The solubility of BaSO_4 in pure water (in g | vedclass

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(D) The dissociation of $BaSO_4$ is given by: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$Let the solubility be $S \, mol \, L^{-1}$.The

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$2.3 	imes 10^{-3}$

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(D) The dissociation of $BaSO_4$ is given by: $BaSO_4(s) ightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$Let the solubility be $S \, mol \, L^{-1}$.Then,$K_{sp} = [Ba^{2+}][SO_4^{2-}] = S 	imes S = S^2$.Given $K_{sp} = 1.0 	imes 10^{-10}$,we have $S^2 = 1.0 	imes 10^{-10}$,so $S = 1.0 	imes 10^{-5} \, mol \, L^{-1}$.To convert solubility from $mol \, L^{-1}$ to $g \, L^{-1}$,multiply by the molar mass of $BaSO_4$ $(233 \, g \, mol^{-1})$:Solubility in $g \, L^{-1} = (1.0 	imes 10^{-5} \, mol \, L^{-1}) 	imes (233 \, g \, mol^{-1}) = 233 	imes 10^{-5} \, g \, L^{-1} = 2.33 	imes 10^{-3} \, g \, L^{-1}$.Thus,the value is closest to $2.3 	imes 10^{-3} \, g \, L^{-1}$.

The solubility of $BaSO_4$ in pure water (in $g \, L^{-1}$) is closest to
[Given: $K_{sp}$ for $BaSO_4$ is $1.0 \times 10^{-10}$ at $25^{\circ} C$. Molecular weight of $BaSO_4$ is $233 \, g \, mol^{-1}$]

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The solubility of $BaSO_4$ in pure water (in $g \, L^{-1}$) is closest to[Given: $K_{sp}$ for $BaSO_4$ is $1.0 \times 10^{-10}$ at $25^{\circ} C$. Molecular weight of $BaSO_4$ is $233 \, g \, mol^{-1}$]