\(\underset{\operatorname{mol} L^{-1}}{A X_{2}} \rightleftharpoons A_{s}^{2+}+2 \underset{2 S}{X}^{-}\)
Solubility product of \(A X_{2}\)
\(K_{s p}=\left[A^{2+}\right]\left[X^{-}\right]^{2}=[S] \times[2 S]^{2}=4 S^{3}\)
\(\because K_{s p}\) of \(A X_{2}=3.2 \times 10^{-11}\)
\(\therefore 3.2 \times 10^{-11}=4 s^{3}\)
or, \(S^{3}=0.8 \times 10^{-11}=8 \times 10^{-12}\)
or, \(S=\sqrt[3]{8 \times 10^{-12}}\)
\(=2 \times 10^{-4} \,mol / L\)
Solubility \(=2 \times 10^{-4}\, mol / L\)
(પાણી માટે $K_f =1.86\, K\, kg\,mol^{-1}$ છે )