# Complex-Ion Equilibria

## Complex-Ion Equilibria

The lone pairs of electrons in water molecules make water an excellent ligand. Metal ions in any body of wter 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 NH3 is present, its molecules compete with water for the coordination site.

Ag(H2O)6+ + NH3 ® Ag(NH3)(H2O)5+
Ag(NH3) (H2O)5+ + NH3 ® Ag(NH3)2(H2O)4+
And these are simply represented by the following with the equilibrium constant as indicated. Ag+ + NH3 ® Ag(NH3)+ K1 = 2.0´103
Ag(NH3)+ + NH3 ® Ag(NH3)2+ K2 = 8.0´103
Where K1 and K2 are called stepwise equilibrium constants as opposed to the overall equilibrium constants, b1 and b2. Note, however, that K1 = [Ag(NH3) +] / [Ag+] [NH3]
b1 = K1
K2 = [Ag(NH3) +2] / [Ag(NH3) +][NH3]
b2 = [Ag(NH3)+2] / [Ag+][NH3]2
= K1 K2
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 C0 is, C0 = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]
= [Ag+] (1 + b1 [NH3] + b2 [NH3]2)
And the mole fraction, X, of a species A, X(A) is easily calculated using the following equations. X(Ag+) = [Ag+] / C0
= 1 / (1 + b1 [NH3] + b2 [NH3]2)
X(Ag(NH3)+) = [Ag(NH3) +] / C0
= b1 [NH3] / (1 + b1 [NH3] + b2 [NH3]2)
= b1 [NH3] X(Ag+)
X(Ag(NH3)2+) = [Ag(NH3)2+] / C0
= b2 [NH3]2 / (1 + b1 [NH3] + b2 [NH3]2)
= b2 [NH3]2 X(Ag+)
A plot of mole fractions of various species against the concentration of NH3, [NH3] or pNH3 (= – log [NH3]) has been shown on the black board during our lecture.

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,

b1 = K1 = 2.0´103
b2 = K1 K2 = 1.6´107
[NH3] = 1.00 M
C0 = [Ag+] = 0.10 M
X(Ag+) = 1 / (1 + b1 [NH3] + b2 [NH3]2)
= 1 / 1.6´107
= 6.25x10-8.
X(Ag(NH3)+) = b1 [NH3] X(Ag+)
= 1.25x10-4.
X(Ag(NH3)2+) = b2 [NH3]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.
X(Ag(NH3)+) = b1 [NH3] X(Ag+)
X(Ag(NH3)2+) = b2 [NH3]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(NH3)+) We have b1[NH3] = 1
[NH3] = 1 / b1
By the same arguments, the cross point of X(Ag(NH3)+) line and X(Ag(NH3)2+) line occur when [NH3] = 1 / (b2)1/2

Discussion -
Note that [NH3] referrs 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 discussion giving above opens up an opportunity for doing more work after the exam for you and/or for me alike!

## The cupric complexes

In the case of Cu2+ ions, the equilibria are more complicated, because the number of ligands can be as high as 4 to 6. Consider the following data: Cu2+ + NH3 ® Cu(NH3) 2+             K1 = 1.1´104
Cu(NH3)2+ + NH3 ® Cu(NH3)22+         K2 = 2.7´103
Cu(NH3)2 2+ + NH3 ® Cu(NH3)32+       K3 = 6.3´102
Cu(NH3)3 2+ + NH3 ®Cu(NH3)42+       K4 = 30
At high concentration of NH3 more of Cu(NH3)42+ 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.