The volume of water required to dissolve 0.1 | vedclass

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(D) The solubility equilibrium for $PbCl_2$ is: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $s \ mol/L$. Then $K_{sp}

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$180$

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(D) The solubility equilibrium for $PbCl_2$ is: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $s \ mol/L$. Then $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 3.2 	imes 10^{-8}$, so $4s^3 = 3.2 	imes 10^{-8} \implies s^3 = 0.8 	imes 10^{-8} = 8 	imes 10^{-9}$.Thus, $s = 2 	imes 10^{-3} \ mol/L$.Molar mass of $PbCl_2 = 207 + 2 	imes 35.5 = 278 \ g/mol$.Mass of $PbCl_2$ in $1 \ L$ solution $= s 	imes 	ext{Molar mass} = 2 	imes 10^{-3} 	imes 278 = 0.556 \ g/L$.To dissolve $0.1 \ g$ of $PbCl_2$, the volume required $V = \frac{	ext{mass}}{	ext{solubility in } g/L} = \frac{0.1}{0.556} \approx 0.1798 \ L$.Converting to $mL$, $V \approx 0.1798 	imes 1000 \approx 180 \ mL$.

The volume of water required to dissolve $0.1 \ g$ $PbCl_2$ to get a saturated solution (in $mL$) is (Given $K_{sp}(PbCl_2) = 3.2 \times 10^{-8}$; Atomic mass of $Pb = 207 \ u$, $Cl = 35.5 \ u$)

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