Complex Ion Equilibria
You mix a 200.0-mL sample of a solution that is 1 $.5 ×10^{-3}$ M in $Cu(NO_3)_2$ with a 250.0-mL sample of a solution that is 0.20 M in $NH_3$ . After the solution reaches equilibrium, what concentration of $Cu^{2+}$ (aq) remains?

Question

Complex Ion Equilibria
You mix a 200.0-mL sample of a solution that is 1[latex].5 ×10^{-3}[/latex] M in [latex]Cu(NO_3)_2[/latex] with a 250.0-mL sample of a solution that is 0.20 M in [latex]NH_3[/latex]. After the solution reaches equilibrium, what concentration of [latex]Cu^{2+}[/latex](aq) remains?

Accepted Answer

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

[latex]Cu^{2+}(aq) + 4 NH_3(aq) ightleftharpoons Cu(NH_3)_4^{2+}(aq)[/latex]

[latex]K_f= 1.7 imes 10^{13}[/latex]

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

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

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

[latex]Cu^{2+}(aq) + 4 NH_3(aq) ightleftharpoons Cu(NH_3)_4^{2+}(aq)[/latex]

	[latex][Cu^{2+}][/latex]	[[latex]NH_3][/latex]	[latex]Cu(NH_3)_4^{2+}[/latex]
Initial	[latex]6.7 imes 10^{-4}[/latex]	0.11	0.0
Change
Equil

Since the equilibrium constant is large and the concentration of ammonia is much larger than the concentration of [latex]Cu^{2+}[/latex], you can assume that the reaction will be driven to the right so that most of the [latex]Cu^{2+}[/latex] 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 [latex]Cu^{2+}[/latex] that remains when equilibrium is reached.

[latex]Cu^{2+}(aq) + 4 NH_3(aq) ightleftharpoons Cu(NH_3)_4^{2+}(aq)[/latex]

	[latex][Cu^{2+}][/latex]	[[latex]NH_3][/latex]	[latex]Cu(NH_3)_4^{2+}[/latex]
Initial	[latex]6.7 imes 10^{-4}[/latex]	0.11	0.0
Change	[latex]≈(-6.7 imes 10^{-4})[/latex]	[latex]≈4(-6.7 imes 10^{-4})[/latex]	[latex]≈(+6.7 imes 10^{-4})[/latex]
Equil	x	0.11	[latex]6.7 imes 10^{-4}[/latex]

Substitute the expressions for the equilibrium concentrations into the expression for [latex]K_f[/latex] and solve for x.

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

[latex]=\frac{6.7 imes 10^{-4}}{x(0.11)^4} [/latex]

[latex]x =\frac{6.7 imes 10^{-4}}{K_f(0.11)^4} [/latex]

[latex]x =\frac{6.7 imes 10^{-4}}{1.7 imes 10^{13}(0.11)^4} [/latex]

[latex]= 2.7 imes 10^{-13}[/latex]

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

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

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

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