Ligand exchange is an equilibrium, so it has an equilibrium constant. The stability constant of each complex tells you which ligand will replace which, and why multidentate ligands such as EDTA grab metal ions so strongly.
The stability constant, Kstab, is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules. For Cu2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq), Kstab = [complex] / ([Cu2+][NH3]4), units mol−4 dm12. Water is left out because it is the solvent and its concentration is effectively constant. Values are huge, so they are often quoted as log Kstab. A ligand exchange happens if the new complex has a larger Kstab: the larger the value, the more stable the complex. Multidentate ligands such as EDTA4− form very stable complexes (the chelate effect) because one ligand releases several water molecules, so the number of particles and the entropy increase.
Watch the Kstab panel on the ligand exchange bench as you add ammonia and chloride. A stability constant is a Kc for complex formation, with the solvent water left out.
| Formation equilibrium | Kstab expression | Units | log Kstab |
|---|---|---|---|
| Cu2+ + 4Cl− ⇌ [CuCl4]2− | [CuCl42−] / ([Cu2+][Cl−]4) | mol−4 dm12 | 5.6 |
| Cu2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ | [Cu(NH3)4(H2O)22+] / ([Cu2+][NH3]4) | mol−4 dm12 | 13.1 |
| Cu2+ + EDTA4− ⇌ [Cu(EDTA)]2− | [Cu(EDTA)2−] / ([Cu2+][EDTA4−]) | mol−1 dm3 | 18.8 |
| Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ | [Ag(NH3)2+] / ([Ag+][NH3]2) | mol−2 dm6 | 7.2 |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What is the stability constant, Kstab, of a complex?
Which complex of copper(II) is most stable?
What are the units of Kstab for Ag+ + 2NH3 ⇌ [Ag(NH3)2]+?
Why is water omitted from Kstab expressions for aqueous complexes?
Compare log Kstab values on the figure. A ligand replaces another if it forms a complex with a larger Kstab, provided it is present in a reasonable concentration.
| Start | Add | Exchange? | Reason |
|---|---|---|---|
| [Cu(H2O)6]2+ | excess NH3 | yes, deep blue | log Kstab 13.1 for the ammine complex |
| [CuCl4]2− | excess NH3 | yes | 13.1 > 5.6 |
| [Cu(NH3)4(H2O)2]2+ | EDTA4− | yes, almost completely | 18.8 > 13.1 |
| [Cu(EDTA)]2− | NH3 | no | 13.1 < 18.8 |
Leave water out of Kstab expressions. Units depend on the number of ligands: four monodentate ligands give mol−4 dm12, but one EDTA4− gives mol−1 dm3. A larger Kstab means a more stable complex, and ligand exchange goes towards it. The chelate effect is mainly an entropy effect, not a stronger bond.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
At equilibrium [Cu(NH3)4(H2O)22+] = 0.0500, [NH3] = 0.200 and [Cu2+] = 2.60 × 10−12 mol dm−3. What is Kstab?
Why does EDTA4− form a much more stable complex with Cu2+ than NH3 does?
log Kstab for [CuCl4]2− is 5.6 and for [Cu(NH3)4(H2O)2]2+ is 13.1. What happens when excess aqueous ammonia is added to [CuCl4]2−?
Kstab for [Ag(NH3)2]+ is 1.7 × 107 mol−2 dm6. If [complex] = 0.0100 and [NH3] = 0.100 mol dm−3 at equilibrium, what is [Ag+]?
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