Equivalent mass of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\cdot 5{\mathrm{H}}_2\mathrm{O}$ in the reaction $2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3+{\mathrm{I}}_2\to {\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6+2\mathrm{NaI}$ is.

$2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\to {\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{e}}^{-}$ Equivalent mass of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\cdot 5{\mathrm{H}}_2\mathrm{O}=\frac{2\times \text{ Molar mass of }{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\cdot 5{\mathrm{H}}_2\mathrm{O}}{2 \mathrm{eq} {\mathrm{mol}}^{-1}}=248 \mathrm{g} {\mathrm{eq}}^{-1}$

As an illustration, the equivalent weight of oxygen is 16.0 g/2 = 8.0g. In an acid-base reaction, the mass that provides or interacts with one mole of hydrogen ion (${\mathrm{H}}^{+}$) determines the equivalent mass of an acid or base.