STRONG>Key to Quiz #9 - Friday, May 31


ODD (i.e.top) PROBLEMS:

All of the questions on the quiz concern the following titration experiment: 10.0 mL of 0.08 M acetic acid (Ka = 1.8 x 10-5) is titrated with 0.08 M NaOH. Answer the following questions relating to this titration:
  1. Calculate the pH at 5.0 mL of titrant added. Is it...
    1. 3.75
    2. 7.00
    3. 4.75
    4. 1.40
    5. Insufficient information.
    ANSWER: This is the half-titration point. Therefore, the pH will equal the pKa at this point, or -Log(1.8 x 10-5) = 4.75, C.

  2. What is present in this solution at the equivalence point?
    1. A weak acid.
    2. A weak base.
    3. A strong base.
    4. Pure H2O.
    5. A strong acid.
    ANSWER: Just the weak base acetate; the product of the neutralization of acetic acid. B.

  3. What will happen to the pH if 100 mL of titrant are added?
    1. >14.00.
    2. 12.9
    3. 1.10
    4. 13.6
    5. Insufficient information.
    ANSWER: This titration curve (and every other) has an assymptote at large quantities of titrant added at the pH of the titrant. Therefore, in this problem, the pH will therefore approach -Log (1.00 x 10-14/0.08) = 12.90, B.

  4. Calculate the equivalence point pH. Is it.
    1. 1.40
    2. 8.74.
    3. 7.00
    4. 5.32.
    5. 7.98
    ANSWER: At the eq. pt., we have a weak base solution having a concentration given by 0.08(10.0/15.0) = 0.0533 M. The [OH-] of this solution will be given by the sq. root of (Kb)(0.0533 M) = 5.44 x 10-6. The pH is therefore equal to 8.74.


EVEN (i.e. bottom) PROBLEMS:

  1. Calculate the initial pH. Is it...
    1. 3.92
    2. 2.92
    3. 1.10
    4. 7.00
    5. Insufficient information.
    ANSWER: The standard procedure for calculating the pH of a seak acid solution is employed. The square root of (0.08)(1.8 x 10-5) = 1.2 x 10-3C. The negative Log of this is 2.92, B.

  2. What is present in the solution after the addition of 5.0 mL of titrant?
    1. A weak acid.
    2. A weak base.
    3. A buffer.
    4. 50% acetate; 50% acetic acid.
    5. All of the above.
    ANSWER: This is the half-titration point and the solution is, in fact, 50% acetate; 50% acetic acid. This is inclusive of the answers A-C, though, so the most correct answer is E, all of the above.

  3. At which titration point is a buffer created?
    1. 0 mL titrant added.
    2. 8.0 mL added.
    3. 12.0 mL added.
    4. 15.0 mL added.
    5. None of the above.
    ANSWER: At any titration point AFTER the addition of base has started, and BEFORE the eq. pt., a buffer exists. The equivalence point in this titration is at 10.0 mL. The only titration point meeting both of these criteria is, B, 8.0 mL added.

  4. Calculate the pH at the half-titration point (i.e. half-way to the eq. point). Is it...
    1. 3.75.
    2. 7.00.
    3. 4.75.
    4. 1.40
    5. 8.42.
    ANSWER: At the half-titration point, the pH will equal the pKa at this point, or -Log(1.8 x 10-5) = 4.75, C.
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Last modified: May 31, 1996