25 mLof household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 Nacetic acid in the titration of the liberated iodine 48 mL of 0.25 N Na_(2)S_(2)O_(3) was used to reach the end point the molarity of the household bleach solution is

25 mLof household bleach solution was mixed with 30 mL of 0.50 M KI an

1.	25 mLof household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 Nacetic acid in the titration of the liberated iodine 48 mL of 0.25 N Na_(2)S_(2)O_(3) was used to reach the end point the molarity of the household bleach solution is
Answer» 0.48 M 0.96 M 0.24 M 0.24 M Solution :The chemical equation involved in this problem are: `CA(OCI)_(2)+4H^(+)+2e^(-)rarrCa^(2)+2H_(2)O+CI_(2)` `2KI+CI_(2)rarr2KCI+I_(2)` `2Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2` Nal `Ca(OCI_(2))+4H^(+)+2e^(-)+2Na_(2)S_(2)O_(3)rarrCa^(2+)+2H_(2)O+2KCI+Na_(2)S_(4)O_(6)+2NaI` Thus the normaility of`Ca(OCI)_(2)` solution =0.48 N Now since EQ (i) is `2e^(-)` change `therefore`Eq wt of `Ca(OcI)_(2)="Mol wt"//2` `=0.48//2` thus the molarity of the household BLEACH =0.24 M