Practice problems for using the Henderson-Hasselbalch equation for buffers
Notes:
A. pKa = 14 - pKb
B. These problems are qualitative or semi-quantitative. You should not need a calculator to solve any
of them.
1. Acetic acid, a monoprotic acid, has a pKa of 4.8. What pH buffer solution would it be ideal for
preparing?
2. Given the need to prepare a solution with its pH buffered to 7.4, which of the following monoprotic
acids and bases would be the best to use based on their pKa?
methylamine pKa = 10.62
Formic acid pKa = 3.6
Tris pKa = 8.3
TAPS pKa = 8.5
MES pKa = 6.2
TES pKa = 7.6
3. In a solution buffered to a pH of 7.0, when the pH is changed to 8.0 by adding NaOH, how much
change would there be in the ratio of (conjugate) base to (conjugate) acid?
4. You are given the same volume of two solutions (A and B) buffered with the same buffering agent
to pH 6.5. The buffering agent's total concentration in solution A is 10 mM while in solution B it is
100 mM. If you add and mix in 1/100th of the solution volume of 1 N HCl to each of solution A
and Solution B, which solution will have the lower pH? Why?
5. In a pH 8.0 solution of ethanolamine (HOCH2CH2NH2), which has a pKa of 9.5, which form is in
excess, the neutral form or the cationic form (or neither)? Why?
6. In a pH 4.5 solution of pyruvic acid, a monocarboxylic acid with a pKa of 4.5, which form is in
excess, the anionic form or the neutral form, or neither? Why?
7. Consider a solution of 0.1 M Tris, a primary amine with a pKa of 8.3. The pH of this solution has
been adjusted to 7.3. To the nearest order of magnitude, what is the ratio of the concentrations in this
solution of conjugate acid of Tris to the base form of Tris?
Answers are on the next page.
Answers:
1. 4.8
2. TES
3. It would increase by 10-fold
4. Solution A, because it is less strongly buffered, it resists the change in pH less, so its pH changes
more for any given "challenge" of added protons or hydroxyl ions. Adding acid always lowers the
pH of the buffer, although it may be by only a very small amount. This is true independent of the
buffering agent. In this case, the pH of both solutions starts out the same, and both decrease, but by
different amounts. Hence the one that changes the most will have the lower pH.
5. The cationic form (the conjugate acid), because at all solution pH values below the pKa the
conjugate acid is in excess. The pH stated in the problem, 8.0, is below the stated pKa (9.5).
6. Neither is excess because their concentrations are equal when the pH = pKa.
7. 10. Remember that the ratio is 1 when the pH = pKa and increases by a power of ten for each pH
unit below the pKa. Hence, since pH 7.3 is one pH unit below the pKa the ratio (of acid to base)
increases by 10-fold. Starting from a ratio of 1, that means it would be 10.