For a buffer containing a weak acid $\mathrm{HA}$ and its conjugate base $\mathrm{A^-}$,

$$\mathrm{HA+H_2O\rightleftharpoons H_3O^+ + A^-},$$

the acid-ionization constant is

$$K_a=\frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}.$$

Solving for hydronium gives

$$[\mathrm{H_3O^+}]=K_a\frac{[\mathrm{HA}]}{[\mathrm{A^-}]}.$$

Taking negative base-10 logarithms produces the Henderson-Hasselbalch equation:

$$\boxed{\mathrm{pH}=pK_a+\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}}.$$

This equation shows that buffer pH is determined by two things: the intrinsic acid strength, represented by $pK_a$, and the ratio of conjugate base to acid.

If the concentrations are equal,

$$[\mathrm{A^-}]=[\mathrm{HA}],$$

then the logarithm is zero and

$$\mathrm{pH}=pK_a.$$

If the base-to-acid ratio is 10,

$$\mathrm{pH}=pK_a+1,$$

while a ratio of $0.1$ gives

$$\mathrm{pH}=pK_a-1.$$

Example

A buffer contains $0.150,\mathrm M$ acetic acid and $0.100,\mathrm M$ acetate. With $pK_a=4.74$,

$$\mathrm{pH}=4.74+\log\frac{0.100}{0.150}\approx4.56.$$

For a prepared buffer in which acid and base occupy the same final volume, the concentration ratio can be replaced by the mole ratio because the common volume cancels.

The equation is a rearranged equilibrium expression, not an independent empirical law. It is most useful when both members of the conjugate pair are present in appreciable amounts and the ordinary dilute-solution concentration approximation is adequate. Near exhaustion of one buffer component, the assumptions behind the simple ratio description fail and a full equilibrium treatment is safer.