 4.0 10 4
c. NH4+ + CH3CO2 NH3 + CH3CO2H
+
[ NH 3 ][CH 3CO 2 H ] [ NH 4 ][CH 3CO 2 ]
+
[H + ] [H + ]
K a for NH 4 Kw = Keq = K a for CH 3CO 2 H (K b for NH 3 )(K a for CH 3CO 2 H) Keq = 1.0 10 14 = 3.1 105 (1.8 10 5 ) (1.8 10 5 ) Keq =
1 = 1.0 1014 Kw
d. H3O+ + OH 2 H2O
e. NH4+ + OH NH3 + H2O HNO2 + OH
Keq =
1 = 5.6 104 K b for NH 3
f.
H2O + NO2
K for HNO 2 [ NO 2 ] 4.0 10 4 [H + ] =a = = 4.0 1010 Keq = + 14 Kw 1.0 10 [HNO 2 ][OH ] [H ] 106. a. In the lungs there is a lot of O2, and the equilibrium favors Hb(O2)4. In the cells there is a deficiency of O2, and the equilibrium favors HbH44+. b. CO2 is a weak acid, CO2 + H2O HCO3 + H+. Removing CO2 essentially decreases H+. Hb(O2)4 is then favored, and O2 is not released by hemoglobin in the cells. Breathing into a paper bag increases CO2 in the blood, thus increasing [H+], which shifts the reaction left. c. CO2 builds up in the blood, and it becomes too acidic, driving the equilibrium to the left. Hemoglobin can't bind O2 as strongly in the lungs. Bicarbonate ion acts as a base in water and neutralizes the excess acidity.
234
107.
CHAPTER 7
ACIDS AND BASES
The light bulb is bright because a strong electrolyte is present; that is, a solute is present that dissolves to produce a lot of ions in solution. The pH meter value of 4.6 indicates that a weak acid is present. (If a strong acid were present, the pH would be close to zero.) Of the possible substances, only HCl (strong acid), NaOH (strong base), and NH4Cl are strong electrolytes. Of these three substances, only NH4Cl contains a weak acid (the HCl solution would have a pH close to zero, and the NaOH solution would have a pH close to 14.0). NH4Cl dissociates into NH4+ and Cl- ions when dissolved in water. Cl is the conjugate base of a strong acid, so it has no basic (or acidic properties) in water. NH4+, however, is the conjugate acid of the weak base NH3, so NH4+ is a weak acid and would produce a solution with a pH = 4.6 when the concentration is ~1.0 M. NH4Cl is the solute. From the pH, C7H4ClO2 is a weak base. Use the weak base data to determine Kb for C7H4ClO2 (which we will abbreviate as CB). CB Initial Equil. + H2O
108.
HCB 0 x
+
OH ~0 x
0.20 M 0.20 - x
Because pH = 8.65, pOH = 5.35 and [OH] = 105.35 = 4.5 106 M = x. Kb =
x2 [HCB][OH ] (4.5 10 6 ) 2 = = = 1.0 1010 6 0.20 x [CB ] 0.20 (4.5 10 )
Because CB is a weak base, HCB, chlorobenzoic acid, is a weak acid. Solving the weak acid problem: HCB Initial Equil. Ka = 0.20 M 0.20 x
H+ ~0 x
+
CB 0 x
Kw x2 x2 1.0 10 14 , 1.0 10 4 = = Kb 0.20 x 0.20 1.0 10 10
x = [H+] = 4.5 103 M; pH = 2.35; assumptions good.
109. CaO(s) + H2O(l) Ca(OH)2(aq); Ca(OH)2(aq) Ca2+(aq) + 2 OH(aq)
0.25 g CaO
[OH] =
1 mol CaO 1 mol Ca (OH) 2 2 mol OH 56.08 g 1 mol CaO mol Ca (OH) 2 = 5.9 10 3 M 1 .5 L
pOH = !log(5.9 10 3 ) = 2.23, pH = 14.00 2.23 = 11.77 110. 10.0 g NaOCN
1 mol = 0.154 mol NaOCN 65.01 g
CHAPTER 7
ACIDS AND BASES
1 mol = 0.111 mol H2C2O4 90.04 g
235
10.0 g H2C2O4
Mol NaOCN 0.154 mol (actual) = = 1.39 Mol H 2 C 2 O 4 0.111 mol
The balanced reaction requires a larger 2 : 1 mole ratio. Therefore, NaOCN in the numerator is limiting. Because there is a 2 : 2 mole correspondence between mol NaOCN reacted and mol HNCO produced, 1.54 mol HNCO will be produced. HNCO Initial Equil.
