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Titration: sodium hydroxide against sulfuric acid

A-Level (AQA) · unreviewed

Titration of a base of unknown concentration with a diprotic acid of known concentration.

Practise this with fresh numbers

Worked example

A 25.0 cm³ portion of NaOH(aq) needed 22.54 cm³ of 0.124 mol dm⁻³ H₂SO₄ for neutralisation. Find the concentration of the NaOH.

1. Concentration from moles and volume: solve for n(H₂SO₄)
$n = cV$
= 0.002795 mol
2. Mole ratio from the balanced equation: solve for n(NaOH)
$n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$
= 0.005590 mol
3. Concentration from moles and volume: solve for c(NaOH)
$c = \dfrac{n}{V}$
= 0.2236 mol dm⁻³

Answer: c(NaOH) = 0.224 mol dm⁻³

Method

n(H₂SO₄) = c × V (volume in dm³). The equation H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O gives n(NaOH) = 2n(H₂SO₄). Then c(NaOH) = n ÷ V(NaOH).

Common mistakes

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O: each mole of sulfuric acid neutralises two moles of NaOH.

Formulas used

Concentration from moles and volume
$c = \dfrac{n}{V}$
Mole ratio from the balanced equation
$n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

Practise this with fresh numbers

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