Some Basic Concepts of Chemistry 2 Question 43
43. A mixture of $\mathrm{H}{2} \mathrm{C}{2} \mathrm{O}{4}$ (oxalic acid) and $\mathrm{NaHC}{2} \mathrm{O}{4}$ weighing $2.02 \mathrm{~g}$ was dissolved in water and the solution made up to one litre. Ten millilitres of the solution required $3.0 \mathrm{~mL}$ of $0.1 \mathrm{~N}$ sodium hydroxide solution for complete neutralisation. In another experiment, $10.0 \mathrm{~mL}$ of the same solution, in hot dilute sulphuric acid medium, required $4.0 \mathrm{~mL}$ of $0.1 \mathrm{~N}$ 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.
$(1990,5 \mathrm{M})$
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Solution:
- Let us consider $10 \mathrm{~mL}$ 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 $\mathrm{NaOH}$, 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 $1.0 \mathrm{~L}$ solution, mole of $\mathrm{H}{2} \mathrm{C}{2} \mathrm{O}_{4}=\frac{0.1}{1000} \times 100=0.01$
$$ \text { Mole of } \mathrm{NaHC}{2} \mathrm{O}{4}=\frac{0.1}{1000} \times 100=0.01 $$
$$ \begin{aligned} \Rightarrow \text { Mass of } \mathrm{H}{2} \mathrm{C}{2} \mathrm{O}{4} & =90 \times 0.01=0.9 \mathrm{~g} \ & \text { Mass of } \mathrm{NaHC}{2} \mathrm{O}_{4}=112 \times 0.01=1.12 \mathrm{~g} \end{aligned} $$
