$AB\text{(s)} \rightleftharpoons A^+\text{(aq)} + B^-\text{(aq)}$

At equilibrium: Saturated solution

• Solid in contact with dissolved ions
• Rate dissolving = rate precipitating

Solubility Product Constant (K_sp)

For general ionic solid:

$M_aX_b\text{(s)} \rightleftharpoons aM^{b+(aq) + bX^{a-}(aq)}$

K_sp expression:

$K_{sp} = [M^{b+}]^a[X^{a-}]^b$

Key points:

• Pure solid NOT included (like all heterogeneous K)
• Only dissolved ions
• Temperature dependent
• Called solubility product

Examples:

1. AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

$K_{sp} = [Ag^+][Cl^-]$

1. PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)

$K_{sp} = [Pb^{2+}][I^-]^2$

1. Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq)

$K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$

Molar Solubility

Molar solubility (s): Maximum moles that dissolve per liter

Different from K_sp:

• K_sp = equilibrium constant
• s = concentration (mol/L)

Relationship depends on stoichiometry

Example: AgCl

AgCl(s) ⇌ Ag⁺ + Cl⁻

If solubility = s:

• [Ag⁺] = s
• [Cl⁻] = s

K_sp = s·s = s²

Example: PbI₂

PbI₂(s) ⇌ Pb²⁺ + 2I⁻

If solubility = s:

• [Pb²⁺] = s
• [I⁻] = 2s (stoichiometry!)

K_sp = s·(2s)² = 4s³

Calculating s from K_sp

General strategy:

1. Write dissolution equation
2. Express ion concentrations in terms of s
3. Write K_sp expression
4. Substitute and solve for s

Example: Calculate s for Ag₂CrO₄ given K_sp = 1.1 × 10⁻¹²

Ag₂CrO₄(s) ⇌ 2Ag⁺ + CrO₄²⁻

If s = solubility:

• [Ag⁺] = 2s
• [CrO₄²⁻] = s

K_sp = [Ag⁺]²[CrO₄²⁻] = (2s)²(s) = 4s³

$1.1 \times 10^{-12} = 4s^3$

$s^3 = 2.75 \times 10^{-13}$

$s = 6.5 \times 10^{-5} \text{ M}$

Ion Product (Q_sp)

Like Q for solubility:

$Q_{sp} = [M^{b+}]^a[X^{a-}]^b$

Use current concentrations (not equilibrium)

Predicting precipitation:

Example: Will AgCl precipitate?

Given: K_sp(AgCl) = 1.8 × 10⁻¹⁰ Mix: [Ag⁺] = 1.0 × 10⁻⁴ M, [Cl⁻] = 1.0 × 10⁻⁴ M

Calculate Q_sp:

$Q_{sp} = [Ag^+][Cl^-] = (1.0 \times 10^{-4})(1.0 \times 10^{-4}) = 1.0 \times 10^{-8}$

Compare: Q_sp (1.0 × 10⁻⁸) > K_sp (1.8 × 10⁻¹⁰)

Result: Yes, AgCl precipitates!

Common Ion Effect

Adding ion already in equilibrium:

Shifts equilibrium by Le Chatelier

Example: AgCl in NaCl solution

AgCl(s) ⇌ Ag⁺ + Cl⁻

Add NaCl (source of Cl⁻):

• Increases [Cl⁻]
• Shifts left (less dissolving)
• Decreases solubility

Common ion decreases solubility

Complex Ion Formation

Metal ions can form complex ions:

Increases solubility

Example: AgCl in NH₃

$Ag^+ + 2NH3 \rightleftharpoons Ag(NH3)2^+$

Removes Ag⁺ from solution:

• Shifts dissolution right
• More AgCl dissolves
• Increases solubility

pH and Solubility

For salts of weak acids/bases:

pH affects solubility

Example: CaF₂ in acidic solution

CaF₂(s) ⇌ Ca²⁺ + 2F⁻

In acid: H⁺ + F⁻ → HF

• Removes F⁻
• Shifts right
• More soluble in acid

Rule:

• Salts of weak acids: more soluble in acid
• Salts of weak bases: more soluble in base

