What is the acid ionisation constant Ka?

For a weak acid HA the ionisation equilibrium is HA H^+ + A^-, and its equilibrium constant is the acid ionisation constant:

What is the base ionisation constant Kb?

For a weak base B that reacts with water, B + H_2O BH^+ + OH^-, the base ionisation constant is

SCSA WACE 2022-style practice: Propanoic acid has K_a = 1.3 × 10^-5. (a) Write the ionisation equilibrium and the K_a expression. (b) Calculate the pH of a 0.20\ mol L^-1 solution, justifying any approximation. (c) Calculate the percentage ionisation.

A 6 mark question rewards the expression, the pH with valid approximation, and the percentage ionisation. **(a)** C_2H_5COOH C_2H_5COO^- + H^+, with K_a = [C_2H_5COO^-][H^+][C_2H_5COOH]. **(b)** Let x = [H^+] and assume 0.20 - x ≈ 0.20: $K_a = (x^2)/(0.20) x^2 = (1.3 × 10^-5)(0.20) = 2.6 × 10^-6,$ $x = 1.61 × 10^-3\ mol L^-1, pH = -_10(1.61 × 10^-3) = 2.79.$ The approximation is valid because x is well under 5\% of 0.20. **(c)** Percentage ionisation = xc × 100 = 1.61 × 10^-30.20 × 100 = 0.81\%, confirming the acid is only slightly ionised. Markers reward the K_a expression, pH = 2.79 with the approximation check, and 0.81\% ionisation.

§-Syllabus dot point

WAChemistrySyllabus dot point

What is the difference between a strong and a weak acid, and how do Ka, Kb and pKa measure acid and base strength?

Distinguish strong from weak acids and bases by degree of ionisation, and define and use Ka, Kb and pKa to compare strengths

A focused answer to the WACE Year 12 Chemistry dot point on strong versus weak acids and bases, the acid and base ionisation constants Ka and Kb, the meaning of pKa, and how they quantify strength, with 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

Acids and bases differ in strength, which describes how completely they ionise in water, and in concentration, which describes how much is dissolved. These are independent ideas and confusing them is the single biggest error in this topic.

Strong versus weak

A strong acid such as hydrochloric acid ionises essentially completely:

\text{HCl}(aq) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq)

We write a single arrow because effectively all the HCl molecules donate their proton. A weak acid such as ethanoic acid only partially ionises, so we write an equilibrium:

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

At any moment most of the ethanoic acid remains as un-ionised molecules. The same distinction applies to bases: sodium hydroxide is a strong base (fully dissociated), while ammonia is a weak base that establishes an equilibrium with water.

The acid ionisation constant Ka

For a weak acid HA the ionisation equilibrium is $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ , and its equilibrium constant is the acid ionisation constant:

K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

A larger $K_a$ means the equilibrium lies further to the right, so the acid is stronger. Strong acids have very large $K_a$ values, so we do not usually quote them.

The base ionisation constant Kb

For a weak base B that reacts with water, $\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$ , the base ionisation constant is

K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}

A larger $K_b$ means a stronger base. For a conjugate acid-base pair, $K_a \times K_b = K_w$ , so a stronger acid has a weaker conjugate base and vice versa.

Calculating pH of a weak acid

Because a weak acid is only partly ionised, you cannot assume $[\text{H}^+]$ equals the acid concentration. You set up an ICE table and use $K_a$ . For a weak acid of concentration $c$ where the ionised amount $x$ is small compared with $c$ , the approximation $K_a \approx x^2 / c$ gives $[\text{H}^+] = x \approx \sqrt{K_a \cdot c}$ .

pH of a weak acid

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

Step 1: Set up the Ka expression with the approximation

Let $x = [\text{H}^+]$ at equilibrium. Because $K_a$ is small compared to the initial concentration, the approximation $[\text{CH}_3\text{COOH}] \approx 0.10$ is valid, giving a simplified $K_a$ expression:

K_a = \frac{x^2}{0.10} = 1.8 \times 10^{-5}

Step 2: Solve for the hydrogen ion concentration

Rearrange to find $x^2$ , then take the square root to get the hydrogen ion concentration:

x^2 = 1.8 \times 10^{-6}, \quad x = 1.34 \times 10^{-3} \text{ mol L}^{-1}

Step 3: Calculate the pH

Take the negative base-10 logarithm of $[\text{H}^+]$ :

\text{pH} = -\log_{10}(1.34 \times 10^{-3}) = 2.87

Compare this with $0.10$ mol L $^{-1}$ HCl (a strong acid), which would give pH $= 1.0$ . The weak acid is much less acidic at the same concentration because it is only partly ionised.

Final answer: pH $= 2.87$

Percentage ionisation and dilution

A useful companion measure is the percentage ionisation, the fraction of acid molecules that have ionised: $\dfrac{[\text{H}^+]}{c} \times 100$ . For a strong acid this is essentially $100\%$ at any concentration; for a weak acid it is small and, importantly, increases on dilution. Diluting a weak acid shifts its ionisation equilibrium to the right (more molecules of water per acid molecule favour ionisation), so the percentage ionised rises even though $[\text{H}^+]$ itself falls. This is a frequent WACE discussion point: dilution lowers the concentration of hydrogen ions and so raises the pH, yet the proportion of acid that is ionised goes up.

Distinguishing strength experimentally

Two solutions of the same concentration, one a strong and one a weak acid, can be told apart by measurements that respond to ionisation. The strong acid has a lower pH (higher $[\text{H}^+]$ ), conducts electricity better (more ions), and reacts faster with a reactive metal or carbonate because of its higher hydrogen-ion concentration. Crucially, however, both will neutralise the same number of moles of base per mole of acid, because the total amount of acid (its concentration) is identical. Separating "rate and pH" effects (which depend on strength) from "amount of base needed" (which depends on concentration and the number of acidic protons) is exactly the skill being assessed.

Why this matters

$K_a$ and $K_b$ let you rank acids and bases, calculate the pH of weak acid and base solutions, and explain buffer behaviour and the position of titration end points. They turn the qualitative idea of strength into a number you can compute with.

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 20226 marksPropanoic acid has

K_a = 1.3 \times 10^{-5}

. (a) Write the ionisation equilibrium and the

K_a

expression. (b) Calculate the

\text{pH}

of a

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

solution, justifying any approximation. (c) Calculate the percentage ionisation.

Show worked answer →

A 6 mark question rewards the expression, the pH with valid approximation, and the percentage ionisation.

(a) $\text{C}_2\text{H}_5\text{COOH} \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^- + \text{H}^+$ , with $K_a = \dfrac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_2\text{H}_5\text{COOH}]}$ .

(b) Let $x = [\text{H}^+]$ and assume $0.20 - x \approx 0.20$ :

K_a = \frac{x^2}{0.20} \Rightarrow x^2 = (1.3 \times 10^{-5})(0.20) = 2.6 \times 10^{-6},

x = 1.61 \times 10^{-3}\ \text{mol L}^{-1},\qquad \text{pH} = -\log_{10}(1.61 \times 10^{-3}) = 2.79.

The approximation is valid because

x

is well under

5\%

0.20

(c) Percentage ionisation $= \dfrac{x}{c} \times 100 = \dfrac{1.61 \times 10^{-3}}{0.20} \times 100 = 0.81\%$ , confirming the acid is only slightly ionised.

Markers reward the $K_a$ expression, $\text{pH} = 2.79$ with the approximation check, and $0.81\%$ ionisation.

WACE 20205 marksTwo solutions are prepared:

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

hydrochloric acid and

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

ethanoic acid (

K_a = 1.8 \times 10^{-5}

). (a) Calculate the

\text{pH}

of each. (b) Use the result to explain the difference between acid strength and acid concentration.

Show worked answer →

A 5 mark question rewards both pH values and the strength-versus-concentration distinction.

(a) $\text{HCl}$ is strong and fully ionised, so $[\text{H}^+] = 0.010$ and $\text{pH} = -\log_{10}(0.010) = 2.0$ .
For ethanoic acid: $x = \sqrt{K_a c} = \sqrt{(1.8 \times 10^{-5})(1.0)} = 4.24 \times 10^{-3}\ \text{mol L}^{-1}$ , so $\text{pH} = -\log_{10}(4.24 \times 10^{-3}) = 2.37$ .

(b) Even though the ethanoic acid is 100 times more concentrated, its pH (2.37) is higher (less acidic) than the dilute strong acid (2.0). This is because strength describes the degree of ionisation: the strong acid releases nearly all its protons, while the weak acid releases only a small fraction. Concentration (how much acid is dissolved) is a separate quantity, so a concentrated weak acid can be less acidic than a dilute strong acid.

Markers reward $\text{pH} = 2.0$ and $\text{pH} = 2.37$ , and the explicit separation of strength (degree of ionisation) from concentration.