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.
$25mL$ of household bleach solution was mixed with $30mL$ of $0.50$ $MKI$ and $10mL$ of $4N$ acetic acid. In the titration of the liberated 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) 0.48 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 :
$2H^{+}+OC{l}^{-}+2I⟶C{l}^{-}+I_2+H_{2}O\dots$ (i)
$I_2+2S_2{O}_3^{2-}⟶2I+S_4{O}_6^{2-}\dots$ (ii)
Also the $n$-factor of $S_2{O}_3^{2-}$ is one as
$2S_2{O}_3^{2-}⟶S_4{O}_6^{2-}+2e^{-}$
[one ' $e$ ' is produced per unit of $S_2{O}_3^{2-}$ ]
$\Rightarrow$ Molarity of $N{a}_2{S}_2{O}_3=0.25N\times 1=0.25M$
$\Rightarrow mmol$ of $N{a}_2{S}_2{O}_3$ used up $=0.25\times 48=12$
Now from stoichiometry of reaction (ii)
$12mmol$ of $S_2{O}_3^{2-}$ would have reduced $6mmol$ of $I_2$.
From stoichiometry of reaction (i)
$mmol$ of $OC{l}^{-}$reduced $=m$ mol in $I_2$ produced $=6$
$\Rightarrow$ Molarity of household bleach solution
$=\frac{6}{25}=0.24M$
Shortcut Method
Milliequivalent of $N{a}_2{S}_2{O}_3=$ milliequivalent of $OC{l}^{-}$
$=0.25\times 48=12$
Also $n$-factor of $OC{l}^{-}=2[C{l}^{+}⟶C{l}^{-}$, gain of $2e^{-}]$
$\Rightarrow mmol$ of $OC{l}^{-}=\frac{12}{2}=6mmol$. Remaining part is solved in the same manner.