Bleaching powder and bleach solution are produced on a large scale and is used in several house-hold products. The effectiveness of bleach solution is often measured by iodometry. $25 m l$ of household bleach solution was mixed with $30 m l$ of $0.50 M$ $K I$ and $10 m l$ of $4 N$ acetic acid. In the titration of the liberated iodine, $48 m l$ 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:

0.48 M

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0.96 M

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0.24 M

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0.024 M

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Solution

The correct option is D $0.24 M$
$C a (O C l) C l + 2 C H_{3} C O O H \to C l_{2} + (C H_{3} C O O)_{2} C a + H_{2} O$

$C l_{2} + 2 K I \to 2 K C l + I_{2}$

$I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 K C l + I_{2}$

$I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}$

$M e q . o f I_{2} = M e q . o f C l_{2}$

$∴$ meq. of $N a_{2} S_{2} O_{3} = 48 \times 0.25$

Millimoles of $I_{2}$ reacted $= 6$

Millmoles of $C l_{2}$ reacted $= 6$

Millimoles of bleach $= 6$

$25 \times M = 6$

$M = \frac{6}{25} = 0.24 M$

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