Question

An excess of acidified solution of $KI$ is added to a $25$ mL $H_2{O}_2$ solution. The iodine liberated requires $20$ mL of $0.3NN{a}_2{S}_2{O}_3$ solution. Calculate the volume strength of $H_2{O}_2$:

• A. $1.144$
• B. $1.244$
• C. $1.344$
• D. $1.444$

Answer 1

The correct option is C $1.344$

$I_2+2N{a}_2{S}_2{O}_3\to N{a}_2{S}_4{O}_6+2NaI$
$20$ ml of $0.3M(=0.3N)$ $N{a}_2{S}_2{O}_3$ will react with $\frac{\left(20\right)\left(0.3\right)}{\left(1000\right)\left(2\right)}$ moles of $I_2$ which is $0.003$ moles or $0.006$ equivalents. Thus there are $0.006$ equivalents in $25$ ml of $H_2{O}_2$, hence, its normality is $0.24$ N and thus, the volume strength is $0.24\times 5.6=1.344$.