Question

$1.17g$ of an impure sample of oxalic acid was dissolved and made up to $200mL$ with water. $10mL$ of this solution in acid medium required $8.5mL$ of a solution of potassium permanganate containing $3.16gperlitre$ of oxidation. Calculate the percentage purity of oxalic acid.

Answer 1

During the redox reaction

we have,

no. of gram equivalent of oxidising agent = gram equivalent weight of reducing agent

or $N_1{V}_1=N_2{V}_2$

$N=$ $\frac{weightweight}{equivalentweight\ast volume}$

let pure form of oxalic acid weigh = x

so we have,

$\frac{x\ast 1000\ast 10}{63\ast 200}$= $\frac{3.16\ast 8.5}{31.6}$

$x$= 1.071 g

percentage purity = $\frac{pureform}{impureform}$$\ast 100$

percentage purity = 91.53