What is a stability constant (Kstab) in the context of transition elements?

A stability constant, often denoted as Kstab, is a quantitative measure of the stability of a complex ion in solution. In the context of transition elements, it refers to the equilibrium constant for the formation of a complex from a metal ion and its ligands. A higher Kstab value indicates a more stable complex, which is a key concept in A-Level Inorganic Chemistry, especially when studying the chemistry of transition elements.

How is the stability constant (Kstab) calculated for a complex ion?

The stability constant (Kstab) is calculated using the equilibrium expression for the formation of a complex ion. For a general reaction where a metal ion M reacts with ligands L to form a complex ML, the expression is Kstab = [ML]/([M][L]^n), where [ML], [M], and [L] are the equilibrium concentrations of the complex, metal ion, and ligands, respectively. This calculation is crucial for understanding the behavior of transition metal complexes in A-Level Chemistry.

Why are stability constants important in the study of transition elements in A-Level Chemistry?

Stability constants are important because they provide insight into the strength and stability of metal-ligand complexes, which are prevalent in the chemistry of transition elements. Understanding Kstab helps predict the formation and behavior of these complexes in various chemical reactions and environments, which is essential knowledge for Cambridge International A Levels students studying Inorganic Chemistry.

What factors affect the stability constant (Kstab) of a complex ion?

Several factors affect the stability constant (Kstab) of a complex ion, including the nature of the metal ion, the type and number of ligands, the charge and size of the metal ion, and the overall geometry of the complex. Additionally, the solvent and temperature can also influence Kstab. These factors are crucial for students to understand when analyzing the chemistry of transition elements in A-Level Chemistry.

How does the stability constant (Kstab) relate to the selectivity of transition metal complexes?

The stability constant (Kstab) is directly related to the selectivity of transition metal complexes. A higher Kstab value indicates a more stable and selective complex formation with specific ligands. This selectivity is important in various applications, such as catalysis and separation processes, which are often explored in A-Level Inorganic Chemistry. Understanding this relationship helps students grasp how transition metals can be used effectively in chemical reactions and industrial applications.

Why is "Including water in the Kstab expression." a common mistake in Stability constants, Kstab?

Why it happens: Treating it like any other species. How to avoid it: Water is the solvent — leave it out of Kstab (and Ka, etc.). Source: 9701 Examiner Reports 2022-2024

Why is "Forgetting the power on [ligand]." a common mistake in Stability constants, Kstab?

Why it happens: Ignoring the number of ligands added. How to avoid it: Raise [ligand] to the number added (e.g. [NH₃]⁴).

Why is "Predicting exchange toward the lower-Kstab complex." a common mistake in Stability constants, Kstab?

Why it happens: Confusing the direction. How to avoid it: Substitution goes toward the HIGHER Kstab (more stable) complex.

Why is "Omitting units (or assuming none) for Kstab." a common mistake in Stability constants, Kstab?

Why it happens: Forgetting they depend on the equation. How to avoid it: Work the units out from the expression (often (mol dm⁻³) to a negative power).

Chemistry of transition elements(A-Level Inorganic Chemistry)Cambridge International A Level Chemistry 9701

Stability constants, Kstab

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Stability Constants, Kstab — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Writing the Kstab expression for complex formation, interpreting its magnitude, predicting ligand-exchange reactions, and the chelate effect.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

28.5 — Define the stability constant Kstab of a complex ion.
28.5 — Write the expression for, and deduce the units of, a stability constant.
28.5 — Use Kstab values to predict ligand-substitution (exchange) reactions.

What Kstab measures and how to write it

The equilibrium constant for replacing water ligands; larger Kstab = more stable complex.

When a ligand is added to an aqueous metal ion, it replaces the water ligands in an equilibrium. The stability constant (Kstab) is the equilibrium constant for this complex formation:

$[\text{Cu(H}_2\text{O)}_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}$

$K_{stab} = \frac{[[\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu(H}_2\text{O)}_6]^{2+}][\text{NH}_3]^4}$

As with Ka and Kw, water (the solvent) is omitted. A large Kstab means the equilibrium lies far to the right, i.e. a very stable complex.

The units depend on the equation: here the four [NH₃] in the denominator leave units of mol⁻⁴ dm¹².

Kstab = equilibrium constant for complex formation.
Products over reactants; omit water.
Larger Kstab = more stable complex.

See the full worked example for stability constants, kstab →

Predicting ligand exchange

A ligand whose complex has a higher Kstab displaces one with a lower Kstab.

Comparing Kstab values predicts ligand-substitution reactions: the metal ends up in the complex with the higher Kstab (the more stable one).

