Question

What is the number of moles of ${\mathrm{KMnO}}_4$ that will be needed to react with one mole of sulphite ions in acidic solution?

Answer 1

The reaction is as follows:

• $2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+3{\mathrm{H}}_2\mathrm{O}+5\mathrm{O}$
• $[{\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}]\times 2$
• $[{\mathrm{SO}}_3^{2-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{SO}}_4^{2-}+2\mathrm{H}+2{\mathrm{e}}^{-}]\times 5$
• $2{\mathrm{MnO}}_4^{-}+5{\mathrm{SO}}_3^{2-}+6{\mathrm{H}}^{+}\to 2{\mathrm{Mn}}^{2+}+5{\mathrm{SO}}_4^{2-}+3{\mathrm{H}}_2\mathrm{O}$ (from the above reactions)
• Since $5\mathrm{moles}$ of sulphite ion $\left({\mathrm{SO}}_3^{2-}\right)$ reacts with $=2\mathrm{moles}\mathrm{of}{\mathrm{MnO}}_4^{-}$
• Hence, $1\mathrm{mole}$of sulphite ion reacts with $=\frac{2}{5}\mathrm{mol}\mathrm{of}{\mathrm{MnO}}_4^{-}$.