Question

$25mL$ of household bleach solution was mixed with $30mL$ of $0.50MKI$ and $10mL$ of $4N$ acetic acid. In the titration of the liberation iodine, $48mL$ of $0.25NN{a}_2{S}_2{O}_3$ was used to reach the end point. The molarity of the household bleach solution is:

• A. $0.48M$
• B. $0.96M$
• C. $0.24M$
• D. $0.024M$

Answer 1

Answer: (C)

$0.24M$

Hydrogen peroxide is household bleach.

$\underset{\left(25mL\right)}{H_2{O}_2}+\underset{\underset{30mL}{0.5M}}{2I^{-}}+\underset{\underset{10mL}{4N}}{2H^{+}}\to I_2+2H_{2}O$

$I_2+\underset{\underset{48mL}{0.25N}}{2N{a}_2{S}_2{O}_3}\to 2NaI+N{a}_2{S}_4{O}_6$

Normality and molarity of $C{H}_{3}COOH\left(H^{+}\right)$ and $N{a}_2{S}_2{O}_3$ will be equal.

Number of moles of $I_2=\frac{1}{2}\times$ Number of moles of $N{a}_2{S}_2{O}_3$

$=\frac{1}{2}\left[\frac{MV}{1000}\right]=\frac{1}{2}\left[\frac{0.25\times 48}{1000}\right]=6\times {10}^{-3}$

Number of moles of $H_2{O}_2=$ Number of moles of $I_2$
$=6\times {10}^{-3}$

Molarity of $H_2{O}_2=\frac{n_{H_2{O}_2}}{V}\times 1000$

$=\frac{6\times {10}^{-3}}{25}\times 1000$

$=0.24$.