25 mL of household bleach contains `CaCIOCI_3` was mixed with 30 mL of 0.50 MkI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N `Na_2S_2O_3` was used to reach the end point. The molarity of the household bleach solution is a)0.48M b)0.96M c)0.24M d)0.024M

Options

A. `0.48M`
B. `0.96M`
C. `0.24M`
D. `0.024M`

Answer

C

Explanation

Consider the reaction `Ocl^(-) + H_(2)O + 2I^(-) rarr Cl^(-) +I_(2) + 2OH^(-)`
According to this reaction, 25mL of `CaOCl_(2)` reacts with 30mL of KI
`I_(2) + 2Na_(2)S_(2)O_(3) rarr Na_(2) S_(4) O_(6) + 2NaI`
Given that 48 mL of 0.25N `Na_(2)S_(2)O_(3)` was used to reach the end point. So, the number of millimoles of `I_(2)` produced =`48 xx 0.25//2=6`
According to the reaction, Number of millimoles of bleaching powder = Number of millimoles of `I_(2)= (1//2)xx` Number of millimoles of `Na_(2)S_(2)O_(3)=6`
So, the molarity of hypochlorate `= ("Number of millimoles of bleaching powder")/("Volume of solution")= (6"millimol")/(25mL) =0.24M`