Titration calculation

Quantitative chemistry

GCSE Chemistry (8462) · 4.3.4 & 4.4.2.5 · Higher

ORIGINAL

Concordant mean titre, moles of the known solution, the equation's ratio, then divide by the unknown's volume (chemistry only).

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Question

25.0 cm³ of sodium hydroxide solution is titrated with 0.150 mol/dm³ sulfuric acid. The titres are 21.40 cm³ (rough), 20.05 cm³ and 19.95 cm³.

\mathrm{H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)}

. Calculate the concentration of the sodium hydroxide in mol/dm³ and in g/dm³.

A_r

: Na = 23, O = 16, H = 1.

Every step worked, with the reasoning.

1
$\text{mean titre} = \dfrac{20.05 + 19.95}{2} = 20.00\ \text{cm}^3$
Leave out the rough titre and average the concordant titres — those within 0.10 cm³ of each other.
2
$n(\mathrm{H_2SO_4}) = 0.150 \times \dfrac{20.00}{1000} = 0.00300\ \text{mol}$
Moles = concentration × volume in dm³.
3
$n(\mathrm{NaOH}) = 2 \times 0.00300 = 0.00600\ \text{mol}$
The equation's 1 : 2 ratio — each H₂SO₄ neutralises two NaOH. Common error: missing the × 2, which gives 0.120 mol/dm³, half the right answer.
4
$c(\mathrm{NaOH}) = \dfrac{0.00600}{0.0250} = 0.240\ \text{mol/dm}^3$
Divide by the sodium hydroxide's own volume in dm³: 25.0 cm³ = 0.0250 dm³.
5
$0.240 \times 40 = 9.60\ \text{g/dm}^3$
g/dm³ = mol/dm³ × $M_r$ , and $M_r(\mathrm{NaOH}) = 23 + 16 + 1 = 40$ .

Answer: $0.240\ \text{mol/dm}^3$ , which is $9.60\ \text{g/dm}^3$ (3 s.f.)