Answer
$c(H_{2}C_{2}O_{4})=0.29464M$
Work Step by Step
Considering stoichiometric coefficients, we can conclude that
$n(H_{2}C_{2}O_{4}):n(MnO_{4}^{-})=5:2$
Therefore, $n(H_{2}C_{2}O_{4}) = \frac{5}{2}\times n(MnO_{4}^{-}) = \frac{5}{2}\times n(KMnO_{4})=\frac{5}{2}\times c(KMnO_{4})\times V(KMnO_{4}) = 2.5\times 0.127\frac{mol}{dm^3}\times 0.0232dm^3=0.007366mol$
Molarity of oxalic acid can now be easily obtained:
$c(H_{2}C_{2}O_{4})=\frac{n(H_{2}C_{2}O_{4})}{V(H_{2}C_{2}O_{4})}=\frac{0.007366mol}{0.025dm^3}=0.29464M$
