Define & Write a Stability Constant for a Complex
- When transition element ions are in aqueous solutions, they will automatically become hydrated
- Water molecules will surround the ion and act as ligands by forming dative covalent bonds to the central metal ion
- When there are other potential ligands present in solution, there is a competing equilibrium in ligand exchange and the most stable complex will be formed
- For example, a Co(II) ion in solution will form a [Co(H2O)6]2+ complex
- Adding ammonia results in the stepwise substitution of the water ligands by ammonia ligands until a stable complex of [Co(NH3)4(H2O)2]2+ is formed
[Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O
- For the substitution reaction above, there are four stepwise constants:
[Cu(H2O)6]2+ + NH3 ⇌ [Cu(NH3)(H2O)5]2+ + H2O K1
[Cu(NH3)(H2O)5]2+ + NH3 ⇌ [Cu(NH3)2(H2O)4]2+ + H2O K2
[Cu(NH3)2(H2O)4]2+ + NH3 ⇌ [Cu(NH3)3(H2O)3]2+ + H2O K3
[Cu(NH3)3(H2O)3]2+ + NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + H2O K4
- These stepwise constants are summarised in the overall stability constant, Kstab
- The stability constant is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules
Expression of Kstab
- The expression for Kstab can be deduced in a similar way as the expression for the equilibrium constant (Kc)
- For example, the equilibrium expression for the substitution of water ligands by ammonia ligands in the Co(II) complex is:
[Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O
Kstab =
- The concentration of water is not included in the expression as the water is in excess
- Therefore, any water produced in the reaction is negligible compared to the water that is already present
- The units of the Kstab can be deduced from the expression in a similar way to the units of Kc
- The stability constants can be used to compare the stability of ligands relative to the aqueous metal ion where the ligand is water
- The larger the Kstab value, the more stable the complex formed is