# Complex Equilibrium

The lone pairs of electrons in water molecules make water an excellent ligand. Metal ions in any body of water in the environment usually are coordinated to 6 water molecules, e.g. Ag(H2O)6+, Cu(H2O)62+, and Cr(H2O)62+, etc. In the literature, these ions are represented by Ag+(aq), Cu2+(aq), and Cr3+(aq) and very often notations (aq) is also omitted.

When a stronger coordinating ligand such as $$NH_3$$ is present, its molecules compete with water for the coordination site.

$Ag(H_2O)_6^+ + NH_3 \rightleftharpoons Ag(NH_3)(H_2O)_5^+$

$Ag(NH_3)(H_2O)_5^+ + NH_3 \rightleftharpoons Ag(NH_3)_2(H_2O)_4^+$

And these are simply represented by the following with the equilibrium constant as indicated.

$Ag^+ + NH_3\rightleftharpoons Ag(NH_3)^+ \;\;\;K_1 = 2.0 \times 10^3$

$Ag(NH_3)^+ + NH_3 \rightleftharpoons Ag(NH_3)_2^+ \;\;\;K_2 = 8.0 \times 10^3$

Where $$K_1$$ and $$K_2$$ are called stepwise equilibrium constants as opposed to the overall equilibrium constants, $$\beta_1$$ and $$\beta_2$$. Note, however, that

$K_1 = \dfrac{[Ag(NH_3)^+]}{[Ag^+] [NH_3]}$

$\beta_1 = K_1$

$K_2 = \dfrac{[Ag(NH_3)^{+2}]}{[Ag(NH_3)^+][NH_3]}$

$\beta_2 = \dfrac{[Ag(NH_3)^{+2}]}{[Ag^+][NH_3]^2} = K_1 K_2$

A mole fraction ($$X$$) distribution diagram for various complexes of silver can be used to show their variations. The total concentration of silver-ion-containing species $$C_0$$ is,

$C_0 = [Ag^+] + [Ag(NH_3)^+] + [Ag(NH_3)_2^+]$

$= [Ag^+] \left(1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2\right)$

And the mole fraction, $$X$$, of a species $$A$$, $$X(A)$$ is easily calculated using the following equations.

$X(Ag^+) = \dfrac{[Ag^+]}{C_0}$

$= \dfrac{1}{1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2}$

$X(Ag(NH_3)+) = \dfrac{[Ag(NH_3)^+]}{C_0}$

= \beta_1 [NH3] / (1 + \beta_1 [NH3] + \beta_2 [NH3]2)
= \beta_1 [NH3] X(Ag+)
X(Ag(NH3)2+) = [Ag(NH3)2+] / C0
= \beta_2 [NH3]2 / (1 + \beta_1 [NH3] + \beta_2 [NH3]2)
= \beta_2 [NH3]2 X(Ag+)

Example 1

Evaluate X(Ag+), X(Ag(NH3)+), and X(Ag(NH3)2+) for a solution containing 0.20 M Ag+ ions mixed with an equal volume of 2.00 M NH3 solution.

SOLUTION

Assume the volume doubles when the solutions are mixed. We have,

$\beta_1 = K_1 = 2.0 \times 10^3$

$\beta_2 = K_1 K_2 = 1.6 \times 10^7$

$[NH_3] = 1.00\; M$

$C_0 = [Ag^+] = 0.10\; M$

$X(Ag^+) = \dfrac{1}{1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2}$

$= 1 / 1.6 \times 10^7 = 6.25 \times 10^{-8}$

$X(Ag(NH_3)^+) = \beta_1 [NH_3] X(Ag^+)$

$= 1.25 \times 10^{-4}$

$X(Ag(NH_3)_2^+) = \beta_2 [NH_3]_2 X(Ag^+)$

$= 1$

DISCUSSION
More precisely [NH3] = 0.80 M, but the results are not changed. Since [NH3] is >> [Ag+], the complex with the highest number of ligands is the dominating species.

Example 2

What are the concentrations of NH3 when

• [Ag+] = [Ag(NH3)+],
• [Ag(NH3)+] = [Ag(NH3)2+],

These are cross points for the mole fraction distribution lines.

SOLUTION
Early on, we have derived these relationships.

$X(Ag(NH_3)^+) = \beta_1 [NH_3] X(Ag^+)$

$X(Ag(NH_3)_2^+) = \beta_2 [NH_3]^2 X(Ag^+)$

When the mole fractions are equal, their concentrations are also equal. Thus, when the mole fraction of Ag+ is the same as the mole fraction of Ag(NH3)+,

$X(Ag^+) = X(Ag(NH_3)^+)$

We have

$\beta_1[NH_3] = 1$
$[NH_3] = \dfrac{1}{\beta_1}$

By the same arguments, the cross point of X(Ag(NH3)+) line and X(Ag(NH3)2+) line occur when

[NH3] = 1 / (\beta_2)1/2

DISCUSSION

Note that [NH3] refers to the concentration of the free ligand, not the total concentration of NH3. For a chemical engineering application, much more details must be revealed in order to understand the complexity of coordination equilibrium. Here is a case for a simulation model either implemented using spread sheet or programming.

### The Cupric Complexes

In the case of $$Cu^{2+}$$ ions, the equilibria are more complicated, because the coordination number of ligands can be as high as 6. Consider the following data:

$Cu^{2+} + NH_3 \rightarrow Cu(NH_3)^{2+}$

with $$K_1 = 1.1 \times 10^4 \label{1}$$

$Cu(NH_3)^{2+} + NH_3 \rightarrow Cu(NH_3)_2^{2+}$

with $$K_2 = 2.7 \times 10^3 \label{2}$$

$Cu(NH_3)_2^{2+} + NH_3 \rightarrow Cu(NH_3)_3^{2+}$

with $$K_3 = 6.3 \times 10^2 \label{3}$$

$Cu(NH_3)_3^{2+} + NH_3 \rightarrow Cu(NH_3)_4^{2+}$

with  $$K_4 = 30 \label{4}$$

At high concentration of $$NH_3$$ more of $$Cu(NH_3)_4^{2+}$$ complexes are formed. The higher the number of ammonia is coordinated to the cupric ion, the deeper is the color blue. What we have done for the silver ion complexes can be applied to the cupric complexes, and the calculation is a little more complicated. However, the theory and method are the same, and you may applied the discussion on silver ion to that of the cupric ion cases.