91Ó°ÊÓ

We have a \(0.138 M\) solution of \({{\mathop{\rm CH}\nolimits} _3}N{H_2}\).

let us calculate the molar solubility of \({\mathop{\rm Pb}\nolimits} {\left( {OH} \right)_2}\)

Now, let us calculate the concentration of OH^-

\(\begin{array}{*{20}{c}}{{K_b} = \frac{{\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{H}}_3} + } \right] \cdot \left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{H}}_2}} \right]}}}\\{4.4 \cdot {{10}^{ - 4}} = \frac{{x \cdot x}}{{0.138 - x}}}\\{{x^2} = 6.072 \times {{10}^{ - 5}} - 4.4 \times {{10}^{ - 4}}x}\\{0 = {x^2} + 4.4 \times {{10}^{ - 4}}x - 6.072 \times {{10}^{ - 5}}}\\{x = 7.58 \times {{10}^{ - 3}}}\\{\left[ {O{H^ - }} \right] = x = 7.58 \cdot {{10}^{ - 3}}}\end{array}\)

Finally, we will find the molar solubility of \({\mathop{\rm Pb}\nolimits} {\left( {OH} \right)_2}\)

\({\rm{Pb}}{({\rm{OH}})_2}({\rm{s}}) \to {\rm{P}}{{\rm{b}}^{2 + }}({\rm{aq}}) + 2{\rm{O}}{{\rm{H}}^ - }({\rm{aq}})\)

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {P{b^{2 + }}} \right] \cdot {{\left[ {O{H^ - }} \right]}^2}}\\{\left[ {P{b^{2 + }}} \right] = \frac{{{K_{sp}}}}{{{{\left[ {O{H^{ - 1}}} \right]}^2}}}}\\{ = \frac{{1.2 \times {{10}^{ - 15}}}}{{{{\left( {7.58 \times {{10}^{ - 3}}} \right)}^2}}}}\\{ = 2.09 \times {{10}^{ - 11}}{\rm{M}}}\end{array}\)