Calculate the solubility of ${\mathrm{A}}_2{\mathrm{X}}_3$ in pure water, assuming that neither kind of ion reacts with water. The solubility product of ${\mathrm{A}}_2{\mathrm{X}}_3,{\mathrm{K}}_{\mathrm{sp}}=1.1\times {10}^{-23}$

${\mathrm{A}}_2{\mathrm{X}}_3⟶2{\mathrm{A}}^{3+}+3{\mathrm{X}}^{2-}$

${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{A}}^{3+}\right]}^2{\left[{\mathrm{X}}^{2-}\right]}^3=1.1\times {10}^{-23}$

$\text{ If }\mathrm{S}=\text{ solubility of }{\mathrm{A}}_2{\mathrm{X}}_3\text{, then }$

$\left[{\mathrm{A}}^{3+}\right]=2\mathrm{S}; \left[{\mathrm{X}}^{2-}\right]=3\mathrm{S}$

therefore, ${\mathrm{K}}_{\mathrm{sp}}=\left(2\mathrm{S}{)}^2\right(3\mathrm{S}{)}^3$

$=\left(108\right){\mathrm{S}}^5=1.1\times {10}^{-23}$

Thus, ${\mathrm{S}}^5=1\times {10}^{-25}$

$\mathrm{S}=1.0\times {10}^{-5} \mathrm{mol} \mathrm{L}$