A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution `(Q_(sp))` becomes greater than its solubility product. If the solubility of `BaSO_(4)` in water is `8 xx 10^(-4)` mol `dm^(-3)`. Calculate its solubility in `0.01` mol `dm^(-3)` of `H_(2)SO_(4)`. Express your answer in scientific notation `x xx 10^(-y)`. Write the value of y.

Explanation

The solubility equilibrium is
`BaSO_4(s) hArr Ba^(2+) +SO_4^(2-)`
Solubility of `BaSO_4 =8xx10^(-4) "mol dm"^(-3)`
`therefore [Ba^(2+)]=8xx10^(-4) "mol dm"^(-3)`
`[SO_4^(2-)]=8xx10^(-4) "mol dm"^(-3)`
`K_(sp)=[Ba^(2+)][SO_4^(2-)]`
`therefore K_(sp)=(8xx10^(-4))(8xx10^(-4))`
`=64xx10^(-8)`
`H_2SO_4` ionizes completely as : `H_2SO_4 hArr 2H^(+) +SO_4^(2-)`
`therefore [SO_4^(2-)]` produced from 0.01 mol `dm^(-3)` of `H_2SO_4`, then
`[Ba^(2+)]=x, [SO_4^(2-)]`=x+0.01
Since `K_(sp)` is constant for a given salt ,
`K_(sp)=[Ba^(2+)][SO_4^(2-)]`
`=x(x+0.01)=64xx10^(-8)`
or `x^2+0.01 x -64xx10^(-8)` =0
Solving for x , we get `x=6xx10^(-4) "mol dm"^(-3)`