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 miximum concentreation of iodide ion at which one of them gets precipitated almost completely. What % of that metal ion is precipitated ? `(K_(SP)of AgI=8.5xx10^(-17)` and `K_(SP)` of `Hg_(2)I_(2)=2.5xx10^(-26))`

Explanation

First determine, which ion starts precipitating first
`[I^(Θ)]_("Minimum for AgI") = (K_(sp) of Ag1)/([Ag^(o+)])`
`= (8.5 xx 10^(-17))/(0.1) = 8.5 xx 10^(-16)M`
`[I^(Θ)]_("Minimum for" Hg_(2)I_(2)) = sqrt((K_(sp) "of" Hg_(2)I_(2))/([Hg_(2)^(2+)]))`
`= sqrt((2.5 xx 10^(-26))/(0.1)) = 5 xx 10^(-13)M`
This means that `Ag1` will be precipitated first as `[I^(Θ)]` required for `Ag1` is less.
But when `[I^(Θ)]` reaches `5 xx 10^(-13)M`, then precipitation of `Hg_(2)I_(2)` also starts.
So [`Ag^(o+)]` left at that stage is given as:
`[Ag^(o+)] = (K_(sp)of AgI)/([I^(Θ)]_(Hg_(2)I_(2))) = (8.5 xx 10^(-17))/(5.0 xx 10^(-13)) = 1.7 xx 10^(-4)M`
`%` of `Ag^(o+)` left un-precipitated `= (1.7 xx 10^(-4))/(0.1) xx 100 = 0.17%`
`rArr %` age of `Ag^(o+)` precipitated `= 99.83%`
This means when `Hg_(2)I_(2)` starts precipitating, `Ag^(o+)` is almost precipitated.