Iodine titrations: Compounds containing iodine are widely used in titrations, commonly known as iodine titration. It is of two kinds:
(i) Iodometric titrations
(ii) Iodimetric titrations.
(i) Iodometric titrations: It is nothing but an indirect method of estimating the iodine. In this type of titration, an oxidising agent is made to react with excess of `KI`, in acidix medium or , basic medium in which `I^(-)` oxidises into `I_(2)`. Now the liberated `I_(2)` can be titrated with `Na_(2)S_(2)O_(3)` solution.
`KI overset("Oxidising Agent")rarr I_(2) overset("Na_(2)S_(2)O_(3)//H^(+))rarr I^(-)+Na_(2)S_(4)O_(6)`
Although solid `I_(2)` is black and insoluble in water, but it converts into soluble `I_(3)` ions `{:(I_(2)(s)+I^(-) hArr, I_(3)^(-)),("Black","dark brown"):}`
Strach is used as indicator near the end point or equivalence point. Even small amount of `I_(2)` molecules, gives blue colour with strach. The completion of the reaction can be detected when blue colour disappears at the and point. In iodimetric titration, the strength of reducing agent is determined by reacting it with `I_(2)`.
When `214g` of `KIO_(3)` reacts with excess of `KI` in presence of `H^(+)`, then it produces `I_(2)`. Now `I_(2)` is completely reacted with `1M Na_(2)S_(2)O_(3)` solution in basic medium, where it converts into `SO_(4)^(-2)` ions. then what volume of `Na_(2)S_(2)O_(3)` is needed to react the end point of the reaction?

Options

A. 500ml
B. 800ml
C. 1500ml
D. 750ml

Answer

D

Explanation

`IO_(3)^(-) +5I^(-) +2H^(+) rarr 3I_(2) +3H_(2)O`
214 g `KIO_(3) = 1` mole of `KIO_(3)`
= 3 mole of `I_(2)` (in the balanced chemical reaction)
= 6 eq. of `I_(2) (n f=2) = 6000` m eq. of `I_(2)`
Let the vol of `Na_(2)S_(2)O_(3) =v mL`, then
`(V xx 1 xx 8) = 6000, v = 750 mL`