Solubility and Complex-Ion Equilibria

The solubility product constant (K_sp) describes the equilibrium between an ionic solid and its dissolved ions. For the general dissolution A_aB_b(s) → aAⁿ⁺(aq) + bB^m−(aq):

K_sp = [Aⁿ⁺]^a[B^m−]^b

The solid does not appear in the expression because its activity is defined as 1. Each ion concentration is raised to the power of its coefficient in the balanced equation.

Examples: AgCl → Ag⁺ + Cl⁻ gives K_sp = [Ag⁺][Cl⁻]. Ca₃(PO₄)₂ → 3 Ca²⁺ + 2 PO₄³⁻ gives K_sp = [Ca²⁺]³[PO₄³⁻]². The exponents dramatically affect the relationship between K_sp and solubility, so you cannot compare the solubility of salts with different formulas by K_sp alone.

Molar solubility (s) is the number of moles of a sparingly soluble salt that dissolve per liter to form a saturated solution. To calculate it from K_sp:

1. Write the dissolution equation and the K_sp expression.
2. Let s = molar solubility. Express each ion concentration in terms of s using the stoichiometric coefficients (e.g., for PbI₂: [Pb²⁺] = s, [I⁻] = 2s).
3. Substitute into K_sp and solve for s.

For a 1:1 salt like AgCl: K_sp = s², so s = √K_sp. For a 1:2 salt like PbI₂: K_sp = (s)(2s)² = 4s³, so s = (K_sp/4)^1/3. The algebra becomes more complex for salts with higher coefficients, but the approach is the same.

Molar solubility can be converted to grams per liter by multiplying by the molar mass, which is often more practical for laboratory work.

The reverse calculation — finding K_sp from experimentally measured solubility — follows the same logic in reverse:

1. Convert the measured solubility to molar solubility (mol/L) if given in g/L.
2. Use stoichiometric ratios to find the equilibrium concentration of each ion.
3. Substitute the ion concentrations into the K_sp expression and evaluate.

For example, if the solubility of CaF₂ is measured as 0.017 g/L, convert to molar solubility: s = 0.017 / 78.08 = 2.2 × 10⁻⁴ M. The ions are [Ca²⁺] = s and [F⁻] = 2s, so K_sp = (s)(2s)² = 4s³ = 4(2.2 × 10⁻⁴)³ = 4.3 × 10⁻¹¹.

This approach assumes the only source of ions is the dissolving solid. If other electrolytes are present, their contributions must be accounted for separately.

To predict whether a precipitate forms when two solutions are mixed, calculate the ion product Q — the same mathematical form as K_sp, but using the initial ion concentrations after mixing (before any reaction):

• Q > K_sp — the solution is supersaturated and a precipitate will form, driving ion concentrations down until Q = K_sp.
• Q = K_sp — the solution is exactly saturated; no precipitate forms, but the system is at equilibrium.
• Q < K_sp — the solution is unsaturated; no precipitate forms and more solute could still dissolve.

A critical step is accounting for dilution: when equal volumes are mixed, each ion’s concentration is halved. Always compute the post-mixing concentrations before evaluating Q.

The common ion effect reduces the solubility of a sparingly soluble salt when one of its ions is already present in solution from another source. This is a direct application of Le Châtelier’s principle: the added common ion shifts the dissolution equilibrium to the left, favoring the solid.

For example, AgCl has K_sp = 1.8 × 10⁻¹⁰. In pure water, s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M. In 0.10 M NaCl, the Cl⁻ is already 0.10 M, so K_sp = [Ag⁺](0.10), giving [Ag⁺] = 1.8 × 10⁻⁹ M — about 7,500 times less soluble than in pure water.

This effect is exploited in analytical chemistry: adding excess precipitating reagent ensures nearly complete removal of the target ion from solution.

Selective precipitation separates ions from a mixture by exploiting differences in their precipitation thresholds. As a precipitating agent is added gradually, the first salt to reach Q = K_sp precipitates first; that order depends on K_sp, salt stoichiometry, and the dissolved-ion concentrations.

The strategy: for each ion that could form a precipitate, calculate the concentration of precipitating agent needed to just begin precipitation (set Q = K_sp and solve). The ion requiring the lowest precipitant concentration precipitates first. Comparing K_sp values alone is safe only for salts with the same stoichiometric form when the relevant ion concentrations are equal.

For example, a solution containing comparable concentrations of Ag⁺ and Cu²⁺ can be separated by slowly adding Cl⁻. AgCl (K_sp = 1.8 × 10⁻¹⁰) reaches its precipitation threshold while CuCl₂ remains soluble, so Ag⁺ can be removed while Cu²⁺ remains in solution.

Selective precipitation is widely used in qualitative analysis schemes to identify unknown cations by systematically precipitating groups of ions with specific reagents.

