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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 79.75g of CuSO_(4) sample containing inert impurity is reacted with KI, the liberated I_(2) is reacted with 50mL (1M) Na_(2)S_(2)O_(3) in basic medium, where it oxidises into SO_(4) ions, and I_(2) reduces into I^(-), then what will be the % purity of CuSO_(4) in sample?

Answer»

`60%`
`80%`
`50%`
`95%`

Solution :m EQ of `CuSO_(4)` reacted = m eq of `Na_(2)S_(2)O_(3)` reacted (n-f=8)
`(w)/(159.5) = (50 xx 1 xx 8)/(1000)`
wt of `CuSO_(4) = (7.975 xx 8)g`
`%` purity of `CuSO_(4) = (7.975 xx 8 xx 100)/(79.75) = 80%`


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