Saved Bookmarks
| 1. |
Calculate the solubility of Ag_(2)CrO_(4) in 0.1M AgNO_(3)K_(sp) of Ag_(3)CrO_(4)=1xx10^(-22) |
|
Answer» Solution :`Ag_(2)CrO_(4)hArr2Ag^(+)+CrO_(2)^(2-)` `2xx0.1+x` `K_(SP)=[Ag^(+)]^(2)[CrO_(4)^(2-)]` `1XX10^(-12)=(2xx0.1)^(2)[CrO_(4)^(2-)]` `[CrO_(4)^(2-)]=(1xx10^(-12))/((2xx0.1)^(2))=(1xx10^(-12))/0.04=0.025xx10^(-10)` |
|