Buffers are solutions that resist large changes in pH when small amounts of acid or base are added. They matter in chemistry, biology, medicine, and environmental science because many reactions only work well within a narrow pH range. Blood, enzyme systems, lakes, and laboratory solutions all rely on buffer behavior.
A buffer usually contains a weak acid and its conjugate base, or a weak base and its conjugate acid.
Key Facts
- Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA])
- For a weak base buffer: pOH = pKb + log([BH+]/[B])
- When [A-] = [HA], pH = pKa because log(1) = 0.
- A buffer works best when pH is within about 1 unit of pKa.
- Adding acid consumes conjugate base: H+ + A- -> HA.
- Adding base consumes weak acid: OH- + HA -> A- + H2O.
Vocabulary
- Buffer
- A buffer is a solution that resists changes in pH when small amounts of acid or base are added.
- Conjugate acid-base pair
- A conjugate acid-base pair consists of two species that differ by one proton, such as HA and A-.
- pKa
- pKa is a measure of acid strength equal to -log(Ka), and it helps predict the pH range where a buffer works best.
- Buffer capacity
- Buffer capacity is the amount of acid or base a buffer can absorb before its pH changes greatly.
- Henderson-Hasselbalch equation
- The Henderson-Hasselbalch equation relates buffer pH to pKa and the ratio of conjugate base to weak acid.
Common Mistakes to Avoid
- Using strong acid and strong base as a buffer is wrong because a buffer requires a weak acid-base conjugate pair that can react reversibly with added H+ or OH-.
- Substituting concentrations in the wrong order is wrong because pH = pKa + log([A-]/[HA]), so reversing the ratio changes the sign of the logarithm.
- Assuming dilution changes buffer pH a lot is wrong because dilution lowers both [A-] and [HA] by the same factor, leaving their ratio nearly unchanged.
- Ignoring neutralization before using Henderson-Hasselbalch is wrong because added strong acid or base first reacts with the buffer components and changes their amounts.
Practice Questions
- 1 A buffer contains 0.20 M acetic acid and 0.30 M acetate ion. If pKa for acetic acid is 4.76, calculate the pH using pH = pKa + log([A-]/[HA]).
- 2 A 1.00 L buffer contains 0.50 mol HA and 0.50 mol A-. If 0.10 mol HCl is added, assume it reacts completely with A-. What are the new moles of HA and A-, and what is the pH if pKa = 7.20?
- 3 Explain why a buffer made with equal amounts of carbonic acid and bicarbonate resists both added acid and added base better than pure water.