Iodine titrations: Compounds containing iodine are widely used in titrations, commonly known as iodine titration. It is of two kinds:
(i) lodometric titrations (ii) lodimetric titrations.
(i) lodometric 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 acidic medium or, basic medium in which `I_2` oxidises into L. Now the liberated `I_2` can be titrated with `Na_(2)S_(2)O_3` . solution.
`KI overset("Oxidising Agent")rarrI_(2)overset(Na_(2).S_(2)O_(3)//H^(+))rarrI^(-)+Na_(2)S_(4)O_6`
Although solid `I_2` is black and insoluble in water, but it converts into soluble `I_3` ions
`underset("Black")(I_2(s))+IhArrunderset("Dark brown")(I_3^(-))`
Starch is used as indicator near the end point or equivalence point. Even small amount of `I_2` molecules, gives blue colour with starch. 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 79.75 g of `CuSO_4` sample containing inert impurity is reacted with KI, the liberated I, is reacted with 50 mL (1M) `Na_(2)S_(2)O_3` in basic medium, where it oxidises into `SO_(4)^(2-)` ions, and `I_2` reduces into `I^(-)` then what will be the % purity of `CuSO_4` in sample ? <

Options

A. `60%`
B. `80%`
C. `50%`
D. `95%`

Answer

B

Explanation

m eq.of `CuSO_4` reacted = m eq. of `Na_(2)S_(2)O_(2)` reacted (n-f-8) `w/(159.5)=(50xx1xx8)/1000`
wt. of `CuSO_(4) = (7.975 xx 8) ` g % purity of
`CuSO_(4)=( 7.975 xx 8 xx100)/(79.75)=80%`