The solubility of BaSO_4 in water is 2.33 tim | vedclass

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(B) The solubility of $BaSO_4$ in $g/L$ is given as $2.33 	imes 10^{-3} \ g/L$. First,convert the solubility into $mol/L$ $(S)$: $S = \frac{	ext{sol

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$1 	imes 10^{-10}$

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(B) The solubility of $BaSO_4$ in $g/L$ is given as $2.33 	imes 10^{-3} \ g/L$. First,convert the solubility into $mol/L$ $(S)$: $S = \frac{	ext{solubility in } g/L}{	ext{molar mass}} = \frac{2.33 	imes 10^{-3}}{233} = 1 	imes 10^{-5} \ mol/L$. Since $BaSO_4$ is a $1:1$ electrolyte,it dissociates as: $BaSO_4(s) ightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$. The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S 	imes S = S^2$. Substituting the value of $S$: $K_{sp} = (1 	imes 10^{-5})^2 = 1 	imes 10^{-10}$.

The solubility of $BaSO_4$ in water is $2.33 \times 10^{-3} \ g/L$. Its solubility product will be (molecular weight of $BaSO_4 = 233$).

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