Chemistry 1412
EXAM # 2
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CHEM 1412 Exam #2
(Chapters 15, 16, and 18)
Name:___________________
Score:
PART I - (3 points each) - Please write your correct answer next to each question number. DO NOT
CIRCLE
_____1. In the reaction HSO4–(aq) + OH–(aq) SO42–(aq) + H2O(l), the conjugate acid-base
pairs are
_____2. Which one of these salts will form an acidic solution upon dissolving in water?
A. LiBr
B. NaF
C. FeBr3
D. KOH
E. NaCN
_____3. Determine the pH of a KOH solution made by mixing 0.251 g KOH with enough water to
make 1.00 × 102 mL of solution.
A. 1.35
B. 2.35
C. 7.00
D. 11.65
E. 12.65
_____4. Acid strength increases in the series: HCN < HF < HSO4–. Which of these species is the
strongest base?
A. H2SO4
B. SO42-
C. F-
D. CN-
E. HSO4-
_____5. Which one of these equations represents the reaction of a weak base with a strong acid?
A.
B.
C.
D.
E.
H+(aq) + OH–(aq) → H2O(aq)
H+(aq) + CH3NH2(aq) → CH3NH3+(aq)
OH–(aq) + HCN(aq) → H2O(aq) + CN–(aq)
HCN(aq) + CH3NH2(aq) → CH3NH3+(aq) + CN–(aq)
None of the above
_____6. Calculate the pH of a 0.14 M HNO2 solution that is 5.7% ionized.
A. 0.85
B. 1.70
C. 2.10
D. 11.90
E. 13.10
_____7. The Ksp for Ag3PO4 is 1.8 × 10–18. Calculate the molar solubility of Ag3PO4.
A. 1.6 × 10–5 M
B. 2.1 × 10–5 M
C. 3.7 × 10–5 M
D. 7.2 × 10–1 M
E. 0.18 M
_____ 8. Calculate the minimum concentration of Mg2+ that must be added to 0.10 M NaF in
order to initiate a precipitate of magnesium fluoride. (For MgF2 , Ksp = 6.9 × 10–9.)
A. 1.4 × 107M
B. 6.9 × 10–9M
C. 6.9 × 10–8M
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D. 1.7 × 10–7M
E. 6.9×10–7
.____ 9. Which one of the following is a buffer solution?
A. 0.40 M HCl and 0.10 KCl
C. 1.0 M NH3 and 1.0 M NH4NO3
E. All of the above
B. 1.0 M HNO3 and 1.0 M NaNO3
D. 0.40 M HCl and 0.10 KOH
____10. For which type of titration will the pH be acidic at the equivalence point?
_
A. Strong acid vs. strong base
B. Strong acid vs. weak base
C. Weak acid vs. strong base
D. All of the above
E. None of the above
_____11. Which response includes all the following processes that are accompanied by an
increase in entropy?
1) 2SO2(g) + O2(g) → SO3(g)
2) H2O(l) → H2O(s)
3) Br2(l) → Br2(g)
4) H2O2(l) → H2O(l) + 1/2O2(g)
A. 1, 2, 3, 4
B. 1,2
C. 2, 3, 4
D. 3, 4
E. 1,4
_____12. If the molar solubility is represented by X, which of the following expressions
is incorrect?
A. AgClO3, KSP=X2
D. Ag3PO4, Ksp=9X4
B. Al2S3, Ksp= 108X5
E. PbF2, Ksp=4X3
C. SrSO4, Ksp=X2
____13. Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from
further corrosion. 4Al(s) + 3O2(g) → 2Al2O3(s)
Using the thermodynamic data provided below, calculate ΔS° in J/K.mol for this reaction.
S°(J/K·mol)
Al(s)
28.3
O2(g)
205.0
Al2O3(s)
`A. 182.3
B. 131.5
50.99
C. –182.3
D. –626.2
E. –802.9
____ 14. Ozone (O3) in the atmosphere can reaction with nitric oxide (NO):
O3(g) + NO(g) → NO2(g) + O2(g).
