Complex Ion Equilibria: Formation and Calculation of Equilibrium Concentrations

Complex Ion Equilibria

Introduction to Complex Ions

Complex ions are an important class of chemical species in aqueous solution, especially in the context of transition metal chemistry. Understanding their formation and equilibrium behavior is essential for predicting solubility, reactivity, and analytical applications.

• Complex ion: A polyatomic cation or anion composed of a central metal ion bonded to surrounding molecules or ions, known as ligands.

• Coordination compound: A compound containing one or more complex ions.

• Ligand: An ion or molecule that binds to a central metal ion in a complex ion. Ligands can be neutral (e.g., NH3) or anionic (e.g., Cl-).

• Coordination number: The number of ligand donor atoms directly bonded to the central metal ion. Common values are 2, 4, and 6.

Examples of Complex Ions:

• [Ag(NH3)2]+ (coordination number 2)

• [Cu(NH3)4]2+ (coordination number 4)

• [Co(NH3)6]2+ (coordination number 6)

Stepwise Formation of Complex Ions

The formation of complex ions often occurs in a stepwise manner, with each ligand binding in a separate equilibrium step. Each step has its own equilibrium constant.

• Step 1: $\mathrm{Ag^+ + NH_3 \rightleftharpoons Ag(NH_3)^+}$    $K_1 = 2.1 \times 10^3$

• Step 2: $\mathrm{Ag(NH_3)^+ + NH_3 \rightleftharpoons Ag(NH_3)_2^+}$    $K_2 = 8.2 \times 10^3$

• Overall: $\mathrm{Ag^+ + 2NH_3 \rightleftharpoons Ag(NH_3)_2^+}$

The overall equilibrium constant ($K_{12}$) is the product of the stepwise constants:

$K_{12} = K_1 \times K_2$

$K_{12} = (2.1 \times 10^3) \times (8.2 \times 10^3) = 1.7 \times 10^7$

Calculating Equilibrium Concentrations: Worked Example

Problem: Calculate the equilibrium concentrations of Ag+ and Ag(NH3)2+ in a solution with initial [Ag+] = $4.0 \times 10^{-3}$ M and [NH3] = 0.500 M.

• Step 1: Set up the ICE table for the overall reaction:

	Ag+ (aq)	NH3 (aq)	Ag(NH3)2+ (aq)
Initial (I)	4.0 × 10−3 M	0.500 M	0
Change (C)	−x	−2x	+x
Equilibrium (E)	4.0 × 10−3 − x	0.500 − 2x	x

• Step 2: Write the equilibrium expression for the overall reaction:

$K_{12} = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$

Substitute the equilibrium values:

$1.7 \times 10^7 = \frac{x}{(4.0 \times 10^{-3} - x)(0.500 - 2x)^2}$

• Step 3: Approximate and solve for x (since K is very large, x ≈ initial [Ag+]):

Assume $x \approx 4.0 \times 10^{-3}$ M$and $0.500 - 2x \approx 0.492$ M.

$1.7 \times 10^7 = \frac{4.0 \times 10^{-3}}{x \times (0.492)^2}$

Solve for x (which is [Ag+] at equilibrium):

$x = [Ag^+] = 9.6 \times 10^{-10}$ M

• Step 4: Calculate [Ag(NH3)2+] at equilibrium:

$[Ag(NH_3)_2^+] = 4.0 \times 10^{-3}$ M

• Step 5: Calculate [NH3] at equilibrium:

$[NH_3] = 0.492$ M

Summary Table: Equilibrium Concentrations

Species	Equilibrium Concentration (M)
Ag+	9.6 × 10−10
NH3	0.492
Ag(NH3)2+	4.0 × 10−3

Key Points and Applications

• When the equilibrium constant for complex ion formation is very large, the reaction proceeds nearly to completion, and the free metal ion concentration becomes extremely small.

• Complex ion equilibria are important in analytical chemistry (e.g., selective precipitation, qualitative analysis) and in biological systems (e.g., hemoglobin binding iron).

• Calculations can often be simplified by using the overall formation constant ($K_f$) rather than stepwise constants when $K$ is very large.

Example Application: The formation of [Ag(NH3)2]+ is used in the Tollens' test for aldehydes, where the complex ion helps keep silver ions in solution until they are reduced to metallic silver.

Additional info: In more advanced courses, the stability of complex ions is discussed in terms of ligand field theory and the chelate effect, which further explain why certain ligands form more stable complexes with specific metal ions.