A solution contains a mixture of `Ag^+(0.1M)` and `Hg^(2+)` (0.1M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated?
`K_(sp)(AgI)=8.5xx10^(-17),K_(sp)(HgI_2)=2.5xx10^(-26)`

Explanation

In solution `Ag^+` and `Hg^(2+)` react with `I^-` ion to form Agl and `Hgl_2` respectively. Agl and `Hgl_2` on partial dissociation, leads to the following equilibrium
`Agl(s) `Hgl_2(s) For Agl, `K_(sp)=[Ag^+][I^-]` and for `HgI_2`
`K_(sp)=[Hg^(2+)][I^-]^2`
In the solution, the concentration of `Ag^+` is 0.1 (M). The minimum concentration of `I^-` necessary to precipitate Agl,
`[I^-]lt(K_(sp)(Agl))/([Ag^+])` or `[I^-]=(8.5xx10^(-17))/(0.1)=8.5xx10^(-16)(M)`
Similarly minimum concentration of `I^-` required to precipitate `Hgl_2," "[I^-]^2lt(K_(sp)(HgI_2))/([Hg^(2+)])` or
`[I^-]=sqrt((2.5xx10^(-6))/(0.1))=5xx10^(-3)(M)`
Hence, Agl starts precipitating when `[I^-]` in solution is
`lt8.5xx10^(-16)(M)`. On the other hand `Hgl_2` starts precipitating when `[I^-]` in solution is `lt5xx10^(-13)(M)`
So, when the `[I^-]` in solution is `5xx10^(-13)(M)`, `Ag^+` ions At this concentration of `I^-` ions
`[Ag+]=(K_(sp)(AgI))/([I^-])=(8.5xx10^(-17))/(5xx10^(-13))=1.7xx10^(-4)(M)`
so, the percentage of `Ag^+` ions will precipitate
`=(0.1-1.7xx10^(-4))/(0.1)xx100=99.83%`