Calculate the ΔG° for this reaction at 25°C. (ΔH° = –199 kJ/mol, ΔS° = –4.1 J/K·mol)
A. 1020 kJ/mol
D. –1.42 × 103 kJ/mol
B. –1.22 × 103 kJ/mol
E. –198 kJ/mol
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C. 2.00 × 103 kJ/mol
____15. Calculate ΔG° for the reaction 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g).
ΔG°f (kJ/mol)
–237.2
H2O(l)
HNO3(l)
–79.9
86.7
NO(g)
NO2(g)
51.8
A. 8.7 kJ/mol
D. –192 kJ/mol
C. –414 kJ/mol
B. 192 kJ/mol
E. –155 kJ/mol
____ 16. Consider a reaction for which ΔH = -90.84 kJ/mol; ΔS = 108.23 J/K-mol
Which of the following statements is true?
A. The reaction is spontaneous at all temperatures
B. The reaction becomes spontaneous at low temperatures below 119 K
C. The reaction becomes spontaneous at high temperatures above 119 K
D. The reaction becomes spontaneous at low temperatures below 840 K
E. The reaction becomes spontaneous at high temperatures above 840 K
____17. Which of the following in each of the following pairs would you expect to have
the lower standard entropy (S°)?
i. Kr(g)
i. Br2(l)
i. O2 at 298K, 1atm
or
or
or
ii. Ne(g)
ii. I2(s)
ii. O2 at 298K, 3 atm
A. i, i, i
D. i, ii, ii
B. ii, ii, ii
E. i, ii, i
C. ii, i, i
____18. Determine the equilibrium constant (Kp) at 25°C for the reaction
CO(g) + H2O(g)  CO2(g) + H2(g) ΔG° = –28.5 kJ/mol.
A. 2.9 × 10–60
B. 1.0 × 10–4
C. 9.9 × 104
D. 3.4 × 1059
E. 1.2
____19. The pH of a 0.020 M solution of an unknown weak acid, HA, is 5.0. The Ka of the acid is:
A. 10-5
B. 5×10-4
C. 10-10
____20. Which of the following statements is correct:
A. The entropy of an element in its stable form is zero
B. The free energy is a minimum at equilibrium
C. The free energy increases in a spontaneous reaction
D. All endothermic reactions are nonspontaneous
E. The entropy always increases in all spontaneous reactions
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D. 5×10-9
E. 0.10
PART II - ( 8 points each) Please show all your work.
21. Consider a 0.021 M NaCN solution; [Ka of HCN = 4.9 × 10–10]
a. What is the Kb of CN-?
b. Write the associated chemical EQUATION for hydrolysis by CNc. Calculate the pH of a 0.021 M NaCN solution.
22. A 500.0 mL buffer solution consists of 0.10 M NaOCl and 0.20 M HOCl.
a. What is the pH of this solution? [Ka(HOCl) = 3.2 × 10–8].
b. Calculate the pH of the buffer upon addition of 100.0 mL of 0.10 M NaOH.
23. The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC.
a. What is the molar solubility for lead(II) iodide at this temperature?
b. What is the solubility product for lead(II) iodide at this temperature?
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24. Consider the titration of 40 mL of 0.15 M HCl with 0.10 M Ba(OH)2?
a. Calculate the pH upon addition of 20.0 mL of the base
b. Calculate the pH upon addition of 35.0 mL of the base
25. The equilibrium constant for the reaction AgBr(s) Ag+(aq) + Br– (aq) is Ksp = 7.7 × 10–13 at 25°C.
a. Calculate ΔGo for the reaction.
b. Calculate ΔG for the reaction when [Ag+] = 1.0 × 10–2 M and [Br–] = 1.0 × 10–3 M.
c. Is the reaction spontaneous or nonspontaneous at these concentrations?
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BONUS QUESTION - (10 points)
Consider the titration of 100 mL of 0.10 M HCN (Ka = 4.9 × 10–10) with 0.10 M NaOH. Calculate the pH
during the following stages of the titration:
.
A. After adding 40.0 ml of 0.10 M NaOH
B. At the half equivalence point
C. At the equivalence point
1. A
15. A
2. C 3. E
16. A 17.B
4. D 5. B 6. C 7. A
18. C 19. D 20. B
8. E
9. C
10. B 11. D 12. D 13. D 14. E
PART II - ( 8 points each) Please show all your work.
