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pH of a weak base

UKChO Round 1 · UKChO Round 2 · IChO · unreviewed

[OH⁻] ≈ √(K_b c), pOH = −log₁₀[OH⁻], then pH = pK_w − pOH.

Practise this with fresh numbers

Worked example

A weak base has K_b = 1.2 × 10⁻⁴. What is the pH of a 0.37 mol dm⁻³ solution? (pK_w = 14.00)

1. Weak base (approximate): solve for [OH⁻]
$[\mathrm{OH^-}] = \sqrt{K_\mathrm{b}\,c}$
= 0.006663 mol dm⁻³
2. pOH: solve for pOH
$\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}]$
= 2.176
3. pH + pOH = pKw: solve for pH
$\mathrm{pH} = \mathrm{p}K_\mathrm{w} - \mathrm{pOH}$
= 11.82

Answer: pH = 11.8

Method

[OH⁻] ≈ √(K_b c), pOH = −log₁₀[OH⁻], then pH = pK_w − pOH.

Common mistakes

That is the pOH. pH = pKw − pOH.

Formulas used

Weak base (approximate)
$[\mathrm{OH^-}] = \sqrt{K_\mathrm{b}\,c}$
pOH
$\mathrm{pOH} = -\log_{10}\dfrac{[\mathrm{OH^-}]}{c^\circ}$
pH + pOH = pKw
$\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_\mathrm{w}$

Practise this with fresh numbers

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