Name ___________________________________________________
Student ID # ___________________________
CHEMISTRY 122 [Tyvoll]
EXAM III
November 30, 2007
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R = 0.0821 Latm/molK
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R = 8.314 J/molK
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N = 6.022 x 1023
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h = 6.63 x 10-34 J s
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c = 3.0 x108 m/s
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1 kJ = 103 J
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1 L = 1000 cm3
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ΔG0 = ΔH0 − TΔS0
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ΔG0 = − RT ln KC
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KP = KC (RT)Δn(gas)
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KW = 1.00 x 10-14
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pH = pKa + log{[base]/[acid]}
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Points Missed:
Part I
Grade (100 Points Possible)
Possibly Useful Information:
Part II
__________
Chemistry 122 (Tyvoll)
EXAMINATION III KEY
November 30, 2007
Part I. Multiple Choice (3 points each -- Total points = 45)
1. Which one of the following sparingly soluble salts is the least soluble?
Answer 2. Co(OH)2 Ksp = 1.3 x 10-15 (all are 1”2 electrolytes, so smallest Ksp corresponds
to least soluble)
2. The Ksp for CuBr is 5.0 x 10−9. The correct molar solubility for CuBr is
Answer 4. 7.1 x 10−5 M (Ksp= S2 so S = (Ksp)1/2 = (5.0 x 10−9)1/2 )
3. The molar solubility of Fe(OH)2 is 1.0 x 10−5 M in water at 25 0C. What is the Ksp for
Fe(OH)2 under these conditions? Fe(OH)2 (s) ' Fe2+ (aq) + 2 OH− (aq)
Answer 1. 4.0 x 10−15 (Ksp = (S)(2S)2 = 4S3 = 4 (1.0 x 10−5)3 = 4.0 x 10−15 )
4. Two aqueous solutions, containing Pb2+ and I− respectively, are mixed. If C(Pb2+) is
1.8 x 10−4 M and C(I−) is 3.7 x 10−3 M, which one of the following statements is true?
For PbI2 , Ksp = 7.9 x 10−9 Qip= C(Pb2+)  C(I−)2 = (1.8 x 10−4)(3.7 x 10−3)2 = 2.5 x 10-9
Answer 1. the solution is unsaturated and no precipitation of PbI2 should occur (Qip< Ksp)
5. Which one of the following processes involves an increase in entropy?
Answer 4. H2O (l) → H2O (g) (the gas is less ordered than the liquid, so entropy increases)
6. Which one of the following pairs of species is a conjugate acid-base pair?
Answer 3. H2SO4, HSO4− (H2SO4 H+ + HSO4−)
7. Which combination of conditions describes a reaction that is always reactant-favored?
Answer 3. ΔH0 positive; ΔS0 negative (ΔG0 = ΔH0 − TΔS0 so when ΔH0 is positive and ΔS0
is negative, ΔG0 is necessarily positive, i.e. reactant-favored)
8. The following reaction is endothermic: 3 O2 (g) ' 2 O3 (g). This reaction is
Answer 2. non-spontaneous at all temperatures. (ΔH0 is positive (endothermic) and ΔS0 is
negative (more mols gas to fewer mols gas) )
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9. Consider the reaction in which 2 CO (g) + O2 (g) ' 2 CO2 (g). Calculate the Kelvin
temperature at which this reaction, with ΔH0 = −565.97 kJ and ΔS0 = −173.01 J/K, makes
the transition between reactant-favored and product-favored.
Answer 1. 3271 K (T = ΔH0/ΔS0 = −565.97 kJ/−0.17301 J/K )
10. If [H3O+] = 5.3 x 10−9 M in an aqueous solution, the solution is most correctly described as
being
Answer 3. basic (pH = −log (5.3 x 10−9) = 8.28 > 7.00, so solution is basic)
11. For the base ionization (hydrolysis) of the conjugate base of hydrogen cyanide, HCN, the
correct Kb expression is Kb =
Answer 2. [HCN][OH-]/[CN1-] (CN1- + H2O ' HCN + OH-)
12. The reason that HI is a stronger acid than HCl is that
Answer 1. bond dissociation energy decreases as you go down the periodic table. (and, the
weaker the H-X binary acid bond, the stronger the acid, see bottom p. 621, Hill)
13. The hydroxide ion concentration in a solution is 7.6 x 10−12 M. The pH of this solution is
Answer 3. 2.88 (pOH = −log (7.6 x 10−12) = 11.12, and pH = 14.00 −11.12 = 2.88)
14. What is the concentration of OH− in a solution that is 0.00025 M HNO3, a strong electrolyte?
Answer 5. 4.0 x 10−11 M (KW = [H3O+][OH−] = 1.00 x 10-14 = (0.00025)  [OH−] )
15. Which is necessarily true for a spontaneous process?
Answer 2. ΔG < 0 (when ΔG is negative (< 0), then the process is necessarily spontaneous
as written)
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Part II - Total points = 55. Answer each of the following questions. SHOW ALL WORK !
