91Ó°ÊÓ

\({\rm{Ksp}} = \left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right){\left( {{\rm{O}}{{\rm{H}}^ - }} \right)^2} = 1.2 \cdot {10^{ - 15}}\)

The volume of \({\rm{PbC}}{{\rm{l}}_2}\)solution is 500 mL(0.500 L).

The concentration of \({\rm{PbC}}{{\rm{l}}_2}\) is 0.050 M.

Let's calculate how many grams of \({\rm{Pb}}{({\rm{OH}})_2}\)will dissolve in a given solution.

Since we have 0.050 M \({\rm{PbC}}{{\rm{l}}_2}\)we can conclude that we have 0.050 M \({\rm{P}}{{\rm{b}}^{2 + }}\)ions in that solution.

When we add \({\rm{Pb}}{({\rm{OH}})_2}\)in that solution, the concentration of the \({\rm{P}}{{\rm{b}}^{2 + }}\)ions will be

\(\left( {P{b^{2 + }}} \right)({\rm{\;total\;}}) = \left( {P{b^{2 + }}} \right)\left( {{\rm{\;from\;}}PbC{l_2}} \right) + \left( {P{b^{2 + }}} \right)\left( {{\rm{\;from\;Pb}}{{(OH)}_2}} \right)\)

\( = 0.050{\rm{M}} + \left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right)\left( {{\rm{\;from\;Pb}}{{({\rm{OH}})}_2}} \right)\)

\({\rm{\;Let\;}}\left( {P{b^{2 + }}} \right)\left( {{\rm{\;from\;}}Pb{{(OH)}_2}} \right) = x,{\rm{\;therefore\;}}\)

\( = 0.050M + x\)

The solubility of product \(({\rm{Ksp}}){\rm{\;of\;Pb}}{({\rm{OH}})_2}\) will be,

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right)({\rm{\;total\;}}) \cdot {{\left( {O{H^ - }} \right)}^2}}\\{1.2 \cdot {{10}^{ - 15}} = (x + 0.050) \cdot {{(2x)}^2}}\end{array}\)

\({\rm{\;Since\;}}{{\rm{K}}_ - }sp{\rm{\;is lower than\;}}{10^{ - 4}}{\rm{,\;}}\)

\({\rm{\;we will assume that\;}}x + 0.05 \approx 0.05\)

\(\begin{array}{*{20}{c}}{1.2 \cdot {{10}^{ - 15}} = (0.050) \cdot {{(2x)}^2}}\\{{{(2x)}^2} = 2.4 \cdot {{10}^{ - 14}}}\\{2x = 1.55 \cdot {{10}^{ - 7}}}\end{array}\)

\(x = 7.746 \cdot {10^{ - 8}}{\rm{M}}\)

The molar solubility for \({\rm{Pb}}{({\rm{OH}})_2}\)is \(x = 7.746 \cdot {10^{ - 8}}{\rm{M}}\)

Since the molar solubility is the number of moles of solute which can be dissolved in the liter of the solution, we can calculate the number of moles of the \({\rm{Pb}}{({\rm{OH}})_2}\)which can be dissolved in $500 mL of the solution.

\(\begin{array}{*{20}{c}}{{n_{{\rm{Pb}}{{(OH)}_2}}} = 7.746 \cdot {{10}^{ - 8}}{\rm{M}} \cdot 0.500{\rm{L}}}\\{ = 3.873 \cdot {{10}^{ - 8}}{\rm{mol}}}\end{array}\)

So the mass is

\(\begin{array}{*{20}{c}}{{m_{{\rm{Pb}}{{({\rm{OH}})}_2}}} = {n_{{\rm{Pb}}{{({\rm{OH}})}_2}}} \cdot 241.21{\rm{g}}/{\rm{mol}}}\\{ = 3.873 \cdot {{10}^{ - 8}}{\rm{mol}} \cdot 241.21{\rm{g}}/{\rm{mol}}}\end{array}\)

=\(9.34 \cdot {10^{ - 6}}\)g

So, the solution is \(9.34 \cdot {10^{ - 6}}\)grams of \({\rm{Pb}}{({\rm{OH}})_2}\)will dissolve in 500 mL of a 0.050 M \({\rm{PbC}}{{\rm{l}}_2}\)solution.