If a g is the mass of NaHC₂O₄ required to neutralize 100 mL of 0.2 M NaOH and b g that required to reduce 100 mL of 0.02 M KMnO₄ in acidic medium, then:

eq. wt. of NaHC₂O₄ as an acid =112
eq. of NaHC₂O₄ = eq. of NaOH
=100x10^-3x0.2
=0.02
a =0.02 x112 = 2.24 g
eq. wt. of NaHC₂O₄ as reductant
$\left({\mathrm{C}}_2{\mathrm{O}}_4^{2-}⟶2{\mathrm{CO}}_2+2{\mathrm{e}}^{-}\right)=\frac{112}{2}=56$
eq. of NaHC₂O₄ as reductant = eq. of
KMnO₄ = 100 x 10^-2 x 0.02 x 5 = 0.01
b = 0.01 x 56 = 0.56 g
Hence, $\mathrm{a}:\mathrm{b}=\frac{2.24}{0.56}=4:1\text{ or }\mathrm{a}=4\mathrm{b}$