Stability Constant, Kstab (Cambridge International A-Level Chemistry)

Stability Constant Expressions

Define & Write a Stability Constant for a Complex

• When transition element ions are in aqueous solutions, they will automatically become hydrated

  • Water molecules will surround the ion and act as ligands by forming dative covalent bonds to the central metal ion

• When there are other potential ligands present in the solution, there is a competing equilibrium in ligand exchange and the most stable complex will be formed

• For example, a Co(II) ion in solution will form a [Co(H₂O)₆]²⁺ complex

• Adding ammonia results in the stepwise substitution of the water ligands by ammonia ligands until a stable complex of [Co(NH₃)₄(H₂O)₂]²⁺ is formed

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

• For the substitution reaction above, there are four stepwise constants:

[Cu(H₂O)₆]²⁺+ NH₃⇌ [Cu(NH₃)(H₂O)₅]²⁺+ H₂O K₁

[Cu(NH₃)(H₂O)₅]²⁺+ NH₃⇌ [Cu(NH₃)₂(H₂O)₄]²⁺+ H₂O K₂

[Cu(NH₃)₂(H₂O)₄]²⁺+ NH₃⇌ [Cu(NH₃)₃(H₂O)₃]²⁺+ H₂O K₃

[Cu(NH₃)₃(H₂O)₃]²⁺+ NH₃⇌ [Cu(NH₃)₄(H₂O)₂]²⁺+ H₂O K₄

• These stepwise constants are summarised in the overall stability constant, K_stab
• The stability constant is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules

Expression of K_stab

• The expression for K_stab can be deduced in a similar way as the expression for the equilibrium constant (K_c)
• For example, the equilibrium expression for the substitution of water ligands by ammonia ligands in the Co(II) complex is:

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

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• The concentration of water is not included in the expression as the water is in excess
• Therefore, any water produced in the reaction is negligible compared to the water that is already present
• The units of the K_stab can be deduced from the expression in a similar way to the units of K_c
• The stability constants can be used to compare the stability of ligands relative to the aqueous metal ion where the ligand is water
• The larger the K_stab value, the more stable the complex formed is

Stability Constant Calculations

Calculations Involving Stability Constants

• If the concentrations of the transition element complex and the reacting ligands are known, the expression for the stability constant (K_stab) can be used to determine which complex is more stable
• The greater the value of K_stab the more stable the complex is

Worked Example The addition of concentrated hydrochloric acid to copper(II) sulfate solution forms an aqueous solution containing [CuCl₄]^2– and [Cu(H₂O)₆]²⁺ complex ions. The overall ligand exchange involved is a series of stepwise reactions as successive ligands are replaced.

The second step in exchanging water ligands with chloride ligands is:

[Cu(H₂O)₅Cl]⁺ (aq) + Cl^– (aq) ⇌ Cu(H₂O)₄Cl₂ (aq) + H₂O (l)

When a 0.15 mol dm^–3 solution of [Cu(H₂O)₅Cl]⁺ (aq) is mixed with 0.15 mol dm^–3 hydrochloric acid, the equilibrium mixture of Cu(H₂O)₄Cl₂ (aq) was found to be 0.10 mol dm^–3.

1. Use this data to calculate K_stab for this step. Include the units for K_stab.
2. Use your answer to (1) to suggest the position of the equilibrium for this step. Explain your answer.

Answer 1

• **Step 1: **Calculate the equilibrium concentration of each ion:

	[Cu(H2O)5Cl]+(aq)	Cl–(aq)	Cu(H2O)4Cl2(aq)
Initial concentration / mol dm-3	0.15	0.15	0
Change in concentration	- 0.1.0	-0.10	+ 0.10
Equilibrium concentration / mol dm-3	0.05	0.05	0.10

• Step 2: Write the K_stab expression for the reaction:

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• Step 3: Substitute the equilibrium concentrations into the K_stab expression and evaluate:

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  • K_stab = 40

• Step 4: Determine the units:

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  • K_stab = dm³ mol^-1

Answer 2:

• The value of K_stab is 40 dm³ mol^-1

• This is a large value, which suggests:

  • The products are favoured
  • Therefore, the position of the equilibrium for this step is to the right / products