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

The concentration of various specices in the equilibrium mixture by represented as :
`BaSO_4(s)hArrBa^(2)+(aq)+SO_(4)^(2-)(aq)`
`"Intial concentration"" " 1" "0" "0`
Concentration
`"at Equilibrium in water"" " (1-S)" "S" "S`
`"at equilibrium in the"" " (1-S)" "S" "(S+0.01)`
presence of `H_(2)SO_(4)`
`K_(sp) " for " BaSO_(4) " in water " =[Ba^(2+)][SO_(4)^(2-)]=(S)xx(S)=S^(2)`
`"But " S=8xx10^(-4) " mole dm"^(-3)`. Therefore, `K_(sp)=(8xx10^(-4))^(2)=64xx10^(-8)`.
`K_(sp)" for "BaSO_(4)` solution in presence of `H_2SO_4=[S]xx[S+0.01]`
`K_(sp)` of the solution will not change as long as temperature is constant.
therefore` (S)xx(S+0.01)=64xx10^(-8)`
or `S^(2)+0.01S-64xx10^(-8)=0`
`S=(-0.01pmsqrt((0.01)^(2)+(4xx64xx10^(-8))))/(2)=(-0.01pmsqrt(10^(-4)(256xx10^(-8))))/(2)`
`=(-0.01pmsqrt(10^(-4)(1+256xx10^(-2))))/(2)=(-0.01pm10^(-2)sqrt(1+0.256))/(2)`
`(-0.01pm10^(-2)sqrt(1.256))/(2)=(-10^(-2)+(1.12xx10^(-2)))/(2)`
`=((-1+1.12)xx10^(-2))/(2)=(0.12)/(2)xx10^(-2)=6xx10^(-4) " mol dm"^(-3)`