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The solubility of $BaSO_4$ in water is $2.42 \times 10^{-3}\ g L^{-1}$ at 298 K. The value of its solubility product ($K_{sp}$) will be (Given molar mass of $BaSO_4$ = 233 g $mol^{-1}$)
Explanation
Convert g/L to mol/L, then $K_{sp} = s^2$ for a 1:1 salt.
Detailed Solution
Solubility of $BaSO_4$, $s = \frac{2.42\times10^{-3}}{233}\ (mol\ L^{-1}) = 1.04\times10^{-5}\ mol\ L^{-1}$
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$, each with concentration s
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = s^2 = (1.04\times10^{-5})^2$
$= 1.08\times10^{-10}\ mol^2 L^{-2}$