Competitive Exams: Chemistry MCQs (Practice-Test 5 of 31)

  1. For the following reaction of hypochlorous acid (HOCl) with water, HOCl + H2O ⇔ H3O + + OCl-what would be the effect of adding sodium hypochlorite (NaOCl) to the reaction at equilibrium?

    1. The concentrations of both HOCl and H3O + would increase.

    2. The concentrations of both HOCl and H3O + would decrease.

    3. The concentration of HOCl would increase and the concentration of H3O + would decrease.

    4. The concentration of HOCl would decrease and the concentration of H3O + would increase.

    5. There would be no change because sodium hypochlorite is a salt without any acidic or basic properties.

  2. Calculate the molar hydronium ion concentration in a solution containing 0.23 M hypochlorous acid (HOCl), a monoprotic weak acid used in bleach solutions. For HOCl, Ka = 2.9 × 10 − 8.

    1. N/A

    2. N/A

    3. N/A

    4. N/A

  3. Given three separate solutions containing equal concentrations of formic acid (Ka = 1.7 × 10 − 4), phenol (Ka = 1.3 × 10 − 10), and acetic acid (Ka = 1.8 × 10 − 5), select the response below that has the acids arranged in order of increasing percent dissociation at equilibrium.

    1. formic < phenol < acetic

    2. formic < acetic < phenol

    3. acetic < formic < phenol

    4. phenol < acetic < formic

    5. No response is correct.

  4. From the following choices, select the one that would be the most basic (least acidic).

    1. 0.1 M hydrochloric acid (a strong acid)

    2. 0.1 M acetic acid (a weak acid)

    3. 0.1 M sodium acetate (the salt of a weak acid)

    4. 0.1 M ammonium chloride (the salt of a weak base)

    5. pure water

  5. If 10 mL a 1.0 × 10 − 4 M solution of a strong acid were added to 100 mL each of one solution containing 1.8 × 10 − 5 M hydrochloric acid and a second solution containing 1.0 M acetic acid (Ka = 1.8 × 10 − 5) plus 1.0 M sodium acetate, it is expected that the:

    1. pH of both solutions would remain unchanged.

    2. change in pH would be very large in both solutions.

    3. change in pH would be the same in both solutions.

    4. change in pH would be larger in the solution containing acetic acid and sodium acetate.

    5. change in pH would be larger in the HCl solution.

  6. Addition of a strong acid to a solution of acetic acid at equilibrium (HOAc + H2O ⇔ H3O + + OAc-) would cause the:

    1. acetate ion concentration to decrease.

    2. acetate ion concentration to increase.

    3. pH to increase.

    4. hydroxide ion concentration to increase.

    5. None of the above is correct.

  7. Calculate to a first approximation the molar concentration of hydronium ion in a 0.171 M solution of benzoic acid (HOBz, a monoprotic weak acid with Ka = 6.5 × 10 − 5).

    1. N/A

    2. N/A

    3. N/A

    4. N/A

  8. Given that the acid dissociation constant for benzoic acid (HOBz) is Ka = 6.5 × 10 − 5, calculate the basic dissociation constant, Kb, of the benzoate ion (OBz-).

    1. N/A

    2. N/A

    3. N/A

    4. N/A

  9. Benzoic acid, C6H5CO2H, is a weak acid (Ka = 6.3 × 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.

    1. N/A

    2. N/A

    3. N/A

    4. N/A

  10. Which of the following weak acid dissociation constants would result in the smallest degree of dissociation?

    1. Ka = 1.0 × 10 − 2

    2. Ka = 1.0 × 10 − 3

    3. Ka = 1.0 × 10 − 4

    4. Ka = 1.0 × 10 − 5

  11. Addition of sodium acetate to an acetic acid solution at equilibrium will cause:

    1. no change in H3O + concentration.

    2. H3O + concentration to decrease.

    3. H3O + concentration to increase.

    4. concentrations of all species to increase.

    5. a decrease in hydroxide concentrations.

  12. What is the H3O + concentration in a 0.17 M solution of a weak acid, HA, with a dissociation constant of 3.21 × 10 − 6.

    1. N/A

    2. N/A

    3. N/A

    4. N/A

  13. Calculate the hydronium ion concentration in a solution that contains 0.21 M acetic acid and 0.17 M sodium acetate. For acetic acid, Ka = 1.8 × 10 − 5.

    1. N/A

    2. N/A

    3. N/A

    4. N/A

  14. What is the concentration of the HPO42-ion in a 0.078 M solution of phosphoric acid (H3PO4)? [Ka1 = 7.5 × 10 − 3; Ka2 = 6.3 × 10 − 8; Ka3 = 4.8 × 10 − 13]

    1. N/A

    2. N/A

    3. N/A

    4. N/A

  15. Why is it necessary to take the acid-base properties of water into account when computing the hydronium ion concentration of very dilute solutions of strong acids?

    1. The hydroxide ion produced from the dissociation of water reacts with most of the hydronium ion produced from the acid.

    2. The dissociation constant for water is larger in dilute rather than in concentrated solutions of acids.

    3. The acids do not dissociate completely in dilute solutions.

    4. The amount of hydronium ion produced by the dissociation of water is significant compared to that produced by the acid.

    5. The conjugate base of the strong acid reacts with the hydroxide ion produced from the dissociation of water.