Question

A mixture of $H _{2} C _{2} O _{4}$ (oxalic acid) and $NaHC _{2} O _{4}$ weighing $2.02\, g$ was dissolved in water and the solution made up to one litre. Ten millilitres of the solution required $3.0 \,mL$ of $0.1 \,N$ sodium hydroxide solution for complete neutralisation. In another experiment, $10.0\, mL$ of the same solution, in hot dilute sulphuric acid medium, required $4.0 \,mL$ of $0.1 \,N$ potassium permanganate solution for complete reaction. Calculate the amount of $H_{2} C _{2} O _{4}$ and $NaHC _{2} O _{4}$ in the mixture.

Answer 1

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