Information: | |
gas
constant: R = 0.08206 L·atm/mole·K
= 8.314 J/mole·K Avogadro's number: NA = 6.022 ´ 1023/mol Q-dependence of free energy change: DG = DGo + RT ln(Q) normal freezing point of water: 0°C = 273K water ionization constant (25°C): Kw = 1.00 ´ 10-14 pressure: 1 atm = 760 mmHg = 760 torr log conversion: ln(x) = 2.303 log(x) quadratic equation: ax2 + bx + c = 0, |
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Multiple choice: select one answer for each question:
1. Calculate K for HCN(aq) + OH-(aq) CN-(aq) + H2O(l) from the reactions
HCN(aq)
H+(aq)
+ CN-(aq)
, K = 4.9 ´10-10 H2O(aq) H+(aq) + OH-(aq) , K = 1.0 ´10-14 |
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a) |
4.9 ´ 104 |
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b) | 2.0 ´ 104 | |
c) | 4.9 ´ 10-24 | |
d) | 2.0 ´ 10-5 | |
e) | 2.0 ´ 10-25 | |
2. CaCO3 has Ksp= 4.8 ´10-9 . The molar solubility of calcium carbonate in water is
a) | 4.8 ´ 10-9 M | |
b) |
7.3 ´ 10-4 M |
|
c) | 6.9 ´ 10-5 M | |
d) | 7.3 ´ 10-7 M | |
e) | 6.7 ´ 10-3 M | |
3. In which of the following reactions is DS° negative?
1.
I2(g) ®
I2(s) 2. H2O(l) ® H2O(s) 3. CO2(g) + CaO(s) ® CaCO3(s) |
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a) | 1 only | |
b) | 2 only | |
c) | 3 only | |
d) | 1 and 2 only | |
e) | 1, 2 and 3 | |
4. Find the quotient [Ag+ ]/[Ag(CN)2- ] from the complex ion formation reaction
Ag+(aq) + 2 CN-(aq) Ag(CN)2- (aq), Kf = 3 ´1020
when the equilibrium concentration of CN- is 0.10 M.
a) | 3 ´ 10-19 | ||
b) | 3 ´ 10-4 | ||
c) | 3 ´ 10+2 | ||
d) | 3 ´ 10-6 | ||
e) | 3 ´ 10-22 | ||
5. The pH after mixing 0.500000 mole of HCl and 0.499996 mole of NaOH in 1000 mL of water is:
a) | 5.88 | |
b) | 5.13 | |
c) | 5.40 | |
d) | 4.92 | |
e) | 5.28 | |
6. Spontaneous
chemical reactions at constant T and P always have __________.
1. DS
> 0 ,
2. DH
< 0,
3. DG
< 0
a) | 1 only | |
b) | 3 only | |
c) | 1 and 2 only | |
d) | 2 and 3 only | |
e) | 1, 2 and 3 | |
7. The pH of a solution that contains 4.8 mole of benzoic acid C6H5COOH (pKa = 4.20) and 3.6 mole of sodium benzoate C6H5COONa in 927 mL of water is:
a) | 3.91 | |
b) | 4.32 | |
c) | 4.15 | |
d) | 4.08 | |
e) | 4.24 | |
8. When DSo < 0 and DHo < 0 a chemical reaction favors
|
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a) | products at low temperature and products at high temperature | |
b) | products at low temperature and reactants at high temperature | |
c) | reactants at low temperature and products at high temperature | |
d) | reactants at low temperature and reactants at high temperature | |
e) | none of the above | |
9. Calculate the equilibrium constant for the reaction at 298K:
|
3C(s) + 4H2(g) C3H8(g), DH° = -103.8 kJ, DS° = -269 J/K | |
a) | 1.0 ´ 1017 | |
b) | 2.1 ´ 1032 | |
c) | 1.0 | |
d) | 3.7 ´ 1019 | |
e) | 1.4 ´ 104 | |
10. Calculate the solubility of Cd(OH)2 (Ksp =1.2 ´10-14) in a solution buffered at pH = 7.00.
a) | 0.3 M | |
b) | 4.8 M | |
c) | 0.6 M | |
d) | 1.2 M | |
e) | 5.0 ´10-8 M | |
Questions 11-15: Titrate 72 mL of 0.200M formic acid HCOOH (Ka = 1.80 ´ 10-4) with 0.200M NaOH. |
11. The pH at the starting point is
a) | 1.80 | |
b) | 3.42 | |
c) | 2.85 | |
d) | 3.16 | |
e) | 2.22 | |
12. The volume of NaOH needed to just reach the buffer point is:
a) | 90 mL | |
b) | 54 mL | |
c) | 120 mL | |
d) | 36 mL | |
e) | 108 mL | |
13. The pH at the buffer point is
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a) | 3.74 | |
b) | 4.36 | |
c) | 4.68 | |
d) | 7.00 | |
e) | 9.32 | |
14. At pH = 5.50 the quotient [HCOO- ] / [HCOOH] is:
a) | 36 | |
b) | 18 | |
c) | 57 | |
d) | 69 | |
e) | 180 | |
15. The pH at the equivalence point is
a) | 7.00 | ||
b) | 8.37 | ||
c) | 7.94 | ||
d) | 8.62 | ||
e) | 9.43 | ||
Questions 16-20: Titrate 84 mL of 0.125 M acid H2A (Ka1 = 3.7 ´ 10-5 , Ka2 = 8.2 ´ 10-10 ) with 0.125 M NaOH . |
16. The volume of NaOH needed to just reach the first equivalence point is:
a) | 42 mL | |
b) | 84 mL | |
c) | 126 mL | |
d) | 105 mL | |
e) | 168 mL | |
17. The pH at the first equivalence point is
a) | 8.27 | |
b) | 7.00 | |
c) | 4.43 | |
d) | 6.76 | |
e) | 8.14 | |
18. The order of concentrations at the first equivalence point is:
a) | [HA- ] > [A2- ] > [H2A] | |
b) | [H2A] = [HA- ] > [A2- ] | |
c) | [HA- ] > [H2A] > [A2- ] | |
d) | [A2- ] = [HA- ] > [H2A] | |
e) | [HA- ] > [A2- ] = [H2A] | |
19. The order of concentrations at pH = 7.00 is:
a) | [HA- ] > [A2- ] = [H2A] | |
b) | [A2- ] = [HA- ] > [H2A] | |
c) | [HA- ] > [H2A] > [A2- ] | |
d) | [H2A] = [HA- ] > [A2- ] | |
e) | [HA- ] > [A2- ] > [H2A] | |
20. The order of concentrations at the second equivalence point is:
a) | [HA- ] > [H2A] > [A2- ] | |
b) | [A2- ] = [HA- ] > [H2A] | |
c) | [A2- ] > [HA- ] > [H2A] | |
d) | [HA- ] > [A2- ] = [H2A] | |
e) | [H2A] = [HA- ] > [A2- ] | |
KEY: |