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 4N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N $N{a}_2{S}_2{O}_3$ 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

Answer 1

The correct option is C 0.24 M

The involved redox reactions are:
$2H^{+}+OC{l}^{-}+2I^{-}\to C{l}^{-}+I_2+H_{2}O$ ...(i)
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$ ...(ii)
Also the n-factor of $S_2{O}_6^{2-}+2e^{-}$
[one 'e' is produced per unit of $S_2{O}_3^{2-}$]
$\Rightarrow$ Molarity of $N{a}_2{S}_2{O}_3$ = 0.25 N $\times$ 1 = 0.25 M
$\Rightarrow$ m mol of $N{a}_2{S}_2{O}_3$ used up = 0.25 N $\times$ 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 $OC{l}^{-}$ reduced = m mol in $I_2$ produced = 6
$\Rightarrow$ Molarity of household bleach solution
$=\frac{6}{25}$ = 0.24 M