1. (10 points) Hypochlorous acid, HOCl, is a weak inorganic acid used as for water purification.
Ka = 3.0 x 10−8 for hypochlorous acid. For a 0.15 M aqueous hypochlorous acid solution,
(a) (2 points) write the complete ionization equation.
HOCl (aq) + H2O (l) ' H3O+ (aq) + OCl− (aq)
(b) (1 point) write the Ka expression for the ionization of hypochlorous acid.
Ka = 3.0 x 10−8 = {[H3O+][OCl−]}/ [HOCl]
(c) (5 points) calculate [H3O+] (you may "neglect x", but must justify why this is reasonable).
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C
E
[HOCl]
0.15
−x
0.15 − x
[H3O+]
0
+x
x
[OCl−]
0
+x
x
Ka = 3.0 x 10−8 = {[H3O+][C8H7O21-]}/ [HOCl] = (x)(x) / (0.15 − x) (assume x << 0.15)
so, Ka = 3.0 x 10−8 ≈ x2/0.15 D x = [H3O+] ≈ 6.7 x 10-5 M << 0.15 M
so assumption is justified.
(d) calculate the pH of the solution. (1 point)
pH = − log (6.7 x 10-5) = 4.17
(e) calculate the % ionization of hypochlorous acid in this solution. (1 point)
% ionization = x/C x 100 = (6.7 x 10-5 M/0.15 M) x 100 = 0.045 %
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2. (10 points) Calcium fluoride (fluorite), CaF2, is a sparingly soluble salt with Ksp = 3.9 x 10−11 .
(a) Write the solubility equilibrium that describes how this salt dissolves in water, indicating
which species are solids and which are aqueous, using our normal conventions.
CaF2 (s) ' Ca2+ (aq) + 2 F− (aq)
(b) Write the solubility product expression and then calculate the molar solubility (S) of
CaF2, expressing the concentration of Ca2+ in mol/L.
S
S
2S
−11
2+
− 2
2+
Ksp = 3.9 x 10 = [Ca ][ F ]
CaF2 (s) ' Ca (aq) + 2 F− (aq)
Ksp = 3.9 x 10−11 = (S)(2S)2 = 4S3
S3 = 9.75 x 10−12
S = 2.1 x 10−4 M
(c) Calculate the solubility of CaF2 (78.075 g/mol) in g/L.
g CaF2/L = (2.1 x 10−4 mol/L)(78.075 g/mol) = 1.7 x 10-2 g/L
(d) Calculate the molar solubility (S) of calcium fluoride in a solution that is also 0.020 M
with respect to F−. Show all work for credit!
S
S
2S+0.020
−11
2+
− 2
2+
Ksp = 3.9 x 10 = [Ca ][ F ]
CaF2 (s) ' Ca (aq) + 2 F− (aq)
Ksp = 3.9 x 10−11 = (S)(2S+0.020)2 ≅ (S)(0.020)2 = 3.9 x 10−11 (assume 2S<<0.020)
S = 3.9 x 10−11/(0.020)2 = 9.9 x 10−8 M
2S = 2 (9.9 x 10−8) = 2.0 x 10−7 << 0.020)
3. (10 points) A buffer is prepared using lactic acid, HC3H6O2, and sodium lactate, NaC3H6O2 so
that the resulting solution is 0.102 M in HC3H6O2 and 0.135 M in C3H6O2−. Calculate the pH
−4
of this solution if the Ka of HC3H6O2 is 1.38 x 10 . Show all work for credit!
