🧪 Stability Constants (\(K_{stab}\)): Measuring the Strength of Chemical Hugs

Welcome to one of the most exciting topics in Transition Element Chemistry! This chapter, Stability Constants (\(K_{stab}\)), links your knowledge of equilibrium (Topic 7) with the unique chemistry of complex ions (Topic 28).


Why is this important? Transition metals form beautiful, colourful complexes, and \(K_{stab}\) is the tool we use to figure out exactly how stable these complexes are—essential knowledge for industries like drug design, dyeing, and chemical analysis!


1. Understanding Complex Ions: A Quick Review

What are Complex Ions?

A complex ion is formed when a central metal ion (usually a transition metal) is surrounded by species called ligands. These ligands "bond" to the metal using a special type of covalent bond.

  • Central Metal Ion: An ion (like \(\text{Cu}^{2+}\) or \(\text{Fe}^{3+}\)) that acts as a Lewis acid (electron pair acceptor).
  • Ligand: A molecule or ion (like \(\text{H}_2\text{O}\), \(\text{NH}_3\), or \(\text{Cl}^{-}\)) that has at least one lone pair of electrons. It acts as a Lewis base (electron pair donor).
  • Coordinate Bond (Dative Covalent Bond): The bond formed when the ligand donates both electrons to the central metal ion.

Think of it like this: The central metal ion is the host, and the ligands are the guests. The stronger the relationship between the host and the guests, the more stable the complex.


2. Defining the Stability Constant, \(K_{stab}\)

Definition (LO 28.5.1)

The stability constant (\(K_{stab}\)) is the equilibrium constant for the formation of a complex ion in a solvent, starting from its constituent metal ion and ligands.


Writing the \(K_{stab}\) Expression (LO 28.5.2)

When a complex ion forms, it is an equilibrium process. For example, when aqueous copper(II) ions react with ammonia ligands, they form the tetraamminecopper(II) ion:


\[ \text{Cu}^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4]^{2+}(\text{aq}) \]


The stability constant expression is written just like any other equilibrium constant (\(K_c\)): concentration of products divided by concentration of reactants, all raised to the power of their stoichiometric coefficients.


The \(K_{stab}\) Expression:

\[ K_{stab} = \frac{[ \text{Product} ]}{[ \text{Reactants} ]} = \frac{[ [\text{Cu}(\text{NH}_3)_4]^{2+} ]}{[ \text{Cu}^{2+} ] [\text{NH}_3]^4} \]


🔑 Key Rule: Excluding Water (LO 28.5.2)

The formation of a complex usually involves the replacement of solvent molecules (water). For example, the aqueous copper ion is actually \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\). The full reaction is:


\[ [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4]^{2+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \]


However, when writing the \(K_{stab}\) expression, we treat the solvent (\(\text{H}_2\text{O}\)) as having a constant concentration, similar to pure liquids in heterogeneous equilibrium.

Therefore, you must exclude \([\text{H}_2\text{O}]\) from the denominator (LO 28.5.2).


✅ Quick Review: Writing \(K_{stab}\)

For the complex formation: \(\text{M}^{n+} + y\text{L} \rightleftharpoons [\text{ML}_y]^{n+}\)

The expression is always: \[ K_{stab} = \frac{[ \text{Complex} ]}{[ \text{Metal Ion} ] [\text{Ligand}]^y} \]

Never include \([\text{H}_2\text{O}]\) in the expression.

3. Interpreting the Magnitude of \(K_{stab}\)

Stability and the Value of \(K_{stab}\) (LO 28.5.4)

Since \(K_{stab}\) is an equilibrium constant, its size tells us about the position of the equilibrium:

  • Large \(K_{stab}\) (e.g., \(10^8\) or higher): The equilibrium lies far to the right (products side). The complex ion is very stable and forms readily.
  • Small \(K_{stab}\) (e.g., \(10^2\) or lower): The equilibrium lies to the left (reactants side). The complex ion is relatively unstable and dissociates easily.

A large \(K_{stab}\) is due to the formation of a stable complex ion.


Analogy: Imagine two people (Metal and Ligand) meeting.
If \(K_{stab}\) is huge, they are inseparable—a highly stable, long-lasting complex!
If \(K_{stab}\) is small, they separate quickly—an unstable complex.

4. Ligand Exchange Reactions and \(K_{stab}\)

Predicting Exchange Feasibility (LO 28.5.4)

Ligand exchange (or substitution) is a competition between different ligands for the central metal ion. The reaction will favour the formation of the complex with the higher \(K_{stab}\) value, as this is the more thermodynamically stable product.


Step-by-Step Prediction:
  1. Identify the ligands in the reactants and products.
  2. Look up (or be given) the \(K_{stab}\) values for the complex formed by the reactant ligand and the complex formed by the product ligand.
  3. If \(K_{stab}(\text{Product Complex}) > K_{stab}(\text{Reactant Complex})\), the exchange reaction is feasible (likely to happen).

Example: \(\text{Copper(II)}\) ions react with \(\text{ammonia}\) to replace water ligands.

