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Unformatted text preview: Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 1) Remember that this is a puzzle: So first start with original K: ⎡CO3−2 ⎤ ⎡ HS −1 ⎤
⎦ K = ⎣ 2 ⎦⎣
−1
⎡Cl ⎤ ⎡ HOCO2 −1 ⎤
⎣
⎦⎣
⎦
Next plug in the solids: Ksp Ag2S & (Ksp AgCl)2: Ksp of Ag2S (s) = [Ag+1]2[HS‐1][OH‐1] since S‐2 automatically hydrolyzes in water. 2 +1
−1
⎡CO3−2 ⎤ ⎡ HS −1 ⎤
⎣
⎦⎣
⎦ * ⎡ Ag ⎤ * ⎡ HO ⎤ ⎣
⎦⎣
⎦
K=
2
2
−1
⎡Cl −1 ⎤ ⎡ HOCO2 −1 ⎤ ⎡ Ag +1 ⎤ ⎡ HO ⎤
⎦
⎣
⎦⎣
⎦⎣
⎦⎣ This works out perfectly so now we can plug in the K’s we know: K sp ( Ag 2 S ) 1
6 x 10−51
=
K b CO3−2
1.7 x 10−10 1
= 9.97 x 10−28 −14
10
4.8 x 10−11 046
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K= ⎡ K sp ( AgCl ) ⎤
⎣
⎦ *
2 ( )( ) 2 * ( ) Notice that we HAD to use Kb here since the Ksp of Ag2S requires OH‐. This is how you solve it with method 1: Ag 2S s + H 2O l 2 Ag +1 aq + HS −1 aq + HO −1 aq HOCO2 −1 aq + HO −1 aq
2 Ag +1 aq + 2 Cl −1 aq CO3−2 aq + H 2O l 2 AgCl s + 2 H 2O l This works out perfectly (don’t worry about the water not cancelling) so now we can plug in the K’s we know: K= K sp Ag 2 S ⎡ K sp AgCl ⎤
⎣
⎦ *
2 1
6 x 10−51
=
K b CO3−2
1.7 x 10−10 2 * ( 1
= 9.97 x 10−28 −14
10
4.8 x 10−11 ) Notice that we HAD to use Kb here since the Ksp of Ag2S requires HO‐1. Answer is (a) 1.0 x 10‐27 2) Look for the solution that yields the lowest value of s making the semi‐soluble salt more stable. Molar solubility “s” will be lowest when the semi‐soluble salt has to deal with a common ion effect. Answers (b), (c) and (e) are out because they do not have a common ion effect. K2CO3 has the highest concentration of the common ion so it will have the lowest value of s. The s value in K2CO3 mixed with BaCO3 is: K sp = 5.0 x 10−9 = ( s ) ( s + 1.0 ) = ( s ) (1.0 ) ;
/
s = 5.0 x 10−9 Answer is (a) 1 M K2CO3 TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 1 3) Since the system is at EQ that means that (a) and (e) are TRUE! (d) is true because we need heat to convert to gas and we increase in disorder as we go to gas. (b) is correct as well because the standard temperature where water boils is 100 °C. This leaves (c) as the answer which is incorrect because there is more disorder in the gas phase rather than the liquid phase. Answer is (c) S° H2O (l) > S° H2O (g) 4) This is a reaction where you are breaking bonds so the reaction is endothermic (ΔH° = +). Also, the amount of disorder increases in the reaction so ΔS° is +. If you put heat into the reaction it speeds up the reaction since the reaction is endothermic; with this (e) is correct. If the pressure is lowered the reaction slows down since both reactants and products are gases. The only answer left is (a) ΔG°r is“+”. This is not true because there is no “standard” temperature. The reaction does not state a temperature but we are saying that really this statement (a) is the most wrong that we have in all the statements. Answer is (a) ΔG°r is“+” 046
