Explain how a buffer maintains pH using equilibrium and calculate buffer pH.

How a weak acid and its conjugate base resist pH change, the action of buffers when acid or base is added, and calculating buffer pH.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-30

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

You must explain the equilibrium behind buffer action and calculate the pH of a buffer.

What a buffer is

A common example is ethanoic acid mixed with sodium ethanoate. The solution contains plenty of the weak acid $\text{CH}_3\text{COOH}$ and plenty of its conjugate base $\text{CH}_3\text{COO}^-$ .

How a buffer works

The buffer relies on the equilibrium $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$ .

When acid (extra $\text{H}^+$ ) is added, it is consumed by the conjugate base: $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$ . The added hydrogen ions are mopped up, so the pH barely changes.

When base (extra $\text{OH}^-$ ) is added, it is neutralised by the weak acid: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$ . The added hydroxide is removed, so the pH barely changes.

Because both partners are present in reasonable amounts, the buffer can absorb additions of either acid or base. Its capacity is limited; if too much acid or base is added one partner is used up and the buffer fails.

example

pH of a buffer

A buffer contains $0.20\ \text{mol L}^{-1}$ ethanoic acid and $0.30\ \text{mol L}^{-1}$ sodium ethanoate. $K_a = 1.8 \times 10^{-5}$ . Find the pH.

Step 1: $[\text{H}^+] = K_a \times \dfrac{[\text{acid}]}{[\text{base}]} = 1.8 \times 10^{-5} \times \dfrac{0.20}{0.30}$ .

Step 2: $[\text{H}^+] = 1.8 \times 10^{-5} \times 0.667 = 1.2 \times 10^{-5}\ \text{mol L}^{-1}$ .

Step 3: $\text{pH} = -\log_{10}(1.2 \times 10^{-5}) = 4.92$ .

The pH sits near the $pK_a$ of ethanoic acid (4.74), shifted slightly because there is more conjugate base than acid.

In the exam, name both components of the buffer, write the neutralising equations for added acid and added base, and use the $K_a$ and concentration ratio to calculate the pH.

Discuss this in the community ->