Question

The solubility of BaSO4, in water is 2.42 x 103 gL- at
298 K. The value of its solubility product (Ksp) will
be
(Given molar mass of BaSO4 = 233 g mol​

Answer 1

Answer: $1.08\times 10^{-8}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of $BaSO_4$ is given as:

$BaSO_4\leftrightharpoons Ba^{2+}+SO_4^{2-}$

We are given:

Solubility of $BaSO_4$ = $2.42\times 10^{-3}g/L$

Molar mass of $BaSO_4$ = 233 g/mol

Molar solubility of $BaSO_4$ =$\frac{2.42\times 10^{-3}g/L}{233g/mol}=10.38\times 10^{-3}mol/L$

By stoichiometry of the reaction:

1 mole of $BaSO_4$ gives 1 mole of $Ba^{2+}$ and 1 mole of $SO_4^{2-}$.

Expression for the equilibrium constant of $BaSO_4$ will be:

$K_{sp}=[Ba^{2+}]^2[SO_4^{2-}]$

$K_{sp}=[10.38\times 10^{-3}][10.38\times 10^{-3}]=1.08\times 10^{-8}$

Hence, the solubility product of $BaSO_4$ is $1.08\times 10^{-8}$

Answer 2

refer the attachment above