Question

$25$ ml of $H_2{O}_2$ solution were added to excess of acidified $KI$ solution. The iodine so liberated required $20$ ml of $0.1\text{N}$ $N{a}_2{S}_2{O}_3$ for titration. Calculate strength of $H_2{O}_2$ in terms of normality and percentage?

• A. $0.04\text{N},0.136\%$
• B. $0.08\text{N},0.136\%$
• C. $0.04\text{N},0.163\%$
• D. $0.02\text{N},0.163\%$
  

Answer 1

The correct option is B $0.08\text{N},0.136\%$

$I_2+2N{a}_2{S}_2{O}_3^{-}\to N{a}_2{S}_4{O}_6+2NaI$

Milliequivalent of $H_2{O}_2=$ Milliequivalent of $I_2=$Milliequivalent of $N{a}_2{S}_2{O}_3$

$\frac{w}{34/2}\times 1000=20\times 0.1$

${\therefore W}_{H_2{O}_2}=0.034$ gm$/25$ ml

$N_{H_2{O}_2}=\frac{0.034}{34/2}\times \frac{1000}{25}=0.08$

Volume strength $=5.6\times 0.08=0.448$

$\%$ strength $=\frac{17}{56}\times 5.6\times 0.08=0.136$.