Question 17.15: Complex Ion Equilibria You mix a 200.0-mL sample of a soluti...

Write the balanced equation for the complex ion equilibrium that occurs and look up the value of K_f in Table 17.3. Since this is an equilibrium problem, you have to create an ICE table, which requires the initial concentrations of Cu^{2+}and NH_3 in the combined solution. Calculate those concentrations from the given values.

Cu^{2+}(aq) + 4 NH_3(aq) \rightleftharpoons Cu(NH_3)_4^{2+}(aq)

K_f= 1.7 \times 10^{13}

[Cu^{2+}]_{initial}=\frac{0.200\cancel{L}\times \frac{1.5\times 10^{-3} \ mol}{\cancel{L}} }{0.200 L + 0.250 L}=6.7\times 10^{-4} \ M

[NH_3]_{initial}=\frac{0.250\cancel{L}\times \frac{0.20 \ mol}{1\cancel{L}} }{0.200 L + 0.250 L}=0.11 \ M

Construct an ICE table for the reaction and write down the initial concentrations of each species.

Cu^{2+}(aq) + 4 NH_3(aq) \rightleftharpoons Cu(NH_3)_4^{2+}(aq)

Since the equilibrium constant is large and the concentration of ammonia is much larger than the concentration of Cu^{2+}, you can assume that the reaction will be driven to the right so that most of the Cu^{2+} is consumed. Unlike in previous ICE tables, where you let x represent the change in concentration in going to equilibrium, here you let x represent the small amount of Cu^{2+} that remains when equilibrium is reached.

Cu^{2+}(aq) + 4 NH_3(aq) \rightleftharpoons Cu(NH_3)_4^{2+}(aq)

Substitute the expressions for the equilibrium concentrations into the expression for K_f and solve for x.

K_f=\frac{[Cu(NH_3)_4^{2+}]}{[Cu^{2+}][NH_3]_4}

=\frac{6.7 \times 10^{-4}}{x(0.11)^4}

x =\frac{6.7 \times 10^{-4}}{K_f(0.11)^4}

x =\frac{6.7 \times 10^{-4}}{1.7 \times 10^{13}(0.11)^4}

= 2.7 \times 10^{-13}

Confirm that x is indeed small compared to the initial concentration of the metal cation.
The remaining Cu^{2+} is very small because the formation constant is very large.

Since x = 2.7 \times 10^{-13} « 6.7 \times 10^{-4} , the approximation is valid.
The remaining [Cu^{2+}] = 2.7 \times 10^{-13}  M.