Buffer solutions: TCE Chemistry (Tasmania)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

A buffer is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts. It resists pH change because the conjugate base neutralises added acid and the weak acid neutralises added base, so the ratio of the two changes only slightly. The pH is set by $[\text{H}^+] = K_a \times \dfrac{[\text{acid}]}{[\text{base}]}$ , which depends on the ratio rather than the absolute concentrations.

What this dot point is asking

TASC expects you to explain buffer action through equilibrium and Le Chatelier's principle, write the neutralising equations, and calculate buffer pH.

What a buffer is

Buffer solution

A solution that resists significant change in pH when small amounts of acid or base are added. It contains comparable amounts of a weak acid and its conjugate base (or a weak base and its conjugate acid).

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 $\text{CH}_3\text{COO}^-$ . A buffer can be made either by mixing the weak acid with a salt of its conjugate base, or by partially neutralising a weak acid with a strong base.

How a buffer works

The buffer relies on the equilibrium $\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^+(aq) + \text{CH}_3\text{COO}^-(aq)$ , with a large reservoir of both partners.

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

Because both partners are present in large amounts, the buffer absorbs additions of either acid or base.

Buffer capacity

Buffer pH depends on the ratio

[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{base}]} \qquad \text{so} \qquad \text{pH} = pK_a + \log_{10}\frac{[\text{base}]}{[\text{acid}]}

The pH is set by

K_a

and the ratio of weak acid to conjugate base, not by their absolute concentrations. Diluting a buffer changes both equally, so the ratio and the pH stay nearly the same. A buffer works best when

[\text{acid}] \approx [\text{base}]

, giving

\text{pH} \approx pK_a

Buffer capacity is the amount of acid or base a buffer can absorb before the pH changes sharply. It is greatest when the two components are present in roughly equal, high concentrations. Once one partner is largely used up, the buffer fails and the pH swings rapidly.

pH of a buffer

Identify the components and ratio

A buffer contains $0.20\ \text{mol L}^{-1}$ ethanoic acid and $0.30\ \text{mol L}^{-1}$ sodium ethanoate, with $K_a = 1.8 \times 10^{-5}$ . The acid is $\text{CH}_3\text{COOH}$ and the conjugate base is $\text{CH}_3\text{COO}^-$ .

Substitute into the buffer expression

[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{base}]} = 1.8 \times 10^{-5} \times \frac{0.20}{0.30} = 1.2 \times 10^{-5}\ \text{mol L}^{-1}.

Convert to pH

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

The pH sits just above the $pK_a$ of ethanoic acid ( $4.74$ ), because there is more conjugate base than acid, which makes the solution slightly less acidic.

Choosing a buffer for a target pH

To make a buffer at a particular pH, choose a weak acid whose $pK_a$ is close to the desired pH, then adjust the ratio of acid to conjugate base. Because $\text{pH} = pK_a + \log_{10}\dfrac{[\text{base}]}{[\text{acid}]}$ , equal amounts of the two give $\text{pH} = pK_a$ , and shifting the ratio up or down moves the pH by at most about one unit before the buffer becomes ineffective. This is why each buffer system has a useful working range of roughly $pK_a \pm 1$ .

Biological and natural buffers

Buffers are essential in living systems. Human blood is held near pH $7.4$ mainly by the carbonic acid and hydrogencarbonate system, $\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$ , with the lungs and kidneys removing or retaining $\text{CO}_2$ and $\text{HCO}_3^-$ to keep the ratio steady. Even small departures from this pH are dangerous, which shows how tightly biological buffers must work. Ocean water is similarly buffered by the carbonate system, which is why rising atmospheric $\text{CO}_2$ gradually lowers ocean pH as the buffer is progressively consumed.

Thinking a buffer keeps pH perfectly constant

A buffer resists change; it does not prevent it entirely. Adding acid or base does move the pH slightly, and once the buffer capacity is exceeded the pH changes sharply. Avoid writing that the pH stays exactly the same. A second slip is inverting the ratio: $[\text{H}^+]$ uses $\dfrac{[\text{acid}]}{[\text{base}]}$ , while the pH form uses $\log_{10}\dfrac{[\text{base}]}{[\text{acid}]}$ .

In the exam, name both components of the buffer, write the neutralising equations for added acid and added base, link the resistance to Le Chatelier's principle, and use $K_a$ with the concentration ratio to calculate the pH.

Exam-style questions

Questions in the style of TASC 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.

Practice5 marks

A buffer is prepared containing

0.20\ \text{mol L}^{-1}

ethanoic acid,

\text{CH}_3\text{COOH}

, and

0.30\ \text{mol L}^{-1}

sodium ethanoate,

\text{CH}_3\text{COONa}

. (

K_a(\text{CH}_3\text{COOH}) = 1.8 \times 10^{-5}

.) (a) Write equations showing how this buffer responds to added

\text{H}^+

and added

\text{OH}^-

. (b) Calculate the pH of the buffer.

Show worked answer →

(a) Added acid is removed by the conjugate base: $\text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{CH}_3\text{COOH}(aq)$ . Added base is removed by the weak acid: $\text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq) \rightarrow \text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l)$ . (2 marks)

(b) $[\text{H}^+] = K_a \times \dfrac{[\text{acid}]}{[\text{base}]} = 1.8 \times 10^{-5} \times \dfrac{0.20}{0.30} = 1.2 \times 10^{-5}\ \text{mol L}^{-1}$ . (2 marks)

$\text{pH} = -\log_{10}(1.2 \times 10^{-5}) = 4.92$ . (1 mark)

Practice3 marks

Carbonic acid and hydrogencarbonate form the main buffer of human blood:

\text{H}_2\text{CO}_3(aq) \rightleftharpoons \text{H}^+(aq) + \text{HCO}_3^-(aq)

. Explain, using this equilibrium and Le Chatelier's principle, how the blood buffer responds when a small amount of lactic acid (a source of

\text{H}^+

) is produced during exercise.

Show worked answer →

Adding lactic acid increases $[\text{H}^+]$ . By Le Chatelier's principle the equilibrium shifts left to oppose this increase, consuming the added $\text{H}^+$ . (1 mark)

The added $\text{H}^+$ reacts with the conjugate base: $\text{HCO}_3^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{CO}_3(aq)$ . (1 mark)

Because most of the extra $\text{H}^+$ is converted to $\text{H}_2\text{CO}_3$ (which can be removed as $\text{CO}_2$ by the lungs), the blood pH stays close to its normal value of about $7.4$ rather than dropping sharply. (1 mark)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC