CIE AS/A Level Chemistry 28.5 Stability constants, $K_{\text {stab }}$ Study Notes- 2025-2027 Syllabus
CIE AS/A Level Chemistry 28.5 Stability constants, $K_{\text {stab }}$ Study Notes – New Syllabus
CIE AS/A Level Chemistry 28.5 Stability constants, $K_{\text {stab }}$ Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Chemistry latest syllabus with Candidates should be able to:
define the stability constant, Kstab, of a complex as the equilibrium constant for the formation of the complex ion in a solvent (from its constituent ions or molecules)
write an expression for a Kstab of a complex ([H₂O] should not be included)
use Kstab expressions to perform calculations
describe and explain ligand exchanges in terms of Kstab values and understand that a large Kstab is due to the formation of a stable complex ion
Stability Constant, \( \mathrm{K_{stab}} \)
The stability constant, \( \mathrm{K_{stab}} \), of a complex is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.
General Expression
For the formation of a complex:
\( \mathrm{M^{n+} + xL \rightleftharpoons [ML_x]^{n+}} \)
The stability constant is given by:
\( \mathrm{K_{stab} = \dfrac{[ML_x]^{n+}}{[M^{n+}][L]^x}} \)
A large value of \( \mathrm{K_{stab}} \) indicates that the complex ion is very stable in solution.
Meaning of \( \mathrm{K_{stab}} \)
- Shows the tendency of a metal ion to form a complex
- Allows comparison of stabilities of different complexes
- Larger \( \mathrm{K_{stab}} \) → more stable complex
Example
For the formation of the hexaamminecopper(II) complex:
\( \mathrm{Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}} \)
The stability constant expresses how far this equilibrium lies to the right.
Example
State what is meant by the stability constant of a complex.
▶️ Answer / Explanation
The stability constant is the equilibrium constant for the formation of a complex ion from its constituent ions or molecules in solution.
Example
Explain what information can be obtained from the value of \( \mathrm{K_{stab}} \).
▶️ Answer / Explanation
The value of \( \mathrm{K_{stab}} \) indicates how stable a complex ion is in solution.
A larger \( \mathrm{K_{stab}} \) means the equilibrium lies further towards complex formation, showing greater stability.
Expression for the Stability Constant, \( \mathrm{K_{stab}} \)
- The stability constant, \( \mathrm{K_{stab}} \), is written for the equilibrium in which a complex ion forms from a metal ion and its ligands in solution.
- Water molecules are not included in the expression because water is the solvent.
General Formation Equation
\( \mathrm{M^{n+} + xL \rightleftharpoons [ML_x]^{n+}} \)
Stability Constant Expression
\( \mathrm{K_{stab} = \dfrac{[ML_x]^{n+}}{[M^{n+}][L]^x}} \)
Worked Examples
1. For the complex \( \mathrm{[Cu(NH_3)_4]^{2+}} \):
\( \mathrm{Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}} \)
\( \mathrm{K_{stab} = \dfrac{[\,[Cu(NH_3)_4]^{2+}\,]}{[Cu^{2+}][NH_3]^4}} \)
2. For the complex \( \mathrm{[CoCl_4]^{2-}} \):
\( \mathrm{Co^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-}} \)
\( \mathrm{K_{stab} = \dfrac{[\,[CoCl_4]^{2-}\,]}{[Co^{2+}][Cl^-]^4}} \)
Example
Write an expression for \( \mathrm{K_{stab}} \) for the formation of \( \mathrm{[Ag(NH_3)_2]^+} \).
▶️ Answer / Explanation
\( \mathrm{K_{stab} = \dfrac{[\,[Ag(NH_3)_2]^+\,]}{[Ag^+][NH_3]^2}} \)
Example
Write an expression for \( \mathrm{K_{stab}} \) for the formation of \( \mathrm{[Ni(en)_3]^{2+}} \).
▶️ Answer / Explanation
\( \mathrm{Ni^{2+} + 3en \rightleftharpoons [Ni(en)_3]^{2+}} \)
\( \mathrm{K_{stab} = \dfrac{[\,[Ni(en)_3]^{2+}\,]}{[Ni^{2+}][en]^3}} \)
Calculations Using the Stability Constant, \( \mathrm{K_{stab}} \)
The stability constant, \( \mathrm{K_{stab}} \), can be used to calculate equilibrium concentrations of metal ions, ligands, and complex ions in solution.
General Approach
- Write the balanced complex formation equation
- Write the correct \( \mathrm{K_{stab}} \) expression (exclude water)
- Substitute known values into the expression
- Solve for the unknown concentration
General Expression
\( \mathrm{M^{n+} + xL \rightleftharpoons [ML_x]^{n+}} \)
\( \mathrm{K_{stab} = \dfrac{[ML_x]^{n+}}{[M^{n+}][L]^x}} \)
Example
The stability constant for the formation of \( \mathrm{[Ag(NH_3)_2]^+} \) is \( \mathrm{1.6 \times 10^7} \). Calculate the concentration of free \( \mathrm{Ag^+} \) ions in a solution containing 0.010 mol dm–3 \( \mathrm{[Ag(NH_3)_2]^+} \) and 1.0 mol dm–3 ammonia.
