Question

A mixture of $H_2{C}_2{O}_4$ and $NaH{C}_2{O}_4$ weighing $2.02g$ was dissolved in water and the solution made up to one litre. $10mL$ of this solution required $3.0mL$ of $0.1NNaOH$ solution for complete neutralisation. In another experiment $10mL$ of same solution in hot dilute $H_{2}S{O}_4$ medium required $4mL$ of $0.1NKMn{O}_4$ for complete neutralisation. Calculate the amount of $H_2{C}_2{O}_4$ and $NaH{C}_2{O}_4$ in the mixture.

Answer 1

Let mass of $H_2{C}_2{O}_4$ present in the mixture be $=ag$ in $1litre$
and mass of $NaH{C}_2{O}_4$ present in the mixture be $bg$ in $1litre$
For acid-base reaction
$H_2{C}_2{O}_4+2NaOH\to N{a}_2{C}_2{O}_4+2H_{2}O$
Eq. mass of $H_2{C}_2{O}_4=\frac{Mol.mass}{2}=\frac{90}{2}=45$
$NaH{C}_2{O}_4+NaOH\to N{a}_2{C}_2{O}_4+H_{2}O$
Eq. mass of $NaH{C}_2{O}_4=\frac{Mol.mass}{1}=112$
Now,
Equivalents of $H_2{C}_2{O}_4$ in $10mL$ solution $+$ Equivalent of $NaH{C}_2{O}_4$ in $10mL$ solution $=\frac{3\times 0.1}{1000}$
$\frac{a\times 10}{45\times 1000}+\frac{b\times 10}{112\times 1000}=\frac{3\times 0.1}{1000}$
or $112a+45b=\frac{3\times 0.1\times 45\times 112}{10}=\mathrm{151.2...}\left(i\right)$
For redox reaction
Eq. mass of $H_2{C}_2{O}_4=\frac{90}{2}=45$;
Eq. mass of $NaH{C}_2{O}_4=\frac{112}{2}=56$
(Change in oxidation number of carbon per molecule $=2;C_2^{3+}\to 2C^{4+})$
Now,
Equivalents of $H_2{C}_2{O}_4$ in $10mL$ solution $+$ Equivalent of $NaH{C}_2{O}_4$ in $10mL$ solution $=\frac{4\times 0.1}{1000}$
$\frac{a\times 10}{45\times 1000}+\frac{b\times 10}{56\times 1000}=\frac{4\times 0.1}{1000}$
or $56a+45b=100.8....\left(ii\right)$
Solving equations (i) and (ii),
$a=0.9g$ and $b=1.12g$.