`K_(SP)` of `BaSO_(4)` is `1.5xx10^(-9)` . Calculate is solubility in: (a) Pure water, , (b) `0.10MBaCI_(2)`.

Explanation

`BaSO_(4)hArr underset(S)(Ba^(2+))+underset(S)(SO_(4)^(2-))`
(where S in m/litre solubility of `BaSO_(4)`)
(i) `K_(SP)=SxxS`
`:. S=sqrt(K_(SP))=sqrt(1.5xx10^(-9))`
`= 3.87xx10^(-5)` mol `litre^(-1)`
(ii) In presence of `0.10M BaCI_(2)`, let S mol/litre `BaSO_(4)` is dissolved
`K_(SP)=[Ba^(2+)][SO_(4)^(2-)]`
`1.5xx10^(-9)=(0.1+S)(S)`
`[(0.1+S)=0.1 as S ltltlt0.1]`
`:. S=1.5xx10^(-8)M`