Introducing the stability constant

In aqueous solutions, transition metal ions become hydrated as water molecules surround the ion and act as ligands. Adding other ligands like NH₃ leads to a competing ligand exchange equilibrium, forming the most stable complex.

For example:

[Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

This occurs through four stepwise displacement reactions with equilibrium constants K₁ to K₄.

These stepwise stability constants are summarised by an overall stability constant, (K_stab), which is the equilibrium constant for the overall formation of the complex from the hydrated metal ion and free ligands.

K_stab indicates the stability of the final complex compared to the original hydrated metal ion. The larger K_stab, the more stable the complex.

Defining and deriving Kstab

The stability constant (K_stab) is defined as the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.

For the general ligand substitution reaction:

Mⁿ⁺ + mL ⇌ [ML_m]ⁿ⁺

Where:

• M = central metal ion.
• L = ligand.
• n = charge on complex.
• m = number of ligands.

The stability constant K_stab is given by:

$K_{stab}=\frac{[{[M{L}_m]}^{n+}]}{[M^{n+}]{[L]}^m}$

For example, for the reaction:

[Cu(H₂O)₆]²⁺ + 4Cl^-_(aq) ⇌ [CuCl₄]^2-_(aq) + 6H₂O_(l)

K_stab would be:

$K_{stab}=\frac{[{[CuC{l}_4]}^{2-}]}{[{[Cu{(H_{2}O)}_6]}^{2+}]{[C{l}^{-}]}^4}$

H₂O does not appear in the equilibrium expression because it is in such a large excess that its concentration is regarded as being constant.

The units for K_stab vary, so we must determine the units after each calculation, in the same way as for the units of K_c.

Comparing complex stability

Stability constants (K_stab) are often expressed as log₁₀K_stab values on a logarithmic scale without units. This is because stability constants can vary over a wide range of values, and using a logarithmic scale allows for easier comparison between different complexes. A higher log₁₀K_stab value indicates a more stable complex.

For example, stability constants for some Cu²⁺ complexes:

The higher value for the [Cu(NH₃)₄]²⁺ complex compared to the [CuCl₄]^2- complex tells us that [Cu(NH₃)₄]²⁺ is more stable.

We can use stability constant values to predict ligand displacement reactions; if the log₁₀K_stab value of ligand B is greater than ligand A, then ligand B will displace ligand A from the complex.

For example, adding NH₃ to [CuCl₄]^2- results in the formation of [Cu(NH₃)₄]²⁺, because ammonia has a higher stability constant and displaces the coordinated chloride ions.

This allows us to predict the products of ligand substitution reactions based on the magnitude of the stability constants for the competing ligands.

Worked example 1 - Calculation involving Kstab

Concentrated hydrochloric acid is added to a copper(II) sulfate solution, forming [CuCl₄]^2- and [Cu(H₂O)₆]²⁺ through a series of ligand exchange reactions. The second step is:

[Cu(H₂O)₅Cl]⁺_(aq) + Cl^-_(aq) ⇌ Cu(H₂O)₄Cl_2(aq) + H₂O_(l)

At equilibrium, the mixture contains 0.80 mol dm^-3 Cu(H₂O)₄Cl₂, 0.20 mol dm^-3 [Cu(H₂O)₅Cl]⁺, and 0.10 mol dm^-3 HCl.

Calculate the stability constant K_stab for this step, including units. Give your answer to 2 significant figures.

Step 1: Write the K_stab expression for the reaction

$K_{stab}=\frac{[Cu{(H_{2}O)}_{4}C{l}_2]}{[{[Cu{(H_{2}O)}_{5}Cl]}^{+}][C{l}^{-}]}$

Step 2: Substitution and correct evaluation

$K_{stab}=\frac{0.80}{(0.20\times 0.1)}=40$

Step 3: Determine the units of K_stab

The units for K_stab are $\frac{{\text{mol}}^1{\text{ dm}}^{-3}}{({\text{mol}}^1{\text{ dm}}^{-3})({\text{mol}}^1{\text{ dm}}^{-3})}={\text{ dm}}^3{\text{ mol}}^{-1}$

So, the stability constant (K_stab) for this step is 40 dm³ mol^-1.

Worked example 2 - Calculation involving Kstab

Ammonia solution is added to a solution containing silver(I) ions, forming [Ag(NH₃)₂]⁺ through a series of ligand exchange reactions. The first step is:

Ag⁺_(aq) + NH_3(aq) ⇌ [AgNH₃]⁺_(aq)

When a 2.50 mol dm^-3 solution of silver(I) ions is mixed with a 2.50 mol dm^-3 solution of ammonia, the equilibrium concentration of the [AgNH₃]⁺ complex is 2.46 mol dm^-3.

Calculate the stability constant K_stab for this step, including units. Give your answer to 3 significant figures.

Step 1: Calculate equilibrium concentrations

Step 2: Write the K_stab expression for the reaction

$K_{stab}=\frac{[{[AgN{H}_3]}^{+}]}{[A{g}^{+}][N{H}_3]}$

Step 3: Substitution and correct evaluation

$K_{stab}=\frac{2.46}{(0.04\times 0.04)}=\mathrm{1,540}$

Step 4: Determine the units of K_stab

The units for K_stab are $\frac{{\text{mol}}^1{\text{ dm}}^{-3}}{({\text{mol}}^1{\text{ dm}}^{-3})({\text{mol}}^1{\text{ dm}}^{-3})}={\text{ dm}}^3{\text{ mol}}^{-1}$

So, the stability constant (K_stab) for this step is 1,540 dm³ mol^-1