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Chemistry · Ch 9 — Equilibrium

The Henderson-Hasselbalch Equation and Buffer Solutions

9.12

The Henderson-Hasselbalch Equation and Buffer Solutions

A buffer solution is one that resists an appreciable change in pH when a small amount of a strong

acid or a strong base is added to it, or when it is moderately diluted. The most common type — an

acidic buffer — is prepared by mixing a weak acid with a salt of its conjugate base (for example,

CH3COOH\text{CH}_3\text{COOH} with CH3COONa\text{CH}_3\text{COONa}); a basic buffer is the analogous mixture

of a weak base with a salt of its conjugate acid (for example, NH4OH\text{NH}_4\text{OH} with

NH4Cl\text{NH}_4\text{Cl}).

Deriving the Henderson-Hasselbalch equation. Starting from the ionization equilibrium of the weak

acid, HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-, the equilibrium expression is

Ka=[H+][A−]/[HA]K_a = [\text{H}^+][\text{A}^-]/[\text{HA}]. Solving for [H+][\text{H}^+],

[H+]=Ka×[HA]/[A−][\text{H}^+] = K_a \times [\text{HA}]/[\text{A}^-]. Taking −log⁡10-\log_{10} of both sides and using the

definitions pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+] and pKa=−log⁡Ka\text{p}K_a = -\log K_a,

pH=pKa+log⁡[A−][HA]=pKa+log⁡[salt][acid]\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}

This is the Henderson-Hasselbalch equation. Its derivation relies on one simplifying assumption:

because the buffer already contains a substantial, deliberately-added concentration of the salt (the

conjugate base A−\text{A}^-), the weak acid's own, otherwise-small ionization is suppressed by the

common ion effect, so the equilibrium concentrations [HA][\text{HA}] and [A−][\text{A}^-] can safely be

taken as equal to the initial, as-mixed concentrations of the acid and the salt, without needing to

solve an ICE table.

Why a buffer resists pH change. The buffer contains a large reservoir of both the weak acid (which

can neutralize any added OH−\text{OH}^-) and its conjugate base (which can neutralize any added

H+\text{H}^+). Adding a small amount of strong acid converts a small amount of A−\text{A}^- into

HA\text{HA}; adding a small amount of strong base converts a small amount of HA\text{HA} into

A−\text{A}^-. In either case, only the ratio [A−]/[HA][\text{A}^-]/[\text{HA}] changes, and only slightly,

so by the Henderson-Hasselbalch equation the pH — which depends on the logarithm of that ratio — …