Question

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

• A. 1.3
• B. 1.699
• C. 1
• D. 2

Answer 1

Answer: (A)

Oxalic acid $=xmol/L$
Oxalic acid $KMn{O}_4$
$M_1{V}_1=M_2{V}_2$
$40mL\times x=16mL\times 0.05$
$x=\frac{16\times 0.05}{40}=\frac{1}{50}$
$x=\frac{1}{50}M$
Now convert molarity into normality
$N\times$ eq. wt. $=M\times$ mol. wt. of oxalic acid
$N\times 45=\frac{1}{50}\times 90$
$N=\frac{1}{25}$
This normality represents the hydrogen ion concentration.
So, $\left[H^{+}\right]=\frac{1}{25}$
$pH=\log \frac{1}{\left[H^{+}\right]}=\log 25=1.3$