Question

Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.
$25\, mL$ 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 $0.25\, N \,Na_2S_2O_3$ was used to reach the end point. The molarity of the household bleach solution is
(a) 0.48 M
(b) 0.96 M
(c) 0.24 M
(d) 0.024 M

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

Answer 1

Answer: (C)

The involved redox reactions are :
$2 H ^{+}+ OCl ^{-}+2 I \longrightarrow Cl ^{-}+ I _{2}+ H _{2} O \ldots$ (i)
$I _{2}+2 S _{2} O _{3}^{2-} \longrightarrow 2 I + S _{4} O _{6}^{2-} \ldots $ (ii)
Also the $n$-factor of $S _{2} O _{3}^{2-}$ is one as
$2 S _{2} O _{3}^{2-} \longrightarrow S _{4} O _{6}^{2-}+2 e^{-}$
[one ' $e$ ' is produced per unit of $S _{2} O _{3}^{2-}$ ]
$\Rightarrow$ Molarity of $Na _{2} S _{2} O _{3}=0.25 \,N \times 1=0.25\, M$
$\Rightarrow m \,mol$ of $Na _{2} S _{2} O _{3}$ used up $=0.25 \times 48=12$
Now from stoichiometry of reaction (ii)
$12\, m \,mol$ of $S _{2} O _{3}^{2-}$ would have reduced $6\, m\, mol$ of $I _{2}$.
From stoichiometry of reaction (i)
$m \,mol$ of $OCl ^{-}$reduced $= m$ mol in $I _{2}$ produced $=6$
$\Rightarrow$ Molarity of household bleach solution
$=\frac{6}{25}=0.24 \,M$
Shortcut Method
Milliequivalent of $Na _{2} S _{2} O _{3}=$ milliequivalent of $OCl ^{-}$
$=0.25 \times 48=12$
Also $n$-factor of $OCl ^{-}=2\left[ Cl ^{+} \longrightarrow Cl ^{-}\right.$, gain of $\left.2 e^{-}\right]$
$\Rightarrow \,m \,mol$ of $OCl ^{-}=\frac{12}{2}=6 \,m\, mol$. Remaining part is solved in the same manner.