Revise Stability constants, Kstab for Cambridge 9701 (28.5)

The formation of a complex ion from an aqueous metal ion and ligands is a reversible process that establishes an equilibrium. The stability constant, Kstab, is the equilibrium constant for this formation reaction. It provides a measure of the strength of the bonding between the metal ion and its ligands and, consequently, the stability of the resulting complex.

$M^{a+}(aq) + nL^{b-}(aq) \rightleftharpoons [ML_n]^{(a-nb)+}(aq) \\ K_{stab} = \frac{[[ML_n]^{(a-nb)+}]}{[M^{a+}][L^{b-}]^n}$

Ligands do not usually attach to a metal ion all at once. Instead, they are added sequentially in a series of equilibrium steps. Each step has its own stability constant, known as a stepwise stability constant, denoted K₁, K₂, etc. The overall stability constant, often given the symbol βn (beta), represents the equilibrium for the total formation from the bare metal ion and is the product of all the individual stepwise constants.

$Step 1: Cu^{2+}(aq) + NH_3(aq) \rightleftharpoons [Cu(NH_3)]^{2+}(aq) \quad K_1 = \frac{[[Cu(NH_3)]^{2+}]}{[Cu^{2+}][NH_3]} \\ Step 2: [Cu(NH_3)]^{2+}(aq) + NH_3(aq) \rightleftharpoons [Cu(NH_3)_2]^{2+}(aq) \quad K_2 = \frac{[[Cu(NH_3)_2]^{2+}]}{[[Cu(NH_3)]^{2+}][NH_3]} \\ ...and so on for K₃ and K₄. \\ Overall: Cu^{2+}(aq) + 4NH_3(aq) \rightleftharpoons [Cu(NH_3)_4]^{2+}(aq) \\ \beta_4 = K_{stab} = K_1 \times K_2 \times K_3 \times K_4$

Ligand substitution reactions are competitive equilibria. By comparing the Kstab values for two different complexes of the same metal ion, we can predict which complex is more stable and therefore which direction the equilibrium will lie. A ligand that forms a complex with a significantly higher Kstab will displace a ligand that forms a less stable complex. This is particularly evident with polydentate ligands like EDTA or ethylenediamine (en), which exhibit the 'chelate effect' and form exceptionally stable complexes due to a large positive entropy change upon their formation.