SCSA WACE 2022-style practice: A buffer is made by dissolving 0.20\ mol of ethanoic acid (K_a = 1.8 × 10^-5) and 0.20\ mol of sodium ethanoate in water to make 1.0\ L of solution. (a) Calculate the pH of the buffer. (b) Calculate the new pH after 0.020\ mol of solid NaOH is added (assume no volume change). (c) State why the pH change is small.

A 7 mark question rewards both pH calculations and the reasoning. **(a)** Rearrange K_a = [H^+][CH_3COO^-][CH_3COOH] for [H^+]: $[H^+] = K_a × [CH_3COOH][CH_3COO^-] = 1.8 × 10^-5 × (0.20)/(0.20) = 1.8 × 10^-5\ mol L^-1.$ $pH = -_10(1.8 × 10^-5) = 4.74.$ **(b)** The added OH^- converts 0.020\ mol of acid into ethanoate: acid = 0.20 - 0.020 = 0.18\ mol; ethanoate = 0.20 + 0.020 = 0.22\ mol. $[H^+] = 1.8 × 10^-5 × (0.18)/(0.22) = 1.47 × 10^-5, pH = 4.83.$ **(c)** The change is only 0.09 of a pH unit because the large reservoir of weak acid neutralises the added base (CH_3COOH + OH^- → CH_3COO^- + H_2O), so the ratio of acid to base shifts only slightly. Markers reward pH = 4.74, the updated mole amounts, pH = 4.83, and the reservoir reasoning.

§-Syllabus dot point

WAChemistrySyllabus dot point

How can a solution resist a change in pH when small amounts of acid or base are added?

Explain how buffer solutions resist changes in pH using Le Chatelier's principle and conjugate acid-base equilibria

A focused answer to the WACE Year 12 Chemistry dot point on buffers, how a conjugate acid-base mixture resists pH change, with the equilibrium reasoning, a worked example and common exam mistakes.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

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What this dot point is asking

A buffer solution resists changes in pH when small quantities of strong acid or strong base are added, and also when the solution is diluted. It is built from a conjugate acid-base pair present in comparable amounts: most often a weak acid together with a salt of its conjugate base (for example ethanoic acid, $\text{CH}_3\text{COOH}$ , with sodium ethanoate, $\text{CH}_3\text{COONa}$ ), or a weak base with its conjugate acid (ammonia with ammonium chloride).

How a buffer works

Take an ethanoic acid / ethanoate buffer. The relevant equilibrium is

\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)

The buffer contains a large reservoir of both the weak acid molecule ( $\text{CH}_3\text{COOH}$ ) and its conjugate base ( $\text{CH}_3\text{COO}^-$ ).

When acid ( $\text{H}^+$ ) is added, the extra $\text{H}^+$ reacts with the conjugate base: $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$ . By Le Chatelier's principle, the added $\text{H}^+$ pushes the equilibrium to the left, consuming most of the added protons. The conjugate base mops up the acid, so $[\text{H}^+]$ and the pH barely change.

When base ( $\text{OH}^-$ ) is added, the hydroxide reacts with the weak acid: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$ . Removing $\text{H}^+$ (as the $\text{OH}^-$ consumes it) shifts the equilibrium to the right to replace it, so again the pH changes only slightly.

The buffer works because it holds a large store of both partners. Each can react with an incoming disturbance, and the equilibrium re-establishes a hydrogen ion concentration close to the original.

Buffer capacity and limits

A buffer only works for small additions. If you add enough strong acid or base to use up one of the reservoirs, the buffering collapses and pH changes rapidly. The amount of acid or base a buffer can absorb before this happens is its buffer capacity, which is greatest when the weak acid and its conjugate base are present in roughly equal concentrations.

Where buffers matter

Blood is buffered close to pH 7.4, largely by the carbonic acid / hydrogen carbonate system: $\text{H}_2\text{CO}_3 \rightleftharpoons \text{HCO}_3^- + \text{H}^+$ . This keeps the pH within the narrow range enzymes need. Buffers are also used in fermentation, shampoos, and many laboratory and industrial processes where a stable pH is required.

Identifying a buffer mixture

Question

Explain which of these is a buffer: (a) $\text{HCl}$ and $\text{NaCl}$ ; (b) $\text{CH}_3\text{COOH}$ and $\text{CH}_3\text{COONa}$ .

Solution

(a) is not a buffer. $\text{HCl}$ is a strong acid, so $\text{Cl}^-$ is the conjugate base of a strong acid and is far too weak to accept protons. There is no weak acid / conjugate base equilibrium to absorb added base.

(b) is a buffer. $\text{CH}_3\text{COOH}$ is a weak acid and $\text{CH}_3\text{COO}^-$ (from the salt) is its conjugate base. Both partners are present in appreciable amounts: the ethanoate neutralises added $\text{H}^+$ and the ethanoic acid neutralises added $\text{OH}^-$ , so the pH stays nearly constant.

When answering buffer questions in the WACE examination, always name both species and write the equation showing how each one reacts, then link the response to Le Chatelier's principle to earn the reasoning marks.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksA buffer is made by dissolving

0.20\ \text{mol}

of ethanoic acid (

K_a = 1.8 \times 10^{-5}

) and

0.20\ \text{mol}

of sodium ethanoate in water to make

1.0\ \text{L}

of solution. (a) Calculate the

\text{pH}

of the buffer. (b) Calculate the new

\text{pH}

after

0.020\ \text{mol}

of solid

\text{NaOH}

is added (assume no volume change). (c) State why the

\text{pH}

change is small.

Show worked answer →

A 7 mark question rewards both pH calculations and the reasoning.

(a) Rearrange $K_a = \dfrac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$ for $[\text{H}^+]$ :

[\text{H}^+] = K_a \times \frac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]} = 1.8 \times 10^{-5} \times \frac{0.20}{0.20} = 1.8 \times 10^{-5}\ \text{mol L}^{-1}.

\text{pH} = -\log_{10}(1.8 \times 10^{-5}) = 4.74.

(b) The added $\text{OH}^-$ converts $0.020\ \text{mol}$ of acid into ethanoate:
acid $= 0.20 - 0.020 = 0.18\ \text{mol}$ ; ethanoate $= 0.20 + 0.020 = 0.22\ \text{mol}$ .

[\text{H}^+] = 1.8 \times 10^{-5} \times \frac{0.18}{0.22} = 1.47 \times 10^{-5},\qquad \text{pH} = 4.83.

(c) The change is only $0.09$ of a pH unit because the large reservoir of weak acid neutralises the added base ( $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$ ), so the ratio of acid to base shifts only slightly.

Markers reward $\text{pH} = 4.74$ , the updated mole amounts, $\text{pH} = 4.83$ , and the reservoir reasoning.

WACE 20204 marksExplain, using equations and Le Chatelier's principle, how a buffer made of ammonia and ammonium chloride resists a change in

\text{pH}

when a small amount of hydrochloric acid is added.

Show worked answer →

A 4 mark explain answer needs the equilibrium, the neutralising reaction, and the Le Chatelier link.

The buffer contains the weak base $\text{NH}_3$ and its conjugate acid $\text{NH}_4^+$ , with the equilibrium

\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq).

Added

\text{H}^+

from the

\text{HCl}

reacts with the ammonia reservoir:

\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+

. Because the large store of

\text{NH}_3

consumes nearly all the added protons,

[\text{H}^+]

rises only slightly. By Le Chatelier's principle the small increase in

\text{H}^+

is opposed as the equilibrium adjusts, so the

\text{pH}

stays close to its original value.

Markers reward naming both species, the $\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+$ reaction, and the Le Chatelier reasoning.