New Exam 2

Chemistry 212-Exam 2	Name:
April 7, 2000		please print

Answers:

1. a	7. b	13. c
2. d	8. b	14. b
3. a	9. c	15. a
4. d	10. c	16. b
5. c	11. b
6. a	12. c

Solutions:

Instructions: Read the whole examination before starting to work on individual problems. Work the easiest problems first. Budget your time to insure that you finish the examination. Don’t waste time on problems that you find hard to finish until you have worked the others. Remember the use of programmable calculators is forbidden and would be considered a serious honor code violation.

1. Which reactant in the reaction below is a base (this one first in the answer) and which reactant is an acid?

S²^-(aq) + HBrO(aq) « HS^-(aq) + BrO^-(aq)

a. S²^-, HBrO	c. HBrO, BrO^-
b. S²^-, BrO^-	d. HBrO, S²^-

Solution1: On the left-hand side, HBrO is giving up a proton and S²^-is accepting it, making them an acid and base respectively.

2. What is the conjugate base and base dissociation constant of H₂O₂ if it’s acid dissociation constant is 2.400x10^-¹²?

a. , 4.17x10¹¹	c. , 4.17x10¹¹
b. , 4.17x10^-³	d. , 4.17x10^-³

Solution2: Since H₂O is acting as an acid, it’s conjugate base will have one less proton. Determine K_b by dividing K_a into K_w.

3. A solution with a hydronium ion concentration of 7.12x10^-⁶ is ___ and the hydroxide concentration is ___.

a. acidic, 1.40x10^‑9 M	c. basic, 1.40x10^‑9
b. basic, 1.40x10⁺⁵	d. acidic, 1.40x10⁺⁵

Solution3: Since the hydronium concentration is greater than 1.00x10^-⁷ M, it is acidic. The hydroxide concentration will be low (it can be calculated from K_w = [H₃O⁺][OH^-]).

4. Which of the acids below has the weakest conjugate base?

a. CH₃COOH	c. Cl₂CHCOOH
b. ClCH₂COOH	d. CCl₃COOH

Solution4: The electronegativity of chlorine pulls electron density away from the proton making it more acidic. The acid with the greatest number of chlorine is the most acidic. The weakest conjugate base is derived from the strongest conjugate acid.

5. Which mixture of products will be produced in significant amounts upon mixing of their conjugates? In decreasing strength their relative acidities are: , H₃PO₄, HCN, H₂O.

a. H₃PO₄ + CN^-	c. HCN +
b. +	d. HCN + OH^-

Solution5: Since these are products, the conjugate acid (of the base listed) and base (of the acid listed) must be stronger than the acid and base found in the products. HCN is a weaker than H₃PO₄ and CN^- is a stronger base than .

6. Which of the acids listed below is the strongest?

a. H₂S, K_a = 9.10x10^-⁸	c. , K_a = 4.8x10^-¹¹
b. HBrO, K_a = 2.0x10^-⁹	d. HClO, K_a = 3.0x10^-⁸

Solution6: The acid with the largest equilibrium constant is the strongest.

7. The pH of a solution of KOH was 10.402. What was the concentration of KOH?

a. 1.09x10^-⁴ M	c. 5.81x10^-⁴ M
b. 2.52x10^-⁴M	d. 2.26x10^-⁴ M

Solution7: The concentration of KOH will equal the [OH^-]

8. Which of the compounds below will not produce a neutral solution when dissolved into water?

a. CaBr₂	c. LiBr
b. Sr(OCl)₂	d. Sr(ClO₄)₂

Solution8: Look for anions or cations that are not derived from a strong acid or base. OCl^- is the conjugate base of HOCl a weak acid. This is thus the salt of a weak acid. The solution will be basic.

9. A solution of F^- had a pH of 8.19. What was the [OH^-]?

a. 2.0x10^-⁶ M	c. 1.6x10^-⁶ M
b. 8.19 M	d. 6.4x10^-⁹ M

Solution9: Determine the pOH and then the hydroxide concentration. We know that answers b and d aren’t reasonable. We need to do the calculations to determine between a and c.

10. Which solution below would have the greatest buffering capacity? Each solution was prepared from a weak base and the salt of its conjugate.

a. 0.093 M C₅H₅N and 0.093 M C₅H₆NNO₃, K_b = 1.4x10^-⁹

b. 0.055 M C₅H₅N and 0.055 M C₅H₆NF, K_b = 1.4x10^-⁹

c. 0.132 M (CH₃)₂NH and 0.132 M (CH₃)₂NH₂Cl, K_b = 5.4x10^-⁴

d. 0.089 M (CH₃)₂NH and 0.089 M (CH₃)₂NH₂Br, K_b = 5.4x10^-⁴

Solution10: The highest buffering capacity will be found with solutions of highest concentration and largest volume! This problem is focussing on the concentrations of the buffer components.

11. Calculate the pOH of 119.1 mL of a buffer initially consisting of 0.1327 M C₂H₅NH₂ and 0.1089 M C₂H₅NH₃NO₃ after addition of 0.0022 mol HCl. Assume no volume change occurs after addition of the acid. K_b (C₂H₅NH₂) = 5.6x10^-⁴.

a. 10.7	c. 3.0
b. 3.3	d. 3.2

Solution11: Determine the moles of conjugate base and acid after addition of HCl and then determine the pOH.

12. What is the [OH^-] of 0.158 M methylamine? CH₃NH₂ + H₂O « + OH^- K_b = 3.70x10^-⁴

a. 2.45x10^-¹²	c. 7.47x10^-³
b. 4.08x10^-³	d. 1.34x10^-¹²

Solution12: This is a weak acid; assume that the [OH^-] from the base is greater than from water to get:

Improve using the method of successive approximations.

13. What is the equilibrium [OH^-] in 0.1413 M H₂CO₃? H₂CO₃ + H₂O « H₃O⁺ + K_a = 4.30x10^-⁷

a. 7.07x10^-¹⁴ M	c. 4.06x10^-¹¹ M
b. 2.46x10^-⁴ M	d. 6.84x10^-⁶ M

Solution13: This is a weak acid. Determine the hydronium ion concentration, and then the hydroxide concentration. We should not that the hydroxide concentration will be relatively small. There is only one answer that fits the concentration level we would expect (answer c).

14. A solution consisting of 3.905x10^-³ M was found to be 79.456% ionized. What is the K_a?

+ H₂O « + H₃O⁺

a. 1.01x10^-²	c. 7.95x10^-¹
b. 1.20x10^-²	d. 1.51x10^-²

Solution14: Determine the sulfate and hydronium ion concentrations. Substitute into equilibrium expression to determine the equilibrium constant.

15. What is the pOH of 0.113 M HOC₂H₄NH₃NO₃? K_b for HOC₂H₄NH₂is 3.20x10^-⁵.

a. 8.8	c. 5.2
b. 12.0	d. 10.3

Solution15: This is the salt of a weak base, which is a weak acid. Determine K_a, pH and finally pOH.

16. When 25.00 mL of a 0.100 M solution of a weak acid was titrated by 0.100 M NaOH, the titration curve below resulted. What is the K_a of this weak acid?

a. 7.88x10^-⁷

b. 1.57x10^-⁸

c. 3.96x10^-⁵

d. 5.60x10^-¹¹

Solution16: The end point is after 25.00 mL of base; after 12.5 mL the pH = pK_a = 7.8. K_a = 10^-^7.8= 1.58x10^-⁸