Topic 8.7 pH and pKa: use the Henderson-Hasselbalch equation to relate the pH of a buffer to the pKa and the ratio of conjugate base to weak acid, and explain buffer capacity.

What this topic is asking

The College Board (Topic 8.7) wants you to use the Henderson-Hasselbalch equation to relate the pH of a buffer to the pKa and the ratio of conjugate base to weak acid, and to explain buffer capacity. This makes the buffer behavior of Topic 8.4 quantitative.

The Henderson-Hasselbalch equation

This equation is derived from the $K_a$ expression by taking negative logarithms. It shows that the pH of a buffer is set primarily by the $\text{p}K_a$ of the weak acid, adjusted by the logarithm of the conjugate-base-to-acid ratio. Because the ratio enters logarithmically, even a large change in the ratio shifts the pH only modestly, which is the mathematical reason a buffer resists pH change.

pH equals pKa when the ratio is one

This special case is both a useful check and the basis of the titration method for finding $\text{p}K_a$. It also marks the center of the buffer's effective range: a buffer works well roughly within one pH unit of its $\text{p}K_a$, where the ratio stays between about 1:10 and 10:1.

Designing a buffer and buffer capacity

To make a buffer at a target pH, choose a weak acid whose $\text{p}K_a$ is close to that pH (within about one unit), then adjust the ratio $\dfrac{[\text{A}^-]}{[\text{HA}]}$ to hit the target exactly using the equation.

Buffer capacity is the amount of strong acid or base a buffer can neutralize before its pH changes appreciably. It is greatest when $[\text{A}^-] = [\text{HA}]$ (pH near $\text{p}K_a$), because the buffer can then absorb added acid and base about equally, and it increases with the total concentration of the buffer components, since a larger reservoir absorbs more. Doubling both concentrations (keeping the ratio fixed) leaves the pH unchanged but doubles the capacity.

Try this

Q1. A buffer has $\text{p}K_a = 5.0$, $[\text{A}^-] = 0.20$ M and $[\text{HA}] = 0.20$ M. Calculate the pH. [2 points]

• Cue. $\text{pH} = 5.0 + \log(1) = 5.0$.

Q2. Explain how to increase a buffer's capacity without changing its pH. [2 points]

• Cue. Increase the concentrations of both the acid and conjugate base by the same factor; the ratio (and so the pH) is unchanged, but the larger reservoir gives greater capacity.