What is the acid dissociation constant?

A weak acid HA ionises only partly in water:

What is calculating the pH of a weak acid?

Set up an equilibrium with the initial acid concentration c. 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 c - x ≈ c.

TASC TCE 2023-style practice: Benzoic acid, C_6H_5COOH, is a weak acid with K_a = 6.46 × 10^-5. Calculate the pH of a solution with an initial concentration of 1.40\ mol L^-1. Clearly state any assumptions made.

Write the dissociation: C_6H_5COOH C_6H_5COO^- + H^+, with K_a = [C_6H_5COO^-][H^+][C_6H_5COOH]. Assumptions: (1) [H^+] = [C_6H_5COO^-] = x; (2) the acid is only weakly dissociated, so [C_6H_5COOH] ≈ 1.40\ mol L^-1 (x is negligible beside 1.40). (1 mark) Then K_a = x^2 / 1.40, so x^2 = 6.46 × 10^-5 × 1.40 = 9.04 × 10^-5, giving x = [H^+] = 9.51 × 10^-3\ mol L^-1. (2 marks) pH = -_10(9.51 × 10^-3) = 2.02. (1 mark)

TASC TCE 2022-style practice: Ethanoic acid is weak: CH_3COOH(aq) CH_3COO^-(aq) + H^+(aq), K_a = 1.76 × 10^-5. A solution has [CH_3COO^-] = 1.2 × 10^-3, [H^+] = 8.3 × 10^-6 and [CH_3COOH] = 1.5 × 10^-3\ mol L^-1. Show that this solution is not at equilibrium and predict the direction of net change.

Calculate the reaction quotient Q from current concentrations and compare with K_a. $Q = [CH_3COO^-][H^+][CH_3COOH] = (1.2 × 10^-3)(8.3 × 10^-6)1.5 × 10^-3 = 6.6 × 10^-6.$ (1 mark) Q\ (6.6 × 10^-6) < K_a\ (1.76 × 10^-5), so the system is not at equilibrium. Because Q < K_a, the forward (dissociation) reaction is favoured and proceeds until Q rises to equal K_a. (1 mark)

§-Syllabus dot point

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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, percentage ionisation, and the relationship between Ka and acid strength, with fully worked TASC-style examples.

Generated by Claude Opus 4.89 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

TASC expects you to define $K_a$ , use $K_a$ and $pK_a$ to compare acid strength, and calculate the pH of a weak acid solution, stating your assumptions.

The acid dissociation constant

A weak acid $\text{HA}$ 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, $pK_a$ is more convenient: a smaller $pK_a$ means a stronger acid, just as a smaller pH means more acidic. Like all equilibrium constants, $K_a$ depends only on temperature, not on concentration.

Calculating the pH of a weak acid

Set up an equilibrium with the initial acid concentration $c$ . 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 $c - x \approx c$ .

pH of ethanoic acid

Set up the equilibrium

Find the pH of $0.10\ \text{mol L}^{-1}$ ethanoic acid, $K_a = 1.8 \times 10^{-5}$ . Write $K_a = \dfrac{x^2}{c - x}$ where $x = [\text{H}^+]$ .

Apply the weak-acid approximation

Because the acid is weak, $c - x \approx c = 0.10$ , so $K_a \approx \dfrac{x^2}{0.10}$ .

Solve for the hydrogen ion concentration

x^2 = K_a \times c = (1.8 \times 10^{-5})(0.10) = 1.8 \times 10^{-6}, \qquad x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\ \text{mol L}^{-1}.

Convert to pH and check the approximation

$\text{pH} = -\log_{10}(1.34 \times 10^{-3}) = 2.87$ . The ionisation is $\dfrac{1.34 \times 10^{-3}}{0.10} \times 100 = 1.3\%$ , well under $5\%$ , so the approximation is valid. This pH is much higher (less acidic) than a $0.10\ \text{mol L}^{-1}$ strong acid (pH $1.0$ ), because ethanoic acid only partly ionises.

Percentage ionisation

Percentage ionisation $= \dfrac{[\text{H}^+]}{c} \times 100\%$ . Diluting a weak acid increases the percentage ionised (Le Chatelier favours the side with more particles) even though $[\text{H}^+]$ falls. The approximation $c - x \approx c$ is valid when ionisation is below about $5\%$ .

