1 gm sample of `AgNO_(3)` is dissolved in 50 ml of water . It is titrated with 50 ml of Kl Solution : The Agl preciplitated is filtered off . Excess of Kl in filtrate is titrated with M/10 `KlO_(3)` in presence of 6 M HCl till all `l^(-)` converted into ICl. It requires 50 ml of M/10 `KlO_(3) `
Soluition : 20 ml of the same stock solution of Kl requires 30 ml of M/10 `KlO_(3)` under similar conditions. Calculate % of `AgNO_(3)` in the sample . The reaction is `KlO_(2) + 2Kl + 6HCl to 3lCl + 3KCl + 3H_(2)O` .

Explanation

`AgNO_(3) + Kl to Agl downarrow + KNO_(3)`
Ag present in `AgNO_(3)` is removed as Agl by adding 50 ml of Kl solution of which 20 ml requires 30 ml of M/10 `KlO_(3)`
Meq. of Kl in 20 ml = Meq. of `KlO_(3) = 30 xx 1/10 xx 4`
(For `KlO_(3) , l^(5+) + 4e^(-) to l^(+) l_(-) to l^(2+) + 2e^(-) ` )
`:." Wq, wt. of " Kl = M/2 `
`:. ` Meq. of Kl in 50 ml added to `AgNO_(3) = (30 xx 4 xx 50)/(10 xx 20) = 30 `
Meq. of Kl left unused by `AgNO_(3) = 50 xx 1/10 xx 4 = 20`
` :. ` Meq. of Kl used for `AgNO_(3) = 30 - 20 = 10`
` :. ` Meq. of `AgNO_(3) = 10`
`W_(AgNO_(3))/(170) xx 2 xx 1000 = 10`
`:. W_(AgNO_(3)) = 0.85 gm = 85 %`
`("Mole of " AgNO_(3))/("Mole of " K_(t)) = 1/l`
` :. ` If equivalent weight of `Kl = M/2`
Equivalent weight of `AgNO_(3) = M/2`