Buffers and the Henderson–Hasselbalch equation

A buffer holds the pH roughly steady when small amounts of acid or base are added. It does this with a weak acid HA and its conjugate base A⁻, both present in comparable amounts: A⁻ mops up added H⁺, and HA mops up added OH⁻.

Where the equation comes from

Start from the acid dissociation constant and take $-\log_{10}$ of both sides:

This is the Henderson–Hasselbalch equation. So far it is exact. The approximation comes when you substitute the amounts you added for the equilibrium concentrations. That only works when HA and A⁻ are both much more concentrated than H⁺ and OH⁻.

Because HA and A⁻ share the same volume, you can use moles in the ratio, and dilution does not change the pH.

Worked example: a titration

25.0 mL of 0.100 mol L⁻¹ ethanoic acid ($\mathrm{p}K_\mathrm{a} = 4.76$) is titrated with 0.100 mol L⁻¹ NaOH.

• At 12.50 mL, half the acid has been neutralised, so $n(\mathrm{HA}) = n(\mathrm{A^-})$ and $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 4.76$. This is the half-equivalence point.
• At 20.00 mL, there is 2.00 mmol of A⁻ and 0.50 mmol of HA left, so $\mathrm{pH} = 4.76 + \log_{10} 4 = 5.36$.

How good is the buffer?

Buffer capacity $\beta$ is the amount of strong base needed per litre to raise the pH by one unit. Ignoring water's own contribution, for total concentration $C = [\mathrm{HA}] + [\mathrm{A^-}]$ and fraction deprotonated $\alpha$:

This is largest when $\alpha = \tfrac12$, that is when pH = pKₐ. It falls to about a third of that maximum at pKₐ ± 1. That is where the rule "choose an acid with pKₐ within one unit of the target pH" comes from.