Weak acids and Ka: TCE Chemistry (Tasmania)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

A weak acid only partly ionises, so it sits at equilibrium. The acid dissociation constant $K_a$ measures the extent of that ionisation: a larger $K_a$ (smaller $pK_a$ ) means a stronger acid. You find the pH of a weak acid by setting up and solving its $K_a$ equilibrium expression, using the approximation that $x$ is small compared with the initial concentration.

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.

Acid dissociation constant

K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \qquad pK_a = -\log_{10}K_a

A larger

K_a

(smaller

pK_a

) means more ionisation and a stronger acid.

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.

Strength and Ka

Strong acids ionise completely and have very large $K_a$ . Weak acids have $K_a \ll 1$ and ionise only slightly. The stronger the acid, the weaker its conjugate base ( $K_a \times K_b = K_w$ for a conjugate pair). Lower $pK_a$ equals stronger acid.

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.

Using pH equal to minus log of the acid concentration

For a weak acid you cannot set $[\text{H}^+]$ equal to the formal acid concentration, because most of the acid stays as undissociated molecules. You must work through the $K_a$ expression to find the small $[\text{H}^+]$ . Treating a weak acid like a strong acid badly overestimates the acidity. Also do not forget to take the square root after solving for $x^2$ .

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 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.

Practice4 marks

Benzoic 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)

Practice2 marks

For 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)

Practice2 marks

Ethanoic 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)

Practise this

Sources & how we know this

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