A student titrates 25.0 cm³ of sodium hydroxide solution with sulfuric acid of concentration 0.120 mol/dm³. The mean titre is 18.75 cm³.
$$2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$
(a) Calculate the concentration of the sodium hydroxide solution in mol/dm³. [3]
(b) Convert your answer to (a) into g/dm³. [$M_r$ of NaOH = 40] [1]
- (a) moles H₂SO₄ = 0.120 × 18.75 ÷ 1000 = 0.00225 (mol)(1)
- (a) moles NaOH = 2 × 0.00225 = 0.00450 (mol) (allow ECF: 2 × their moles of H₂SO₄)(1)
- (a) concentration = 0.00450 ÷ 0.0250 = 0.180 (mol/dm³) (allow 0.18; allow ECF: their moles of NaOH ÷ 0.0250, e.g. 0.090 if the 2 : 1 ratio is missed scores 2; correct answer with no working scores 3)(1)
- (b) 0.180 × 40 = 7.20 (g/dm³) (accept 7.2; allow ECF from (a))(1)
- Answer: (a) 0.180 mol/dm³ (b) 7.20 g/dm³
- Examiner tip: Remember the 2 : 1 ratio from the equation and convert cm³ to dm³ (÷1000) — both are frequent slips.