Question

A mixture of $H_2{C}_2{O}_4$ and $NaH{C}_2{O}_4$ weighing $2.02$ g was dissolved in water and the solution made upto one litre. $10$ mL of this solution required $3$ mL of $0.1NNaOH$ solution for complete neutralization. In another experiment, $10$ mL of the same in hot dilute $H_{2}S{O}_4$ medium required $4$ mL of $0.1NKMn{O}_4$ for complete neutralization. Calculate the mass of $NaH{C}_2{O}_4$ in mixture (as nearest integer).

Answer 1

Let the amount of $H_2{C}_2{O}_4$ and $NaH{C}_2{O}_4$ be $a$ and $b$ respectively.
$a+b=2.02$ $...\left(i\right)$
For $1$ L mixture undergoing acid-case reaction,
Meq. of $H_2{C}_2{O}_4+$ Meq. of $NaH{C}_2{O}_4=$ Meq. of $NaOH$ used for $1000$ mL

$\left(\frac{a}{45}\right)\times 1000+\left(\frac{b}{112}\right)\times 1000=3\times 0.1\times \frac{1000}{10}$ $...\left(ii\right)$

[$\therefore$ Eq. mass of $H_2{C}_2{O}_4=\frac{M}{2}$ (as acid) and Eq. mass of $NaH{C}_2{O}_4=\frac{M}{1}$ (as acid salt)]
For redox changes, for $1000$ mL mixture,

Meq. of $H_2{C}_2{O}_4+$ Meq. of $NaH{C}_2{O}_4=$ Meq. of $KMn{O}_4$ used for $1000$ mL

$\frac{a}{45}\times 1000+\frac{b}{\frac{112}{2}}\times 1000=4\times 0.1\times \frac{1000}{10}=40$

$\therefore \frac{1000a}{45}+\frac{2000b}{112}=40$ $...\left(iii\right)$

[$\therefore$ Redox changes are $(C^{3+}{)}_2⟶2C^{4+}+2e$]
$M{n}^{7+}+5e⟶M{n}^{2+}$
($\therefore$ Eq. mass of $H_2{C}_2{O}_4=\frac{M}{2}$ and Eq. mass of $NaHC{O}_4=\frac{M}{2}$)
Solving Equations $\left(i\right)$ and $\left(ii\right)$ or $\left(ii\right)$ and $\left(iii\right)$ or $\left(i\right)$ and $\left(iii\right)$, we get
$a=0.90$ g, $b=1.12$ g
So, the nearest integer value is $1$ g.