InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`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. |
|
Answer» 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. |
|