Worked. [Cu(NH₃)₄(H₂O)₂]²⁺ has a much larger Kstab than [Cu(H₂O)₆]²⁺. So adding ammonia to aqueous copper(II) displaces water: $[\text{Cu(H}_2\text{O)}_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}$ and the pale-blue solution turns deep blue.

A ligand with a still larger Kstab will in turn displace ammonia — for instance EDTA (lg Kstab ≈ 18) easily displaces ammonia (lg Kstab ≈ 8). Comparing the lg Kstab values shows the EDTA complex is about 10¹⁰ times more stable.

Exchange goes toward the higher-Kstab complex.
NH₃ displaces water (deep blue Cu complex).
EDTA (very high Kstab) displaces NH₃.

See the full worked example for stability constants, kstab →

The chelate effect

Multidentate ligands give especially stable complexes because they increase entropy.

Complexes of multidentate ligands (e.g. EDTA, a hexadentate ligand, or bidentate 'en') are unusually stable — they have very high Kstab. This is the chelate effect.

The reason is largely entropy. When one EDTA molecule replaces, say, six water (or other monodentate) ligands: $[\text{M(H}_2\text{O)}_6]^{n+} + \text{EDTA}^{4-} \rightarrow [\text{M(EDTA)}]^{(n-4)+} + 6\text{H}_2\text{O}$ the number of free particles increases (one EDTA in → six water molecules out), giving a large positive ΔS. With ΔH roughly similar (similar bonds made and broken), the positive entropy term makes ΔG more negative and Kstab very large.

Multidentate ligands → very stable complexes (high Kstab).
Driven by a positive entropy change (more free particles).
EDTA displaces monodentate ligands like NH₃/H₂O.

See the full worked example for stability constants, kstab →

How it’s examined

Paper 4 asks you to write a Kstab expression (and its units), and to use Kstab values to predict and explain a ligand-substitution reaction (often with a colour change). Examiner reports flag including water in the expression, omitting the ligand power, and predicting exchange toward the lower Kstab.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 28.5; 9701 Examiner Reports 2022-2024; 9701/41 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-20.

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Step-by-step worked examples — Stability constants, Kstab

Step-by-step solutions to past-paper-style questions on stability constants, kstab, written exactly the way a tutor would explain them at the board.

1What Kstab measures (2 marks)
Getting started• definition
Question
State what the stability constant Kstab measures and what a large value indicates.
Step-by-step solution
1. Step 1
  Kstab is the equilibrium constant for forming a complex ion from the aqueous metal ion (with its water ligands) and the new ligands.
2. Step 2
  A large Kstab means the complex is very stable (the equilibrium lies far to the product side).
Answer
The equilibrium constant for complex formation; a large Kstab = a more stable complex.
Examiner tip
Kstab is a ligand-exchange equilibrium constant — water ligands are replaced.
2Writing a Kstab expression (3 marks)
Getting started• expression
Question
Write the Kstab expression for $[\text{Cu(H}_2\text{O)}_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}$ .
Step-by-step solution
1. Step 1
  Products over reactants; water (the solvent) is omitted.
  $K_{stab} = \frac{[[\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu(H}_2\text{O)}_6]^{2+}][\text{NH}_3]^4}$
Answer
$K_{stab} = \dfrac{[[\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu(H}_2\text{O)}_6]^{2+}][\text{NH}_3]^4}$ .
Examiner tip
Leave water (the solvent) out of the expression; raise [NH₃] to the power 4.
3Units of Kstab (2 marks)
Building confidence• units
Question
Deduce the units of Kstab for the formation above (4 NH₃ added).
Step-by-step solution
1. Step 1
  Concentrations cancel except the four [NH₃] in the denominator.
  $K_{stab}: \frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^4}$
2. Step 2
  Net (mol dm⁻³)⁻⁴.
  $\text{units} = \text{mol}^{-4}\,\text{dm}^{12}$
Answer
mol⁻⁴ dm¹² (= (mol dm⁻³)⁻⁴).
4Predicting ligand exchange (4 marks)
Building confidence• ligand exchange
Question
The Kstab of $[\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ is much larger than that of $[\text{Cu(H}_2\text{O)}_6]^{2+}$ . Predict what happens when ammonia is added to aqueous copper(II) sulfate, and explain.
Step-by-step solution
1. Step 1
  A larger Kstab means the ammonia complex is more stable than the aqua complex.
2. Step 2
  So ammonia displaces water ligands (ligand substitution).
  $[\text{Cu(H}_2\text{O)}_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}$
3. Step 3
  Observation: the pale blue solution turns deep blue.
Answer
NH₃ (higher Kstab complex) displaces water → deep-blue [Cu(NH₃)₄(H₂O)₂]²⁺.
Examiner tip
Ligand exchange proceeds toward the complex with the HIGHER stability constant.
5Comparing two ligands by Kstab (4 marks)
Stretch• comparison
Question
For a metal ion, the EDTA complex has lg Kstab ≈ 18 and the ammonia complex lg Kstab ≈ 8. Predict and explain what happens if EDTA is added to the ammonia complex.
Step-by-step solution
1. Step 1
  EDTA's complex has a far larger Kstab (lg 18 vs 8 → ~10¹⁰ times more stable).
2. Step 2
  EDTA displaces the ammonia ligands, forming the much more stable EDTA complex.
3. Step 3
  Driving force: EDTA is multidentate, so it releases many small ligand molecules (a favourable entropy increase) and gives a very high Kstab.
Answer
EDTA displaces NH₃ (much higher Kstab); the multidentate ligand gives a very stable complex (entropy-favoured).
Examiner tip
A multidentate ligand like EDTA replaces several monodentate ligands, increasing the number of free particles — a positive entropy change that boosts Kstab (the 'chelate effect').
6Using Kstab to compare complexes (3 marks)
Stretch• comparison
Question
Two complexes of the same metal ion have Kstab = 1.0 × 10⁴ and 1.0 × 10¹³. State which ligand binds more strongly and explain how you could test it.
Step-by-step solution
1. Step 1
  The complex with Kstab = 10¹³ is far more stable, so that ligand binds more strongly.
2. Step 2
  Adding the stronger ligand to the weaker complex should cause ligand exchange (a colour change), confirming the order.
Answer
The Kstab = 10¹³ ligand binds more strongly; adding it to the weaker complex causes substitution (colour change).
Examiner tip
Higher Kstab = stronger binding = displaces the lower-Kstab ligand.

