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9701 · 28.5

Stability constants, Kstab — practice questions

Practice and worked examples for 9701 Stability constants, Kstab. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A solution at equilibrium was found to contain the complex ion [Ag(NH₃)₂]⁺ at a concentration of 0.095 mol dm⁻³. The concentrations of the constituent ions were [Ag⁺] = 5.6 × 10⁻⁹ mol dm⁻³ and [NH₃] = 0.12 mol dm⁻³. Calculate the stability constant, Kstab, for [Ag(NH₃)₂]⁺ and state its units.

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Step 1: Write the balanced equation for the formation of the complex ion. Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq)

Step 2: Write the expression for Kstab. Kstab=[[Ag(NH3)2]+][Ag+][NH3]2K_{stab} = \frac{[[Ag(NH_3)_2]^+]}{[Ag^+][NH_3]^2}

Step 3: Substitute the given equilibrium concentrations into the expression. Kstab=(0.095)(5.6×10−9)×(0.12)2K_{stab} = \frac{(0.095)}{(5.6 \times 10^{-9}) \times (0.12)^2} Kstab=(0.095)(5.6×10−9)×(0.0144)K_{stab} = \frac{(0.095)}{(5.6 \times 10^{-9}) \times (0.0144)} Kstab=(0.095)8.064×10−11K_{stab} = \frac{(0.095)}{8.064 \times 10^{-11}} Kstab=1.18×109K_{stab} = 1.18 \times 10^9 (to 3 s.f.)

Step 4: Determine the units. Units = mol dm−3(mol dm−3)(mol dm−3)2=1(mol dm−3)2=mol−2 dm6\frac{mol \ dm^{-3}}{(mol \ dm^{-3})(mol \ dm^{-3})^2} = \frac{1}{(mol \ dm^{-3})^2} = mol^{-2} \ dm^6

Final Answer: Kstab=1.18×109 mol−2 dm6K_{stab} = 1.18 \times 10^9 \ mol^{-2} \ dm^6

Worked example 2

Consider the following stability constants:

  1. For [Co(NH₃)₆]³⁺: Kstab=4.5×1033 mol−6 dm18K_{stab} = 4.5 \times 10^{33} \ mol^{-6} \ dm^{18}
  2. For [Co(en)₃]³⁺: Kstab=4.0×1048 mol−3 dm9K_{stab} = 4.0 \times 10^{48} \ mol^{-3} \ dm^{9} (en = ethylenediamine, a bidentate ligand)

If a solution of [Co(NH₃)₆]³⁺ has ethylenediamine added to it, predict the outcome of the reaction and justify your answer.

Show solution outline

Step 1: Compare the stability constants for the two complexes. Kstab([Co(en)3]3+)=4.0×1048K_{stab}([Co(en)_3]^{3+}) = 4.0 \times 10^{48} Kstab([Co(NH3)6]3+)=4.5×1033K_{stab}([Co(NH_3)_6]^{3+}) = 4.5 \times 10^{33}

Step 2: Analyse the comparison. The stability constant for the [Co(en)₃]³⁺ complex is approximately 10¹⁵ times larger than that for the [Co(NH₃)₆]³⁺ complex. This indicates that the [Co(en)₃]³⁺ complex is vastly more stable.

Step 3: Predict the outcome. When ethylenediamine is added to the solution of [Co(NH₃)₆]³⁺, a ligand substitution reaction will occur. The equilibrium will shift strongly to the right to favour the formation of the much more stable [Co(en)₃]³⁺ complex. The ammonia ligands will be displaced by the ethylenediamine ligands.

Justification: The reaction favours the formation of the more stable product. Since Kstab([Co(en)3]3+)≫Kstab([Co(NH3)6]3+)K_{stab}([Co(en)_3]^{3+}) \gg K_{stab}([Co(NH_3)_6]^{3+}), the complex with the bidentate ligand 'en' is thermodynamically much more stable, driven by the chelate effect (a large positive entropy change). The reaction is: [Co(NH3)6]3+(aq)+3en(aq)→[Co(en)3]3+(aq)+6NH3(aq)[Co(NH_3)_6]^{3+}(aq) + 3en(aq) \rightarrow [Co(en)_3]^{3+}(aq) + 6NH_3(aq)

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