When `15mL` of `0.05M AgNO_(3)` is mixed with `45.0mL` of `0.03M K_(2)CrO_(4)`, predict whether precipitation of `Ag_(2)CrO_(4)` occurs or not? `K_(sp)` of `Ag_(2)CrO_(4) = 1.9 xx 10^(-12)`

Explanation

First find the concentrations of `Ag^(o+)` and `CrO_(4)6(2-)` ions in the resulting mixture.
`[Ag^(o+)] = (15 xx 0.05)/(15+45) = 1.25 xx 10^(-2)M`
`[CrO_(4)^(2-)] = (45 xx 0.03)/(15+45) = 2.25 xx 10^(-2)M`
The ionic product for `Ag_(2)CrO_(4)` is given as follows:
`Ag_(2)CrO_(4) hArr 2Ag^(oplus)+CrO_(4)^(2-)`
Ionic product `= [Ag^(o+)]^(2) [CrO_(4)^(2-)]`
`= (1.25 xx 10^(-2))^(2) (2.15 xx 10^(-2))`
`= 3.51 xx 10^(-6) gt K_(sp)`
Hence, precipiation occurs.