Some Basic Concepts of Chemistry - Result Question 40
43. A mixture of $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$ (oxalic acid) and $\mathrm{NaHC}_2 \mathrm{O}_4$ 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 $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$ and $\mathrm{NaHC}_2 \mathrm{O}_4$ in the mixture.
(1990,5 M)
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Solution:
- Let us consider 10 mL of the stock solution contain $x$ millimol oxalic acid $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$ and $y$ millimol of $\mathrm{NaHC}_2 \mathrm{O}_4$.
When titrated against NaOH , basicity of oxalic acid is 2 while that of $\mathrm{NaHC}_2 \mathrm{O}_4$ is 1 .
$\Rightarrow \quad 2 x+y=3 \times 0.1=0.3$
When titrated against acidic $\mathrm{KMnO}_4, n$-factors of both oxalic acid and $\mathrm{NaHC}_2 \mathrm{O}_4$ would be 2 .
$\Rightarrow \quad 2 x+2 y=4 \times 0.1=0.4$
Solving equations (i) and (ii) gives
$y=0.1, x=0.1$
$\Rightarrow$ In 1.0 L solution, mole of $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4=\frac{0.1}{1000} \times 100=0.01$
Mole of $\mathrm{NaHC}_2 \mathrm{O}_4=\frac{0.1}{1000} \times 100=0.01$
$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 & =90 \times 0.01=0.9 \mathrm{~g} \\ \text { Mass of } \mathrm{NaHC}_2 \mathrm{O}_4 & =112 \times 0.01=1.12 \mathrm{~g} \end{aligned}$
BETA
