A solution contains a mixture of `Ag^(+)(0.10M)` and `Hg_(2)^(2+)(0.10M)` which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. `(K_(SP)of AgI=8.5xx10^(-17)` and `K_(SP)` of `Hg_(2)I_(2)=2.5xx10^(-26))`

Explanation

`K_(sp)(AgI) = 8.5 xx 10^(-17) = [Ag^(+)][I^(-)]`
`[I^(-)]` required to start prepitation og AgI = `8.5 xx 10^(-17)/0.10 = 8.5 xx 10^(-16)M`
`K_(sp) = (HgI_(2)) = 2.5 xx 10^(-26) = [Hg^(2+)][I^(-)]^(2)`
`[I^(-)]`required to start precipitation of `HgI_(2) = sqrt(2.5 xx 10^(-26)/0.10) = 5 xx 10^(-13) M`
The above calculation indicates that lower `[I^(-)]` is required for precipitation of AgI. When `[I^(-)]` reaches to `5 xx 10^(-3)` ,AgI gets precipitated almost completely.
When `HgI_(2)` starts precipitating.
`[Ag^(+)] = 8.5 xx 10^(-17)/5xx10^(-13) = 1.70 xx 10^(-4)M`
`% Ag^(+) remai n i n g = 1.70 xx 10^(-4) xx 100/0.10 = 0.17`
`% Ag^(+)` p r ecip i ta t e d = 100 - 0.17 = 99.83