H+ ~0 x
+
NCO 0 x
Ka = 1.2 10 4
0.154 mol/0.100 L 1.54 ! x
Ka = 1.2 10 4 =
x2 x2 , x = [H+] = 1.4 10 2 M 1.54 x 1.54
pH = log(1.4 10 2 ) = 1.85; assumptions good. 30.0 mg papH + Cl 1000 mL 1g 1 mol papH + Cl 1 mol papH + mL soln L 1000 mg 378.85 g mol papH + Cl = 0.0792 M papH+ Initial 0.0792 M Equil. 0.0792 ! x Ka = 2.5 10 6 =
111.
pap + H+ 0 x ~0 x
Ka =
Kw K b, pap
=
2.1 10 14 = 2.5 10 6 8.33 10 9
x2 x2 , x = [H+] = 4.4 10 4 M 0.0792 x 0.0792
pH = -log(4.4 10 4 ) = 3.36; assumptions good. 112. For this problem we will abbreviate CH2=CHCO2H as Hacr and CH2=CHCO2 as acr. a. Initial Equil. Hacr 0.10 M 0.10 x
H+ ~0 x
+
acr0 x
236
Ka =
CHAPTER 7
ACIDS AND BASES
x2 x2 , 5.6 105 , x = [H+] = 2.4 103 M; pH = 2.62 0.10 x 0.10
Assumptions good. b. Percent dissociation = 2.4 10 3 100 = 2.4% 0.10
c. For 0.010% dissociation: [acr] = 1.0 104(0.10) = 1.0 105 M Ka = [ H + ] (1.0 10 5 ) [H + ][acr ] , 5.6 105 = , [H+] = 0.56 M [Hacr] 0.10 (1.0 10 5 )
d. acr is a weak base and the major source of OH in this solution. acr+ H2O
Hacr 0 x
+
OH ~0 x
Kb =
Kw 1.0 10 14 = Ka 5.6 10 5
Initial 0.050 M Equil. 0.050 x Kb = [OH ][ Hacr] [acr ] , 1.8 1010 =
Kb = 1.8 1010
x2 x2 0.050 x 0.050
x = [OH] = 3.0 106 M; pOH = 5.52; pH = 8.48; assumptions good.
113.
a. Initial Equil. Ka =
Fe(H2O)63+ + H2O 0.10 M 0.10 x
[H 3O + ][Fe( H 2 O) 5 (OH ) 2 + ] [Fe(H 2 O) 6 ]
3+
Fe(H2O)5(OH)2+ 0 x
+
H3O+ ~0 x
,
6.0 103 =
x2 x2 0.10 x 0.10
x = 2.4 102 M; assumption is poor (24% error).
Using successive approximations: x2 = 6.0 103, x = 0.021 0.10 0.024
x2 x2 = 6.0 103, x = 0.022; = 6.0 103, x = 0.022 0.10 0.021 0.10 0.022 x = [H+] = 0.022 M; pH = 1.66
CHAPTER 7
b.
ACIDS AND BASES
= 0.0010 [H + ] (0.0010) ; Ka = 6.0 103 = 0.9990 0.9990
237
[Fe(H 2 O) 5 (OH) 2 + ] [Fe( H 2 O) 6 ]
3+
Solving: [H+] = 6.0 M; pH = log(6.0) = 0.78 c. Because of the lower charge, Fe2+(aq) will not be as strong an acid as Fe3+(aq). A solution of iron(II) nitrate will be less acidic (have a higher pH) than a solution with the same concentration of iron(III) nitrate. 114. At a pH = 0.00, the [H+] = 100.00 = 1.0 M. Begin with 1.0 L 2.0 mol/L NaOH = 2.0 mol OH. We will need 2.0 mol HCl to neutralize the OH plus an additional 1.0 mol excess to reduce to a pH of 0.00. We need 3.0 mol HCl total to achieve pH = 0.00. 0.50 M HA, Ka = 1.0 103; 0.20 M HB, Ka = 1.0 1010; 0.10 M HC, Ka = 1.0 1012 Major source of H+ is HA because its Ka value is significantly larger than other Ka values. HA Initial Equil. Ka = 0.50 M 0.50 x
115.
H+ ~0 x
+
A 0 x
x2 x2 0.022 , 1.0 103 , x = 0.022 M = [H+], 100 = 4.4% error 0.50 x 0.50 0.50
Assumption good. Let's check out the assumption that only HA is an important source of H+. (0.022) [B ] , [B] = 9.1 1010 M For HB: 1.0 1010 = (0.20) At most, HB will produce an additional 9.1 1010 M H+. Even less will be produced by HC. Thus our original assumption was good. [H+] = 0.022 M. 116. [HA]0 =
1.0 mol = 0.50 mol/L; solve using the Ka equilibrium reaction. 2 .0 L
HA Initial Equil. Ka = 0.50 M 0.50 x
H+ ~0 x
+
A 0 x
x2 [H + ][ A ] = ; in this problem, [HA] = 0.45 M so: [HA] 0.50 x
[HA] = 0.45 M = 0.50 M x, x = 0.05 M; Ka = 117.
(0.05) 2 = 6 103 0.45
Since NH3 is so concentrated, we need to calculate the OH contribution from the weak base NH3.
238
NH3 + H2O Initial 15.0 M Equil. 15.0 x Kb = 1.8 105 =
CHAPTER 7
ACIDS AND BASES
NH4+ 0 x
+
OH
Kb = 1.8 105
0.0100 M (Assume no volume change.) 0.0100 + x
x(0.0100 + x) x(0.0100) , x = 0.027; assumption is horrible 15.0 x 15.0 (x is 270% of 0.0100).
Using the quadratic formula: (1.8 105)(15.0 x) = (0.0100)x + x2, x2 + (0.0100)x 2.7 104 = 0
x = 1.2 102 M, [OH] = (1.2 102) + 0.0100 = 0.022 M
118.
