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Reaction with Metals

25 mL of hous...

Question

25 mL of household 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 $N a_{2} S_{2} O_{3}$ 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 C 0.24 M
$C a O C l_{2} (a q) + 2 K I \to I_{2} + C a (O H)_{2} + K C l$
25 mL 30 mL
(M)nder 0.5(M)
$I_{2} + 2 N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + 2 N a l$
48 mL
$0.25 (N) = 0.25 M$
So, number of millimoles of $I_{2}$ produced $= 48 \times \frac{0.25}{2} = 24 \times 0.25 = 6$
In reaction;
Number of millimoles of bleaching power $(n_{C a O C l_{2}}) = n_{I_{2} - p r o d u c e d} = \frac{1}{2} \times n_{N a_{2} S_{2} O_{3}} u s e d = 6$
So, $(M) = \frac{n_{C a O C l_{2}} (m i l l i m o l e s)}{V (i n m L)} = \frac{6 m i l l i m o l e s}{25 m L} = 0.24 M$

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