Discussion Forum

Bleaching Powder and bleach solutions 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 $Na_{2}S_{2}O_{3}$, was used to reach the endpoint. The molarity of the household bleach solution is

Answer:

Explanation:

The involved redox reactions are :

 $2H^{+}+OCl^{-}+2I^{-} \rightarrow Cl^{-}+I_{2}+H_{2}O$ ....(i)

$I_{2}+2S_{2}O_{3}^{2-} \rightarrow 2I^{-}+S_{4}O_{6}^{2-}$ ......(ii)

 Also   the n factor of$S-{2}O_{3}^{2-}$ is one as

 $2S_{2}O_{3}^{2-} \rightarrow S_{4}O_{3}^{2-}+2e^{-}$

[obe '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 6mmol of $I_{2}$

From stoichiometry of reaction (i)

m mol of $OCl^{-}$ reduced =m mol of $I_{2}$ produced =6

 $\Rightarrow$ Molarity of household  bleach solution= $\frac{6}{25}=0.24 M$