Ksp(BaSO4)= 1.5*10-9. find solubility in a)pure water b)0.1 M BaCl2 solution

(i) Let the solubility of BaSO₄ in pure water be x mol/L
$$\begin{gathered} BaS{O}_4 \to B{a}^{2+} + S{O}_4^{2-} \\ x x \\ Now, \\ K_{sp} = \left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right] \\ = x \times x \\ 1.5 \times {10}^{-9} =x^2 \\ x =3.87\times {10}^{-5} mol/L \end{gathered}$$

(ii) Let y be the solubility of BaSO₄ in 0.1 BaCl₂ solution.
Therefore,
$$\begin{gathered} \left[B{a}^{2+}\right] = 0.1 +y \approx 0.1 mol/L \\ \left[S{O}_4^{2-}\right] = y mol/L \\ Now, \\ K_{sp} = \left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right] \\ 1.5 x {10}^{-9} = 0.1 \times y \\ y=1.5 x {10}^{-8} mol/L \end{gathered}$$