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pH of a weak acid

A-Level (AQA) · C3L6 · UKChO Round 1 · IChO · unreviewed

[H⁺] ≈ √(Kₐc) when dissociation is small (check: [H⁺]/c < 5 %), then pH = −log₁₀[H⁺].

Practise this with fresh numbers

Worked example

Calculate the pH of a 0.37 mol dm⁻³ solution of a weak acid HA (Kₐ = 1.2 × 10⁻⁴).

1. Weak acid (approximate): solve for [H⁺]
$[\mathrm{H^+}] = \sqrt{K_\mathrm{a}\,c}$
= 0.006663 mol dm⁻³
2. pH: solve for pH
$\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$
= 2.176

Answer: pH = 2.18

Method

[H⁺] ≈ √(Kₐc) when dissociation is small (check: [H⁺]/c < 5 %), then pH = −log₁₀[H⁺].

Common mistakes

You forgot the square root: [H⁺] ≈ √(Kₐc).
That treats HA as a strong acid ([H⁺] = c). A weak acid is only slightly dissociated.

Formulas used

Weak acid (approximate)
$[\mathrm{H^+}] = \sqrt{K_\mathrm{a}\,c}$
pH
$\mathrm{pH} = -\log_{10}\dfrac{[\mathrm{H^+}]}{c^\circ}$

Practise this with fresh numbers

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