21. a. Kb=10-14/4.9 × 10–10 = 2.0 × 10-5
b. CN-(aq) + H2O(l) HCN(aq) +OH-(aq)
c. [OH-] = sqrt(2.0 × 10-5×0.021) = 6.5 × 10-4M
pOH = -log(6.5 × 10-4) = 3.2; pH = 14-3.2 = 10.8
22.
a. pH = -log(3.2 × 10–8) + log(0.10/0.20) = 7.2
b. OH- +HOCl  OCl- + H2O; moles of OH- added = 0.100 L × 0.10M = 0.010 moles
Moles of HOCl before neutralization = 0.500 L × 0.20M = 0.10 moles
Moles of HOCl after neutralization = 0.10 moles – 0.010 moles = 0.090 moles
Moles of OCl- before neutralization = 0.500 L × 0.10M = 0.050 moles
Moles of OCl- after neutralization = 0.050 moles +0.010 moles = 0.060 moles
pH = -log(3.2 × 10–8) + log(0.060/0.090) = = 7.3
23.
a. 0.064g/100 mL = 0.64 g/L = (0.64/461)mol/L = 1.4 × 10-3M = molar solubility, X
b. PbI2(s)Pb+2(aq) + 2I-(aq); Ksp = 4X3 = 4×(1.4 × 10-3)3 = 1.1×10-8
24.
a. H+ + OH-  H2O
H+
Initial moles
0.0060
Change in moles
-0.0040
Final moles
0.0020
+
Excess [H ] = 0.0020 moles/0.060 L = 0.033 M
pH = -log(0.033) =1.5
OH0.0040
-0.0040
0
b. a. H+ + OH-  H2O
Initial moles
Change in moles
Final moles
H+
0.0060
-0.0060
0
OH0.0070
-0.0060
0.0010
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Excess [OH-] = 0.0010 moles/0.075 L = 0.013 M
pOH = -log(0.0133) =1.9; pH = 14-1.9 = 12.1
25.
a. ΔGo=-RTln(Ksp) = -8.314 JK-1mol-1×298K× ln(7.7 × 10–13) = 69100 J
b. ΔG = ΔGo + RTln(Q); 69100 J +8.314 JK-1mol-1×298K× ln(Q)
Q=[Ag+]×[Br–]=1.0 × 10–2 ×1.0 × 10–3= 1.0 × 10–5
ΔG = ΔGo + RTln(Q) = 69100 J +8.314 JK-1mol-1×298K× ln(1.0 × 10–5) = 40600 J
c. Not spontaneous as ΔG > 0; Another way to arrive at this : since Q>Ksp, precipitation and not solubility is
favored.
BONUS QUESTION - (10 points)
HCN(aq) + OH-(aq)  CN-(aq) +H2O(l)
A.
HCN
OH
CNInitial Moles
0.010
0.0040
0
Change in Moles
-0.0040
-0.0040
0.0040
Final Moles
0.0060
0
0.0040
Solution is a buffer since it has 0.0060 moles of HCN and 0.0040 moles of it conjugate base, CN-.
pH = -log(4.9 × 10–10) + log(0.0040/0.0060) = 9.1
B. pH = pKa at half-equivalence point = -log(4.9 × 10–10) = 9.3
C. At equivalence point, moles of HCN used = mole of OH- used = moles of CN- formed
HCN
OH
CNInitial Moles
0.010
0.010
0.010
Change in Moles
-0.010
-0.010
0.010
Final Moles
0
0
0.010
Solution only contains 0.010 moles of the weak base, CN-.
Kb of CN- = 10-14/(4.9 × 10–10) = (2.0 × 10–5)
CN-(aq) + H2O(l) HCN(aq) +OH-(aq)
L of NaOH added = moles of NaOH used/M of NaOH used = 0.010 moles/0.10M = 0.10 L
Therefore final volume of solution = L of HCN + L of NaOH = 0.10 L + 0.10 L = 0.20 L
[CN-] at equivalence point = 0.010 moles/0.20 L = 0.050 M
[OH-] formed at equivalence point due to hydrolysis by CN- = sqrt(2.0 × 10–5×0.050) =0.0010 M
pOH = -log(0.0010) = 3.0; pH = 14 – pOH = 11.0
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