[H+]0 = (1.0 102) + (1.0 102) = 2.0 102 M from strong acids HCl and H2SO4. HSO4 is a good weak acid (Ka = 0.012). However, HCN is a poor weak acid (Ka = 6.2 1010) and can be ignored. Calculating the H+ contribution from HSO4-: HSO4 Initial Equil. Ka =
H+ 0.020 M 0.020 + x
+
SO42 0 x
Ka = 0.012
0.010 M 0.010 x
x(0.020 + x) x(0.020) , 0.012 , x = 0.0060; assumption poor (60% error). 0.010 x 0.010
Using the quadratic formula: x2 + (0.032)x 1.2 104 = 0, x = 3.4 103 M [H+] = 0.020 + x = 0.020 + (3.4 103) = 0.023 M; pH = 1.64 119. a. The initial concentrations are halved since equal volumes of the two solutions are mixed. HC2H3O2 Initial Equil. 0.100 M 0.100 x
H+
+
C2H3O2 0 x
5.00 104 M 5.00 104 + x
Ka = 1.8 105 =
x(5.00 10 4 + x) x(5.00 10 4 ) 0.100 x 0.100
x = 3.6 103; assumption is horrible. Using the quadratic formula: x2 + (5.18 104)x 1.8 106 = 0 x = 1.1 103 M; [H+] = 5.00 104 + x = 1.6 103 M; pH = 2.80
b. x = [C2H3O2] = 1.1 103 M 120. a. NH4(HCO3) NH4+ + HCO3
CHAPTER 7
ACIDS AND BASES
K 1.0 10 14 1.0 10 14 = 5.6 1010; K b , HCO = w = = 2.3 108 5 7 3 K a1 1.8 10 4.3 10
239
K a , NH + =
4
Solution is basic since HCO3 is a stronger base than NH4+ is as an acid. The acidic properties of HCO3 were ignored because K a 2 is very small (4.8 1011). b. NaH2PO4 Na+ + H2PO4; ignore Na+. K 1.0 10 14 = 1.3 1012 K a , H PO = 6.2 108; K b , H PO = w = 2 2 4 2 4 K a1 7.5 10 3 Solution is acidic since Ka > Kb. c. Na2HPO4 2 Na+ + HPO42; ignore Na+. Ka = 4.8 1013; K b , HPO = Kw 1.0 10 14 = = 1.6 107 Ka2 6.2 10 8
3,
HPO 4
2
4
2
Solution is basic since Kb > Ka. d. NH4(H2PO4) NH4+ + H2PO4 NH4+ is weak acid and H2PO4 is also acidic (see part b). Solution with both ions present will be acidic. e. NH4(HCO2) NH4+ + HCO2; from Appendix 5, K a , HCO 2 H = 1.8 104. K a , NH + = 5.6 1010; K b , HCO =
4 2
Kw 1.0 10 14 = = 5.6 1011 Ka 1.8 10 4
Solution is acidic since NH4+ is a stronger acid than HCO2 is a base. 121. Because the values of K a1 and K a 2 are fairly close to each other, we should consider the amount of H+ produced by the K a1 and K a 2 reactions. H3C6H5O7 Initial Equil.
H2C6H5O7
+
H+ ~0 x
K a1 = 8.4 104
0 0.15 M x 0.15 - x 2 2 x x 8.4 104 = , x = 1.1 102; 0.15 x 0.15
assumption fails the 5% rule.
Solving more exactly using the method of successive approximations: 8.4 104 =
x2 , x = 1.1 102 M (consistent answer) (0.15 1.1 10 2 )
Now lets solve for the H+ contribution from the K a 2 reaction.
240
H2C6H5O7 Initial Equil. 1.8 105 = 1.1 102 M 1.1 102 x
CHAPTER 7
ACIDS AND BASES
K a 2 = 1.8 105
HC6H5O72 0 x
+
H+ 1.1 102 M 1.1 102 + x
x (1.1 10 2 + x) x (1.1 10 2 ) , x = 1.8 105 M; assumption good 2 2 (1.1 10 x) 1.1 10 (0.2% error).
