An aqueous solution of 0.10 g KIO_(3) was treated with an excess of KI solution the solution was acidified with HCI the liberated I_(2) consumed 45.0 mL ofthisuplhate solutoin to decolourise the blue starch iodine complex calculate the molarity of sodium thiossuphate solution

An aqueous solution of 0.10 g KIO_(3) was treated with an excess of KI

1.	An aqueous solution of 0.10 g KIO_(3) was treated with an excess of KI solution the solution was acidified with HCI the liberated I_(2) consumed 45.0 mL ofthisuplhate solutoin to decolourise the blue starch iodine complex calculate the molarity of sodium thiossuphate solution
Answer» SOLUTION :The reaction involved are `2IO_(3)^(-)+12 H^(+)+10 I^(-) rarr 6 I_(2)+6 H_(2)O` `2S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(-)]xx6` `2IO_(3)^(-)+12 H^(+)+12 H^(+)+12 S_(2)O_(3)^(2-)rarr6S_(4)O_(6)^(2-)+2 I^(-)+6H_(2)O` Now of moles of `KIO_(3)=(0.1)/(214)` Now 2 moles of `KIO_(3)` react with `Na_(2)S_(2)_(3)`=12 moles `therefore (0.1)/(214)` mole of `KIO_(3)` will react with `Na_(2)S_(2)O_(3)=12/2xx(0.1)/(214)` mole Now `12/2xx(0.1)/(214)` mole of `Na_(2)S_(2)O_(3)` s present in 45 ML of the soluton `therefore` MOLARITY of `Na_(2)S_(2)O_(3)` solution =`12/2xx(0.1)/(214xx1000/45=0.0623 M`