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

The `[I^(-)]` needed for precipitation of `Ag^(+)` and `Hg_(2)^(2+)` are derived as:
For AgI: `[Ag^(+)][I^(-)]=K_(SP_(AgI))`
`(0.1)[I^(-)]=8.5xx10^(-17`
`:. [I^(-)]=8.5xx10^(-16)M`
For `Hg_(2)I_(2): [Hg_(2)^(2+)][I^(-)]^(2)=2.5xx10^(-26)`
`(0.1)[I^(-)]^(2)=2.5xx10^(-26)`
`:. [I^(-)]=5xx10^(-13)M`
Since `[I^(-)]` required for precipitation of AgI is less and thus AgI being to Precipitate first. Also it will continue upto addition of `[I^(-)]=5xx10^(-13)` when `H_(2)I_(2)` begins to precipitate and thus, Maximum `[I^(-)]` for AgI precipitation `= 5xx10^(-13)M` Now at this concentration of `I^(-), [Ag^(+)]` left solution is
`[Ag^(+)]_("left")[I^(-)]=K_(SP_(AgI))`
`:. [Ag^(+)]_("left")=(8.5xx10^(-17))/(5.0xx10^(-13))=1.7xx10^(-4)M`
`:' 0.1M Ag^(+)` will be left `= 1.7xx10^(-4)M Ag^(+)` in solution
`:. 100 M Ag^(+)` will be left `= 0.17%M Ag^(+)` in solution
`:. %` of Ag precipitatated `= 99.83%`