Beyond the common ion effect, several other factors influence the solubility of ionic compounds in water:

• pH effects: salts containing the anion of a weak acid (e.g., CaCO₃, Mg(OH)₂) are more soluble in acidic solution. The added H⁺ reacts with the anion (CO₃²⁻ + H⁺ → HCO₃⁻), removing it from the equilibrium and shifting dissolution to the right.
• Complex ion formation: metal ions that form stable complexes with ligands (e.g., Ag⁺ with NH₃) become more soluble because the free metal-ion concentration decreases, driving more solid to dissolve.
• Temperature: for most ionic solids, solubility increases with temperature, though a few salts (e.g., Ce₂(SO₄)₃) show the opposite trend.

When solving solubility problems, always check whether pH or complexation effects apply before using the simple K_sp expression.

A complex ion is a central metal cation bonded to one or more molecules or ions called ligands (e.g., NH₃, CN⁻, S₂O₃²⁻). Complex-ion formation is itself an equilibrium, described by the formation constant K_f. For the general formation Mⁿ⁺(aq) + j L(aq) ⇌ ML_j^z+(aq):

K_f = [ML_j^z+] / ([Mⁿ⁺][L]^j)

For example, for Ag⁺ + 2 NH₃ ⇌ Ag(NH₃)₂⁺: K_f = [Ag(NH₃)₂⁺] / ([Ag⁺][NH₃]²). Typical K_f values are very large (10⁷ to 10²⁵ or more), meaning complex formation is strongly product-favored. The reverse (dissociation) reaction has the reciprocal constant: K_d = 1/K_f.

Free-metal concentration in excess ligand: when K_f is large and the ligand is in excess, treat the complexation as complete first (stoichiometry step: each mole of metal consumes j moles of ligand), then let the complex dissociate back by a tiny x. The equilibrium free-metal concentration follows from [Mⁿ⁺] = [ML_j] / (K_f[L]^j) — typically vanishingly small (10⁻¹⁵ M or less).

Complex-ion formation increases solubility. When a sparingly soluble salt MX dissolves in the presence of a complexing ligand, the two equilibria couple, and their constants multiply:

• MX(s) ⇌ M⁺ + X⁻  (K_sp)
• M⁺ + j L ⇌ ML_j  (K_f)
• Net: MX(s) + j L ⇌ ML_j + X⁻  (K_net = K_sp × K_f)

Even when K_sp is tiny, a huge K_f can make K_net large enough for substantial dissolution. The classic application is photographic fixing: AgBr (K_sp = 5.0 × 10⁻¹³) dissolves in sodium thiosulfate solution because Ag(S₂O₃)₂³⁻ has K_f = 4.7 × 10¹³, giving K_net ≈ 24. Similarly, AgCl dissolves in ammonia via Ag(NH₃)₂⁺, and Al(OH)₃ dissolves in strong base via Al(OH)₄⁻.

For a 1:1 salt dissolving via the net reaction, if s mol/L dissolves then [ML_j] = [X⁻] = s and K_net = s²/[L]^j, so s = √(K_net × [L]^j) when the ligand stays approximately fixed; if dissolution consumes a significant fraction of the ligand, account for the 1:j stoichiometry in an ICE table (each mole dissolved consumes j moles of ligand) and solve the resulting quadratic.

Watch the charges when writing K_f expressions: the complex's net charge is the metal charge plus the total ligand charge (e.g., Cu²⁺ + 4 CN⁻ gives Cu(CN)₄²⁻, net −2).

Follow this workflow for solubility equilibrium problems:

1. Write the dissolution equation and the K_sp expression. Each ion concentration is raised to its stoichiometric coefficient.
2. For molar solubility from K_sp: let s = molar solubility. Express each ion concentration in terms of s using the stoichiometry (e.g., for Ca₃(PO₄)₂: [Ca²⁺] = 3s, [PO₄³⁻] = 2s).
3. For will-it-precipitate questions: calculate Q_sp from the actual ion concentrations after mixing. If Q_sp > K_sp, a precipitate forms.
4. For common-ion problems: the initial concentration of the common ion is not zero—include it in the ICE table.
5. For selective precipitation: compute the precipitant concentration at which each salt reaches Q = K_sp; the salt with the lowest threshold precipitates first. (Comparing K_sp values directly works only for salts of the same formula type whose target ions are present at comparable concentrations; each threshold depends on both K_sp and the ion’s existing concentration.)

Common mistakes: forgetting to raise ion concentrations to their stoichiometric powers in the K_sp expression (e.g., writing [Ag⁺] instead of [Ag⁺]² for Ag₂CrO₄), ignoring dilution when two solutions are mixed (volumes add, concentrations drop), assuming molar solubility equals K_sp directly without setting up the algebra, and neglecting the common-ion effect when a shared ion is already present in solution.