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A solution contains a mixture of Ag^(+)(...

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))`

Text Solution

Verified by Experts

the `[I^(-)]` needed for precipitaion of `Ag^(+)` and `Hg_(2)^(2+)` are derived as:
For `Agl:[Ag^(+)][l^(-)]=K_(spAg)`
`(0.1)[l^(-)]=8.5xx10^(-16)M`....(1)
`[Hg_(2)^(2+)][l^(2-)]=2.5xx10^(-26)`
`(0.1)[l^(-)]=2.5xx10^(-26)`
`[l^(-)]=5xx10^(-13)M`....(2)
Since `[l^(-)]` required for precipitation of `Agl` is less and thus `Agl` begins to precipitate first,Also it will continue upto addition of `[l^(-)]=5xx10^(-3)` when `Hg_(2)l_(2)` begins to precipitate and thus
Maximum `[l^(-)]` for `Agl` precipitaion `=5xx10^(-13)M`
Now at this concentration of `l^(-),[Ag^(+)]` left in solution is `[Ag^(+)]_("left")[l^(-)]=(K_(sp))_(Agl)`
`[Ag^(+)]_("left")=(8.5xx10^(-17))/(5.0xx10^(-13))=1.7xx10^(-4)M`
`0.1M Ag^(+)` will have `=1.7xx10^(-4)M Ag^(+)` in solution
`%` of `Ag^(+)` precipitated =`(0.1-1.7xx10^(-4))/(0.1)xx100=99.83%`
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