In a neutralisation reaction, 25 mL of 0.2 M sulfuric acid reacts with excess sodium hydroxide. Find the volume (in mL) of 0.1 M NaOH required to completely neutralise the acid.

Step-by-step solution

1. Convert 25 mL to litres: $0.025,\mathrm{L}$
2. Calculate moles of ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$: $0.2,\mathrm{M}\times 0.025,\mathrm{L}=0.005,\mathrm{mol}$
3. Each ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ gives 2 ${\mathrm{H}}^{+}$; moles of ${\mathrm{H}}^{+}$: $0.005\times 2=0.010,\mathrm{mol}$
4. Neutralisation requires 1 mol $\mathrm{NaOH}$ per 1 mol ${\mathrm{H}}^{+}$; thus moles of $\mathrm{NaOH}$ needed = $0.010,\mathrm{mol}$
5. Volume of 0.1 M $\mathrm{NaOH}$: $\frac{0.010}{0.1}=0.10,\mathrm{L}$
6. Convert to mL: $0.10,\mathrm{L}\times 1000=100,\mathrm{mL}$

Concept tested: Stoichiometry of neutralisation

Published 4 October 2026 · How this solution was prepared and checked