pBlock ElementsII 2 Question 26
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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
======= ####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
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $0.48 \mathrm{M}$
(b) $0.96 \mathrm{M}$
(c) $0.24 \mathrm{M}$
(d) $0.024 \mathrm{M}$
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Solution:
- The involved redox reactions are :
$$ \begin{aligned} 2 \mathrm{H}^{+}+\mathrm{OCl}^{-}+2 \mathrm{I}^{-} & \longrightarrow \mathrm{Cl}^{-}+\mathrm{I}{2}+\mathrm{H}{2} \mathrm{O} \ \mathrm{I}{2}+2 \mathrm{~S}{2} \mathrm{O}{3}^{2-} & \longrightarrow 2 \mathrm{I}^{-}+\mathrm{S}{4} \mathrm{O}_{6}^{2-} \end{aligned} $$
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)
$12 \mathrm{~m} \mathrm{~mol} \mathrm{of} \mathrm{S}{2} \mathrm{O}{3}^{2-}$ would have reduced $6 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{I}_{2}$.
From stoichiometry of reaction (i)
$\mathrm{m} \mathrm{mol} \mathrm{of} \mathrm{OCl}^{-}$reduced $=\mathrm{m} \mathrm{mol} \mathrm{in} \mathrm{I}_{2}$ 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 $\mathrm{OCl}^{-}=\frac{12}{2}=6 \mathrm{mmol}$. Remaining part is solved in
the same manner.
