che141b_test[3] memo_sept 2012 (1)

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DEPARTMENT OF CHEMISTRY CHEMISTRY IA (CHE141B) SEMESTER TEST 3 Time allowed: 90 minutes Total marks: 50 1. Define or describe 1.1. Common ion effect (2) The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. √√ 1.2. A Brønsted-Lowry acid (2) A Brønsted-Lowry acid is a proton (H + ) donor. √√ 1.3. A Lewis base (2) A Lewis base is an electron pair donor. √√ 1.4. A buffer solution (2) A buffer solution is an aqueous solution consisting of a mixture of a weak acid 1

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Page 1: CHE141B_test[3] Memo_Sept 2012 (1)

DEPARTMENT OF CHEMISTRYCHEMISTRY IA (CHE141B)

SEMESTER TEST 3

Time allowed: 90 minutes Total marks: 50

1. Define or describe

1.1. Common ion effect (2)

The common ion effect is the shift in equilibrium caused by the addition of a

compound having an ion in common with the dissolved substance. √√1.2. A Brønsted-Lowry acid (2)

A Brønsted-Lowry acid is a proton (H+) donor. √√1.3. A Lewis base (2)

A Lewis base is an electron pair donor. √√1.4. A buffer solution (2)

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. √√

2. Distinguish between homogenous and heterogeneous equilibrium. (4)

A homogeneous equilibrium is one in which everything in the equilibrium mixture is present in the same phase, whereas for heterogeneous equilibrium, different phases of either the reactants and/or products will be observed. √√√√

3. Why for a polyprotic acid, Ka2 is much less than Ka1? (2)

Ka2 << Ka1, since it is always easier to remove the first proton from a neutral

polyprotic acid than to remove the second proton from a negatively charged ion.

√√

4. Give the conjugate acid for each of the following:

4.1. SO42- HSO4

- √ (1)

4.2. HSO3- H2SO3 √ (1)

4.3. HPO42- H2PO4

- √ (1)

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Page 2: CHE141B_test[3] Memo_Sept 2012 (1)

5. In a study of the chemistry of glass etching, an inorganic chemist examines the

reaction between sand (SiO2) and hydrogen fluoride at a temperature above the

boiling point of water:

SiO2(s) + 4HF(g) SiF4(g) + 2H2O(g)

Predict the effect on [SiF4] when

5.1. H2O(g) is removed; increases √ (1)

5.2. HF is added; increases √ (1)

5.3. H2O(g) is added; decreases √ (1)

5.4. some sand is removed; decreases √ (1)

5.5. a catalyst is added. no effect √ (1)

Answer increases, decreases or no change.

6. Decomposition of CO2(g) at elevated temperature is given by the following

reaction:

2CO2(g) 2CO(g) + O2(g)

At 3000 K, 2.00 mol of CO2 is placed into a 1.00 L container and allowed to come

to equilibrium. At equilibrium, 0.90 mol CO2 remains.

6.1. Calculate the equilibrium concentrations of CO and O2. (4)

2CO2(g) 2CO(g) + O2(g)

Initial 2.00 0 0

Change -2x +2x +x

Equilibrium 2.00 – 2x 2x x √

2.00 – 2x = 0.90

2x = 1.10

x = 0.55 √

[CO] = 1.10 M √ and [O2] = 0.55 M √

6.2. Calculate the value for Kc at this temperature. (2)

Kc = [CO] 2 [O 2] = (1.10) 2 (0.55) = 0.82 √√

[CO2]2 (0.90)2

7. At 100oC the equilibrium constant for the reaction

2

Page 3: CHE141B_test[3] Memo_Sept 2012 (1)

COCl2(g) CO(g) + Cl2(g)

has the value of Kc = 2.19 x 10-10. Are the following mixtures of COCl2, CO, and

Cl2 at 100oC at equilibrium? If not, indicate the direction that the reaction must

proceed to achieve equilibrium.

7.1. [COCl2] = 2.00 x 10-3 M, [CO] = 3.3 x 10-6 M, and [Cl2] = 6.62 x 10-6 M; (4)

Q = [CO] [Cl2] = 3.3 x 10 -6 x 6.62 x 10 -6 = 1.1 x 10-8 √√

[COCl2] 2.00 x 10-3

Q > Kc, the reaction will proceed to the left. √√

7.2. [COCl2] = 4.50 x 10-2 M, [CO] = 1.1 x 10-7 M, and [Cl2] = 2.25 x 10-6 M. (4)

Q = [CO] [Cl2] = 1.1 x 10 -7 x 2.25 x 10 -6 = 5.5 x 10-12 √√

[COCl2] 4.50 x 10-2

Q < Kc, the reaction will proceed to the right. √√

8. What are the concentrations of hydronium (H3O+) and hydroxide (OH‐) ions in

0.075 M Ca(OH)2 solution? (4)

Ca(OH)2 → Ca2+ + 2OH-

[OH-] = 2 x 0.075 = 0.15 M √√

Kw = [H3O+][OH-]

[H3O+] = 1.0 x 10 -14 = 6.7 x 10-14 M √√

0.15

9. Benzoic acid, C6H5COOH, is a weak acid (Ka = 6.3 x 10-5). Calculate the initial

concentration (in M) of benzoic acid that is required to produce an aqueous

solution of benzoic acid that has a pH of 2.54. (5)

= 10-2.54 = 2.884 x 10-3 M √

√√

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10. Calculate the pH of a solution that contains 0.55 M acetic acid and 0.195 M potassium acetate. For acetic acid, Ka = 1.8 x 10-5. (5)

√√

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