Calculate the volume (in mL) of 0.15 M sodium thiosulphate solution required to titrate 12.0 mL of a 0.04 M iron(III) chloride solution in a standard redox titration where ${\mathrm{Fe}}^{3+}+{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\to {\mathrm{Fe}}^{2+}+{\mathrm{S}}_4{\mathrm{O}}_6^{2-}$ (1:1).

Step-by-step solution

1. Calculate moles of ${\mathrm{Fe}}^{3+}$: $n=\text{volume(L)}\times \text{concentration(M)}=0.012\times 0.04=4.8\times 10^{-4},\text{mol}$.
2. Use 1:1 stoichiometry; moles of ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ needed = $4.8\times 10^{-4},\text{mol}$.
3. Compute volume of 0.15 M thiosulphate: $V=\frac{n}{C}=\frac{4.8\times 10^{-4}}{0.15}=3.2\times 10^{-3},\text{L}=3.2,\text{mL}$.
4. Thus the required volume is 3.2 mL.

Concept tested: Redox titration stoichiometry

Published 13 October 2026 · How this solution was prepared and checked