The solubility product of PbCl_2 at 20 , ^oC | vedclass

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(B) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ⇌ Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $S \, mol/L$.Then,$[Pb^{2+}] = S$ and $[Cl^-

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$3.34 	imes 10^{-2}$

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(B) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ⇌ Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $S \, mol/L$.Then,$[Pb^{2+}] = S$ and $[Cl^-] = 2S$.The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 1.5 	imes 10^{-4}$,we have $4S^3 = 1.5 	imes 10^{-4}$.$S^3 = \frac{1.5 	imes 10^{-4}}{4} = 0.375 	imes 10^{-4} = 3.75 	imes 10^{-5}$.$S = \sqrt[3]{3.75 	imes 10^{-5}} = \sqrt[3]{37.5 	imes 10^{-6}} \approx 3.34 	imes 10^{-2} \, mol/L$.

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

The solubility product of $PbCl_2$ at $20 \, ^oC$ is $1.5 \times 10^{-4}$. Calculate the solubility.

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