Question

$H_2{O}_2$ solution of $0.2$ N is prepared and after $5$ hours it is titrated with $KMn{O}_4$ solution. If the volume of the $H_2{O}_2$ solution taken is half of the volume of $KMn{O}_4$ used (in acidic medium), find the molarity of $KMn{O}_4$ solution?

• A. $0.4$ M
• B. $0.01$ M
• C. $0.02$ M
• D. $0.2$ M

Answer 1

Answer: (B)

$5$ hours corresponds to half life period for hydrogen peroxide in which its strength will reduce to half.
In $5$ hours, the normality of hydrogen peroxide solution will decrease from $0.2$ N to $0.1$ N.
$0.1$ N hydrogen peroxide corresponds to $0.05$ M hydrogen peroxide as its normality is twice its molarity.
$2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$.
Thus $2$ moles of potassium permanganate reacts with $5$ moles of hydrogen peroxide.
Hence, the number of moles of potassium permanganate that will react with $1$ L of $0.05$ M hydrogen peroxide solution are $0.05\times \frac{2}{5}=0.02$ moles.
Since double volume of potassium permanganate is take, the molarity of potassium permanganate solution is $\frac{0.02}{2}mol/L=0.01$ M.