Saved Bookmarks
| 1. |
The `K_(sp)` of `BaSO_(4)` is `1.6 xx 10^(-9)`. Find the solubility of `BaSO_(4)` in `gL^(-1)` in a. Pure water b. `0.1M Ba(NO_(3))_(2)` |
|
Answer» Correct Answer - A::B::C i. `{:(BaSO_(4) hArr,Ba^(2+)+,SO_(4)^(2-),,),(,S,2S,,):}` `S_(H)_(2)O) = sqrt(K_(sp)) = (1.6 xx 10^(-9))^(1//2) = 4 xx 10^(-5)M` `("Mw of" BaSO_(4) = 137 + 32 + 64 = 233 g)` `S_(H_(2)O)"in" gL^(-1) = 4 xx 10^(-5) xx 233` `= 9.32 xx 10^(-3)gL^(-1)` ii. In pressure of `0.1M Ba(NO_(3))_(2)`, due to common ion effect, the solubility is supressed. For unti-univalent or di-divalent salt, `S_("new") = (K_(sp))/((C)^(n))` (where `C` is the concentration of common ion added and `n` is the number of common ion in the salt `(BaSO_(4))`. `S_("new") = (1.6 xx 10^(-9))/((0.1)^(1)) = 1.6 xx 10^(-8)M` `S_("new") "is" gL^(-1) = 1.6 xx 10^(-8)xx233` `= 3.72 xx 10^(-6)gL^(-1)`. |
|