If the solubility product of BaSO_4 is 1.5 ti | vedclass

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(B) The dissociation of $BaSO_4$ is represented as: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$.Let the solubility be $S \ mol/L$.Then,

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$3.9 	imes 10^{-5}$

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(B) The dissociation of $BaSO_4$ is represented as: $BaSO_4(s) ightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$.Let the solubility be $S \ mol/L$.Then,$[Ba^{2+}] = S$ and $[SO_4^{2-}] = S$.The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S 	imes S = S^2$.Given $K_{sp} = 1.5 	imes 10^{-9}$.Therefore,$S^2 = 1.5 	imes 10^{-9}$.$S = \sqrt{1.5 	imes 10^{-9}} = \sqrt{15 	imes 10^{-10}} \approx 3.87 	imes 10^{-5} \ mol/L$.Rounding to two significant figures,$S \approx 3.9 	imes 10^{-5} \ mol/L$.

If the solubility product of $BaSO_4$ is $1.5 \times 10^{-9}$ in water,its solubility in moles per litre is

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