At most, 1.8 105 M H+ will be added from the K a 2 reaction. [H+]total = (1.1 102) + (1.8 105) = 1.1 102 M Note that the H+ contribution from the K a 2 reaction was negligible compared to the H+ contribution from the K a1 reaction even though the two Ka values only differed by a factor of 50. Therefore, the H+ contribution from the K a 3 reaction will also be negligible since K a 3 < Ka2 . Solving: pH = log(1.1 102) = 1.96
Challenge Problems
122. 1.000 L 1.00 10 4 mol HA = 1.00 104 mol HA L
25.0% dissociation gives: mol H+ = 0.250 (1.00 104) = 2.50 105 mol mol A = 0.250 (1.00 104) = 2.50 105 mol mol HA = 0.750 (1.00 104) = 7.50 105 mol 2.50 10 5 2.50 10 5 V V [H + ][A ] 4 = 1.00 10 = Ka = 5 [ HA] 7.50 10 V 1.00 104 = (2.50 10 5 ) 2 (2.50 10 5 ) 2 , V= = 0.0833 L = 83.3 mL (7.50 10 5 )(V) (1.00 10 4 )(7.50 10 5 )
The volume goes from 1000. mL to 83.3 mL, so 917 mL of water evaporated. 123. a. HCO3 + HCO3 H2CO3 + CO32
CHAPTER 7
Keq =
ACIDS AND BASES
[H 2 CO 3 ][CO 3 ] [HCO 3 ][HCO 3 ]
2
241
K [H + ] 4.8 10 11 = a2 = = 1.1 104 + 7 K a1 [H ] 4.3 10
b. [H2CO3] = [CO32] since the reaction in part a is the principal equilibrium reaction. c. H2CO3 2 H+ + CO32 Keq = [H + ]2 [CO 3 ] = K a1 K a 2 [H 2 CO 3 ]
2
Because [H2CO3] = [CO32] from part b, [H+]2 = K a1 K a 2 . [H+] = (K a1 K a 2 )1/ 2 , or taking the log of both sides: pH = pK a1 + pK a 2 2
d. [H+] = [(4.3 107) (4.8 1011)]1/2, [H+] = 4.5 109 M; pH = 8.35 124. a. NaHCO3(aq) Na+(aq) + HCO3(aq); NaHSO4 Na+(aq) + HSO4(aq) Na+ has no acidic (or basic) properties. HCO3 is a weak acid with Ka = 4.8 1011. HCO3 is also the conjugate base of the weak acid H2CO3, which makes it a weak base. HCO3 is amphoteric; the dominant equilibrium of the best acid reacting with the best base present in a bicarbonate solution is: HCO3(aq) + HCO3(aq)
H2CO3(aq) + CO32(aq)
Because the best acid and best base present are the same species, adding more HCO3 adds both the acid and the base to the equilibrium at the same time. The H2CO3 and CO32 concentrations are increased by the same proportions as more HCO3 is added. The proportional increase is determined only by the Ka value for HCO3 and the Kb value for HCO3. Thus bicarbonate solutions are concentration independent. For HSO4 solutions, the dominant equilibrium of the best acid reacting with the best base present is: HSO4(aq) + H2O(l)
SO42(aq)
+ H3O+(aq)
This is just the Ka reaction for HSO4. HSO4 is the conjugate base of the strong acid H2SO4, so HSO4 is a much worse base than water. Water is the best base present in bisulfate solutions. When more HSO4 is added, more H3O+ will be produced, resulting in a more acidic pH. The pH of HSO4 solutions does depend on the concentration of HSO4 present. b. The dominant equilibrium reaction is: HCO3(aq) + HCO3(aq)
H2CO3(aq) + CO32(aq)
242
CHAPTER 7
ACIDS AND BASES
From this reaction, the equilibrium concentrations of H2CO3 and CO32 must be equal to each other. If we add the K a1 reaction for H2CO3 to the K a 2 reaction for HCO3, the result is: 2 [H + ]2 [CO 3 ] K = K a1 K a 2 = H2CO3 2 H+ + CO32 [H 2 CO 3 ] 2 +2 Because [H2CO3] = [CO3 ]: [H ] = K a1 K a 2 [H+] = ( K a1 K a 2 )1/2 or taking the log of both sides: pH = pH = c. Initial Equil. pK a1 + pK a 2 2 HSO4 0.010 M 0.010 x = pK a1 + pK a 2 2
log(4.3 10 7 ) log(4.8 10 11 ) 6.37 + 10.32 , pH = = 8.35 2 2
H+ + SO42 ~0 x 0 x
Ka = 1.2 10 2
1.2 10 2 =
x2 ; solving using the quadratic equation: 0.010 x
x = [H+] = 6.5 103 M; pH = 2.19
125.
HC2H3O2 Initial 1.00 M Equil. 1.00 x 1.8 10 5 =
H+ +
~0 x
C2H3O2 0 x
Ka = 1.8 10 5
x2 x2 , x = [H+] = 4.24 10 3 M (using one extra sig. fig.) 1.00 x 1.00
pH = !log(4.24 10 3 ) = 2.37; assumptions good. We want to double the pH to 2(2.37) = 4.74 by addition of the strong base NaOH. As is true with all strong bases, they are great at accepting protons. In fact, they are so good that we can assume they accept protons 100% of the time. The best acid present will react the strong base. This is HC2H3O2. The initial reaction that occurs when the strong base is added is: HC2H3O2 + OH C2H3O2 + H2O Note that this reaction has the net effect of converting HC2H3O2 into its conjugate base, C2H3O2. For a pH = 4.74, lets calculate the ratio of [C2H3O2]/[HC2H3O2] necessary to achieve this pH.
CHAPTER 7
ACIDS AND BASES
243
HC2H3O2
H
+
+ C2H3O2
[H + ][C 2 H 3 O 2 ] Ka = [ HC 2 H 3 O 2 ]
When pH = 4.74, [H+] = 10 4.74 = 1.8 10 5 . (1.8 10 5 )[C 2 H 3O 2 ] [C 2 H 3O 2 ] , = 1.0 [HC 2 H 3O 2 ] [HC 2 H 3O 2 ]
Ka = 1.8 10 5 =
For a solution having pH = 4.74, we need to have equal concentrations (equal moles) of C2H3O2 and HC2H3O2. Therefore, we need to add an amount of NaOH that will convert onehalf of the HC2H3O2 into C2H3O2. This amount is 0.50 M NaOH. HC2H3O2 + OH C2H3O2 + H2O Before 1.00 M Change !0.50 After 0.50 M completion 0.50 M 0 !0.50 +0.50 0 0.50 M
From the preceding stoichiometry problem, adding enough NaOH(s) to produce a 0.50 M OH solution will convert one-half the HC2H3O2 into C2H3O2; this results in a solution with pH = 4.74. Mass NaOH = 1.00 L 0.50 mol NaOH 40.00 g NaOH = 20. g NaOH L mol
0.08206 L atm 298 K K mol = 125 g/mol 1.00 atm
126.
