9701 · 28.5
Stability constants, Kstab — FAQ
Frequently asked questions for 9701 Stability constants, Kstab. Direct answers first, then deeper explanation — then practise with marking.
Why do Kstab values have different, and often complicated, units?
The units of any equilibrium constant depend on the stoichiometry of the reaction. For Kstab = [[MLn]] / ([M][L]ⁿ), the units are (concentration) / (concentration × concentrationⁿ) = concentration¹⁻⁽¹⁺ⁿ⁾. Since 'n' (the number of ligands) can vary, the overall power on the concentration unit changes, leading to different units for each type of complex.
Is Kstab just another name for Kc?
Essentially, yes. Kstab is a specific application of the general equilibrium constant, Kc. The subscript 'stab' is used to specify that the equilibrium in question is the formation of a stable complex ion. You treat it mathematically in exactly the same way as any other Kc.
Why are complexes with polydentate ligands (chelates) so much more stable?
This is due to the 'chelate effect'. When one polydentate ligand replaces several monodentate ligands, the total number of independent particles (molecules/ions) in the solution increases. For example, one 'en' molecule replaces two H₂O molecules. This increase in the disorder of the system corresponds to a large positive entropy change (ΔS). According to the equation ΔG = ΔH - TΔS, a large positive ΔS makes ΔG much more negative, indicating a more spontaneous and favourable reaction, resulting in a more stable complex.
Do stepwise stability constants (K₁, K₂, etc.) decrease as more ligands are added?
Yes, generally K₁ > K₂ > K₃ > ... . It becomes progressively harder to add another ligand as the complex becomes more crowded (steric hindrance) and as the positive charge of the central metal ion is increasingly neutralised by the electron pairs from the ligands, making it less attractive to further ligands.