`1.0L` of solution which was in equilibrium with solid mixture of `AgC1` and `AgC1` and `Ag_(2)CrO_(4)` was found to contain `1xx 10^(-4) mol of Ag^(o+)` ions, `1.0 xx 10^(-6) mol` of `C1^(Θ)` ions and `8.0 xx 10^(-4)` moles of `CrO_(4)^(2-)` ions. `Ag^(o+)` ions added slowely to the above mixture (keeping volume constant) till `8.0 xx 10^(-7)` mol of `AgC1` got precipitated. How many moles of `Ag_(2)CrO_(4)` were also precipitated?

Explanation

We have to assume here that an equilibrium is achieved when `AgC1` is precipitated.
First find `K_(sp)` values of `AgC1` and `AgCrO_(4)`
`K_(sp)` of `AgC1 = [Ag^(o+)] [C1^(Θ)]`
`=( 1xx 10^(-4)) (1xx10^(-6)) = 1.0 xx 10^(-10)`
`K_(sp) of Ag_(2)CrO_(4) = [Ag^(o+)]^(2) [crO_(4)^(2-)]`
`=(1xx10(-4))^(2) (8 xx 10^(-4)) = 8.0 xx 10^(-12)`
Since `8 xx 10^(-7) mol` of `AgC1` are precipitated per litre of solution, so moles of `C1^(Θ)` ions left in the solution
`= 1 xx 10^(-6) - 8 xx 10^(-7) =2 xx 10^(-7)M`
`rArr [C1^(Θ)]_("left") = 2 xx 10^(-7)M`
`[Ag^(o+)]_("left") = (K_(sp) of AgC1)/([C1^(Θ)]) = (1xx10^(-10))/(2xx10^(-7)) = 5 xx 10^(-4)M`
For this much amount of `Ag^(o+)` ion,
`[CrO_(4)^(2-)]` left `=(K_(sp)` of `AgCrO_(4))/([Ag^(o+)]^(2))`
`= (8 xx 10^(-12))/((5xx10^(-4))^(2)) =3.2 xx 10^(-5)M`
Hence, moles of `CrO_(4)^(2-)` precipated (as `Ag_(2)CrO_(4))`
`= (8xx10^(-4)-3.2 xx 10^(-5)) = 7.68 xx 10^(-4)mol L^(-1)`