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Equilibrium Multiple Choice Review Questions for AP Chemistry (page 2)

based on 6 ratings
By — McGraw-Hill Professional
Updated on Feb 9, 2011

Answers and Explanations

  1. B—An acid, any acid, will give a pH below 7; thus, answers C–E are eliminated. A 0.1-molar solution of a strong acid would have a pH of 1. Acetic acid is not a strong acid, so answer A is eliminated.
  2. A—The K nearest 10–8 will give a pH near 8. The answer must involve the H2PO4 ion.
  3. C—The generic Ka is:
  4. Ka = [H+][A]/[HA] = 9 × 10–4 = x2/1.0 – x

    x = 3 × 10–2 M and the percent dissociation is (3 × 10–2/1.0) × 100% = 3%

    You will need to be able to do calculations at this level without a calculator.

  5. D—The two substances are not a conjugate acid–base pair, so this is not a buffer. Both compounds are salts of a strong base and a weak acid; such salts are basic (pH > 7).
  6. C—The two substances constitute a conjugate acid–base pair, so this is a buffer. The pH should be near –log Ka1. This is about 2 (acid).
  7. E—The two substances constitute a conjugate acid–base pair, so this is a buffer. The pOH should be near –log Kb. This is about 4. The pH would be about 14 – 4 = 10.
  8. B—Any time an acid is added, the pH will drop. The reaction of the weak base with the acid produces the conjugate acid of the weak base. The combination of the weak base and its conjugate is a buffer, so the pH will not change very much until all the base is used. After all the base has reacted, the pH will drop much more rapidly. The equivalence point of a weak base–strong acid titration is always below 7 (only strong base–strong acid titrations will give a pH of 7 at the equivalence point). The value of pOH is equal to pKb half-way to the equivalence point.
  9. E—If pH = 4.0, then [H+] = 1 × 10–4 = [A], and [HA] = 0.30 – 1 × 10–4. The generic Ka is [H+][A]/[HA], and when the values are entered into this equation: (1 × 10–4)2/0.30 = 3.3 × 10–8. Since you can estimate the answer, no actual calculations need be done.
  10. A—This is an acid-dissociation constant, thus the solution must he acidic (pH < 7). The pH of a 0.010 M strong acid would be 2.0. This is not a strong acid, so the pH must be above 2.
  11. A—A is the salt of a strong acid and a weak base; it is acidic. B and E are salts of a strong acid and a strong base; they are neutral. C and D are salts of a weak acid and a strong base; they are basic. The lowest pH would be the acidic choice.
  12. E—Sodium nitrite is a salt of a weak acid and a strong base. Ions from strong bases, Na+ in this case, do not undergo hydrolysis, and do not affect the pH. Ions from weak acids, NO2 in this case, undergo hydrolysis to produce basic solutions.
  13. B—The presence of a strong acid, HNO3, would make this the most acidic (lowest pH).
  14. E—The weak acid and the weak base partially cancel each other to give a nearly neutral solution.
  15. C—Both A and C are buffers, because they have conjugate acid–base pairs of either a weak acid (A) or a weak base (C). The weak acid buffer would have a pH below 7, and the weak base buffer would have a pH above 7.
  16. A—See the answer to question 14.
  17. A—The equilibrium constant expression is: Kb = 4.0 × 10–10 = [OH][C6H5NH3+ ]/[C6H5NH2]. This expression becomes: (x)(x)/(1.0 – x) = 4.0 × 10–10, which simplifies to: x2/1.0 = 4.0 × 10–10. Taking the square root of each side gives: x = 2.0 × 10–5 = [OH]. Since you can estimate the answer, no actual calculations need be done.
  18. E—The solubility-product constant expression is: Ksp = [Zn2+][IO3]2 = 4 × 10–6. This may be rearranged to: [IO3]2 = 4 × 10–6/[Zn2+]. Inserting the desired zinc ion concentration gives: [IO3]2 = 4 × 10–6/(1 × 10–6) = 4. Taking the square root of each side leaves a desired IO3 concentration of 2 M. 2 mol of KIO3 must be added to 1.00 L of solution to produce this concentration. Since you can estimate the answer, no actual calculations need be done.
  19. A—When dealing with gaseous equilibria, volume changes are important when there is a difference in the total number of moles of gas on opposite sides of the equilibrium arrow. All the answers, except A, have differing numbers of moles of gas on opposite sides of the equilibrium arrow.
  20. C—Hydrolysis of any ion begins with the interaction of that ion with water. Thus, both the ion and water must be on the left side of the equilibrium arrow, and hence in the denominator of the equilibrium-constant expression (water, as with all solvents, will be left out of the expression). The oxalate ion is the conjugate base of a weak acid. As a base it will produce OH in solution along with the conjugate acid (HC2O4) of the base. The equilibrium reaction is: C2O42–(aq) + H2O(l) OH(aq) + HC2O4(aq).
  21. D—The low value for the equilibrium constant means that the equilibrium lies to the left. For this to be true, the weaker acid and the weaker base must be on the left side.
  22. E—The equilibrium given is actually the sum of the following three equilibria:
  23. Summing these equations means you need to multiply the equilibrium constants:

