91Ó°ÊÓ

Let us calculate the equilibrium concentration of Cu²⁺in a solution initially with 0.050 M Cu²⁺and 1.00 M NH₃.

The reaction of formation of\({\left( {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right]^{2 + }}\)

Since Cu²⁺ is limiting reagent, 0.050 M Cu²⁺will react with\((0.050{\rm{M}} \cdot 4)\)\(0.2{\rm{MN}}{{\rm{H}}_3}\), to produce 0.050 M\({\left( {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right)^{2 + }}\)

Therefore, we will be left with

\(\begin{array}{*{20}{c}}{\left[ {{\rm{C}}{{\rm{u}}^{2 + }}} \right) = 0{\rm{M}}}\\{\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 1{\rm{M}} - 0.2{\rm{M}} = 0.8{\rm{M}}}\\{\left( {{{\left( {{\rm{Cu}}{{\left( {{\rm{NI}}{{\rm{I}}_3}} \right)}_1}} \right)}^{2 + }}} \right) = 0.050{\rm{M}}}\end{array}\)

The reaction ofdissociation of\({\left( {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right)^{2 + }}\)

The dissociation constant of\({\left( {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right)^{2 + }}\)is

\(\begin{array}{*{20}{c}}{{K_d} = \frac{{\left( {C{u^{2 + }}} \right) \cdot {{\left( {N{H_3}} \right)}^4}}}{{\left( {{{\left( {Cu{{\left( {N{H_3}} \right)}_4}} \right)}^{2 + }}} \right)}}}\\{5.88 \cdot {{10}^{ - 14}} = \frac{{x \cdot {{(0.8 + 4x)}^4}}}{{0.050 - x}}}\end{array}\)

Since K_d is smaller than 10^-4,

We will assume that \(0.050 - {\rm{x}} \approx 0.050\)

And that\(0.8 + 4{\rm{x}} \approx 0.8\)

\(\begin{array}{*{20}{c}}{5.88 \cdot {{10}^{ - 14}} = \frac{{x \cdot {{(0.8)}^4}}}{{0.050}}}\\{0.4096x = 2.94 \cdot {{10}^{ - 15}}}\\{x = 7.18 \cdot {{10}^{ - 15}}{\rm{M}}}\\{\left( {C{u^{2 + }}} \right) = 7.18 \cdot {{10}^{ - 15}}{\rm{M}}}\end{array}\)