Question

Oxalic acid, $H_2{C}_2{O}_4$ forms two series of salt, ${H{C}_2{O}_4}^{-}$ and ${C_2{O}_4}^{-}$. If $0.9$ g of oxalic acid is in $100$ mL solution, ${H{C}_2{O}_4}^{-}$ and ${C_2{O}_4}^{-}$ have normality respectively :

• A. 0.1 N, 0.1 N
• B. 0.1 N, 0.2 N
• C. 0.2 N, 0.2 N
• D. 0.2 N, 0.1 N

Answer 1

The correct option is B 0.1 N, 0.2 N

Molarity of oxalic acid $=\frac{0.9}{90}\times \frac{1000}{100}=0.1M$
The dissociation reactions are as shown below.
$H_2{C}_2{O}_4\rightleftharpoons H^{+}+{H{C}_2{O}_4}^{-}$
${H{C}_2{O}_4}^{-}\rightleftharpoons H^{+}+{C_2{O}_4}^{2-}$.
For first dissociation, it acts as monobasic acid and for second dissociation, it acts as dibasic acid.
The normality of ${H{C}_2{O}_4}^{-}$ is equal to its molarity and is equal to $0.1$ N.
The normality of ${C_2{O}_4}^{-}$ is equal to two times its molarity. It is equal to $0.2$ N.