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. Calculate the molarity of the household bleach solution

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answer is 0.24.

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Detailed Solution

$C a O C l_{2} + 2 H^{+} \to C a^{2 +} + H_{2} O + C l_{2} \begin{array}{l} C l_{2} + 2 K I \to 2 K C I + I_{2} \\ I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6} \end{array} \begin{array}{l} N a_{2} S_{2} O_{3} M . m o l e s = M V_{M L} \\ = 0.25 \times 48 \\ = 12 \\ C a O C l_{2} M . M o l e s = 6 \\ C a O C l_{2} m o l a r i t y = \frac{6}{25} = 0.24 \end{array}$