Question

To a 25 ml $H_2{O}_2$ solution, excess of acidified solution of KI was added. The iodine liberated required 20 ml of 0.3 N $N{a}_2{S}_2{O}_3$ solution. The volume strength of $H_2{O}_2$ solution is:

• A. 1.344 g/L
• B. 3.244 g/L
• C. 5.4 g.L
• D. 4.08 g/L

Answer 1

The correct option is A 1.344 g/L

Given,

$25ml$ $H_2{O}_2$ was reacted with excess of $KI$

Also $20ml$ $0.3N$ $N{a}_2{S}_2{O}_3$ was required to liberate $I_2$

The reaction involved in this liberation is

$2KI+H_2{O}_2⟶I_2+2KOH$

Let, Normality of $25ml$ $H_2{O}_2$ be $xN$

$\therefore$ From above titration

$N_1=$ Normality of $H_2{O}_2$

$V_1=$ Volume of $H_2{O}_2$

$N_2=$ Normality of $N{a}_2{S}_2{O}_3$ or $I_2$

$V_2=$ Volume of $N{a}_2{S}_2{O}_3$ or $I_2$

$\Rightarrow N_1{V}_1=N_2{V}_2$

$\Rightarrow x\times 25=0.3\times 20$

$\Rightarrow x=\frac{0.3\times 20}{25}$

$=0.24$

$\therefore$ Normality of $H_2{O}_2$ solution=$0.24N$

Now, We know Volume strength=$Normality\times Equivalent$ $weight$

Equivalent weight of $H_2{O}_2$ in terms of oxygen=$5.6L$

$\Rightarrow$ Volume strength=$0.24\times 5.6$

$=1.344g{L}^{-1}$