An aqueous solution containing 0.10 g `KIO_(3)` (formula weight`=214.0`) was treated with an excess of KI solution the solution was acidified with HCl. The liberated `I_(2)` consumed 45.0 " mL of " thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thosulphate solution.

Explanation

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`