91Ó°ÊÓ

We have a buffer solution with 0.100 M NH₃and 0.400 NH₄⁺.

Calculate the molar solubility of Al (OH)₃

\(\begin{array}{*{20}{c}}{{K_b} = \frac{{\left[ {N{H_4} + } \right] \cdot \left[ {O{H^ - }} \right]}}{{\left[ {N{H_3}} \right]}}}\\{\left[ {O{H^ - }} \right] = \frac{{{K_b} \cdot \left[ {N{H_3}} \right]}}{{\left[ {N{H_4} + } \right]}}}\\{ = \frac{{1.8 \cdot {{10}^{ - 5}} \cdot 0.100}}{{0.400}} = 4.5 \cdot {{10}^{ - 6}}}\end{array}\)

The reaction of Al (OH)₃

\({\rm{Al}}{({\rm{OH}})_3}({\rm{s}}) \to {\rm{A}}{{\rm{l}}^{3 + }}({\rm{aq}}) + 3{\rm{O}}{{\rm{H}}^ - }({\rm{aq}})\)

The solubility product of\({\rm{Al}}{({\rm{OH}})_3}{\rm{\;is\;}}{K_{sp}} = 2 \cdot {10^{ - 32}}\)

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {A{l^{3 + }}} \right] \cdot {{\left[ {O{H^ - }} \right]}^3}}\\{\left[ {A{l^{3 + }}} \right] = \frac{{{K_{sp}}}}{{{{\left[ {O{H^ - }} \right]}^3}}}}\\{ = \frac{{2 \cdot {{10}^{ - 32}}}}{{{{\left( {4.5 \cdot {{10}^{ - 6}}} \right)}^3}}}}\\{ = 2.05 \cdot {{10}^{ - 16}}{\rm{M}}}\end{array}\)