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*Revision Materials* 1 Atomic Structure 2 Atoms, molecules and stoichiometry 3 Chemical Bonding 4 States of matter 5 Chemical energetics 6 Electrochemistry 7 Equilibria 8 Reaction kinetics 9 The Periodic Table, chemical periodicity 10 Group 2 11 Group 17 12 Nitrogen and sulfur 13 Organic 14 Hydrocarbons 15 Halogen compounds 16 Hydroxy compounds 17 Carbonyl compounds 18 Carboxylic acids and derivatives 19 Nitrogen compounds 20 Polymerisation 21 Organic synthesis 22 Analytical techniques 23 Chemical energetics 24 Electrochemistry 25 Equilibria 26 Reaction kinetics 27 Group 2 28 Chemistry of transition elements 29 Organic 30 Hydrocarbons 31 Halogen compounds 32 Hydroxy compounds 33 Carboxylic acids and derivatives 34 Nitrogen compounds 35 Polymerisation 36 Organic synthesis 37 Analytical techniques

28 Chemistry of transition elements

28.1 General physical and chemical properties of the first row of transition elements, titanium to copper 28.2 General characteristic chemical properties of the first set of transition elements, titanium to copper 28.3 Colour of complexes 28.4 Stereoisomerism in transition element complexes 28.5 Stability constants, Kstab

Stability Constants, Kstab

Specification Reference Inorganic Chemistry, Chemistry of transition elements 28.5

Quick Notes

  • A stability constant (Kstab) is the equilibrium constant for the formation of a complex ion from its metal ion and ligands.
  • Large Kstab values indicate greater stability of the complex.
  • Example for [Cu(NH₃)₄(H₂O)₂]²⁺ formation:
    Kstab = [Cu(NH₃)₄²⁺] / ([Cu²⁺][NH₃]⁴)
  • Ligand exchange is more likely if the new complex has a higher Kstab than the original.
  • [H₂O] is not included in Kstab expressions because it is the solvent.

Full Notes

The stability constant (Kstab) is the equilibrium constant for a reaction where a complex ion forms from a metal ion and its ligands.

A larger Kstab value means the complex is more stable and forms more readily.

Writing the Expression for Kstab

Kstab is written just like any equilibrium expression:

Kstab = [complex ion] ÷ ([metal ion] × [ligand]n)

Example:
In aqueous solution, copper(II) exists as [Cu(H₂O)₆]²⁺ and can react with ammonia:

CIE A-Level Chemistry copper(II) complex ion before ligand exchange.

The Kstab expression for the new complex ion formed would be:

CIE A-Level Chemistry copper(II) complex ion after ammonia ligand exchange showing Kstab equilibrium expression Kstab = [[Cu(NH₃)₄(H₂O)₂]²⁺] / ([[Cu(H₂O)₆]²⁺] × [NH₃]⁴.
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Matt’s Exam Tip

Remember if water is a solvent, concentration of H2O isn’t included in an equilibrium expression. Meaning, for Kstab, [H2O] isn’t included in the expression.

What Does Kstab Tell Us?

Examples:

Ligand Exchange and Kstab

Ligand exchange involves one ligand replacing another around a metal ion.

This happens if the new complex has a higher Kstab than the original one.

Example:
[Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

Stronger-field ligands like NH₃, CN⁻, and EDTA⁴⁻ tend to form more stable complexes.

Example Calculation with Kstab

Worked Example

Calculate Kstab for [Cu(NH₃)₄(H₂O)₂]²⁺ given:

  • [[Cu(H₂O)₆]²⁺] = 0.010 mol dm⁻³
  • [NH₃] = 0.100 mol dm⁻³
  • [[Cu(NH₃)₄(H₂O)₂]²⁺] = 0.040 mol dm⁻³

Kstab = 0.040 / (0.010 × 0.100⁴)
= 0.040 / (1.0 × 10⁻¹⁰)
= 4.0 × 10⁸ mol⁻⁴ dm¹²

This large value shows the complex is highly stable.

Summary