1. CE 541
Complex Formation

2. Complex ions consist of one or more central ions (usually metals) associate with one or more ions or molecules (ligands). Ligands act to stabilize the central ion (metal) and keep it in solution.

3. Mononuclear Complexes
Consist of one single central ion to which is bound a number of neutral or anionic ligands
The number of ligands attached to the central ion is called coordination number
If we take Cu2+ as a central ion and NH3 as ligand

4. Table 4 shows more formation constants for various ligands and complexes.
When formation constants are available, the concentration of the complex ion can be performed. The significance of complexes can be illustrated graphically using Logarithmic Concentration Diagrams.

5. pNH3 (-log NH3)
if concentration of Cu2+ is fixed, equations (1) to (4) become straight lines on the Logarithmic Concentration Diagram.
Equation (1) becomes
from Table 4
log K1 = 3.99

7. Similarly,

9. Mixed Ligand Complexes
Natural waters contain several ligands (Cl-, NH3, S2-, OH-) competing for complex formation with metal ions
How to solve the problem:
All stepwise equilibrium for each ligand and the metal must be considered
Assume the unassociated (uncomplexed) ion concentration
Determine concentration of each complex for each ligand

10. Problem 4.66
Draw a logarithmic concentration diagram illustrating the effect of chloride concentration on the relative concentrations of various chloride complexes of mercury, assuming [Hg2+] = 10-7 mol/l. Which complex predominates when the chloride concentration equals 0.1 mg/l, 1 mg/l, 10 mg/l, and 100 mg/l.

13. Solubility of Salts “All salts are soluble to some degree”
AgCl  Ag+ + Cl-
AgCl is known to be insoluble. With insoluble salts the saturation value is reached very quickly.
The equilibrium that exists between the solids state of a compound and its ion can be represented by the following equilibrium:

14. Since AgCl is in solid state, it can be considered as constant (say Ks)
Ksp = solubility product constant

15. For more complex substances:

16. Logarithmic Concentration Diagrams for Simple Solubility Determination
The diagram was draw based on:
Solubility product constants (Table 2.5)
The following solubility product relationships:
[M2+][CO32-]=Ksp
log[M2+] = pCO3 + log Ksp

17. The result is straight line with slope = 1
[CO32-] is represented by a line with slope = -1 and as a function of pCO3
Pb2+ is the least soluble and Mg2+ is the most soluble
Intersection of CO32- line with cation line (say Pb) indicates the solubility condition for that cation (Pb)
[M2+] = [CO32-] = solubility
Example
Determine the solubility of Ca2+ in water containing 10-3 M 25 C.
Solution
pCO3 = -log 10-3 = 3
from the figure
at pCO3 = 3, log[Ca2+] = -5.3
so, solubility = S = = 5  10-6 M

18. Complex Solubility Relationships
“Solubility relationships are generally complex”
ionic strength affects the ion activity (using molar concentration will not give exact solution)
Cations and anions may form complexes with other materials in solution (reduce the effective concentration of cations and anions)
Other ions may form salts with less solubility than the one under consideration
Other equilibria beside solubility product affect the concentration of the ions present

19. (1) Effect of pH on Solubility of Hydroxide salts

20. The relationship can be drawn as shown in the figure
The relationship can be drawn as shown in the figure. Slope of the line = -Z

21. (2) Effect of Weak Acids and Bases pH
Affects solubility of metal hydroxides, and
Affects other equilibria which affect solubility (such as that of weak acids)
Take CaCO3 as an example

23. How can we solve the problem? Use equations (1) to (4) in addition to:
If NaOH was added to adjust the pH and if CaCl2 was added to the solution
At what [Ca2+] will the solution be saturated with respect to CaCO3
How the saturation value will be affected by pH and CT
How can we solve the problem?
Use equations (1) to (4) in addition to:
Equations (1) to (6) can be solved by trial and error. The relationship between CT, pH, and saturation value of [Ca2+] can be represented graphically.

24. Study Examples 21 to 23

25. (3) Effect of Complexes
“Complex formation affects the solubility of salts”
Take Zn as an example
Base is added to increase pH to form insoluble Zn(OH)2
If excess base was added, Zn forms complexes with OH- (soluble)
This behavior can be expressed by the following:
(a) Solubility product

26. (b) Complex formation
(c) Water ionization

27. Equilibrium Equations for the Above

28. How to construct the diagram (solubility of Zn(OH)2 as a function of pH)
For a given pH, find OH- from equation (6)
Use it in Equation (1) to find Zn2+
Use Zn2+ in equation (2) to find [ZnOH+]
Continue the approach to find all complexes
Use equation (7) to find CT

29. logC-pH equations for each Zn species can be developed
take log of equation (1)

30. Similarly, taking the log of equation (2)
following the same approach, similar equations for the remaining Zn-hydroxide complexes can be developed

31. See Table 4-5 for other metals Study Examples 24 to 25

32. The solubility of metals can be complicated by the presence of ligands other than hydroxides (ammonia as an example)
Now, how ammonia will affect the solubility of Zn(OH)2. There are four (4) Zn–NH3 complexes as shown in Table 4-4.

34. Assuming that the concentration of Zn-NH3 complexes is small compared to [NH4+] and [NH3], then
Use these equations together with those of Zn-OH to construct the diagram of Zn(OH)2 solubility in the presence of ammonia (NH3)

35. Oxidation – Reduction Reactions
“Oxidation - reduction reactions tend towards a state of equilibrium”
Oxidation-reduction reactions are complex
Understanding the equilibrium of such reactions will help to indicate whether certain reactions can occur under certain conditions
Example
Use half reactions for the reduction of NO3- to NH4 (Table 2-4)

36. The Redox potential can be indicated by pE
If pE is large, then system oxidizing. So, half reactions are driven to the left
If pE is small, then system is reducing. So, half reactions are driven to the right

37. Logarithmic Concentration Diagram
It can be used to illustrate the equilibrium concentrations for NH4+ and NO3- and a function of pE or E. The diagram was 25 C using:
pE value can be found in Table 2-4

38. At pH = 7
[H+] = 10-7
Thus

39. pE-pH Diagrams As shown
pE is a function of [H+] which means it is a function of pH
The general form of equation to be used in constructing the pE-pH diagrams is:
(reduced) represents species on the reduced side
(oxidized) represents species on the oxidized side

40. Then
[H+] was separated from the log term
In order to draw pE-pH diagram for nitrogen system, the following relationships will be needed: