Strong and weak acids and bases, Ka and Kb: WACE Year 12 Chemistry

Syllabus dot point

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

8 min readUpdated 2 Jun 2026

Quick answer

Strong acids and bases ionise essentially completely in water; weak ones ionise only partially and sit at equilibrium. The acid ionisation constant $K_a$ (and base constant $K_b$ ) measures how far that equilibrium lies to the right. A larger $K_a$ (or smaller $\text{p}K_a = -\log K_a$ ) means a stronger acid. Strength is not the same as concentration.

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.

Strength is about degree of ionisation

A strong acid is fully ionised regardless of concentration; a weak acid is only partly ionised. A dilute strong acid can have a higher pH than a concentrated weak acid. Never equate strong with concentrated.

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.

Ka and pKa

$K_a$ is the equilibrium constant for the ionisation of a weak acid in water. $\text{p}K_a = -\log_{10} K_a$ . A larger $K_a$ , or equivalently a smaller $\text{p}K_a$ , indicates a stronger acid.

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.

Confusing strength with concentration

The headline error is treating strong and concentrated as synonyms. A second is calculating the pH of a weak acid as if it were strong (using $[\text{H}^+] = c$ instead of $\sqrt{K_a c}$ ), which badly overestimates the acidity. A third is writing a full arrow for a weak acid equation; weak acids and bases must always be shown with the equilibrium arrows.

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-style6 marks

Propanoic 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 2020-style5 marks

Two 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.

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA