Calculate and interpret complex stability constants
Calculate Kstab or an unknown equilibrium concentration and interpret the result without confusing stability with reaction rate.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 28. Use the first-row transition elements titanium to copper and the species explicitly named in the syllabus. Where redox potentials or stability constants are needed, use the values supplied in the question. Colour descriptions can depend on conditions and concentration. Written practice is self-checked
Before you begin
- Use electron configurations, oxidation numbers, ionic equations, equilibrium expressions and the AS redox and bonding models.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which ligand concentration belongs in a Kstab calculation?
What you will learn
- Substitute equilibrium concentrations into Kstab
- Rearrange a Kstab expression to find an unknown concentration
- Interpret a stability constant in the context of the stated complex and conditions
Use equilibrium concentrations in a calculation
Kstab is calculated from the equilibrium expression using equilibrium concentrations. For M²⁺ + 2L ⇌ ML₂²⁺, Kstab = [ML₂²⁺]/([M²⁺][L]²). Convert units consistently and use any convention supplied by the question.
A common error is to use the amount of ligand added instead of the free equilibrium ligand concentration. If the question gives equilibrium concentrations, substitute those values directly.
Interpret magnitude with the same reaction written
For a fixed formation reaction, a larger Kstab indicates a larger product-to-reactant ratio at equilibrium and a more stable complex relative to its constituent species under the stated conditions.
Kstab does not give how quickly equilibrium is reached. Nor should constants for differently written reactions be compared without adjusting for the equation and conditions.
Pause and explain
Does a large Kstab imply a fast complex-formation rate?
Worked example
For M²⁺ + 2L ⇌ ML₂²⁺, Kstab = 50, [M²⁺] = 0.020 mol dm⁻³ and [L] = 0.10 mol dm⁻³. Find [ML₂²⁺].
Show the worked solution
- Rearrange [ML₂²⁺] = Kstab[M²⁺][L]².
- Substitute all equilibrium values.
- Check units and magnitude.
Answer [ML₂²⁺] = 50 × 0.020 × 0.10² = 0.010 mol dm⁻³.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For M²⁺ + 2L ⇌ ML₂²⁺, calculate Kstab when [ML₂²⁺] = 0.0060, [M²⁺] = 0.030 and [L] = 0.20 mol dm⁻³.
Give me a hint
Use product concentration divided by metal and squared free-ligand concentrations.
Compare my reasoning
- Calculate [L]² = 0.040.
- Calculate the denominator 0.030 × 0.040.
- Divide 0.0060 by that denominator.
Kstab = 5.0 using the stated concentration convention.
Look for these in your work
- Ligand exponent applied
- Denominator calculated
- Value reported
A complex has Kstab = 8.0 × 10⁵ for a stated formation equation. What does this say qualitatively about the equilibrium?
Give me a hint
Interpret products relative to reactants for that exact equation.
Compare my reasoning
- The product-to-reactant ratio is large at equilibrium under the stated convention.
- Formation is favoured relative to the constituent species.
- No conclusion about reaction speed follows.
The stated equilibrium strongly favours the complex relative to its constituents; the value does not by itself specify the rate of formation.
Look for these in your work
- Equilibrium interpretation correct
- Equation scope retained
- Rate distinction stated
Two sources quote Kstab values for [M(L)₂] and [M(L)₃] formations. Why should you not compare the numbers without checking the equations?
Give me a hint
The stoichiometry and concentration powers differ between formation reactions.
Compare my reasoning
- Write each balanced formation equation.
- Check the ligand coefficients and K expressions.
- Compare only under a consistent convention and stated conditions.
The constants refer to different reactions with different ligand stoichiometries and expressions. Check equations, conditions and concentration conventions before making a stability comparison.
Look for these in your work
- Reaction equations checked
- Stoichiometric differences recognised
- Comparison qualified
Common mistakes
- Using the amount of ligand initially added as free equilibrium [L]
- Forgetting the ligand exponent
- Calling a large Kstab a fast reaction
- Comparing constants for differently balanced formation reactions directly
What can you explain now?
Which ligand concentration belongs in a Kstab calculation?
Compare with the explanation
The equilibrium concentration of free ligand
The expression describes equilibrium composition.
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Make it stick
What does Kstab tell you about equilibrium composition, and what does it not tell you about rate?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.