Complex Ion Formation

1
Complex Ion Formation
Read Chapter 6-6
I will introduce this topic by considering
dissolving PbI2 in solution.
PbI2(s) Pb2 2I-
Ksp 7.9x10-9
If I-1.0x10-3 and Pb21.0x10-4 initially,
what will be the concentrations after equilibrium
is reached?
I-2 Pb2 1.0x10-10 lt Ksp , so no
precipitate.
What happens if I start to add NaI to the
solution?
2
Complex Ion Formation Continued
There are additional equilibria to be considered
PbI2(s) Pb2 2I- Pb2
I- PbI PbI
I- PbI2(aq) PbI2(aq)
I- PbI-3 PbI-3
I- PbI-4
Ksp 7.9x10-9
K1 K2 K3 K4
Write down equilibrium expressions for each of
these reactions.
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Equilibrium expressions
PbI2(s) Pb2 2I- Pb2
I- PbI PbI
I- PbI2(aq) PbI2(aq)
I- PbI-3 PbI-3
I- PbI-4
Ksp 7.9x10-9
1 2 3 4 5
K1 K2 K3 K4
4
Complex Ion Formation Continued
PbI2(s) Pb2 2I- Pb2
I- PbI PbI
I- PbI2(aq) PbI2(aq)
I- PbI-3 PbI-3
I- PbI-24
Ksp 7.9x10-9
1 2 3 4 5
K1 K2 K3 K4
Add Eqs. (2) and (3) and write down the
equilibrium constant for this new equation. You
should find that it is K1K2.
Add Eqs. (2), (3) and (4) and write down the
equilibrium constant for this reaction. You
should find that it is K1 K2 K3 .
5
If we multiply these two expressions together, we
get
This the equilibrium constant for the reaction
Pb2 2I- PbI2(aq)
6
Complex Ion Formation Continued
PbI2(s) Pb2 2I- Pb2
I- PbI Pb2
2I- PbI2(aq) Pb2
3I- PbI-3 Pb2 4I-
PbI-24
Ksp 7.9x10-9
1 2 3 4 5
b1K1 b2 K1 K2 b3 K1 K2 K3 b4 K1 K2 K3 K4
The bi are overall or cumulative formation
constants. If one knows I- then one knows all
the concentrations, given that we are at
equilibrium.
7
Complex Ion Formation Continued
PbI2(s) Pb2 2I- Pb2
I- PbI Pb2
2I- PbI2(aq) Pb2
3I- PbI-3 Pb2 4I-
PbI-24
Ksp 7.9x10-9
1 2 3 4 5
b1K1 b2 K1 K2 b3 K1 K2 K3 b4 K1 K2 K3 K4
Pb2 Ksp/I-2
6
PbI b1 Pb2 I- b1 Ksp I--1
7a 7b
PbI2(aq) b2 Pb2 I-2 b2Ksp 8
8
Complex Ion Formation Continued
PbI2(s) Pb2 2I- Pb2
I- PbI Pb2
2I- PbI2(aq) Pb2
3I- PbI-3 Pb2 4I-
PbI-24
Ksp 7.9x10-9
1 2 3 4 5
b1 K1 b2 K1 K2 b3 K1 K2 K3 b4 K1 K2 K3 K4
6 7 8 9
Pb2 Ksp/I-2 PbI K1 I- Pb2 K1
Ksp/ I- PbI2(aq) b2 Pb2 I-2 b2
Ksp PbI-3 b3 Pb2 I-3 b3 KspI-
At low iodide concentrations what species
dominates?

9
Graph of Pb Species
Pb2 Ksp I--2 PbI
b1Ksp I--1 PbI2(aq) b2 Ksp I-0
PbI-3 b3Ksp I-1
Log-Log plots are easier.
y b mx
logPb2 log(Ksp) 2log
I- 10 logPbI log( b1Ksp) 1log
I- 11 logPbI2(aq) log( b2 Ksp)
12 logPbI-3 log( b3Ksp) 1log
I- 13
10
y b mx
logPb2 log(Ksp) 2log
I- 10 logPbI log( b1 Ksp) 1log
I- 11 logPbI2(aq) log( b2 Ksp)
12 logPbI-3 log( b3 Ksp) 1log
I- 13
I II III IV
What is the value of I- when red and orange
lines intersect?
log of Pb species
?
log I-
11
Pb2 Ksp I--2 PbI
b1 Ksp I--1 PbI2(aq) b2 Ksp I-0
PbI-3 b3 Ksp I-1
KspI--2 b1 Ksp I--1
1/b1 I-
or log I- pK1
What is the value of I- when red and orange
lines intersect?
log of Pb species
log I-
12
Pb2 Ksp I--2 PbI
b1 Ksp I--1 PbI2(aq) b2 Ksp I-0
PbI-3 b3Ksp I-1
b2 KspI-0 b3 Ksp I-1
b2/b3 I- 1/K3
or log I- pK3
What is the value of I- when green and blue
lines intersect?
log of Pb species
log I-
13
Mathcad treatment of this problem
In this class we will use Excel to solve complex
equilibria concentrations. We will not use
Mathcad. See 2-10 if you need Excel help.
Excel treatment of this problem
Sometimes I will show Mathcad results. Here is
what they look like.
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(No Transcript)
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Key Points
Metal-ligand complexes are important species that
must be considered in predicting equilibrium
concentrations. Appendix I includes formation
constants. Take a look at this appendix. Given
Ksp and formation constants and the concentration
of the ligand one can calculate the concentration
of the metal ion and all of its complexes.