Question

$25 \, mL$ of household bleach solution was mixed with $30 \, mL$ of $0.50 \, M$ $KI$ and $10 \, mL$ of $4 \, N$ acetic acid. In the titration of the liberated iodine, $48 \, mL$ of $0.25 \, N$ $Na_{2}S_{2}O_{3}$ was used to reach the end point. The molarity of the household bleach solution is

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

Answer 1

Answer: (A)

$\underset{\underset{25 ml}{\text{Bleach}}}{H_{2} O_{2}}+\underset{\underset{30 ml}{0 . 5 M}}{2 I^{-}}+\underset{\underset{4 N \left(\right. 10 ml \left.\right)}{\left(\right. \text{from} \left(CH\right)_{3} COOH \left.\right)}}{2 H^{+}} \rightarrow I_{2 \, }+2H_{2}O$ ...(i)
$I_{2}+\underset{0 . 25 N \left(\right. 48 ml \left.\right)}{2 \left(Na\right)_{2} S_{2} O_{3}} \rightarrow \left(Na\right)_{2}S_{4}O_{6}+2NaI$ ...(ii)
m. mol of $I_{2}=\frac{1}{2}$ (m. Moles of $Na_{2}S_{2}O_{3}$ )
$=\frac{1}{2}\times \left(\right.0.25\times 48\left.\right)=6m.mol$
Using equation (i)
1 m. Mol of $I_{2}\equiv 1$ m. Mol of $H_{2}O_{2}$
m.mol of $H_{2}O_{2}=6$ m.mol
Molarity of $H_{2}O_{2}=\frac{6}{25}=0.24\,M$