Question

The concentration of oxalic acid is $x$ mol litre${}^{-1}$. $40mL$ of this solution reacts with $16mL$ of $0.05M$ acidified $KMn{O}_4$. What is the $pH$ of ${}^{\prime }{x}^{\prime }M$ oxalic acid solution? (Assume that oxalic acid dissociates completely.)

• A. $3$
• B. $1.5$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Oxalic acid $\left(H_2{C}_2{O}_4\right)$ acts as an reducing agent here.
The corresponding reaction would be:
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

So, n-factor of $KMn{O}_4=+5$
n -factor of $H_2{C}_2{O}_4=+2$
Now,
$\text{Milliequiv of}{C}_2{O}_4^{2-}=\text{milliequiv. of}KMn{O}_4$
$\Rightarrow 2\left(40\right)x=5\times 16\times 0.05\Rightarrow x=0.05M\therefore \left[H^{+}\right]=2x=0.1MpH=-log\left[H^{+}\right]=1$