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28.5 Stability constants, K stab

Syllabus
9701–2028–2029
Topic
28.5
Level
A2

Learning objectives

28.5.1Stability constant, Kstab, of a complex• Define the stability constant, Kstab, of a complex as the equilibrium constant for the formation of the complex ion in a solvent (from its constituent ions or molecules)28.5.2Write an expression for a Kstab of a complex• Write an expression for a Kstab of a complex ([H₂O] should not be included)28.5.3Kstab expressions to perform calculations• Use Kstab expressions to perform calculations28.5.4Ligand exchanges• Describe/explain: ligand exchanges in terms of K stab values and understand that a large Kstab is due to the formation of a stable complex ion

Kstab is the equilibrium constant for formation of a complex in a solvent

MXz+(solv)+n L⇌[MLXn]Xq\ce{M^{z+}(solv) + nL <=> [ML_n]^{q}}

The stability constant, Kstab, is the equilibrium constant for forming a complex ion in a specified solvent from its constituent metal ion and ligand ions or molecules.

A larger Kstab means the formation equilibrium lies further toward the complex under the stated conditions, so the complex is thermodynamically more stable relative to the separated constituents used in that definition.

Kstab is not a rate constant. A complex can have a large formation constant yet exchange ligands slowly or quickly; equilibrium stability and kinetic inertness are different ideas.

Build a Kstab expression from the balanced formation equilibrium

Kstab=[MLn][M][L]nK_{stab}=\frac{[ML_n]}{[M][L]^n}

[Cu(HX2O)X6]X2++4 NHX3⇌[Cu(NHX3)X4(HX2O)X2]X2++4 HX2O\ce{[Cu(H2O)6]^2+ + 4NH3 <=> [Cu(NH3)4(H2O)2]^2+ + 4H2O}

Kstab=[Cu(NHX3)X4(HX2O)X2X2+][Cu(HX2O)X6X2+][NHX3]4K_{stab}=\frac{[\ce{Cu(NH3)4(H2O)2^2+}]}{[\ce{Cu(H2O)6^2+}][\ce{NH3}]^4}

Write the balanced formation equation first. Put the complex concentration in the numerator and each dissolved constituent concentration in the denominator, raised to its stoichiometric coefficient. Omit H₂O when water is the solvent because its activity is effectively constant.

Derive units from the final expression rather than memorising one unit. In the copper example, one concentration divided by five concentration factors gives (mol dm⁻³)⁻⁴ = dm¹² mol⁻⁴.

A coefficient becomes a power, not a multiplier in front of a concentration. Do not include [H₂O] merely because water appears in the balanced aqueous exchange equation.

Use equilibrium changes before substituting into Kstab

[Cu(HX2O)X5Cl]X++ClX−⇌[Cu(HX2O)X4ClX2]+HX2O\ce{[Cu(H2O)5Cl]+ + Cl- <=> [Cu(H2O)4Cl2] + H2O}

concentration / mol dm⁻³ [Cu(H₂O)₅Cl]⁺ Cl⁻ [Cu(H₂O)₄Cl₂]
initial 0.15 0.15 0
change −0.10 −0.10 +0.10
equilibrium 0.05 0.05 0.10

Kstab=0.10(0.05)(0.05)=40 dm3 mol−1K_{stab}=\frac{0.10}{(0.05)(0.05)}=40\ \mathrm{dm^3\ mol^{-1}}

Use the balanced coefficients to construct concentration changes, calculate every equilibrium concentration, write the expression with water omitted, substitute equilibrium—not initial—values, then derive units and interpret the magnitude.

Here Kstab = 40, so products are favoured for this step under the stated conditions, although appreciable reactant remains. A large value does not mean the equilibrium goes literally to completion.

If the supplied amounts and equilibrium value do not fit a simple one-to-one change, follow the actual stoichiometric coefficients. Reject any calculated equilibrium concentration that is negative or exceeds what atom balance permits.

Ligand exchange favours the complex with the larger conditional stability

MLXa+b LX′⇌MLXb′+a L\ce{ML_a + bL' <=> ML'_b + aL}

Kexchange=Kstab(MLb′)Kstab(MLa),log⁡Kexchange=log⁡Kstab,new−log⁡Kstab,oldK_{exchange}=\frac{K_{stab}(ML'_b)}{K_{stab}(ML_a)},\qquad \log K_{exchange}=\log K_{stab,new}-\log K_{stab,old}

The ratio applies when the two overall Kstab values use the same free metal reference and consistent standard-state convention. If Kexchange is much greater than 1, the new complex is favoured under comparable ligand activities.

Co(II) ligand environment log₁₀ Kstab Relative conclusion
chloride complex 5.6 less stable in this comparison
ammonia complex 13.1 more stable in this comparison
exchange 13.1 − 5.6 = 7.5 Kexchange ≈ 10⁷⋅⁵ ≈ 3.2 × 10⁷, favouring the ammonia complex

Adding sufficient NH₃ therefore favours replacement of chloride by ammonia because formation of the ammine complex has the larger Kstab. The actual equilibrium mixture still depends on incoming and outgoing ligand concentrations through the reaction quotient.

Do not rank equilibrium concentrations from Kstab alone when ligand concentrations differ greatly. Kstab compares intrinsic equilibrium tendency under its definition; composition also reflects the amounts and activities present.