What is the molar solubility of \({\bf{Tl}}{\left( {{\bf{OH}}} \right)_{\bf{3}}}\) in a \({\bf{0}}.{\bf{10}}{\rm{ }}{\bf{M}}\) solution of \({\bf{N}}{{\bf{H}}_{\bf{3}}}\) ?

We have a \(0.10 M\) solution of \({{\mathop{\rm NH}\nolimits} _3}\).

let us calculate the molar solubility of \({\mathop{\rm Tl}\nolimits} {\left( {OH} \right)_3}\)

Now, let us calculate the concentration of OH^-

\(\begin{array}{*{20}{c}}{{K_b} = \frac{{\left[ {{\rm{N}}{{\rm{H}}_4}^ + } \right] \cdot \left[ {O{{\rm{H}}^ - }} \right]}}{{\left[ {N{H_3}} \right]}}}\\{1.8 \times {{10}^{ - 5}} = \frac{{x \cdot x}}{{0.10 - x}}}\\{{x^2} = 1.8 \times {{10}^{ - 6}} - 1.8 \times {{10}^{ - 5}}x}\\{0 = {x^2} + 1.8 \times {{10}^{ - 5}}x - 1.8 \times {{10}^{ - 6}}}\\{x = 1.33 \times {{10}^{ - 3}}}\\{\left[ {O{H^ - }} \right] = x = 1.33 \times {{10}^{ - 3}}}\end{array}\)

Finally, we will find the molar solubility of \({\mathop{\rm Tl}\nolimits} {\left( {OH} \right)_3}\)

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

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {T{l^{3 + }}} \right] \cdot {{\left[ {O{H^ - }} \right]}^3}}\\{\left[ {T{l^{3 + }}} \right] = \frac{{{K_{sp}}}}{{\left[ {O{H^{ - 13}}} \right.}}}\\{ = \frac{{6.3 \times {{10}^{ - 46}}}}{{{{\left( {1.33 \times {{10}^{ - 3}}} \right)}^3}}}}\\{ = 2.68 \times {{10}^{ - 37}}{\rm{M}}}\end{array}\)