Question

$25mL{H}_2{O}_2$ were added to excess of acidified solution of $KI$. The iodine so liberated required $20mL$ of $0.1N$ sodium thiosulphate for titration. Calculate the strength in terms of normality percentage and volume.

Answer 1

Let $N$ be normality of $H_2{O}_2$

$20$ml of $I_2=25$ ml of$0.1NN{a}_2{S}_2{O}_3=20$ ml of $I_2$ $=25$ml of $N{H}_2{O}_2$

$\to 20$ml of $I_2=25$ml of $N{H}_2{O}_2$

$20\ast 0.1=25N$

$N=\frac{2}{25}=0.08$

So strength in terms of normality

Normality * Equivalent Mass

$=0.08\ast 17$

$=1.36$ g/L

In terms of percentage $=\frac{1.36}{1000}\ast 100=0.136$

For $100$ml of solution $0.136$g of $H_2{O}_2$, volume of oxygen given $=22.4L$

So, $0136g$ of $H_2{O}_2=\frac{22400}{68}\ast 0.00136$

$=0.448ml$ of oxygen

Strength$=0.8N$

Percentage $=0.136$%

Volume : $0.448$ml