The solubility of BaSO_4 in water is 2.33 tim | vedclass

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(B) The solubility $S$ in $mol/L$ is calculated as: $S = \frac{	ext{solubility in } g/L}{	ext{molar mass}} = \frac{2.33 	imes 10^{-3}}{233} = 1 	i

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$1 	imes 10^{-10}$

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(B) The solubility $S$ in $mol/L$ is calculated as: $S = \frac{	ext{solubility in } g/L}{	ext{molar mass}} = \frac{2.33 	imes 10^{-3}}{233} = 1 	imes 10^{-5} \ mol/L$. For the salt $BaSO_4$,the dissociation is $BaSO_4 ightleftharpoons Ba^{2+} + SO_4^{2-}$. The solubility product $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S 	imes S = S^2$. 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 is....... (Molar mass of $BaSO_4 = 233 \ g/mol$)

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