Question

A mixture of $H_2{C}_2{O}_4$ (oxalic acid) and $NaH{C}_2{O}_4$ weighing $2.02g$ was dissolved in water and the solution made up to one litre. Ten millilitres of the solution required $3.0mL$ of $0.1N$ sodium hydroxide solution for complete neutralisation. In another experiment, $10.0mL$ of the same solution, in hot dilute sulphuric acid medium, required $4.0mL$ of $0.1N$ potassium permanganate solution for complete reaction. Calculate the amount of $H_2{C}_2{O}_4$ and $NaH{C}_2{O}_4$ in the mixture.

Answer 1

Let us consider $10mL$ of the stock solution contain $x$ millimol oxalic acid $H_2{C}_2{O}_4$ and $y$ millimol of $NaH{C}_2{O}_4$.
When titrated against $NaOH$, basicity of oxalic acid is 2 while that of $NaH{C}_2{O}_4$ is $1$ .
$\Rightarrow 2x+y=3\times 0.1=0.3$
When titrated against acidic $KMn{O}_4,n$-factors of both oxalic acid and $NaH{C}_2{O}_4$ would be 2 .
$\Rightarrow 2x+2y=4\times 0.1=0.4$
Solving equations (i) and (ii) gives
$y=0.1,x=0.1$
$\Rightarrow$ In $1.0L$ solution, mole of $H_2{C}_2{O}_4=\frac{0.1}{1000}\times 100=0.01$
Mole of $NaH{C}_2{O}_4=\frac{0.1}{1000}\times 100=0.01$
$\Rightarrow$ Mass of $H_2{C}_2{O}_4=90\times 0.01=0.9g$
Mass of $NaH{C}_2{O}_4=112\times 0.01=1.12g$