Calculate `[Ag^(o+)]` in a solution made by dissolving both `AgCrO_(4)` and `AgC_(2)O_(4)` untill saturation is reached with respect to both salts. Given `K_(sp) of Ag_(2)CrO_(4)` and `Ag_(2)C_(2)O_(4)` are `9.0 xx 10^(-12)` and `6.0 xx 10^(-12)`, respectively.

Explanation

Assume no hydrolysis of `C_(2)O_(4)^(2-)`. The solubilities of the two salts are too similar to make simplifying assumptions about which provides `Ag^(+)` to the solution.
`Ag_(2)CrO_(4)hArr 2Ag^(o+) +CrO_(4)^(2-)`
`K_(sp) = 9.0 xx10^(-12) =[Ag^(o+)]^(2) [CrO_(4)^(2-)]`
`Ag_(2)C_(2)O_(4)hArr 2Ag^(o+) +C_(2)O_(4)^(2-)`
`K_(sp) = 6.0 xx 10^(-12) = [Ag^(o+)]^(2) [C_(2)O_(4)^(2-)]`
By electroneutality,
`[{:("Total +ve charge Total - ve charge"),("Total charge = charge" xx "conc"):}]`
`[Ag^(o+)] = 2[CrO_(4)^(2-)] +2 [C_(2)O_(4)^(2-)]` or
`0.5 [Ag^(o+)] = [CrO_(4)^(2-)] +[C_(2)O_(4)^(2-)]`
`= (9.0xx10^(-12))/([Ag^(o+)](2)) +(6.0xx10^(-12))/([Ag^(o+)]^(2))`
`0.5[Ag^(o+)] = 9.0 xx 10^(-12) + 6.0 xx10^(-12)`
`= 1.5 xx10^(-11)`
`:. [Ag^(o+)] = 3.1 xx10^(-4)M`