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Exa 5) For ΔG°r to be positive or nonnegative means that the reaction must be anti‐spontaneous. The formation of a solid in (a) indicates the value of ΔG°r to be “–”. (c) is the mixing of a large extent strong acid. (d) is acombustion reaction which is spontaneous and (e) is forming a more stable compound than what was in the reactants so that is definitely spontaneous. The reaction in (b) indicates that you readily break up a very stable salt in Ag2O! You of course do not. This is again one of those best answer situations that the H‐Dog really loves! Watch out. Answers is (b) Ag2O (s) → 2 Ag+1 (aq) + O‐2 (aq) 6) Ca(OH)2 breaks up into → Ca+2 (aq) and 2 OH‐1 (aq). The Ksp expression is then [Ca+2] [OH‐1]2 = Ksp which then turns into (s) * [OH‐1]2 = 3.0 x 10‐5. The problem states the solubility (s = 2.5 x 10‐
4) so you can plug s into the equation to get [OH‐1]. With the concentration of OH‐1 you can determine the pH of the solution: 3 x 10−5
⎡OH −1 ⎤ =
= 0.346 M ;
⎣
⎦
2.5 x 10−4 − log [ 0.346] = 0.461 = pOH ; pH = 14 − 0.461 = 13.54 ≈ 13.5
Answer is (c) 13.5 7) Remember that for any (HO)2SO2 this is always true: [H3O+1] = [Mi + x] [HSO4‐1] = [Mi – x] [SO4‐2] = [x] So if the solution is not extremely dilute or not too concentrated then: ⎡ H 3O +1 ⎤ = ⎡ HSO4 −1 ⎤ + 2 ⎡ SO4 −2 ⎤ ;
⎣
⎦⎣
⎦
⎣
⎦ [ M i + x] = [ M i − x] + 2 [ x] = [ M i − x + 2 x] = [ M i + x]
Answer is (c) [HSO4‐1] + 2 [SO4‐2] = [H3O+1] TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 2 8) This is a puzzle to solve for K. Begin with the starting K expression: 2 ⎡ Z −1 ⎤
* ⎣ ⎦2
K=
2
+2
−1 2
⎡ Ba ⎤ [ HZ ] ⎡OH ⎤ ⎡ Z −1 ⎤
⎣
⎦
⎣
⎦⎣⎦
1 ⎛
1
=⎜
⎜ K b Z −1
⎝ () 2 ⎞
1
⎟*
= K b Z −1
⎟ K sp ( BaZ 2 )
⎠ () −2 * K sp ( BaZ 2 ) −1 Answer is (b) Ksp(BaZ2)‐1 * (Kb(Z‐1))‐2 9) (a) Graphite is certainly not harder than diamonds…right Ladies! (b) A diamond is held closer together (Lower S°) than a piece of graphite because of the molecular bonding network in diamonds. The molecular bonding network is stronger than any covalent bonding that exists in graphite. (c) The bond order in Cd is 1.0 and the bond order in Cgr is 1.5 from the resonance so Cgr has a stronger set of covalent bonds. (e) Cgr certainly exhibits resonance because of its strong covalent bonds. Strong covalent bonding usually means the molecule has a higher bond order than 1.0. Resonance can exist in a compound that can have a higher bond order than 1.0. (d) is the answer because the conversion is real exothermic since the Cd molecule has weaker covalent bonds than the covalent bonding in Cgr. When you go from weaker bonds to stronger bonds, the process is exothermic. Below are some extra calculations to show you why the conversion is exothermic: ΔH° Cd = 1.895 kJ/mol ΔH° Cgr = 0.000 kJ/mol Cd → Cgr
ΔH° = 0.000 – 1.895 = –1.895 kJ/mol Answer is (d) conversion of Cd to Cgr is exo 10) In this question we have to first look at the elements that can exist at 75.0 °C. I can see that (c) and (b) can be crossed out since they form solid and gas. Next we can look for the reaction with the least enthalpic drive (ΔH° = + or as close to positive as we can get). The least enthalpic reactions have to break the most bonds since breaking bonds takes energy. The answer is (a) since (e) and (d) are not breaking the very strong O2 molecule. Answer is (a) H2 (g) + ½ O2 (g) → H2O (l) 046
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Exa TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 3 11) Firstset up the ICE table so you know where all the numbers go: Ag(CN)2‐1 (aq) +
