91Ó°ÊÓ

We have a buffer solution containing equal concentrations of NH₃ and NH₄⁺.

Calculate the molar solubility of Sn (OH)₂

\(\begin{array}{*{20}{c}}{{K_b} = \frac{{\left[ {{\rm{N}}{{\rm{H}}_4}^ + } \right] \cdot \left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {N{H_3}} \right]}}}\\{{\rm{\;Since\;}}\left[ {{\rm{N}}{{\rm{H}}_ - }3} \right] = \left[ {N{H_ - }{4^ + }} \right],{\rm{we get\;}}}\\{\left[ {O{H^ - }} \right] = {K_b}}\\{ = 1.8 \cdot {{10}^{ - 5}}}\end{array}\)

The reaction of Sn (OH)₂

\(Sn{({\rm{OH}})_2}({\rm{s}}) \to {\rm{S}}{{\rm{n}}^{2 + }}({\rm{aq}}) + 2{\rm{O}}{{\rm{H}}^ - }({\rm{aq}})\)

The solubility product of\(Sn{({\rm{OH}})_2}{\rm{\;is\;}}{K_{sp}} = 3 \cdot {10^{ - 27}}\)

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {S{n^{2 + }}} \right] \cdot {{\left[ {O{H^ - }} \right]}^2}}\\{\left[ {S{n^{2 + }}} \right] = \frac{{{K_{sp}}}}{{{{\left[ {O{H^ - }} \right]}^2}}}}\\{ = \frac{{3 \cdot {{10}^{ - 27}}}}{{{{\left( {1.8 \cdot {{10}^{ - 5}}} \right)}^2}}}}\\{ = 9.26 \cdot {{10}^{18}}{\rm{M}}}\end{array}\)