0.3 g of KI is dissolved in 25 " mL of " water. After adding to this solution double its volume of concentration HCl, a solution of KIO_(3) is graduaaly added with stirring. Iodine is liberated as first but redissolved. It is observed that 24.1 " mL of " iodate solution is just sufficient to dissolve the iodine. If the iodate solution contains 0.8 g per 100 mL formulate the reaction that has taken place.

0.3 g of KI is dissolved in 25 " mL of " water. After adding to this s

1.	0.3 g of KI is dissolved in 25 " mL of " water. After adding to this solution double its volume of concentration HCl, a solution of KIO_(3) is graduaaly added with stirring. Iodine is liberated as first but redissolved. It is observed that 24.1 " mL of " iodate solution is just sufficient to dissolve the iodine. If the iodate solution contains 0.8 g per 100 mL formulate the reaction that has taken place.
Answer» Solution :0.3 of KI CONTAINS `(0.3)/(166)=1.807xx10^(-3)` " mol of "iodide ion, 24.1 " mL of " iodate SALUTION contains `(24.1)/(100)xx(0.8)/(214)=9.01xx10^(-4)` mol. It is clear from these figures that 2 moles of `I^(ɵ)` ions react with 1 MOLE of iodate ion. In thus PROCESS `I^(ɵ)` is oxidised and `IO_(3)^(ɵ)` is reduced to form some common species in which the oxidation number of iondine is x (say). Thus. `2(x-(-1)=(5-x)` i.e., `2x+2=5-x` `therefore3x=3impliesx=+1` In presence of excess HCl, this would correspond to the formation of ICL. `therefore` the reaction is formulated as follows, `2I^(ɵ)+IO_(3)^(ɵ)+3Cl^(ɵ)+6H^(o+)to3ICl+3H_(2)O`