A sparingly soluble salt gets precipitated only when the prudct of concentration of its ions in th e solution `(Q_(sp))` becomes greater than its solubility product. If solubility of `BaSO_(4)` in water is `8xx10^(-4)" mol dm"^(-3)`. Calculater its solubility in `0.01" mol dm"^(-3) " of " H_(2)SO_(4)`.

Explanation

Solubility of `BaSO_(4) ` in water `(S) = 8 xx 10^(-4) ` mol `dm^(-3)` ltbgt `BaSO_(4) hArr Ba^(2) + SO_(4)^(2-)`
`:. K_(sp) ` for `BaSO_(4) = [Ba^(2+)][SO_(4)^(2-)]=(8xx10^(-4))(8xx10^(-4))=64xx10^(-8)`
As `H_(2)SO_(4)` ionizes completely as `H_(2)SO_(4) rarr 2H^(+) + SO_(4)^(2-), [SO_(4)^(2-)]` produced from 0.01 M `H_(2)SO_(2-)]` produced from 0.01 M `H_(2)SO_(4)=0.01 M `. Thus, if S is the solubility of `BaSO_(4) ` in `H_(2)SO_(4) ` , then
`K_(ap) = [Ba^(2+)][SO_(4)^(2-)] = S (S+ 0.01 ) = 64 xx 10^(-8) `(calculated above )
or `S^(2)+0.01 S - 64 xx 10^(-8) = 0 `
`:. S= (-0.01 pm sqrt((0.01)^(2)+4xx64xx10^(-8)))/(2) = (-0.01 pm sqrt(10^(-4)+256xx10^(-8)))/(2)`
`=(-0.01 pm sqrt(10^(-4)(1+256xx10^(-2))))/(2) = (-0.01 pm 10^(-2) sqrt(1+256xx10^(-2)))/(2)`
`=(-0.01 pm 10^(-2)sqrt(1.256))/(2) = (-10^(-2)+1.12xx10^(-2))/(2)`
`=((-1+1.12)xx10^(-2))/(2)`
`=((-1+1.12)xx10^(-2))/(2) = (0.12)/(2) xx 10^(-2) = 6 xx 10^(-4) ` mol `dm^(-3)`.