The solubility product of `BaSO_4` is `1.1 xx 10^(-10)` at 298 K temp. calculate its water solubility.

Explanation

`BaSO_(4(s)) hArr Ba_((aq))^(2+) + SO_(4(aq))^(2-)`
If `S=BaSO_4` Solubility of `BaSO_4` in mol `L^(-1)`
`therefore K_(sp)=S^2=1.1xx10^(-10)`
`therefore S=sqrt(1.1xx10^(-10))`
`=1.049xx10^(-5) "mol L"^(-1)`