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

Let us first calculate `[I^-]` to be precipitate AgI and `H_(g_(2))I_(2)`
`K_(sp) AgI=[Ag^+] [I^-]`
`8.5 xx 10^(-17)=(0.1) [I^-]`
`[I^-]` to precipitate `AgI=(8.5 xx 10^(-17))/(0.1) =8.5 xx 10^(-16)M`
`K_(sp) (Hg_(2)I_(2))=[Hg_(2)^(2+)] [I^(-)]^(2)`
`2.5 xx 10^(-26)=(0.1) [I^(-)]^2`
`[I^-]` to precipitate `Hg_(2)I_(2)=5.0 xx 10^(-13)M`
`[I^-]` to precipitate AgI is smaller, therefore, AgI will start precipitating first. On further addition of `I^-` more Agi will precipitate and when `[I^-] ge 5.0 xx 10^(-13)M,Hg_2I_2` will start precipitating. The maximum concentration of `Ag^+` at this stage will thus be calculated as
`K_(sp) (AgI)=[Ag^+] [I^-]`
`8.5 xx 10^(-17)=[Ag^(+)] (5.0 xx 10^(-13))`
`[Ag^(+)]=(8.5 xx 10^(-17))/(5.0 xx 10^(-13))=1.7 xx 10^(-4)M`
percentage of `Ag^(+)` remained unprecipitated `=(1.7 xx 10^(-4))/(0.1) xx 100`
=0.17%
Thus, percentage of `Ag^(+)` precipitated =99.83%