    Ksum = KspKK' = Ksp/Ka2Ka1

          = 1.6 × 10–24/[(1 × 10–19) (9.5 × 10–8)]
  24. D—As the reaction moves to the left, the H2O behaves as a base (accepts H+). When the reaction moves to the right, HPO42– behaves as a base.
  25. D—The equilibrium constant for the two successive ionizations will be the product of the two equilibrium constants given. Thus, K = Ka1Ka2 = (1.5 × 10–3)(2.0 × 10–6).
  26. B—The addition or removal of some solid, as long as some remains some present, will not change the equilibrium. An increase in volume will cause the equilibrium to shift towards the side with more moles of gas (right). Raising the temperature of an endothermic process will shift the equilibrium to the right. Any shift to the right will increase the amounts of the products.
  27. E—Assuming 1 mol of each reactant is used, the equilibrium quantities would be: [C2H4] = 1 – x, [O2] = 1 – 3x, [CO2] = 2x, and [H2] = 2x. Unless the value of x is known, it is not possible to relate the actual concentrations of any reactant to any product.
  28. B—Using the following table:
  29. The presence of 0.20 mol of CO (0.20 M) at equilibrium means that 2x = 0.20 and that x = 0.10. Using this value for x, the bottom line of the table becomes:

    The equilibrium expression is: K = [CO]2[H2]2/[CH4][CO2]. Entering the equilibrium values into the equilibrium expression gives: K = (0.20)2(0.20)2/(0.20)(0.30)

  30. A—The addition of a product will cause the equilibrium to shift to the left. The amounts of all the reactants will increase, and the amounts of all the products will decrease (the O2 will not go below its earlier equilibrium value since excess was added). The value of K is constant, unless the temperature is changed. The rates of the forward and reverse reactions are equal at equilibrium.
  31. C—The only way to change the value of K is to change the temperature. For an exothermic process (ΔH < 0), K is increased by a decrease in temperature.
  32. B—The low value of K means that the equilibrium lies to the left. The equilibrium always lies away from the stronger acid and the stronger base.
  33. B—Nitric acid, being an acid, will react with a base. In addition to obvious bases containing OH, the salts of weak acids are also bases. All of the anions, except CN, are from strong acids.
  34. A—The equilibrium constant expression for the dissolving of manganese(II) hydroxide is:
  35. Ksp = [Mn2+][OH]2 = 1.6 × 10–13

    If s is used to indicate the molar solubility, the equilibrium expression becomes:

    Ksp = (s)(2s)2 = 4s3 = 1.6 × 10–13

    This rearranges to:

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