Stability constants, Kstab

Chemistry A2 · Chemistry of transition elements. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

When a ligand replaces the water molecules around a metal ion in solution, an equilibrium is set up. The stability constant, Kstab, is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.

The expression is written like Kc, with the complex on top and the starting ion and the free ligand below. Water is the solvent and is in large excess, so [H2O] is left out.

A large Kstab means the equilibrium lies far to the right and the complex is stable. In a ligand exchange, a ligand that forms a complex with a larger Kstab replaces one that forms a complex with a smaller Kstab.

Rules to remember

Common mistake

Forgetting the power on the ligand concentration, or writing the expression upside down. The complex formed is always on top, and the ligand concentration is raised to the number of ligands that enter.

Worked example

For [Fe(H2O)6]3+ + SCN− ⇌ [Fe(SCN)(H2O)5]2+ + H2O, the equilibrium concentrations are [Fe(H2O)6]3+ 0.0020 mol dm−3, SCN− 0.0050 mol dm−3 and [Fe(SCN)(H2O)5]2+ 0.0090 mol dm−3. What is Kstab?

  1. Kstab = [[Fe(SCN)(H2O)5]2+] ÷ ([[Fe(H2O)6]3+][SCN−]); water is left out.
  2. Bottom line: 0.0020 × 0.0050 = 1.0 × 10−5.
  3. Kstab = 0.0090 ÷ (1.0 × 10−5) = 900.
  4. Units: (mol dm−3) ÷ (mol dm−3)2 = dm3 mol−1.

Answer: Kstab = 9.0 × 102 dm3 mol−1

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