dRT Molar mass = = P 1.50 g [HA]0 = HA
5.11 g / L
1 mol 125 g = 0.120 M; pH = 1.80, [H+] = 101.80 = 1.6 102 M 0.100 L
H+ + A ~0 x 0 x where x = [H+] = 1.6 102 M
Initial 0.120 M Equil. 0.120 x Ka = 127.
[H + ][A ] (1.6 10 2 ) 2 = = 2.5 103 [HA] 0.120 0.016
Major species: BH+, X, and H2O; because BH+ is the best acid and X is the best base in solution, the principal equilibrium is:
244
BH+ + Initial 0.100 M Equil. 0.100 x K= K a , BH + K a , HX X
CHAPTER 7
ACIDS AND BASES
B+ 0 x
HX 0 x
0.100 M 0.100 x
=
[B][HX ] , where [B] = [HX] and [BH+] = [X] + [BH ][X ]
To solve for the Ka of HX, lets use the equilibrium expression to derive a general expression that relates pH to the pKa for BH+ and to the pKa for HX. K a , BH + K a , HX
=
[HX ]2 [H + ][X ] [HX] [H + ] ; K a , HX = , = [HX] K a , HX [ X ]2 [X ]
2
, [ H + ]2 = K K a , HX a , BH + K a , HX pK a , BH + + pK a , HX Taking the log of both sides: pH = 2 K a , BH + This is a general equation that applies to all BHX type salts. Solving the problem: Kb for B = 1.0 103; Ka for BH+ = pH = 8.00 = 128. 11.00 + pK a , HX 2 Kw = 1.0 1011 Kb
[H + ] [HX]2 = = K a , HX [ X ]2
, pK a , HX = 5.00 and Ka for HX = 105.00 = 1.0 105
Major species: NH4+, C2O42, and H2O; reacting the best acid with the best base: NH4+ Initial Change Equil. K= 0.200 M x 0.200 x + C2O42
NH3 + HC2O4 0 +x x 0 +x x
K=
K a , NH
4
+
K a , HC O
2
4
0.100 M x 0.100 x
K = 9.2 106
( x)( x) = 9.2 106; solving: x = 4.3 104 M (0.200 x)(0.100 x)
Use either Ka expression to solve for [H+]. K a 2 = 6.1 105 = [H + ][C 2 O 4 ] [HC 2 O 4 ]
2
=
[H + ](0.100 4.3 10 4 ) , [H+] = 2.6 107 M; (4.3 10 4 ) pH = 6.59
We get the same answer using the Ka equilibrium for NH4+.
CHAPTER 7
ACIDS AND BASES
[H + ][ NH 3 ] [ NH 4 ]
+
245
[H + ](4.3 10 4 ) , [H+] = 2.6 10 7 M; pH = 6.59 4 (0.200 4.3 10 )
Ka = 5.6 1010 = 129.
=
0.0500 M HCO2H (HA), Ka = 1.77 104; 0.150 M CH3CH2CO2H (HB), Ka = 1.34 105 Because two comparable weak acids are present, each contributes to the total pH. Charge balance: [H+] = [A] + [B] + [OH] = [A] + [B] + Kw/[H+] Mass balance for HA and HB: 0.0500 = [HA] + [A] and 0.150 = [HB] + [B] [H + ][A ] [H + ][B ] = 1.77 10 4 ; = 1.34 105 [HA] [HB] We have five equations and five unknowns. Manipulate the equations to solve. [H+] = [A] + [B] + Kw/[H+]; [H+]2 = [H+][A] + [H+][B] + Kw [H+][A] = (1.77 104)[HA] = (1.77 104) (0.0500 [A]) If [A] << 0.0500, then [H+][A] (1.77 104) (0.0500) = 8.85 106. Similarly, assume [H+][B] (1.34 105)(0.150) = 2.01 106. [H+]2 = 8.85 106 + 2.01 106 + 1.00 1014, [H+] = 3.30 103 mol/L 8.85 10 6 2.68 103 Check assumptions: [H+][A] 8.85 106, [A] 3 3.30 10 Assumed 0.0500 [A] 0.0500. This assumption is borderline (2.68 103 is 5.4% of 0.0500). The HB assumption is good (0.4% error). Using successive approximations to refine the [H+][A] value: [H+] = 3.22 103 M, pH = log(3.22 103) = 2.492
Note: If we treat each acid separately:
H+ from HA = 2.9 103 H+ from HB = 1.4 103
