Question

If x $g$ is the mass of $NaH{C}_2{O}_4$ is required to neutralise 100 mL of 0.2 (M) $NaOH$ and y $g$ is required to reduce 100 mL of 0.02 (M) $KMn{O}_4$ in acidic medium, then:

• A. $x=y$
• B. $x=2y$
• C. $x=4y$
• D. $x=3y$

Answer 1

The correct option is C $x=4y$

Let the molar wt. of $NaH{C}_2{O}_4=M$

$NaH{C}_2{O}_4+NaOH\to N{a}_2{C}_2{O}_4+H_{2}O$

only one H exchange is taking place so, n factor of both $NaH{C}_2{O}_4\text{and}NaOH$ is 1.

$\therefore \frac{x}{M}=100\times 0.2\times {10}^{-3}=20\times {10}^{-3}$ ...(I)


Case 2:

$5H{C}_2{O}_4^{-}+Mn{O}_4^{-}+11H^{+}\to M{n}^{2+}+10C{O}_2+8H_{2}O$

from here,
Calculating n-factor of $NaH{C}_2{O}_4$

In reactant side O.S. of carbon=3
In product side O.S. of carbon=4

n factor of $NaH{C}_2{O}_4=1\times 2=2$

n factor of $KMn{O}_4=5$, as medium is acidic.


$\therefore \frac{2y}{M}=100\times 0.02\times 5\times {10}^{-3}=10\times {10}^{-3}$

$\therefore \frac{y}{M}=100\times 0.02\times 5\times {10}^{-3}=5\times {10}^{-3}$ ...(ii)

Dividing equation (i) by equation (ii)
$\frac{x}{y}=\frac{20}{5}$
or x = 4y