Skip to main content

Buffer solutions: WACE Year 12 Chemistry

Syllabus dot point

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

8 min read

Quick answer

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. It is made from a weak acid and its conjugate base (or a weak base and its conjugate acid) in similar amounts. Added acid is neutralised by the conjugate base and added base is neutralised by the weak acid, so the pH stays nearly constant.

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, CH3COOH\text{CH}_3\text{COOH}, with sodium ethanoate, CH3COONa\text{CH}_3\text{COONa}), or a weak base with its conjugate acid (ammonia with ammonium chloride).

Buffer solution

A solution that resists changes in pH on the addition of small amounts of acid or base, made from a weak acid and its conjugate base (or a weak base and its conjugate acid) in similar concentrations.

How a buffer works

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

CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)\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 (CH3COOH\text{CH}_3\text{COOH}) and its conjugate base (CH3COO−\text{CH}_3\text{COO}^-).

When acid (H+\text{H}^+) is added, the extra H+\text{H}^+ reacts with the conjugate base: CH3COO−+H+→CH3COOH\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}. By Le Chatelier's principle, the added H+\text{H}^+ pushes the equilibrium to the left, consuming most of the added protons. The conjugate base mops up the acid, so [H+][\text{H}^+] and the pH barely change.

When base (OH−\text{OH}^-) is added, the hydroxide reacts with the weak acid: CH3COOH+OH−→CH3COO−+H2O\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}. Removing H+\text{H}^+ (as the OH−\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.

The two reservoirs

A buffer needs appreciable amounts of both the weak acid (to neutralise added base) and the conjugate base (to neutralise added acid). If either runs low, the buffer fails and the pH changes sharply.

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: H2CO3⇌HCO3−+H+\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) HCl\text{HCl} and NaCl\text{NaCl}; (b) CH3COOH\text{CH}_3\text{COOH} and CH3COONa\text{CH}_3\text{COONa}.

Solution

(a) is not a buffer. HCl\text{HCl} is a strong acid, so Cl−\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. CH3COOH\text{CH}_3\text{COOH} is a weak acid and CH3COO−\text{CH}_3\text{COO}^- (from the salt) is its conjugate base. Both partners are present in appreciable amounts: the ethanoate neutralises added H+\text{H}^+ and the ethanoic acid neutralises added OH−\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.

Thinking strong-acid mixtures buffer

A common error is calling a mixture of a strong acid and its salt (such as HCl\text{HCl} and NaCl\text{NaCl}) a buffer. The conjugate base of a strong acid is negligibly basic, so it cannot resist pH change. Another frequent slip is saying a buffer keeps the pH exactly constant; it only minimises the change, and it fails once a reservoir is exhausted. A third is forgetting to show the equation for both added acid and added base.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks
A buffer is made by dissolving 0.20 mol0.20\ \text{mol} of ethanoic acid (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}) and 0.20 mol0.20\ \text{mol} of sodium ethanoate in water to make 1.0 L1.0\ \text{L} of solution. (a) Calculate the pH\text{pH} of the buffer. (b) Calculate the new pH\text{pH} after 0.020 mol0.020\ \text{mol} of solid NaOH\text{NaOH} is added (assume no volume change). (c) State why the pH\text{pH} change is small.
Show worked answer →

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

(a) Rearrange Ka=[H+][CH3COO−][CH3COOH]K_a = \dfrac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} for [H+][\text{H}^+]:

[H+]=Ka×[CH3COOH][CH3COO−]=1.8×10−5×0.200.20=1.8×10−5 mol L−1.[\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}.

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

(b) The added OH−\text{OH}^- converts 0.020 mol0.020\ \text{mol} of acid into ethanoate:
acid =0.20−0.020=0.18 mol= 0.20 - 0.020 = 0.18\ \text{mol}; ethanoate =0.20+0.020=0.22 mol= 0.20 + 0.020 = 0.22\ \text{mol}.

[H+]=1.8×10−5×0.180.22=1.47×10−5,pH=4.83.[\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 0.090.09 of a pH unit because the large reservoir of weak acid neutralises the added base (CH3COOH+OH−→CH3COO−+H2O\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 pH=4.74\text{pH} = 4.74, the updated mole amounts, pH=4.83\text{pH} = 4.83, and the reservoir reasoning.

WACE 2020-style4 marks
Explain, using equations and Le Chatelier's principle, how a buffer made of ammonia and ammonium chloride resists a change in pH\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 NH3\text{NH}_3 and its conjugate acid NH4+\text{NH}_4^+, with the equilibrium

NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq).\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq).

Added H+\text{H}^+ from the HCl\text{HCl} reacts with the ammonia reservoir: NH3+H+→NH4+\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+. Because the large store of NH3\text{NH}_3 consumes nearly all the added protons, [H+][\text{H}^+] rises only slightly. By Le Chatelier's principle the small increase in H+\text{H}^+ is opposed as the equilibrium adjusts, so the pH\text{pH} stays close to its original value.

Markers reward naming both species, the NH3+H+→NH4+\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+ reaction, and the Le Chatelier reasoning.

Practise this

Sources & how we know this

ExamExplained