_________________________________________________
4.3 103 M = [H+]total This assumes the acids did not suppress each others ionization. They do, and we expect the [H+] to be less than 4.3 103 M. We get such an answer. 130. Initial Equil. HA
H+
+
A 0 1.00 104
Ka = 1.00 106 C = [HA]0 for pH = 4.000 x = [H+] = 1.00 104 M
C C 1.00 104
~0 1.00 104
246
Ka =
CHAPTER 7
(1.00 10 4 ) 2 = 1.00 106; solving: C = 0.0101 M 4 (C 1.00 10 )
ACIDS AND BASES
The solution initially contains 50.0 103 L 0.0101 mol/L = 5.05 104 mol HA. We then dilute to a total volume V in liters. The resulting pH = 5.000, so [H+] = 1.00 105. In the typical weak acid problem, x = [H+], so: HA Initial Equil. Ka =
H+
+
A 0 1.00 105
5.05 104 mol/V (5.05 104/V) (1.00 105)
~0 1.00 105
(1.00 10 5 ) 2 = 1.00 106 (5.05 10 4 /V) (1.00 10 5 )
1.00 104 = (5.05 104/V) 1.00 105 V = 4.59 L; 50.0 mL are present initially, so we need to add 4540 mL of water. 131. Initial Change Equil. HA [HA]0 x [HA]0 x
H+ ~0 +x x
+
A 0 +x x
Ka = 5.00 1010
From the problem: pH = 5.650, so [H+] = x = 105.650 = 2.24 106 M 5.00 1010 = x2 (2.24 10 6 ) 2 = , [HA]0 = 1.00 102 M [HA]0 x ([HA]0 2.24 10 6 )
After the water is added, the pH of the solution is between 6 and 7, so the water contribution to the [H+] must be considered. The general expression for a very dilute weak acid solution is: Ka = [ H + ]2 K w [ H + ]2 K w [HA]0 [H + ]
pH = 6.650; [H+] = 106.650 = 2.24 107 M Let V = volume of water added: 5.00 1010 = (2.24 10 7 ) 2 (1.00 10 14 ) 0.0500 (2.24 10 7 ) 2 (1.00 10 14 ) (1.00 10 2 ) 0.0500 + V 2.24 10 7
Solving, V = 6.16 L of water were added. 132. Major species = Na+, HSO4, NH3, and H2O; reaction: HSO4 + NH3
SO42 + NH4+
CHAPTER 7
ACIDS AND BASES
2 +
247
K=
[SO 4 ][ NH 4 ] [HSO 4 ][ NH 3 ]
=
K a , HSO K a , NH
4
=
4
+
1.2 10 2 = 2.1 107 5.6 10 10
Because K is a large number, let the reaction go to completion, and then solve the back equilibrium problem. HSO4 + NH3 Before After 0.10 M 0 0.10 M 0
SO42 0 0.10 M
+
NH4+ 0 0.10 M
Allow the reaction to attain equilibrium: HSO4 + Initial Change Equil. 0 +x x NH3 0 +x x
SO42
+
NH4+ 0.10 M x 0.10 x
0.10 M x 0.10 x
(0.10 x) 2 (0.10) 2 = 2.1 107, x = 2.2 105 M; assumptions good. 2 2 x x [HSO4] = 2.2 105 M; [NH3] = 2.2 105 M; [SO42] = 0.10 M [NH4+] = 0.10 M
Using one of the Ka equilibrium expressions to solve for [H+]: K a , HSO =
4
[H + ] (0.10) = 1.2 10 2 , [H+] = 2.6 106 M; pH = 5.59 2.2 10 5
133.
0.135 mol CO 2 = 5.40 102 mol CO2/L = 5.40 102 M H2CO3; 0.105 M CO32 2.50 L The best acid (H2CO3) reacts with the best base present (CO32) for the principal equilibrium. H2CO3 + CO32 2 HCO3 K= K a1 , H 2CO 3 K a 2 , H 2CO3
=
4.3 10 7 = 9.0 103 4.8 10 11
Because K >> 1, assume all CO2 (H2CO3) is converted into HCO3; that is, 5.40 102 mol/L CO32 is converted into HCO3. [HCO3] = 2(5.40 102) = 0.108 M; [CO32] = 0.105 - 0.0540 = 0.051 M Note: If we solve for the [H2CO3] using these concentrations, we get [H2CO3] = 2.5 105 M; our assumption that the reaction goes to completion is good (2.5 105 is 0.05% of 0.051). Whenever K >> 1, always assume the reaction goes to completion.
