A LevelChemistryCIERevision Notes28. Chemistry of Transition Elements28.5 Stability Constants, KstabStability Constant, Kstab

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Stability Constant, Kstab (CIE A Level Chemistry)

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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, Kstab
  • 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 Kstab

  • The expression for Kstab can be deduced in a similar way as the expression for the equilibrium constant (Kc)
  • 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 

Kstab = Error converting from MathML to accessible text.

  • 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 Kstab can be deduced from the expression in a similar way to the units of Kc
  • 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 Kstab value, the more stable the complex formed is

Calculations Involving Stability Constants

  • If the concentrations of the transition element complex and the reacting ligands are known, the expression for the stability constant (Kstab) can be used to determine which complex is more stable
  • The greater the value of Kstab 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) ⇌{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true} 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 Kstab for this step. Include the units for Kstab.
  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 Kstab expression for the reaction:
    • Kstab = CuH2O4Cl2 aqCuH2O5Cl+ aq Cl- aq{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}
  • Step 3: Substitute the equilibrium concentrations into the Kstab expression and evaluate:
    • Kstab = [0.10][0.05] [0.05]{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}
    • Kstab = 40
  • Step 4: Determine the units:
    • Kstab = [mol dm-3][mol dm-3] [mol dm-3]{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}
    • Kstab = dm3 mol-1 

Answer 2:

  • The value of Kstab 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

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    Richard

    Author: Richard

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.