Coupled equilibria, solubility products and complex formation

Cancel identical species and multiply compatible constants

For hypothetical isomers A ⇌ B with K₁° = 4.00 and B ⇌ C with K₂° = 0.500, the summed equation is A ⇌ C. Multiplying (a_B/a_A)(a_C/a_B) cancels the same a_B and gives K° = K₁°K₂° = 2.00. Temperature and standard-state conventions must be compatible.

Reversing a formation reaction gives its reciprocal dissociation constant on the same basis. An overall constant for multiple ligand additions is not automatically the constant of each single step. Keep the written stoichiometry attached to every value.

The dissolution equation determines the powers

For a separate hypothetical salt AB₂(s) ⇌ A²⁺(aq) + 2 B⁻(aq), at a fixed stipulated temperature, suppose initially ion-free water is in contact with excess pure solid that remains. The solution is ideal and dilute, c° = 1 mol L⁻¹ and no side equilibria occur. If molar solubility is s, the free-ion concentrations are [A²⁺] = s and [B⁻] = 2s.

The stipulated dimensionless K_sp° = 3.20 × 10⁻¹¹ equals (s/c°)(2s/c°)² = 4(s/c°)³. Thus s = 2.00 × 10⁻⁴ mol L⁻¹. K_sp° is not s, and this stoichiometry does not give s². The two charges balance: 2s positive equivalents and 2s negative equivalents.

Use the free-ion product after mixing

For a dissolution equation, Q_sp > K_sp° indicates that precipitation is thermodynamically favoured under the model. Calculate free-ion activities after mixing and dilution. Supersaturation can persist if nucleation is slow, so the quotient does not promise immediate visible precipitation.

A common ion can lower a simple salt’s solubility. For another fictional AB(s) ⇌ A⁺ + B⁻ model with K_sp° = 1.00 × 10⁻¹⁰ and c° = 1 mol L⁻¹, at the same fixed stipulated temperature with excess pure AB(s) remaining, pure-water s is 1.00 × 10⁻⁵ mol L⁻¹. A separate equilibrium in the same ideal/no-side-reaction scope, with a soluble charge-balanced salt initially supplying 0.0100 mol L⁻¹ B⁻ and excess pure AB(s) remaining, gives s(0.0100+s) = 1.00 × 10⁻¹⁰ in numerical mol L⁻¹ values; s is approximately 1.00 × 10⁻⁸ mol L⁻¹. Activity corrections, protonation and complex formation are excluded here.

Bound metal contributes to total metal without being free metal

A final hypothetical equilibrium is M⁺ + 2 L ⇌ ML₂⁺, where L is neutral and the complex retains charge +1. Use ideal normalized concentrations with c° = 1 mol L⁻¹ and K_f° = 1.00 × 10³. A consistent stipulated equilibrium state has free [M⁺] = 0.00100 mol L⁻¹ and free [L] = 0.100 mol L⁻¹. It gives [ML₂⁺] = 0.0100 mol L⁻¹.

Total dissolved metal is 0.00100 + 0.0100 = 0.0110 mol L⁻¹, and total ligand is 0.100 + 2 × 0.0100 = 0.120 mol L⁻¹. Counterions balance the cation charge. These are supplied equilibrium concentrations, not initial stocks; other complexes and ligand protonation are excluded. Binding free ions can increase total dissolved metal, so a free-ion quotient must not silently substitute the total.

Worked examples

Try it yourself

1. What is the summed A ⇌ C constant for compatible K₁° = 4.00 and K₂° = 0.500?

   Show the answer and reasoning

   2.00.

   Cancel the shared activity and multiply the two constants.

2. For ideal AB₂ dissolution with solubility s, what is K_sp° in the stated convention?

   Show the answer and reasoning

   4(s/c°)³.

   One A²⁺ and two B⁻ give factors s/c° and (2s/c°)².

3. Should a precipitation quotient use the unmixed stock concentrations?

   Show the answer and reasoning

   No.

   It needs the relevant free-ion activities after mixing and dilution.

4. Is total dissolved metal necessarily equal to free-metal concentration?

   Show the answer and reasoning

   No.

   Complexes and other dissolved metal-containing species also contribute to the total.

What to carry forward

• Combine constants only with compatible equations and conventions.
• Use dissolution stoichiometry, charge balance and a remaining solid phase.
• A quotient predicts tendency rather than precipitation speed.
• Keep free species distinct from conserved analytical totals.