248
CHAPTER 7
ACIDS AND BASES
To solve for the [H+] in equilibrium with HCO3 and CO32, use the Ka expression for HCO3. HCO3 H+ + CO32 4.8 1011 =
2
K a 2 = 4.8 1011
[H + ][CO 3 ] [HCO 3 ]
=
[H + ] (0.051) 0.108
[H+] = 1.0 1010; pH = 10.00; assumptions good. 134. H2O
H+ + OH;
B + H2O
HB+ + OH
Kw = [H+][OH];
Kb =
[HB+ ][OH ] [B] material balance: [B]o = [B] + [HB+]
Charge balance: [H+] + [HB+] = [OH]; So: [OH] = [H+] + [HB+] [OH] =
Kw Kw + [HB+] or: [HB+] = [OH] [OH ] [OH ]
[B] = [B]0 [HB+] Kw [B] = [B]0 [OH ] [OH ] Kw [OH ] [OH ] [OH ] Kb = Kw [B]0 [OH ] [OH ] Assuming [B]0 >> [OH ]2 K w [OH ]2 K w [B]0 [OH ]
=
[OH ]2 K w , then: [OH ]
Kb
[OH ]2 K w [OH ]2 1.0 10 14 , 6.1 1011 = [B]0 2.0 10 5
[OH] = 1.1 107; pOH = 6.96; pH = 7.04 (assumption good) 135. Major species: H2O, Na+, and NO2; NO2 is a weak base. NO2 + H2O
HNO2 + OH;
CHAPTER 7
ACIDS AND BASES
249
Because this is a very dilute solution of a weak base, the OH contribution from H2O must be considered. The weak base equations for dilute solutions are analogous to the weak acid equations derived in Section 7.9 of the text. For A type bases (A + H2O Kb =
HA + OH), the general equation is:
[OH ]2 K w [OH ]2 K w [A ]0 [OH ] [OH ]2 K w [OH ]2 K w , then Kb = [OH ] [A ]0 and:
When [A]0 >>
[OH] = (Kb[A]0 + Kw)1/2
1.0 10 14 4 14 Try: [OH ] = 4.0 10 4 (6.0 10 ) + (1.0 10 )
1/ 2
= 1.6 107 M
Checking assumption: 6.0 104 >>
(1.6 10 7 ) 2 (1.0 10 14 ) = 9.8 108 7 1.6 10
Assumption good. [OH] = 1.6 107 M; pOH = 6.80; pH = 7.20 136. Major species: H+, HSO4, and H2O (water is important!) Charge balance: [H+] = [OH] + [HSO4] + 2[SO42] [HSO4]0 = [SO42] + [HSO4] = 1.00 107 M (from the 1.00 107 M H2SO4) Kw = [H+][OH] = 1.0 1014 Material balance: Ka = 1.2 102 = [H+] = [H + ][SO 4 ] [HSO 4 ]
2
; [HSO4] = (1.00 107) [SO42]
Kw Kw + (1.00 107) [SO42] + 2[SO42], [SO42] = [H+] (1.00 107) + [H ] [H + ]
Kw [H + ] [H + ] (1.00 10 7 ) + [H ] [H ][SO 4 ] 1.2 H 102 = = Kw [HSO 4 ] (1.00 10 7 ) [H + ] + + (1.00 10 7 ) [H + ]
+ 2
This is a complicated expression to solve. Because this is such a dilute solution of H2SO4 (1.00 107 M ), the Ka equilibrium expression for HSO4 dictates that [SO42] >> [HSO4]. Lets assume that [SO42] = 1.00 107 M (assume most of the HSO4 dissociates):
250
[H+] =
CHAPTER 7
Kw Kw + (1.00 107) + [SO42] = + 2.00 107 + [H ] [H + ]
ACIDS AND BASES
Solving: [H+] = 2.4 107 M; pH = 6.62 Assumption good: [SO 4 ] [HSO 4 ]
2
=
Ka 1.2 10 2 = = 5.0 104. + 7 [H ] 2.4 10
We do have mostly SO42 at equilibrium. 137. Major species: H2O, NH3, H+, and Cl; The H+ from the strong acid will react with the best base present (NH3). Because strong acids are great at donating protons, the reaction between H+ and NH3 essentially goes to completion, that is, until one or both of the reactants runs out. The reaction is: NH3 + H+ NH4+ Because equal volumes of 1.0 104 M NH3 and 1.0 104 M H+ are mixed, both reactants are in stoichiometric amounts, and both reactants will run out at the same time. After reaction, only NH4+ and Cl remain. Cl has no basic properties since it is the conjugate base of a strong acid. Therefore, the only species with acid-base properties is NH4+, a weak acid. The initial concentration of NH4+ will be exactly one-half of 1.0 104 M since equal volumes of NH3 and HCl were mixed. Now we must solve the weak acid problem involving 5.0 105 M NH4+. K H+ + NH3 Ka = w = 5.6 1010 NH4+ Kb 5 Initial 5.0 10 M ~0 0 x x Equil. 5.0 105 x Ka =
x2 x2 = 5.6 1010, x = 1.7 107 M; check assumptions. (5.0 10 5 x) 5.0 10 5
We cannot neglect [H+] that comes from H2O. As discussed in Section 7.9 of the text, assume 5.0 105 >> ([H+]2 Kw)/[H+]. If this is the case, then: [H+] = (Ka[HA]0 + Kw)1/2 = 1.9 107 M; checking assumption: [ H + ]2 K w = 1.4 10-7 << 5.0 105 (assumption good) [H + ] So: [H+] = 1.9 107 M; pH = 6.72 138. Ca(OH)2 (s) Ca2+(aq) + 2 OH(aq) This is a very dilute solution of Ca(OH)2, so we can't ignore the OH contribution from H2O. From the dissociation of Ca(OH)2 alone, 2[Ca2+] = [OH]. Including the H2O autoionization into H+ and OH, the overall charge balance is:
CHAPTER 7
ACIDS AND BASES
251
2[Ca2+] + [H+] = [OH] 2(3.0 107 M) + Kw/[OH] = [OH], [OH]2 = (6.0 107)[OH] + Kw [OH]2 (6.0 107)[OH] 1.0 1014 = 0; using quadratic formula: [OH] = 6.2 107 M
Marathon Problems
139. To determine the pH of solution A, the Ka value for HX must be determined. Use solution B to determine Kb for X, which can then be used to calculate Ka for HX (Ka = Kw/Kb). Solution B: X Initial Change Equil. Kb = + H2O
HX 0 +x x
+
OH ~0 +x x
Kb =
[HX][OH ] [X ]
0.0500 M x 0.0500 x
x2 ; from the problem, pH = 10.02, so pOH = 3.98 and [OH] = x = 103.98 0.0500 x
(10 3.98 ) 2 = 2.2 107 3.98 0.0500 10
Kb =
Solution A: Ka, HX = K w /K b, X = (1.0 1014)/(2.2 107) = 4.5 108 HX Initial Change Equil.
