Question

$NaH{C}_2{O}_4$ is neutralised by $NaOH$ and can also be oxidised by $KMn{O}_4$ (in acidic medium). Equivalent weight is related to molecular weight $\left(M\right)$ of $Na{H}_2{C}_2{O}_4$ in these reactions as :

• A. $M,M$
• B. $\frac{M}{2},\frac{M}{2}$
• C. $\frac{M}{2},M$
• D. $M,\frac{M}{2}$

Answer 1

Answer: (D)

$M,\frac{M}{2}$

The balanced reactions are as follows:
$NaH{C}_2{O}_4+NaOH⟶{Na}_2{C}_2{O}_4+H_{2}O$
$NaH{C}_2{O}_4\equiv H{C}_2{O}_4^{-}\equiv H^{+}\equiv {OH}^{-}$
So, in this reaction, $NaH{C}_2{O}_4$ gives one mole of $H^{+}$ ion.
$\text{equivalent weight}$ $=\frac{\text{Molar mass}}{\text{Valency}}=\frac{M}{1}=M$
In oxidation process,
$H{C}_2{O}_4^{-}\equiv \underset{\uparrow +6}{C_2{O}_4^{3-}}\equiv 2\underset{\uparrow +8}{C{O}_2}\left(fortwoatoms\right)$
Change in oxidation is 2.
So. its equivalent weight is $\frac{M}{2}$.