Calculate the percentage purity of a sample of bleaching powder from the following data:
14.3 g of a sample of bleaching powder when treated with acetic acid and excess of KI, liberated iodine which required 100 mL 1 M sodium thiosulphate solution:
$C a (O C l)_{2} + 2 C H_{3} C O O H \to C a (C H_{3} C O O)_{2} + H_{2} O + C l_{2}$
$C l_{2} + 2 K I \to 2 K C l + I_{2}$
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$

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Solution

Given,
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$
Again, 100 mL 1 M sodium thiosulphate solution ( $N a_{2} S_{2} O_{3}$ )
= 0.1 mol of $N a_{2} S_{2} O_{3}$
2 moles $N a_{2} S_{2} O_{3}$ reacts with 1 mole of $I_{2}$
Thus, 0.1 moles $N a_{2} S_{2} O_{3}$ will react with 0.05 moles of $I_{2}$ .
Again,
$C l_{2} + 2 K I \to 2 K C l + I_{2}$
So, 1 mole of $I_{2}$ is produced from 1 mole of $C l_{2}$
Thus, 0.05 moles of $I_{2}$ will be produced from 0.05 moles of $C l_{2}$
Given,
$C a (O C l)_{2} + 2 C H_{3} C O O H \to C a (C H_{3} C O O)_{2} + H_{2} O + C l_{2}$
So, 1 mole $C l_{2}$ is obtained fro 1 mole of $C a (O C l)_{2}$ .
Thus 0.05 moles of $C l_{2}$ will be obtained from 0.05 moles of $C a (O C l)_{2}$ .

Now, 0.05 moles of $C a (O C l)_{2}$ $= 0.05 \times 143 g = 7.15 g$

$∴$ Percentage purity of the sample $= \frac{7.15}{14.30} \times 100 = 50 %$

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