Ligand Substitutions & Kstab
Chemistry: Cambridge International A Level, topic 2.4.9
Ligand Subsitution & Kstab
Complex metal ions have ligands which can be substituted. Kstab is defined as the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.

The reaction
- The complex ion [ Cu(H2O)6 ]2+ undergoes a ligand substitution reaction where H2O is replaced by Cl-.
- The reaction is reversible.
- [ Cu(H2O)6 ]2+ + 4Cl− [ CuCl4 ]2- + 6H2O
- The equilibrium can be pushed to the right using concentrated hydrochloric acid (HCl).
- The reaction can be reversed by adding water. This will push the equilibrium to the left.
![[ Cu(H2O)6 ]2+ vs. [ CuCl4 ]2-](https://image-v2.cdn.app.senecalearning.com/2018-11/b6cb988e-c777-43c6-9070-520bb870abec/chlorine-electron%20shell-,w_800.png)
[ Cu(H2O)6 ]2+ vs. [ CuCl4 ]2-
- In the reaction, the copper goes from having six water ligands to having four chloride ligands.
- The co-ordination number changes from six to four in the reaction.
- This is because the chloride ions are larger than the water molecules.
- There is only space for four chloride ions around the copper.

Reversible reaction
- As this reaction is reversible, we can form an equilibrium expression. This is known as Kstab, or the stability constant.
- For the below reaction:
- [ Cu(H2O)6 ]2+ + 4Cl− [ CuCl4 ]2- + 6H2O
- Kstab =
- Looking at the expression, we can see a large Kstab results in the formation of a stable complex ion.
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Colour change
- The colour change observed should be from blue to yellow.
- However, there is normally a mixture of [ Cu(H2O)6 ]2+ and [ CuCl4 ]2- ions.
- This means the observed colour change is from blue to green (blue + yellow = green).
Substituting Water for Chlorine in a Complex Cobalt Ion
Let's go through another example. The chemistry in this reaction with a complex cobalt ion is exactly the same as the reaction with the with a complex copper ion.

The reaction
- The complex ion [ Co(H2O)6 ]2+ undergoes a ligand substitution reaction where H2O is replaced by Cl-.
- The reaction is reversible.
- [ Co(H2O)6 ]2+ + 4Cl− [ CoCl4 ]2- + 6H2O
- The equilibrium can be pushed to the right using concentrated hydrochloric acid (HCl).
- The reaction can be reversed by adding water. This will push the equilibrium to the left.
![[ Co(H2O)6 ]2+ vs. [ CoCl4 ]2-](https://image-v2.cdn.app.senecalearning.com/2018-11/b6cb988e-c777-43c6-9070-520bb870abec/chlorine-electron%20shell-,w_800.png)
[ Co(H2O)6 ]2+ vs. [ CoCl4 ]2-
- In the reaction, the cobalt goes from having six water ligands it to having four chloride ligands.
- The co-ordination number changes from six to four in the reaction.
- This is because the chloride ions are larger than the water molecules.
- There is only space for four chloride ions around the cobalt.

Reversible reaction
- As this reaction is reversible, we can form an equilibrium expression. This is known as Kstab, or the stability constant.
- For the below reaction:
- [ Co(H2O)6 ]2+ + 4Cl− [ CoCl4 ]2- + 6H2O
- Kstab =
- Looking at the expression, we can see a large Kstab results in the formation of a stable complex ion.

Colour change
- The colour change observed should be from pink [ Co(H2O)6 ]2+ ions to blue [ CoCl4 ]2- ions.
Test yourself on Ligand Substitutions & Kstab
What colour change is observed when [ Cu(H2O)6 ]2+ reacts with chloride ions to form [ CuCl4 ]2-?
When [ Cu(H2O)6 ]2+ reacts to form [ CuCl4 ]2-, what happens to the co-ordination number of the complex ion?
Which of the follwoing are true about the formation of [CuCl4]2- from [Cu(H2O)6]2+?
In the reaction [ Cu(H2O)6 ]2+ + 4Cl− [ CuCl4 ]2- + 6H2O, how do you push the equilibrium to the right?
What colour change is observed when [ Co(H2O)6 ]2+ reacts with chloride ions to form [ CoCl4 ]2-?