Pblock Elementsii Result Question 28
Passage
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.
(2012)
28. $25 \mathrm{~mL}$ of household bleach solution was mixed with $30 \mathrm{~mL}$ of $0.50 $ $\mathrm{M} $ $\mathrm{KI}$ and $10 \mathrm{~mL}$ of $4 \mathrm{~N}$ acetic acid. In the titration of the liberated iodine, $48 \mathrm{~mL}$ of $0.25 \mathrm{~N} \mathrm{~Na} _2 \mathrm{~S} _2 \mathrm{O} _3$ was used to reach the end point. The molarity of the household bleach solution is
(a) $0.48 $ $\mathrm{M}$
(b) $0.96 $ $\mathrm{M}$
(c) $0.24 $ $\mathrm{M}$
(d) $0.024 $ $\mathrm{M}$
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Answer:
Correct Answer: 28. ( c )
Solution:
The involved redox reactions are :
$ 2 \mathrm{H}^{+}+\mathrm{OCl}^{-}+2 \mathrm{I}^{-} \longrightarrow \mathrm{Cl}^{-}+\mathrm{I}_2+\mathrm{H}_2 \mathrm{O} $ ….(i)
$\mathrm{I}_2+2 \mathrm{~S}_2 \mathrm{O}_3^{2-} \longrightarrow 2 \mathrm{I}^{-}+\mathrm{S}_4 \mathrm{O}_6^{2-}$ …(ii)
Also the $n$-factor of $\mathrm{S}_2 \mathrm{O}_3^{2-}$ is one as
$2 \mathrm{~S}_2 \mathrm{O}_3^{2-} \longrightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 e^{-}$
[one ’ $e$ ’ is produced per unit of $\mathrm{S}_2 \mathrm{O}_3^{2-}$ ]
$\Rightarrow$ Molarity of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3=0.25 \mathrm{~N} \times 1=0.25 $ $\mathrm{M}$
$\Rightarrow \mathrm{m} $ $\mathrm{mol}$ of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ used up $=0.25 \times 48=12$
Now from stoichiometry of reaction (ii)
$12 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{S}_2 \mathrm{O}_3^{2-}$ would have reduced $6$ $ m$ $mol$ of $\mathrm{I}_2$.
From stoichiometry of reaction (i)
$ m$ $ mol$ of $\mathrm{OCl}^{-}$ reduced $=\mathrm{m} $ $\mathrm{mol}$ in $\mathrm{I}_2$ produced $=6$
$\Rightarrow$ Molarity of household bleach solution $=\frac{6}{25}=0.24 $ $\mathrm{M}$
Shortcut Method
Milliequivalent of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3=$ milliequivalent of $\mathrm{OCl}^{-}$
$=0.25 \times 48=12$
Also $n$-factor of $\mathrm{OCl}^{-}=2\left[\mathrm{Cl}^{+} \longrightarrow \mathrm{Cl}^{-}\right.$, gain of $\left.2 e^{-}\right]$
$\Rightarrow \mathrm{m} $ $\mathrm{mol}$ of $\mathrm{OCl}^{-}=\frac{12}{2}=6 \mathrm{~m}$ mol. Remaining part is solved in the same manner.
BETA