Br‐1 (aq) AgBr (s) + 2 NC‐1 (aq)
2 moles Br −1
=2M
Initial 2.0 M Mi
1L
Change Equilibrium – x 2 – x – x 2 – x + x 1.6 moles + 2x Mi + 2x You are told in the problem that 1.6 moles of AgBr (s) is produced so that means x = 1.6 M since the solution is in 1 L. With this information you can set up to solve for the unknown concentration: ⎡ M i + 2 (1.6 ) ⎤
⎦;
K = 435 = ⎣
2
[ 2 − 1.6]
2 435*0.42 = ⎡ M i + 2 (1.6 ) ⎤ ;
⎣
⎦ 435*0.42 = M i + 3.2 2 046
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M i = 435*0.42 − 3.2 = 5.14 M = Initial Molarity of NC −1 Answer is (e) 5.1 M 12) You can first set up the EQ to be: Ag2S (s) + NH3 (aq) + HO‐1 (aq) Ag2O (s) + S‐2 (aq) + NH4+1 (aq) Initial 1.0 mols 0.5* 2.5
= 1.25M
1L 0.5* 2.5
= 1.25M
1L 0 0 0 Change Equilibrium – x 1.25 – x – x 1.25 – x + x x + x x + x x So K = x2 1.25 − x 2 = 3.3 x 10−4 . Square root both sides then solve for x. x2 1.25 − x 2 = 3.3 x 10−4 ⇒ x
= 0.0182;
1.25 − x x = 0.02275 − 0.0182 x ⇒ 1.0182 x = 0.02275 x = 0.022 M consumed Since 0.022 M consumed and the volume of the system stays at 1 L then the number of moles consumed of Ag2S = 1.0 moles – 0.022 moles = 0.978 left at EQ. Answer is (b) 0.98 mols 13) This is a statement of stability. Since the pH is close to 7 in CaF2 this indicates that F‐1 does not hydrolyze to a large extent in water and with CaCO3 pH close to 10.0 this means that the CO3‐2 hydrolyzes to a large extent. (a) Most fluorides ARE soluble. (b) the carbonate ion has always been stronger than the fluoride ion. (c) the concentration of the acid has to be more concentrated for this to work since CaF2 is so stable. (d) the fact that CaF2 is more stable means that CaF2 is not soluble. (e) is the best answer. Answer is (e) treating CaCO3 with 1 M HF will likely produce CaF2 TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 4 14) To solve problems like this you should use hypothetical units. Say: [OH‐1] pOH pH 1.0 M 0 14 30 M –1.48 15.48 The pH went up about 1.5 pH units meaning that basic strength increased which means that conc. of [HO‐1] increases. By increasing [HO‐1] by a factor of 30 we changed pH by an increase of pH by 1.50 units. Answer is (c) + 1.50 pH units 15) Low S° corresponds to less disorder. This is an application of what you know from CHM2045. (a) strong hydrogen bonds creates less disorder, (b) low FM means that you must be compact so again less disorder, (d) solid phase is the least disordered phase & (e) low T means less disorder once again. (c) indicates a loose molecule, something floppy so S° is highest here. Answer is (c) weak London forces 16) First setup another ICE table to organize your thoughts: CaCO3 (s) + SO4‐2 (aq) CaSO4 • 2 H2O (s) + CO3‐2 (aq) 046
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Initial 1 mol 2 L *0.3 M
= 0.6 M SO4 −2
1L Change – x Equilibrium 0.6 – x Once organized you can solve for missing values: K= 0 0 + x x + x x x
= 6.8 x 10−4 ;
0.6 − x x = 0.000408 − 6.8 x 10 x= −4 x ⇒ 1.00068 x = 0.000408 0.000408
= 0.00041 M = ⎡CO3−2 ⎤
⎣
⎦
1.00068 Since this molarity is so small and the K is small then you are led to believe that (e) is correct. Which it is! Answer is (e) Nonexistent ⎡ SO4 −2 ⎤ [ H 2CO3 ]
