`100.0mL` of a saturated solution of `Ag_(2)SO_(4)` is added to `250.0mL` of saturated solution of `PbCrO_(4)`. Will may precipitate form and if so what? Given `K_(sp)` for `Ag_(2)SO_(4), Ag_(2)CrO_(4), PbCrO_(4)`,and `PbSO_(4)` are `1.4 xx 10^(-5), 2.4 xx 10^(-12), 2.8 xx 10^(-13)`, and `1.6 xx 10^(-8)`, respectively.

Explanation

For `{:(Ag_(2)SO_(4) hArr,2Ag^(o+)+,SO_(4)^(2-),,),(,2S,S,,):}`
`K_(sp) = 4S^(3)` or `S = 3sqrt((K_(sp))/(4)) = 3sqrt((1.4 xx 10^(-5))/(4)) = 1.52 xx 10^(-2)M`
For `PbCrOhArr Pb^(2+) + CrO_(4)^(2-)`
`K_(sp) = S_(1)^(2)` or `S_(1) = sqrt(K_(sp)) = sqrt(2.8 xx 10^(-13)) = 5.29 xx 10^(-7)M`
In solution, conce. of each ion can be given as :
Thus, `[Ag^(o+)] = (2S xx 100)/(350) = (2xx1.52xx10^(-2)xx100)/(350)`
`= 0.869 xx 10^(-12)M`
`[SO_(4)^(2-)] = (S xx 100)/(350) = (1.52 xx 10^(-2) xx 100)/(350) = 0.43 xx 10^(-2)`
`[Pb^(2+)] = (S_(1)xx250)/(350) = (5.29 xx 10^(-7)xx250)/(350) = 3.78 xx 10^(-7)`
`[CrO_(4)^(2-)] = (S_(1)xx250)/(350) = (5.29xx10^(-7)xx250)/(350) = 3.78 xx 10^(-7)`
It is thus evident that
`[Ag^(o+)] = [CrO_(4)^(2-)] = (0.869 xx 10^(-2))^(2) xx (3.78 xx 10^(-7))`
`=2.85 xx 10^(-11) gt K_(sp) Ag_(2)CrO_(4)`
The product is greater than `K_(sp)` of `Ag_(2)CrO_(4)` and thus `Ag_(2)CrO_(4)` will ppt.