Question

An impure sample of sodium oxalate $(N{a}_2{C}_2{O}_4)$ weighing $0.20g$ is dissolved in aqueous solution of $H_{2}S{O}_4$ and solution is titrated at $70^{\circ }C$, requiring $45mL$ of $0.02MKMn{O}_4$ solution. The end point is overrun, and back titration in carried out with $10mL$ of $0.1M$ oxalic acid solution. Find the percentage purity of $N{a}_2{C}_2{O}_4$ in sample.

Answer 1

Answer: 83.75

$m$-eq. of $N{a}_2{C}_2{O}_4=m$-eq. of $KMn{O}_4$ reacted
total $m$-eq. of $KMn{O}_4^{-}$ excess $m$-eq. of $KMn{O}_4$ reacted with $H_2{C}_2{O}_4$
$=45\times 0.02\times 5-10\times 0.1\times 2=2.5$
$1000\times \frac{W}{134}\times 2=2.5$
$W_{N{a}_2{C}_2{O}_4}=0.1675g$
$\%$ purity of $N{a}_2{C}_2{O}_4$ in sample
$=\frac{0.1675}{0.2}\times 100=83.75$