Mass of ${\mathrm{KHC}}_2{\mathrm{O}}_4$ (potassium acid oxalate) required to reduce 100 ml of 0.02 M ${\mathrm{KMnO}}_4$ in acidic medium $\left(\mathrm{to} {\mathrm{Mn}}^{2+}\right)$ is x g and to neutralize 100 ml of 0.05 M $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$ is y g, then

$$\begin{gathered} \mathrm{Eq}. \mathrm{of} {\mathrm{KMnO}}_4 = \mathrm{Eq}. \mathrm{of} {\mathrm{KHC}}_2{\mathrm{O}}_4 \\ \stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}+{\stackrel{+3}{\mathrm{C}}}_2{\mathrm{O}}_4^{-2}\to {\mathrm{Mn}}^{+2}+2\stackrel{+4}{\mathrm{C}}{\mathrm{O}}_2 \\ \therefore 100 \times {10}^{-3} \times 0.02 \times 5 = \frac{\mathrm{x}}{{\displaystyle \frac{128}{2}}} \\ \Rightarrow \mathrm{x} = 0.64 \mathrm{g} \\ \mathrm{Eq}. \mathrm{of} {\mathrm{KHC}}_2{\mathrm{O}}_4 = \mathrm{Eq}. \mathrm{of} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2 \\ \Rightarrow \frac{\mathrm{y}}{{\displaystyle \frac{128}{1}}} = 100 \times {10}^{-3} \times 0.05 \times 2 \\ (\because {\mathrm{KHC}}_2{\mathrm{O}}_4 \mathrm{contains} \mathrm{only} \mathrm{one} \mathrm{replaceable} \mathrm{Hydrogen}, \mathrm{Bascity}=1) \\ \Rightarrow \mathrm{y} = 1.28 \mathrm{g} \therefore 2 \mathrm{x} = \mathrm{y} \end{gathered}$$