Key Formulae — Stability constants, Kstab

The formulae you need to memorise for stability constants, kstab on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Stability constant
$K_{stab} = \frac{[\text{complex}]}{[\text{aqua ion}][\text{ligand}]^n}$
$[complex]$
concentration of the ligand-substituted complex
$n$
number of ligand molecules added (the power)
When to use
Quantifying complex stability; omit water (solvent). Larger Kstab = more stable complex.

Key Definitions and Keywords — Stability constants, Kstab

Definitions to memorise and the exact keywords mark schemes credit for stability constants, kstab answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Stability constant (Kstab)
Examiner keyword
The equilibrium constant for the formation of a complex ion from its aqueous metal ion and ligands (a ligand-exchange equilibrium).
Ligand exchange (substitution)
Examiner keyword
Replacement of one ligand by another; it proceeds toward the complex with the larger Kstab.
Chelate effect
Examiner keyword
Complexes of multidentate ligands (e.g. EDTA) are especially stable (high Kstab) because their formation increases entropy (more free particles).

Common Mistakes and Misconceptions — Stability constants, Kstab

The traps other students keep falling into on stability constants, kstab questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Including water in the Kstab expression.
9701 Examiner Reports 2022-2024
Why it happens
Treating it like any other species.
How to avoid it
Water is the solvent — leave it out of Kstab (and Ka, etc.).
✕Forgetting the power on [ligand].
Why it happens
Ignoring the number of ligands added.
How to avoid it
Raise [ligand] to the number added (e.g. [NH₃]⁴).
✕Predicting exchange toward the lower-Kstab complex.
Why it happens
Confusing the direction.
How to avoid it
Substitution goes toward the HIGHER Kstab (more stable) complex.
✕Omitting units (or assuming none) for Kstab.
Why it happens
Forgetting they depend on the equation.
How to avoid it
Work the units out from the expression (often (mol dm⁻³) to a negative power).

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Stability constants, Kstab — frequently asked questions

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Stability constants, Kstab

Short Study Notes in the form of Slides

Short Study Notes — Stability constants, Kstab

Detailed Notes

What you’ll learn

How it’s examined

1What Kstab measures (2 marks)

2Writing a Kstab expression (3 marks)

3Units of Kstab (2 marks)

4Predicting ligand exchange (4 marks)

5Comparing two ligands by Kstab (4 marks)

6Using Kstab to compare complexes (3 marks)

Stability constant

Stability constant (Kstab)

Ligand exchange (substitution)

Chelate effect

✕Including water in the Kstab expression.

✕Forgetting the power on [ligand].

✕Predicting exchange toward the lower-Kstab complex.

✕Omitting units (or assuming none) for Kstab.

Practice questions

Past paper style quiz

Assess your understanding

Stability constants, Kstab — frequently asked questions