  • Complex 1 (Reactant): \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) (The natural aqueous ion)
  • Complex 2 (Product): \([\text{Cu}(\text{NH}_3)_4]^{2+}\)

Since the ammonia complex has a much larger \(K_{stab}\) than the aqua complex (which is often taken as the reference), the equilibrium strongly favours the formation of the ammonia complex.


The Chelate Effect (LO 28.5.4 Explanation)

One of the most powerful concepts related to stability constants is the Chelate Effect.

What is Chelation?

A chelating ligand is a ligand that can form two or more coordinate bonds with the central metal ion simultaneously. These are polydentate ligands (e.g., bidentate like 1,2-diaminoethane (en) or hexadentate like EDTA$^{4-}$).

Complexes formed with chelating ligands are dramatically more stable than those formed with simple monodentate ligands (like \(\text{H}_2\text{O}\) or \(\text{NH}_3\)). This enhanced stability is the chelate effect.

Why are Chelates so Stable? (The Entropy Factor)

The stability increase is primarily due to a favourable entropy change (\(\Delta S\)). Remember, a reaction is feasible if the Gibbs Free Energy change (\(\Delta G\)) is negative. \(\Delta G = \Delta H - T\Delta S\).

Consider the exchange where 3 bidentate 'en' ligands replace 6 monodentate water ligands:

\[ [\text{Ni}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 3\text{en}(\text{aq}) \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \]

  1. Reactant Side: 1 complex ion + 3 ligand molecules = 4 particles.
  2. Product Side: 1 complex ion + 6 water molecules = 7 particles.

The reaction goes from 4 particles to 7 particles (4 moles to 7 moles in solution). This massive increase in the number of particles leads to a huge increase in disorder (entropy), making \(\Delta S\) large and positive.

Since \(\Delta G = \Delta H - T\Delta S\), a large positive \(\Delta S\) makes \(\Delta G\) more negative, resulting in a much more feasible reaction and therefore a very large \(K_{stab}\).


⚠️ Common Mistake Alert

Do NOT confuse \(K_{stab}\) with solubility product, \(K_{sp}\). They both deal with equilibrium, but \(K_{sp}\) deals with solids dissolving, while \(K_{stab}\) deals with soluble complex ions forming in solution.

5. Using \(K_{stab}\) Expressions in Calculations (LO 28.5.3)

You may be required to use the \(K_{stab}\) expression to calculate concentrations of ions or the value of \(K_{stab}\), given the equilibrium concentrations.

Step-by-Step Calculation Guide

Let's use the reaction: \(\text{Ag}^{+}(\text{aq}) + 2\text{NH}_3(\text{aq}) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^{+}(\text{aq})\).
The \(K_{stab}\) expression is: \[ K_{stab} = \frac{[ [\text{Ag}(\text{NH}_3)_2]^{+} ]}{[ \text{Ag}^{+} ] [\text{NH}_3]^2} \]


Scenario 1: Calculating \(K_{stab}\) (Given equilibrium concentrations)

If at equilibrium, the concentrations are: \([\text{Ag}^{+}] = 1.0 \times 10^{-5} \text{mol dm}^{-3}\), \([\text{NH}_3] = 0.10 \text{mol dm}^{-3}\), and \([[ \text{Ag}(\text{NH}_3)_2]^{+}] = 0.50 \text{mol dm}^{-3}\).

1. Substitute the values: \[ K_{stab} = \frac{0.50}{(1.0 \times 10^{-5})(0.10)^2} \]

2. Calculate: \[ K_{stab} = \frac{0.50}{(1.0 \times 10^{-5})(0.01)} = \frac{0.50}{1.0 \times 10^{-7}} = 5.0 \times 10^6 \text{ mol}^{-2}\text{ dm}^{6} \]

(Note the units: \(\text{mol dm}^{-3}\) in the numerator cancels one unit in the denominator, leaving the units \(\text{mol}^{-2}\text{ dm}^{6}\)).


Scenario 2: Finding an Unknown Concentration (Given \(K_{stab}\))

If the \(K_{stab}\) is large (like \(5.0 \times 10^6\)), we often assume that almost all the metal ion reacts to form the complex. However, if the question asks for the concentration of the *unreacted* metal ion (\(\text{Ag}^{+}\)), we use the established concentrations and rearrange the \(K_{stab}\) formula:


\[ [ \text{Ag}^{+} ] = \frac{[ [\text{Ag}(\text{NH}_3)_2]^{+} ]}{K_{stab} [\text{NH}_3]^2} \]

A large \(K_{stab}\) in the denominator means the concentration of the unreacted metal ion (\([\text{Ag}^{+}]\)) is extremely small—confirming the complex is highly stable.


🔑 Key Takeaways on \(K_{stab}\)
  • \(K_{stab}\) measures the stability of a complex ion.
  • A large \(K_{stab}\) means a highly stable complex.
  • Ligand exchange reactions favor the complex with the higher \(K_{stab}\).
  • The chelate effect (polydentate ligands) leads to massive increases in \(K_{stab}\) due to a favourable entropy increase (more particles produced).