▶️ Answer / Explanation
Formation equation:
\( \mathrm{Ag^+ + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+} \)
Stability constant expression:
\( \mathrm{K_{stab} = \dfrac{[\,[Ag(NH_3)_2]^+\,]}{[Ag^+][NH_3]^2}} \)
Substitute values:
\( \mathrm{1.6 \times 10^7 = \dfrac{0.010}{[Ag^+](1.0)^2}} \)
Rearranging:
\( \mathrm{[Ag^+] = \dfrac{0.010}{1.6 \times 10^7}} \)
\( \mathrm{[Ag^+] = 6.25 \times 10^{-10}\ mol\,dm^{-3}} \)
Example
Nickel(II) ions form a complex with ethane-1,2-diamine:
\( \mathrm{Ni^{2+} + 3en \rightleftharpoons [Ni(en)_3]^{2+}} \)
The value of \( \mathrm{K_{stab}} \) is \( \mathrm{4.2 \times 10^{18}} \). A solution initially contains 0.020 mol dm–3 \( \mathrm{Ni^{2+}} \) and a large excess of ethane-1,2-diamine so its concentration remains effectively constant at 1.0 mol dm–3. Calculate the equilibrium concentration of free \( \mathrm{Ni^{2+}} \).
▶️ Answer / Explanation
Stability constant expression:
\( \mathrm{K_{stab} = \dfrac{[\,[Ni(en)_3]^{2+}\,]}{[Ni^{2+}][en]^3}} \)
Assume almost all nickel(II) ions are complexed:
\( \mathrm{[\,[Ni(en)_3]^{2+}\,] \approx 0.020\ mol\,dm^{-3}} \)
Substitute into the expression:
\( \mathrm{4.2 \times 10^{18} = \dfrac{0.020}{[Ni^{2+}](1.0)^3}} \)
Rearranging:
\( \mathrm{[Ni^{2+}] = \dfrac{0.020}{4.2 \times 10^{18}}} \)
\( \mathrm{[Ni^{2+}] = 4.8 \times 10^{-21}\ mol\,dm^{-3}} \)
This extremely small value shows that the complex is very stable.
Ligand Exchange Explained Using Stability Constants, \( \mathrm{K_{stab}} \)
Ligand exchange reactions can be explained and predicted using stability constant values, \( \mathrm{K_{stab}} \). These values show how strongly a metal ion binds to a particular ligand.
Meaning of \( \mathrm{K_{stab}} \) in Ligand Exchange
\( \mathrm{K_{stab}} \) measures the position of equilibrium for the formation of a complex ion.
Therefore:
- Large \( \mathrm{K_{stab}} \) → equilibrium lies far to the right
- Small \( \mathrm{K_{stab}} \) → little complex is formed
Ligand Exchange and Relative \( \mathrm{K_{stab}} \) Values
- Ligand exchange occurs when a ligand forms a more stable complex with the metal ion than the original ligand.
- This means, incoming ligand has a larger \( \mathrm{K_{stab}} \) than the ligand being replaced
Example: Copper(II) with Water and Ammonia
- In aqueous solution, copper(II) ions form:
\( \mathrm{[Cu(H_2O)_6]^{2+}} \)
Water forms a relatively weak complex with copper(II), so the \( \mathrm{K_{stab}} \) value is moderate.
- When ammonia is added, ligand exchange occurs:
\( \mathrm{[Cu(H_2O)_6]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O} \)
Ammonia forms a more stable complex with copper(II), so the \( \mathrm{K_{stab}} \) for the ammine complex is much larger.
As a result, the equilibrium lies strongly to the right and ammonia replaces water ligands.
Example: Cobalt(II) with Water and Chloride
- Cobalt(II) ions in water form:
\( \mathrm{[Co(H_2O)_6]^{2+}} \)
- At high chloride ion concentration, ligand exchange occurs:
\( \mathrm{[Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O} \)
Although chloride is a weaker ligand than water, a very high \( \mathrm{Cl^-} \) concentration shifts the equilibrium towards formation of the chloro complex.
Why a Large \( \mathrm{K_{stab}} \) Means a Stable Complex
- Strong metal–ligand bonding
- Lower energy of the complex compared with free ions
- Equilibrium strongly favours complex formation
A stable complex is therefore one that is unlikely to dissociate once formed.
Example
Explain why ammonia replaces water ligands in a copper(II) complex.
▶️ Answer / Explanation
Ammonia forms a complex with copper(II) that has a larger stability constant than the aqua complex.
This means the complex is more stable and the equilibrium lies towards ammonia coordination.
Example
A metal ion can form two complexes with ligands L and X. The \( \mathrm{K_{stab}} \) value for the L complex is much larger than that for X. Predict and explain which complex will predominate in solution.
▶️ Answer / Explanation
The complex with ligand L will predominate.
A larger \( \mathrm{K_{stab}} \) means the L complex is more stable and the equilibrium lies strongly towards its formation.