H+ ~0 +x x
+
X 0 +x x
Ka = 4.5 108 =
[H + ][X ] [HX]
0.100 M x 0.100 x
Ka = 4.5 108 =
x2 x2 , x = [H+] = 6.7 105 M 0.100 x 0.100
Assumptions good (x is 0.067% of 0.100); pH = 4.17 Solution C: Major species: H2O, HX (Ka = 4.5 108), Na+, and OH; the OH from the strong base is exceptional at accepting protons. OH will react with the best acid present (HX), and we can assume that OH will react to completion with HX, that is, until one (or both) of the reactants runs out. Because we have added one volume of substance to another, we have diluted both solutions from their initial concentrations. What hasnt changed is the moles of each reactant. So lets work with moles of each reactant initially.
252
Mol HX = 0.0500 L Mol OH = 0.0150 L
CHAPTER 7
0.100 mol HX = 5.00 103 mol HX L
ACIDS AND BASES
0.250 mol NaOH 1 mol OH = 3.75 103 mol OH L mol NaOH
Now lets determine what is remaining in solution after OH reacts completely with HX. Note that OH is the limiting reagent. HX Before 5.00 103 mol 3.75 103 Change After 1.25 103 mol completion + OH 3.75 103 mol 3.75 103 0
X
+
H2O
0 +3.75 103 3.75 103 mol
+3.75 103
After reaction, the solution contains HX, X, Na+ and H2O. The Na+ (like most +1 metal ions) has no effect on the pH of water. However, HX is a weak acid and its conjugate base, X, is a weak base. Since both Ka and Kb reactions refer to these species, we could use either reaction to solve for the pH; we will use the Kb reaction. To solve the equilibrium problem using the Kb reaction, we need to convert to concentration units since Kb is in concentration units of mol/L. [HX] = 1.25 10 3 mol 3.75 10 3 mol = 0.0192 M; [X] = = 0.0577 M (0.0500 + 0.0150) L 0.0650 L
[OH] = 0 (We reacted all of it to completion.) X Initial Change Equil. + H2O
HX
+
OH 0 +x x
Kb = 2.2 107
0.0577 M 0.0192 M x mol/L of X reacts to reach equilibrium x +x 0.0577 x 0.0192 + x (0.0192 + x)x (0.0192) x 0.0577 x 0.0577
Kb = 2.2 107 =
(assuming x is << 0.0192)
x = [OH] =
(2.2 10 7 )(0.0577) = 6.6 107 M; assumptions great (x is 0.0034% of 0.0192 0.0192).
[OH] = 6.6 107 M, pOH = 6.18, pH = 14.00 = 6.18 = 7.82 = pH of solution C The combination is 4-17-7-82. 140. a. Strongest acid from group I = HCl; weakest base (smallest Kb) from group II = NaNO2 0.20 M HCl + 0.20 M NaNO2; major species = H+, Cl, Na+, NO2, and H2O
CHAPTER 7
ACIDS AND BASES
253
Let the H+ react to completion with the NO2; then solve the back equilibrium problem. H+ Before After 0.10 M 0 HNO2 Initial 0.10 M Change x Equil. 0.10 x + NO2 0.10 M 0
HNO2
0 0.10 M NO2 0 +x x (Molarities are halved due to dilution.) Ka = 4.0 104
H+ + 0 +x x
x2 = 4.0 104; solving, x = [H+] = 6.1 103 M; pH = 2.21 0.10 x
b. Weakest acid from group I = (C2H5)3NHCl; best base from group II = KOI; The dominant equilibrium will be the best base reacting with the best acid. OI + (C2H5)3NH+ Initial Equil. 0.10 M 0.10 x 0.10 M 0.10 x
HOI 0 x
+ (C2H5)3N 0 x
K=
K a , (C
2 H 5 ) 3 NH
+
K a , HOI
=
1.0 10 14 1 = 1.25 (carrying extra sig. fig.) 4 4.0 10 2.0 10 11
x2 x = 1.25, = 1.12, x = 0.053 M 2 0.10 x (0.10 x)
So: [HOI] = 0.053 M and [OI] = 0.10 x = 0.047 M; using the Ka equilibrium constant for HOI to solve for [H+]: 2.0 1011 = [H + ](0.047) , [H+] = 2.3 1011 M; pH = 10.64 (0.053) 1.0 10 14 = 2.5 1011 4.0 10 4
c. Ka for (C2H5)3NH+ = Kb for NO2 =
1.0 10 14 = 2.5 1011 4.0 10 4
Because Ka = Kb, mixing (C2H5)3NHCl with NaNO2 will result in a solution with pH = 7.00.