che141b_test[3] memo_sept 2012 (1)
TRANSCRIPT
![Page 1: CHE141B_test[3] Memo_Sept 2012 (1)](https://reader036.vdocuments.mx/reader036/viewer/2022082703/5572133d497959fc0b91e83e/html5/thumbnails/1.jpg)
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)
1
![Page 2: CHE141B_test[3] Memo_Sept 2012 (1)](https://reader036.vdocuments.mx/reader036/viewer/2022082703/5572133d497959fc0b91e83e/html5/thumbnails/2.jpg)
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)](https://reader036.vdocuments.mx/reader036/viewer/2022082703/5572133d497959fc0b91e83e/html5/thumbnails/3.jpg)
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 √
√
√√
3
![Page 4: CHE141B_test[3] Memo_Sept 2012 (1)](https://reader036.vdocuments.mx/reader036/viewer/2022082703/5572133d497959fc0b91e83e/html5/thumbnails/4.jpg)
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)
√
√
√√
4