x2
⎣
⎦
= 0.22 17) I will skip the ICE table and just set up the K: K =
=
2
2
⎡ HCO3−1 ⎤
( 0.35 − 2 x )
⎣
⎦
Square root both sides and solve for x: x2 ( 0.35 − 2 x ) 2 = 0.22 ⇒ x
= 0.469;
0.35 − 2 x x = 0.16415 − 0.938 x ⇒ 1.938 x = 0.16415 0.16415
= 0.085 M = moles of CaCO3 produced since1L container
1.938
Answer is (a) 0.10 mols x= TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 5 18) This is a formation equation so you must form this compound from the elements: 1
1
Na ( s ) + H 2 ( g ) + S8 ( s ) + 2 O2 ( g ) → NaHSO4
2
8 ΔG ° f ( NaHSO4 ) 19
. 88 Solid phase elemental forms are Na (s) and S8 (s) so the answer is 1 + = Answer is (d) 9/8 19) Remember that for any (HO)2SO2 this is always true: [H3O+] = [Mi + x]; [HSO4‐] = [Mi – x]; [SO42‐] = [x]. We are then given a way to relate Mi and x: ⎡ H 3O +1 ⎤ = 3 ⎡ SO4 −2 ⎤
⎣
⎦⎣
⎦ Mi + x =3 x ⇒ Mi = 2 x
By knowing Mi = 2x we can use the Ka of HSO4‐1 to find Mi: ( K a HSO4 −1 ) ⎡ H 3O +1 ⎤ ⎡ SO4 −2 ⎤ [ M i + x ][ x ] [ 2 x + x ][ x ] 3 x 2
⎦⎣
⎦=
=⎣
=
=
= 3 x = 1.2 x 10−2
M i − x]
x
2 x − x] ⎡ HSO4 −1 ⎤
[
[
⎣
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x = 0.004 M meaning that M i = 2 ( 0.004 M ) = 0.008 M Answer is (d) 0.008 M 20) Just compare Ka vs. Kb since the number of moles is equal. In this case we have to compare Ka of NH4+1 and Kb of Ac‐1 since both values are essentially the same the solution is essentially neutral. If Ka > Kb = mildly acidic; if Ka > > > Kb then strongly acidic; the reverse is also true. Answer is (c) essentially neutral 21) Remember that water boils at a standard state temperature of 100 °C so this EQ is only standard at 100 °C so this EQ is not at a standard state temperature. For this problem you are asked to tell which one is true: (a) S° H2O (l) > S° H2O (g) – False because gas has more entropy. (b) VP H2O (l) = P H2O (g) – True because the system is at EQ. (c) ∆H°r is +, ∆S°r is – this is False because the reaction is going to more disorder so ∆S°r is + (d) ΔG°r is zero – False because the temperature is not 100 °C. The degree sign on Gibb’s Free energy, ΔG°r , means a standard reaction and 25.0 °C is not standard. (e) ∆H°r is –, ∆S° r is + this is False because the reaction needs energy to proceed. Answer is (b) VP H2O (l) = P H2O (g) 22) Standard state means how you would find it in nature. The diatomic elements, metal solids, P4 (s), S8 (s) and C (s) are the elemental states. The pressure of something standard is 1 atm or 1 bar (Memorize) and the concentration of a standard solution is 1 M (Memorize). With all this the answer has to be the temperature because standard state temperature only depends on the reaction; for example water boils at a standard state temperature of 100 °C or 373 K but ice freezes at a standard state temperature of 0.00 °C or 273 K. Answer is (e) T = 298 K TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 6 23) This is a precipitation question based on the value of Qsp. Look for the highest value of Qsp, which will give you the best chance to precipitate M2X3. Qsp = [M+3]2 [X‐2]3 so you can start plugging in each answer choice to see which one gives the highest Qsp: (a) Qsp = ⎡ M +3 ⎤ ⎡ X −2 ⎤ = [1 M ] [ 0.2 M ] = 0.008 ⎣
⎦⎣
