If the solubility product of PbS is 8 times 1 | vedclass

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(B) The dissociation of $PbS$ is given by: $PbS(s) \rightleftharpoons Pb^{2+}(aq) + S^{2-}(aq)$.The solubility product expression is $K_{sp} = [Pb^{2+

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$282$

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(B) The dissociation of $PbS$ is given by: $PbS(s) ightleftharpoons Pb^{2+}(aq) + S^{2-}(aq)$.The solubility product expression is $K_{sp} = [Pb^{2+}][S^{2-}] = S 	imes S = S^{2}$.Given $K_{sp} = 8 	imes 10^{-28}$.Therefore,$S = \sqrt{8 	imes 10^{-28}} = \sqrt{4 	imes 2 	imes 10^{-28}} = 2 \sqrt{2} 	imes 10^{-14} \ mol \ L^{-1}$.Using the given value $\sqrt{2} = 1.41$,we get $S = 2 	imes 1.41 	imes 10^{-14} = 2.82 	imes 10^{-14} \ mol \ L^{-1}$.To express this in the form $x 	imes 10^{-16} \ mol \ L^{-1}$,we write $2.82 	imes 10^{-14} = 282 	imes 10^{-16}$.Thus,the value of $x$ is $282$.

If the solubility product of $PbS$ is $8 \times 10^{-28}$,then the solubility of $PbS$ in pure water at $298 \ K$ is $x \times 10^{-16} \ mol \ L^{-1}$. The value of $x$ is $\dots$.
[Given $\sqrt{2} = 1.41$]

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If the solubility product of $PbS$ is $8 \times 10^{-28}$,then the solubility of $PbS$ in pure water at $298 \ K$ is $x \times 10^{-16} \ mol \ L^{-1}$. The value of $x$ is $\dots$.[Given $\sqrt{2} = 1.41$]