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^(+)" and " CrO_(4)^(2-)` ions in the resulting mixture.
`[Ag^(+)] = (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^(+) + CrO_(4)^(2-)`
Ionic product (I.P) ` = [Ag^(+)]^(2) [CrO_(4)^(2-)]`
` = (1.25 10^(-2))^(2) (2.25 xx 10^(-2))`
` = 3.51 xx 10^(-6) gt K_("sp")`
Hence precipitation occurs.