⎦ (b) Qsp = ⎡ M +3 ⎤ ⎡ X −2 ⎤ = [ 0.8 M ] [ 0.4 M ] = 0.041 ⎣
⎦⎣
⎦ (c) Qsp = ⎡ M +3 ⎤ ⎡ X −2 ⎤ = [ 0.4 M ] [ 0.8 M ] = 0.082 ⎣
⎦⎣
⎦ (d) Qsp = ⎡ M +3 ⎤ ⎡ X −2 ⎤ = [ 0.2 M ] [1.0 M ] = 0.040 ⎣
⎦⎣
⎦ (e) Qsp = ⎡ M +3 ⎤ ⎡ X −2 ⎤ = [ 0.30 M ] [ 0.60 M ] = 0.019 ⎣
⎦⎣
⎦ 2 3 2 3 2 3 2 3 2 2 3 2 3 2 3 2 3 3 2 3 The highest number is (c) so that gives us the biggest chance to precipitate M2X3. Answer is (c) 0.4 M MCl3 w. 0.8 M K2X 24) This is just a conceptual question. Since AgCN produces NC‐1 a basic solution is made. To reliably calculate the solutbility strictly from Ksp value you MUST prevent hydrolysis. So we must find a stronger base in solution than NC‐1 in order to prevent it from reacting. Remember that the strongest base and the strongest acid in solution will provide the principal reaction. Answer is (e) 0.6 M K2O which produces 2 HO‐1 which is a stronger base than NC‐1. 25) This is a combination question which combines Hess’s Law & Enthalpy of Formation. You need to first find out the goal reaction, which is the formation of C2H2: 046
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2 C graphite + H 2 g → C2 H 2 g C (graphite) is the same thing as C (s). Once the goal reaction is established which is an enthalpy of formation it is time to create the reaction from the three equations. The first step is to have the right reactants and the right products in the correct stoichiometric coefficients. First we need to multipy formation of CO2 by two which also multiplies the ethalpy of formation by 2: 2* ⎡C graphite + O2 g → CO2 g ⎤ = 2 C graphite + 2 O2 g → 2 CO2 g ⎣
⎦ Then we need to reverse the reaction with water: −1* ⎡ H 2O ( l ) → H 2 ( g ) + 1 / 2 O2 ( g ) ⎤ = H 2 ( g ) + 1 / 2 O2 ( g ) → H 2O ( l ) ⎣
⎦ The resulting equations are: 2 C (graphite) + 2 O2 (g) → 2 CO2 (g) H2 (g) + 1/2 O2 (g) → H2O (l) 2 CO2 (g) + H2O (l) → C2H2 (g) + 5/2 O2 (g) Since you only want the goal reaction, the rest of the reactants and products have to cancel. The sum of the equations is: 2 C ( graphite ) + H 2 ( g ) → C2 H 2 ( g ) The last step is to sum up the enthalpies: 2* − 393.5 kJ + 289.7 kJ + ( −285.8 kJ ) = − 783.1 kJ Answer is (3) –783.1 kJ TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 7 26) First set up the ICE table so you know where all the numbers go: AgBr (s) + 2 NC‐1 (aq)
Ag(CN)2‐1 (aq) + Br‐1 (aq) Initial 0.050 mols
Mi 0 0 Change – x – 2x + x + x Equilibrium 0.050 – x Mi – 2x x 0.040 M ‐1 is produced so that means x = 0.040 M. With this You are told in the problem that 0.040 M Br
information you can set up to solve for the unknown concentration: [0.040][0.040] ;
2
⎡ M i − 2 ( 0.040 ) ⎤
⎣
⎦
2
[0.040][0.040] ⇒ M = [0.040][0.040] + 2
⎤=
K = 0.00230 = ⎡ M i − 2 ( 0.040 ) ⎦
⎣ i 0.00230
0.00230
M i = 0.914 M = Initial Molarity of NC −1 ( 0.040 ) Answer is (a) 0.93 M 046
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Exa TutoringZone's Fall 2010 CHM2046 Exam 2 Practice Problem Solutions 8 ...
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This note was uploaded on 11/09/2010 for the course CHM 2046 taught by Professor Veige/martin during the Fall '07 term at University of Florida.
 Fall '07
 veige/martin
 Chemistry

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