`1 g` sample of `AgNO_(3)` is dissolved in `50 mL` of water, It is titrated with `50 mL` of `KI` solution. The `Agl`percipitated is filtered off. Excess of `KI` filtrate is titrated with `M//10KIO_(3)` in presence of `6 M HCl` till all `I^(-)` converted into `ICI`. It requires `50 mL` of `M//10 KIO_(3)` solution. `20 mL` of the same stock solution of `KI` requires `30 mL` of `M//10KIO_(3)` under similar conditions. Calculate `%` of `AgNO_(3)` in sample. The reaction is
`KIO_(3)+2KI+6HClrarr3ICl+3KCl+3H_(2)O`

Explanation

(A) `AgNO_(3)+KIrarrAgI+KNO_(3)`
1.`Ag` present in `AgNO_(3)` is removed as `AgI` by adding `50 mL KI` of which `20 mL` requires `30 mL` of `M//10KIO_(3)`.
2. The solution contains `KI` unused. The unused `KI` is converted into `ICI` by `KIO_(3)`.
`4e+I^(5+)rarrI^(1+), I^(-)rarr2e`
`therefore` Eq.wt.of `KI=(M)/(2)`
`therefore` Meq.of `KI` in `20 mL=` Meq.of `KIO_(3)=30xx(1)/(10)xx4`
`therefore` Meq.of `KI` in `50 mL` added to `AgNO_(3)=(30xx4xx50)/(10xx20)=30`
Now Meq.of `KI` left unused by
`AgNO_(3)=50xx(1)/(10)xx4=20`
`because` Mole ratio of `AgNO_(3)` and `KI` ractions is `1:1` and thus if eq.wt. of `KI` is `M//2`, then
`therefore` Meq.of `KI` used for `AgNO_(3)=30-20`
`therefore` Meq.of `AgNO_(3)=10`
`therefore (w)/(170//2)xx1000=10`
`therefore w=0.85 g`
`therefore %` purity of `AgNO_(3)` in sample `=(0.85xx100)/(1)=85%`
(B) Alternative method:
`m` moles of `KIO_(3)` used by `20 mL` of `KI` stock solution `=30xx(1)/(10)=3`
`therefore m` mole of `KIO_(3)` used by `50 mL` of `KI` stock solution
`(3xx50)/(20)=7.5`
`therefore m` moles of `KIO_(3)` used by `KI` left in `50 mL` solution after reaction with `AgNO_(3)=(1)/(10)xx50=5`
`therefore` mole ratio of `KIO_(3)` and `KI` is `1:2` in reaction
`therefore m` d moles of `KI` in `50 mL` stock solution `=7.5xx2=15`
`therefore m` moles of `KI` left in `50 mL` solution after reaction with `AgNO_(3)=5xx2=10`
`:. m` moles of `KI` used for `AgNO_(3)=15-10=5`
`:. m ` moles of `AgNO_(3)=5`
( `:.` mole ratio of `KI` and `AgNO_(3)` reactions is `1:1`)
Now for `AgNO_(3)`:
`(w)/(M)xx1000=5`
`:. (w)/(170)xx1000=5`
`:. w_(AgNO_(3))=0.85`
`:. % "of" AgNO_(3)=(0.85xx100)/(1)=85%`