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(D) The molar mass of $PbCl_2 = 207 + 2 	imes 35.5 = 278 \ g/mol$. The solubility equilibrium is $PbCl_2(s) \leftrightarrow Pb^{2+}(aq) + 2Cl^-(aq)$.

Vedclass · Accepted Answer

$0.18$

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(D) The molar mass of $PbCl_2 = 207 + 2 	imes 35.5 = 278 \ g/mol$. The solubility equilibrium is $PbCl_2(s) \leftrightarrow Pb^{2+}(aq) + 2Cl^-(aq)$. The solubility product expression is $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$. Given $K_{sp} = 3.2 	imes 10^{-8}$,we have $4s^3 = 3.2 	imes 10^{-8}$,so $s^3 = 0.8 	imes 10^{-8} = 8 	imes 10^{-9}$. Thus,solubility $s = 2 	imes 10^{-3} \ mol/L$. The number of moles of $PbCl_2$ is $n = \frac{0.1 \ g}{278 \ g/mol} \approx 3.597 	imes 10^{-4} \ mol$. Since $s = \frac{n}{V}$,the volume $V = \frac{n}{s} = \frac{3.597 	imes 10^{-4}}{2 	imes 10^{-3}} \approx 0.1798 \ L \approx 0.18 \ L$.

The minimum volume of water required to dissolve $0.1 \ g$ of lead $(II)$ chloride to get a saturated solution ($K_{sp}$ of $PbCl_2 = 3.2 \times 10^{-8}$; atomic mass of $Pb = 207 \ u$,$Cl = 35.5 \ u$) is......$L$

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