0.102 M
0.135 M
+
HC3H6O2 (aq) + H2O (l) ' H3O (aq) + C3H6O2− (aq)
−4
Ka = 1.38 x 10
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C
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= {[H3O+][C3H6O2−]}/ [HC3H6O2] = (x)(0.135)/(0.102)
[HC3H6O2]
0.102
−x
0.102 − x
[H3O+]
0
+x
x
[C3H6O2−]
0.135
+x
0.135 + x
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So, x = [H3O+] = 1.38 x 10−4 x (0.102/0.135) = 1.04 x 10
−4
M
pH = − log (1.04 x 10−4) = 3.98
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Alternatively, pKa = − log (1.38 x 10−4) = 3.86
pH = pKa + log ([base]/[acid]} = 3.86 + log ([0.135]/[0.102]}
pH = 3.86 + 0.122 = 3.98
4. (15 points) Many reactions are product-favored at some temperatures and reactant-favored at
other temperatures. Consider the reaction at 298 K, in which
SiO2 (g) + 4 HF (g) ' SiF4 (g) + 2 H2O (g)
Table 1. Selected Values of ΔH f0 and Sf0
Compound or Element
SiO2 (g)
HF (g)
SiF4 (g)
H2O (g)
ΔH f0 (kJ/mol)
− 910.94
− 271.1
− 1614.94
− 241.818
Sf0 (J/mol K)
+ 41.84
+ 173.779
+ 282.49
+ 188.825
(a) Calculate ΔHo (in kJ) for this reaction at 298 K. (use attached table values)
ΔHo = [1 (−1614.94 kJ +2 (−241.818 kJ)] − [1 (−910.94 kJ + 4 (−271.1 kJ)] = −103.2 kJ
(b) Calculate ΔSo (in J/K) for this reaction at 298 K. (use attached table values)
ΔSo = [1 (282.49 J/K) +2(188.82 J/K)] − [1(41.84 J/K) + 4(173.78 J/K)] = −76.82 J/K
(c) Calculate the minimum temperature at which the reaction will be product-favored (you
may assume no change in the values of ΔH and ΔS with temperature).
ΔG0 = ΔH0 − TΔS0 ≡ 0 at equilibrium; so ΔH0 = TΔS0 and T = ΔH0/ΔS0
T = ΔH0/ΔS0 = (−103.2 x 103 J)/(−76.82 J/K = 1344 K
(d) Calculate ΔGo (in kJ) for this reaction at 298 K, using ΔHo and ΔSo values from (a) and
(b) above.
ΔG0 = ΔH0 − TΔS0 = −103.2 kJ − (298 K)( −0.07682 kJ/K)
ΔG0 = − 80.3 kJ
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(e) Calculate KC for this reaction at 298 K. Is it reactant-favored or product-favored at this
temperature? Explain briefly.
ΔG0 = − RT ln KC @ − 80.3 x 103 J = − (8.3143 J/mol K)(298 K)ln KC
ln KC = (− 80.3 x 103 J)/[− (8.3143 J/mol K)(298 K)] = +32.4
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KC = exp (+32.4) = 1.1 x 10
5. (10 points) Sodium benzoate, NaC7H5O2, is an organic salt used as a food preservative in soft
drinks. The benzoate ion, C7H5O2−, is the conjugate base of benzoic acid, HC7H5O2, for
which Ka is 6.3 × 10−5. Write the chemical equation for the ionization reaction of the
benzoate ion and then calculate the pH of an aqueous 0.0500 M solution of NaC7H5O2
(C7H5O2−). If you “neglect x”, you must clearly justify why you were able to do so. Show all
work for credit!
(1) NaC7H5O2 (aq) C7H5O2− (aq) + Na+ (aq)
(2) C7H5O2− (aq) + H2O (l) ' H3O+ (aq) + HC7H5O2
Kb = [H3O+][HC7H5O2]/[C7H5O2−] = Kw/Ka(HC7H5O2) = 1.0 x 10-14/6.3 x 10-5 = 1.6 x 10-10
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C
E
[C7H5O2−]
0.0500
−x
0.0500 − x
[HC7H5O2]
0
+x
x
[OH−]
0
+x
x
Assume x << 0.0500, especially since Kb = 1.6 x 10-10 (very reactant-favored)
Kb = 1.6 x 10-10 = {[ HC7H5O2][ OH−]}/[ C7H5O2−] = (x)(x) / (0.0500 − x) ≅ x2/0.0500
x2 ≅ 0.0500 (1.6 x 10-10) = 7.9 x 10-12 @ x = [OH−] ≅ 2.8 x 10-6 M << 0.0500
pOH = −log (2.8 x 10-6) = 5.55
pH = 14.00 − 5.55 = 8.45
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