Define stability constants and write Kstab expressions
Represent complex-ion formation as an equilibrium and construct its stability-constant expression with water omitted.
Section 28 · Chemistry of transition elements
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
What does a large Kstab indicate for the stated formation equilibrium?
What you will learn
- Define Kstab for complex formation from constituent species
- Write Kstab expressions using product-over-reactant concentrations
- Omit water from expressions when it is the solvent
Kstab describes complex formation in solution
The stability constant Kstab is the equilibrium constant for forming a complex ion in a solvent from its constituent ions or molecules. A large Kstab indicates that the equilibrium favours the complex under the stated conditions.
For [M(H₂O)₆]²⁺ + 4NH₃ ⇌ [M(NH₃)₄(H₂O)₂]²⁺ + 4H₂O, water is the solvent and is omitted from the expression: Kstab = [[M(NH₃)₄(H₂O)₂]²⁺] / ([M(H₂O)₆]²⁺[NH₃]⁴). Use brackets to distinguish a complex species from multiplication.
Match exponents to balanced stoichiometry
For a general formation equilibrium M²⁺ + 2L ⇌ ML₂²⁺, Kstab = [ML₂²⁺]/([M²⁺][L]²). The power on each concentration comes from its coefficient in the balanced equation.
Use equilibrium concentrations. Include charges in the species labels, but charges are not exponents. Follow the units or concentration convention given in the question; do not include liquid water as a variable term.
Pause and explain
For M²⁺ + 2L ⇌ ML₂²⁺, what power is used for [L]?
Worked example
Write the Kstab expression for M²⁺ + 2L ⇌ ML₂²⁺.
Show the worked solution
- Put the complex product in the numerator.
- Put free metal and free ligand in the denominator.
- Use the coefficient 2 as the ligand exponent.
Answer Kstab = [ML₂²⁺] / ([M²⁺][L]²).
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Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Write Kstab for [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O.
Give me a hint
Omit water as the solvent and use equilibrium concentrations.
Compare my reasoning
- Place the ammine complex in the numerator.
- Place the aqua complex and free NH₃ in the denominator.
- Raise [NH₃] to the fourth power.
Kstab = [[Cu(NH₃)₄(H₂O)₂]²⁺] / ([Cu(H₂O)₆]²⁺[NH₃]⁴).
Look for these in your work
- Product over reactants
- Water omitted
- NH₃ exponent correct
For M²⁺ + 2L ⇌ ML₂²⁺, the equilibrium concentrations are [ML₂²⁺] = 0.010 mol dm⁻³, [M²⁺] = 0.020 mol dm⁻³ and [L] = 0.10 mol dm⁻³. Calculate Kstab using the stated concentration expression.
Give me a hint
Substitute into the product-over-reactant expression.
Compare my reasoning
- Kstab = 0.010/(0.020 × 0.10²).
- Evaluate the denominator.
- Report the numerical value under the given convention.
Kstab = 50 (using the supplied concentration convention).
Look for these in your work
- Expression set up
- Ligand exponent used
- Numerical value correct
A student writes Kstab = [M²⁺][L]²/[ML₂²⁺] and includes [H₂O]⁴ in the denominator. Identify both corrections.
Give me a hint
Check expression direction and identify water as solvent.
Compare my reasoning
- Formation constant has product over reactants.
- Water is omitted as solvent.
- Keep [L]² from the balanced coefficient.
Use Kstab = [ML₂²⁺]/([M²⁺][L]²). Omit solvent water; retain the ligand exponent of two.
Look for these in your work
- Ratio reversed corrected
- Water omitted
- Stoichiometric exponent preserved
Common mistakes
- Including solvent water in Kstab
- Using initial instead of equilibrium concentrations
- Forgetting a ligand coefficient in the exponent
- Putting reactants over products
What can you explain now?
What does a large Kstab indicate for the stated formation equilibrium?
Compare with the explanation
The equilibrium favours the complex ion
A larger product-to-reactant ratio corresponds to greater complex formation at equilibrium.
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Make it stick
How do the balanced formation equation and solvent convention determine a Kstab expression?
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.