1.

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_2oxidises into L. Now the liberated I_2can 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_6Although solid I_2is black and insoluble in water, but it converts into soluble I_3ions 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_2molecules, 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_2When 79.75 g of CuSO_4sample containing inert impurity is reacted with KI, the liberated I, is reacted with 50 mL (1M) Na_(2)S_(2)O_3in basic medium, where it oxidises into SO_(4)^(2-) ions, and I_2reduces into I^(-)then what will be the % purity of CuSO_4in sample ?

Answer»

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

Solution :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%`


Discussion

No Comment Found