What is calculating the pH of a weak acid?

Set up an equilibrium with the initial acid concentration. Let x be the concentration of H^+ formed. Then [H^+] = [A^-] = x and [HA] = c - x. Because the acid is weak, x is small, so we approximate c - x ≈ c.

TASChemistrySyllabus dot point

How do we measure the strength of a weak acid and find the pH of its solution?

Use Ka and pKa to describe weak acid strength and calculate the pH of a weak acid solution.

The acid dissociation constant Ka and pKa, calculating the pH of a weak acid solution, and the relationship between Ka and acid strength.

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

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

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What this dot point is asking

You must use $K_a$ and $pK_a$ to compare acid strength and calculate the pH of a weak acid solution.

The acid dissociation constant

A weak acid HA ionises only partly in water:

\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)

The equilibrium constant for this ionisation is the acid dissociation constant.

Because $K_a$ values span many orders of magnitude, we often use $pK_a = -\log_{10} K_a$ . A smaller $pK_a$ means a stronger acid, just as a smaller pH means more acidic.

Calculating the pH of a weak acid

Set up an equilibrium with the initial acid concentration. Let $x$ be the concentration of $\text{H}^+$ formed. Then $[\text{H}^+] = [\text{A}^-] = x$ and $[\text{HA}] = c - x$ . Because the acid is weak, $x$ is small, so we approximate $c - x \approx c$ .

example

pH of ethanoic acid

Find the pH of $0.10\ \text{mol L}^{-1}$ ethanoic acid, $K_a = 1.8 \times 10^{-5}$ .

Step 1: $K_a = \dfrac{x^2}{c - x} \approx \dfrac{x^2}{0.10}$ , where $x = [\text{H}^+]$ .

Step 2: $x^2 = K_a \times c = (1.8 \times 10^{-5})(0.10) = 1.8 \times 10^{-6}$ .

Step 3: $x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\ \text{mol L}^{-1}$ .

Step 4: $\text{pH} = -\log_{10}(1.34 \times 10^{-3}) = 2.87$ .

This is higher (less acidic) than a $0.10\ \text{mol L}^{-1}$ strong acid, which would have pH 1.0, because ethanoic acid only partly ionises.

The approximation $c - x \approx c$ is valid when the acid ionises less than about 5 percent, which holds for typical weak acids at moderate concentrations.

In the exam, write the ionisation equation and $K_a$ expression, set up the $x$ terms, justify the approximation, and convert the resulting $[\text{H}^+]$ to pH.

Exam-style practice questions

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

2023 TASC4 marksBenzoic acid, C6H5COOH, is a weak acid with a Ka of 6.46 x 10-5. Calculate the pH of a solution of benzoic acid which has an initial concentration of 1.40 mol L-1. Clearly state any assumptions made in your working.

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Write the dissociation: C6H5COOH reversible C6H5COO- + H+, with Ka = [C6H5COO-][H+] / [C6H5COOH].

Assumptions: (1) [H+] = [C6H5COO-] = x; (2) the acid is only weakly dissociated, so [C6H5COOH] at equilibrium is approximately the initial 1.40 mol L-1 (x is negligible compared with 1.40). (1 mark)

Then Ka = x^2 / 1.40, so x^2 = Ka x 1.40 = 6.46 x 10-5 x 1.40 = 9.044 x 10-5. x = [H+] = 9.510 x 10-3 mol L-1. (2 marks)

pH = -log10(9.510 x 10-3) = 2.02. (1 mark)

2021 TASC2 marksHypochlorous acid: HClO(aq) reversible H+(aq) + ClO-(aq), Ka = 3 x 10-8. Define Ka for hypochlorous acid and use it to state which component of the equilibrium mixture has the greatest concentration.

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Definition. Ka is the acid dissociation constant: Ka = [H+][ClO-] / [HClO], the equilibrium constant for the ionisation of the acid in water. (1 mark)

Interpretation. Ka is very small (3 x 10-8, much less than 1), so the equilibrium lies far to the left and the acid is only slightly ionised. Therefore the undissociated acid, HClO, is present in the greatest concentration; the ions H+ and ClO- are present in only small amounts. (1 mark)

2021 TASC2 marksEthanoic acid is a weak acid: CH3COOH(aq) reversible CH3COO-(aq) + H+(aq), Ka = 1.76 x 10-5. A solution contains [CH3COO-] = 1.2 x 10-3, [H+] = 8.3 x 10-6 and [CH3COOH] = 1.5 x 10-3 mol L-1. Show that this solution is not at equilibrium.

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Calculate the reaction quotient Q using the current concentrations and compare it with Ka.

Q = [CH3COO-][H+] / [CH3COOH] = (1.2 x 10-3)(8.3 x 10-6) / (1.5 x 10-3) = 9.96 x 10-9 / 1.5 x 10-3 = 6.6 x 10-6.

Q (6.6 x 10-6) is less than Ka (1.76 x 10-5), so the system is not at equilibrium. Because Q is below Ka, the forward (dissociation) reaction is favoured and will proceed until Q rises to equal Ka. (2 marks: calculate Q and compare with Ka.)

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