91Ó°ÊÓ

Given,

\(1 - 0.100{\rm{L\;of\;}}0.0100{\rm{MCd}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)

And \(1.150{\rm{L\;of\;}}0.100{\rm{MN}}{{\rm{H}}_3}\)

\({K_f}{\rm{\;for\;Cd}}\left( {{\rm{N}}{{\rm{H}}_3}} \right)_4^{2 + }{\rm{\;is\;}}1.3 \cdot {10^7}\)

Calculate number of moles of \({\rm{Cd}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}{\rm{\;and\;N}}{{\rm{H}}_3}\),

\(\begin{array}{*{20}{c}}{{n_{{\rm{Cd}}{{\left( {N{O_3}} \right)}_2}}} = 0.100{\rm{L}} \cdot 0.0100{\rm{M}}}\\{ = 0.001{\rm{mol}}}\end{array}\)

\({n_{{\rm{N}}{{\rm{H}}_3}}} = 1.15{\rm{L}} \cdot 0.100{\rm{M}} = {\bf{0}}.{\bf{115}}{\rm{mol}}\)

\(0.001{\rm{\;mol of\;Cd}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)dissociates into \(0.001{\rm{molC}}{{\rm{d}}^{2 + }}\) and \(0.002{\rm{molNO}}_3^ - \)

\(0.001{\rm{mol\;of\;C}}{{\rm{d}}^{2 + }}\)will react with \(0.004{\rm{mol\;of\;N}}{{\rm{H}}_3}\)and produce \({\bf{0}}.{\bf{001}}{\rm{\;mol of\;Cd}}{\left( {{\rm{N}}{{\rm{H}}_3}} \right)_4}\)

So, excess of \({\rm{N}}{{\rm{H}}_3}\)is \(0.115{\rm{mol}} - 0.004{\rm{mol}} = {\bf{0}}.{\bf{111}}{\rm{mol}}\).

The initial concentration of \({\rm{N}}{{\rm{H}}_3}{\rm{\;and\;Cd}}{\left( {{\rm{N}}{{\rm{H}}_3}} \right)_4}\),

\(\begin{array}{*{20}{c}}{\left. {\left( {{\rm{N}}{{\rm{H}}_3}} \right){\rm{\;(initial\;}}} \right) = \frac{{0.111{\rm{mol}}}}{V}}\\{ = \frac{{0.111{\rm{mol}}}}{{1.250{\rm{L}}}}}\\{ = 0.0888{\rm{M}}}\end{array}\)

\(\begin{array}{*{20}{c}}{\left( {{\rm{Cd}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right)({\rm{\;initial\;}}) = \frac{{0.001{\rm{mol}}}}{V}}\\{ = \frac{{0.001{\rm{mol}}}}{{1.250{\rm{L}}}}}\\{ = 0.0008{\rm{M}}}\end{array}\)

\(\begin{array}{*{20}{c}}{{K_f} = \frac{{\left( {{\rm{Cd}}\left( {{\rm{N}}{{\rm{H}}_3}} \right)_4^{2 + }} \right)}}{{\left( {{\rm{C}}{{\rm{d}}^{2 + }}} \right) \cdot {{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}^4}}}}\\{1.3 \cdot {{10}^7} = \frac{{0.0008 - x}}{{(x) \cdot {{(0.0888 + 4x)}^4}}}}\end{array}\)

Since \({K_f}\) is larger than \({10^4}\)we will assume \(0.0888 + 4{\rm{x}} = 0.0888\),

\(\begin{array}{*{20}{c}}{1.3 \cdot {{10}^7} = \frac{{0.0008 - x}}{{(x) \cdot {{(0.0888)}^4}}}}\\{0.0008 - x = 808.34x}\\{x = 9.88 \cdot {{10}^{ - 7}}{\rm{M}}}\end{array}\)

The solution for concentration of cadmium is \(9.88 \cdot {10^{ - 7}}{\rm{M}}\) .