Questions: Stability of Complex Ions and Formation Constants

Question 1

A saturated solution of AgCl (Ksp ≈ 1.8 × 10⁻¹⁰) has very low [Ag⁺]. Excess aqueous ammonia is added and AgCl begins to dissolve. Why?

• A. Ammonia reacts with Cl⁻ to form NH₄Cl, removing chloride and pulling the equilibrium toward dissolution
• B. Ammonia forms [Ag(NH₃)₂]⁺, a stable complex with large Kf, drastically lowering [Ag⁺] and shifting the solubility equilibrium toward more dissolved AgCl
• C. Ammonia raises pH, which generally increases the solubility of ionic compounds
• D. Ammonia increases ionic strength, raising the activity of Ag⁺ and pulling more Cl⁻ into solution

Answer: (B)

This is competing equilibria in action. AgCl dissolves according to AgCl(s) ⇌ Ag⁺ + Cl⁻ (Ksp very small). Ammonia forms [Ag(NH₃)₂]⁺ with a large Kf ≈ 1.7 × 10⁷. Complex formation removes free Ag⁺ from solution, dropping [Ag⁺] below the value consistent with the solubility equilibrium. Le Chatelier's principle drives AgCl to dissolve further to replenish Ag⁺ — which is immediately complexed again. This continuous removal of Ag⁺ by complexation drives dissolution far beyond what Ksp alone would allow. Ammonia is not reacting with Cl⁻ (option A), and the pH and ionic-strength effects (options C and D) are not the operative mechanism here.

Question 2

Complex A has Kf = 1 × 10²⁰ and Complex B has Kf = 1 × 10⁵. Equal concentrations of both metal ions compete for a limiting amount of ligand. Which complex forms predominantly?

• A. Complex B, because less stable complexes form faster kinetically
• B. Equal amounts, because Kf only describes equilibrium ratios, not competition between different metals
• C. Complex A, because the larger Kf means its equilibrium position strongly favors complex over free metal, out-competing B for the available ligand
• D. Neither predominates — Kf values describe different reactions and cannot be directly compared

Answer: (C)

Kf is an equilibrium constant that directly compares how completely different metal ions are complexed at equilibrium. Complex A's Kf is 10¹⁵ times larger than B's, meaning A's equilibrium lies overwhelmingly further toward the complex. With limiting ligand, both reactions compete for the same resource, and the reaction with the vastly larger Kf captures essentially all available ligand. Option A incorrectly conflates thermodynamics with kinetics — Kf describes where equilibrium lies, not how fast it is reached. Option B is wrong because Kf values for reactions using the same ligand can absolutely be compared to predict selectivity.

Question 3

The overall formation constant Kf for a complex that forms in n stepwise stages equals the product of all n stepwise formation constants.

• T. True
• F. False

Answer: True

This follows from the standard rule for combining equilibrium constants: when successive reactions are summed to give an overall reaction, their equilibrium constants multiply. Each stepwise constant Ki describes one ligand addition: [MLᵢ]/([MLᵢ₋₁][L]). Summing all n steps to give M + nL ⇌ MLn produces an overall Kf = K₁ × K₂ × ⋯ × Kn. This is why Kf values for complexes with many ligands can reach astronomically large numbers — multiplying several successive stepwise constants compounds the effect even if each individual step is modest.

Question 4

A complex ion with a large Kf will be present in appreciable concentration mainly at high ligand concentrations, because most of the metal remains as the free aquo complex at lower ligand levels.

• T. True
• F. False

Answer: False

This reverses the meaning of Kf. A large Kf means the equilibrium lies far toward the complex — at equilibrium, essentially all metal is complexed and very little remains as free ion, even at moderate ligand concentrations. For [Cu(NH₃)₄]²⁺ with Kf ≈ 10¹³, even modest NH₃ concentrations lock up virtually all Cu²⁺. A small Kf would indicate a weak complex requiring high ligand concentrations to form appreciably. Since Kf = [complex]/([free metal][ligand]ⁿ), a large Kf means a large numerator and tiny denominator at equilibrium.

Question 5

Explain how the formation constant Kf connects to competing equilibria, using the dissolution of an insoluble metal salt in the presence of a complexing ligand as your example.

Model answer: Kf quantifies how completely complexation proceeds at equilibrium. When a complexing ligand is added to a solution in contact with an insoluble salt (e.g., AgCl + NH₃), two equilibria operate through a shared species (Ag⁺): dissolution (AgCl ⇌ Ag⁺ + Cl⁻, governed by Ksp) and complexation (Ag⁺ + 2NH₃ ⇌ [Ag(NH₃)₂]⁺, governed by Kf). A large Kf means complexation removes free Ag⁺ rapidly and nearly completely, driving [Ag⁺] far below the value consistent with the solubility equilibrium. Le Chatelier's principle pushes the dissolution equilibrium right to replenish Ag⁺, which is immediately re-complexed. The overall process couples both equilibria: the net equilibrium constant for AgCl dissolving into the complex equals Ksp × Kf, and when Kf is large enough, this product makes the overall process thermodynamically favorable.

The key is coupling through a shared species: when Kf effectively eliminates free Ag⁺, the solubility equilibrium is forced right to compensate, dissolving an otherwise insoluble solid.