The solubility product `(K_(sp))` of `BaSO_(4)` is `1.5xx10^(-9)`. Calculate the solubility of barium sulphate in pure water and in 0.1 M `BaCl_(2)`.

Answer

A, B, C

Explanation

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)`.