Effect of Ligand Exchange on Stability Constant (Cambridge (CIE) A Level Chemistry): Revision Note
Exam code: 9701
Effect of Ligand Exchange on Stability Constant
The stability constants (Kstab) of ligands are often given on a log10 scale so that it becomes easier to compare them with each other
Ligand exchange in a complex occurs to form a more stable complex with a larger Kstab
The stability constants can be used to explain the substitution of ligands in a copper complex
Ligand substitution in a Co(II) complex
When excess ammonia is added to the [CoCl4]2- complex a brown solution is obtained
Ligand exchange of the [CoCl4]2- complex by ammonia
The chloride ligands are substituted by the ammonia ligands to form the more stable ammonia complex
The formation of the ammonia complex could be explained by comparing the stability of the chloride and ammonia ligands
Stability of chloride and ammonia ligands table
Ligand | Stability (log10 Kstab) |
|---|---|
Cl– | 5.6 |
NH3 | 13.1 |
The stability constant of the ammonia ligand is greater than that of the chloride ligands
The brown ammonia complex is therefore more stable
As a result, the position of the equilibrium is shifted to the right
Worked Example
The numerical values for the stability constants, Kstab, of three silver(I) complexes are given.
Silver(I) complex | Numerical value of Kstab |
|---|---|
[Ag(S2O3)2]3– | 2.9 x 1013 |
[Ag(CN)2]– | 5.3 x 1018 |
[Ag(NH3)2]+ | 1.6 x 107 |
An aqueous solution of Ag+ is added to a solution containing equal concentrations of S2O32– (aq), CN– (aq) and NH3 (aq). The mixture is left to reach equilibrium.
Deduce the relative concentrations of [Ag(S2O3)2]3–, [Ag(CN)2]– and [Ag(NH3)2]+ present in the equilibrium mixture. Explain your answer.
Answer
The highest concentration will be [Ag(CN)2]–
This is because the Kstab value for [Ag(CN)2]– is the largest value
OR[Ag(CN)2]– is the most stable
The lowest concentration will be [Ag(NH3)2]+
This is because the Kstab value for [Ag(NH3)2]+ is the smallest value
OR[Ag(NH3)2]+ is the least stable
An alternative explanation could be to state that higher Kstab values form a more stable complex
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