The solubility of ${\mathrm{BaSO}}_4$ in water is $2.42\times {10}^{-3} \mathrm{g}{ \mathrm{L}}^{-1}$ at 298 K. The value of its solubility product $\left({\mathrm{K}}_{\mathrm{sp}}\right)$ will be (Given molar mass of ${\mathrm{BaSO}}_4=233 \mathrm{g}{ \mathrm{mol}}^{-1}$)

$\text{ Solubility of }{\mathrm{BaSO}}_4,\mathrm{s}=\frac{2.42\times {10}^{-3}}{233} \mathrm{mol}{ \mathrm{L}}^{-1}=1.04\times {10}^{-5} \mathrm{mol}{ \mathrm{L}}^{-1}$

${\mathrm{BaSO}}_4\text{ ionizes completely in the solution as : }$

${\mathrm{BaSO}}_{4\left(\mathrm{s}\right)}\rightleftharpoons {\mathrm{Ba}}_{ }^{2+}+{\mathrm{SO}}_4^{2-}$

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]\left[{\mathrm{SO}}_4^{2-}\right]={\mathrm{s}}^2$

$={\left(1.04\times {10}^{-5}\right)}^2=1.08\times {10}^{-10}{ \mathrm{mol}}^2{ \mathrm{L}}^{-2}$