Question

The concentration of oxalic acid is ' $x$ ' mol $L ^{-1}$. $40 \,mL$ of this solution reacts with $16\, mL$ of $0.05\, M$ acidified $KMnO _{4} .$ 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 $=x \,mol / L$
Oxalic acid $KMnO _{4}$
$M_{1} \, V_{1} =M_{2} \, V_{2} $
$40 mL \times x =16 \,mL \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, $\left[ H ^{+}\right]=\frac{1}{25}$
$pH =\log \frac{1}{\left[ H ^{+}\right]}=\log 25=1.3$