Ka, Kb and the conjugate relationship

For a conjugate acid-base pair, the strength of the acid and the strength of its conjugate base are linked through the ionic product of water:

K_a \times K_b = K_w = 1.0 \times 10^{-14} \text{ at } 25\ ^\circ\text{C}.

A stronger acid (larger $K_a$ ) therefore has a weaker conjugate base (smaller $K_b$ ), and vice versa. This is why the ethanoate ion, the conjugate base of a weak acid, is itself a weak base that makes its solutions slightly basic, while the chloride ion, the conjugate base of the strong acid $\text{HCl}$ , is so weak a base that it has no measurable effect on pH.

Comparing acids fairly

Because $K_a$ is fixed at a given temperature, it is the proper way to compare the intrinsic strength of acids, independent of concentration. pH, by contrast, depends on both strength and concentration, so two acids at the same pH need not be equally strong. When a question asks you to rank acids, use $K_a$ or $pK_a$ : the larger the $K_a$ (or the smaller the $pK_a$ ), the stronger the acid and the more completely it ionises at a given concentration.

In the exam, write the ionisation equation and $K_a$ expression, define the $x$ terms, state and justify the approximation, solve for $[\text{H}^+]$ , and convert to pH, then sanity-check that the percentage ionisation is small.

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.

TCE 20234 marksBenzoic acid,

\text{C}_6\text{H}_5\text{COOH}

, is a weak acid with

K_a = 6.46 \times 10^{-5}

. Calculate the pH of a solution with an initial concentration of

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

. Clearly state any assumptions made.

Show worked answer →

Write the dissociation: $\text{C}_6\text{H}_5\text{COOH} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}^+$ , with $K_a = \dfrac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]}$ .

Assumptions: (1) $[\text{H}^+] = [\text{C}_6\text{H}_5\text{COO}^-] = x$ ; (2) the acid is only weakly dissociated, so $[\text{C}_6\text{H}_5\text{COOH}] \approx 1.40\ \text{mol L}^{-1}$ ( $x$ is negligible beside $1.40$ ). (1 mark)

Then $K_a = x^2 / 1.40$ , so $x^2 = 6.46 \times 10^{-5} \times 1.40 = 9.04 \times 10^{-5}$ , giving $x = [\text{H}^+] = 9.51 \times 10^{-3}\ \text{mol L}^{-1}$ . (2 marks)

$\text{pH} = -\log_{10}(9.51 \times 10^{-3}) = 2.02$ . (1 mark)

TCE 20212 marksFor hypochlorous acid,

\text{HClO}(aq) \rightleftharpoons \text{H}^+(aq) + \text{ClO}^-(aq)

K_a = 3 \times 10^{-8}

. Define

K_a

for this acid and use it to state which component of the equilibrium mixture has the greatest concentration.

Show worked answer →

Definition: $K_a$ is the acid dissociation constant, $K_a = \dfrac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]}$ , the equilibrium constant for ionisation of the acid in water. (1 mark)

$K_a$ is very small ( $3 \times 10^{-8} \ll 1$ ), so the equilibrium lies far to the left and the acid is only slightly ionised. The undissociated acid $\text{HClO}$ is therefore present in the greatest concentration; $\text{H}^+$ and $\text{ClO}^-$ are present only in small amounts. (1 mark)

TCE 20222 marksEthanoic acid is weak:

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

K_a = 1.76 \times 10^{-5}

. A solution has

[\text{CH}_3\text{COO}^-] = 1.2 \times 10^{-3}

[\text{H}^+] = 8.3 \times 10^{-6}

and

[\text{CH}_3\text{COOH}] = 1.5 \times 10^{-3}\ \text{mol L}^{-1}

. Show that this solution is not at equilibrium and predict the direction of net change.

Show worked answer →

Calculate the reaction quotient $Q$ from current concentrations and compare with $K_a$ .

Q = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{(1.2 \times 10^{-3})(8.3 \times 10^{-6})}{1.5 \times 10^{-3}} = 6.6 \times 10^{-6}.

(1 mark)

$Q\ (6.6 \times 10^{-6}) < K_a\ (1.76 \times 10^{-5})$ , so the system is not at equilibrium. Because $Q < K_a$ , the forward (dissociation) reaction is favoured and proceeds until $Q$ rises to equal $K_a$ . (1 mark)