Question

Find out the $\%$ of oxalate ion in given sample of oxalate salt of which $0.3 \, g$ is present in $100 \, mL$ of solution required $90 \, mL$ of $N/20 \, KMnO_{4}$ for complete oxidation.

Answer 1

The redox changes are $Mn ^{+7}+5 e ^{-} \rightarrow Mn ^{2+}$

$\left(C^{3+}\right)_{2} \rightarrow 2 C^{4+}+2 e^{-}$

$\because$ meq. of oxalate ion $=$ meq. of $KMnO _{4}$ $( w / E ) \times 1000=90 \times(1 / 20)$

Or $\left(\frac{ w }{88 / 2}\right) \times 1000=\frac{9}{2}\left(\because E _{ C _{2} O _{4}^{-2}}=\frac{88}{2}\right)$

or $w _{ C _{2} O _{4}^{-2}}=0.198 g$

$\therefore \%$ of oxalate in sample $=(0.198 \times 100) / 0.3=66 \%$