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^(-) to Cl^(-) +I_(2)+2OH^(-)`
According to this reaction 25mL of `CaOCl_2` reacts with 30mL of KI
`I_(2)+2Na_(2)S_2O_3 to Na_(2)S_4O_6+2NaI`
Given that 48mL of 0.25 N `Na_(2)S_2O_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_2S_2O_3=6`
So, the molarity of hypochlorate `=("Number of millimoles of bleaching powder")/("Volume of solution")=("6 millimol")/("25 mL")=0.24M`