Selective Precipitation

Separate ions by precipitation:

Strategy:

1. Calculate K_sp for each compound
2. Find [anion] needed to precipitate each
3. Add reagent slowly
4. First compound precipitates first

Example: Separate Ag⁺ and Pb²⁺ using Cl⁻

Given:

• K_sp(AgCl) = 1.8 × 10⁻¹⁰
• K_sp(PbCl₂) = 1.7 × 10⁻⁵

AgCl precipitates at much lower [Cl⁻]

Can separate by controlling [Cl⁻]

Note: For every 1 Pb²⁺, get 2 I⁻

K_sp expression:

$K_{sp} = [Pb^{2+}][I^-]^2$

$7.9 \times 10^{-9} = (s)(2s)^2$

$7.9 \times 10^{-9} = s \cdot 4s^2$

$7.9 \times 10^{-9} = 4s^3$

$s^3 = \frac{7.9 \times 10^{-9}}{4} = 1.975 \times 10^{-9}$

$s = \sqrt[3]{1.975 \times 10^{-9}}$

$s = 1.25 \times 10^{-3} \text{ M}$

Answer (a): s = 1.3 × 10⁻³ M in pure water

(b) Molar solubility in 0.10 M NaI

NaI provides common ion (I⁻):

• NaI → Na⁺ + I⁻
• [I⁻]₀ = 0.10 M (from NaI)

ICE table:

K_sp expression:

$7.9 \times 10^{-9} = (s)(0.10 + 2s)^2$

Small s approximation:

Since K_sp is very small, s << 0.10

Assume: 0.10 + 2s ≈ 0.10

$7.9 \times 10^{-9} = (s)(0.10)^2$

$7.9 \times 10^{-9} = s(0.010)$

$s = \frac{7.9 \times 10^{-9}}{0.010}$

$s = 7.9 \times 10^{-7} \text{ M}$

Check: 2s = 1.6 × 10⁻⁶ << 0.10 ✓ (approximation valid)

Answer (b): s = 7.9 × 10⁻⁷ M in 0.10 M NaI

Comparison:

Pure water: s = 1.3 × 10⁻³ M 0.10 M NaI: s = 7.9 × 10⁻⁷ M

Common ion effect:

• Solubility decreased by factor of ~1600!
• I⁻ from NaI shifts equilibrium left
• Much less PbI₂ dissolves

$1.8 \times 10^{-10} = (0.010)[Cl^-]$

$[Cl^-] = \frac{1.8 \times 10^{-10}}{0.010}$

$[Cl^-] = 1.8 \times 10^{-8} \text{ M}$

AgCl precipitates when [Cl⁻] ≥ 1.8 × 10⁻⁸ M

For PbCl₂: PbCl₂(s) ⇌ Pb²⁺ + 2Cl⁻

Precipitation when Q_sp = K_sp:

$K_{sp} = [Pb^{2+}][Cl^-]^2$

$1.7 \times 10^{-5} = (0.010)[Cl^-]^2$

$[Cl^-]^2 = \frac{1.7 \times 10^{-5}}{0.010}$

$[Cl^-]^2 = 1.7 \times 10^{-3}$

$[Cl^-] = \sqrt{1.7 \times 10^{-3}}$

$[Cl^-] = 0.041 \text{ M}$

PbCl₂ precipitates when [Cl⁻] ≥ 0.041 M

Compare:

AgCl precipitates at: [Cl⁻] = 1.8 × 10⁻⁸ M PbCl₂ precipitates at: [Cl⁻] = 0.041 M

AgCl requires much less Cl⁻!

Answer: AgCl precipitates first

Separation:

Can separate Ag⁺ and Pb²⁺:

Step 1: Add Cl⁻ slowly to [Cl⁻] = 1.8 × 10⁻⁸ M

• AgCl precipitates
• Pb²⁺ stays in solution

Step 2: Continue adding to [Cl⁻] = 0.041 M

• Now PbCl₂ precipitates
• Most Ag⁺ already removed

Window for separation:

1.8 × 10⁻⁸ M < [Cl⁻] < 0.041 M

In this range:

• AgCl precipitated
• PbCl₂ still dissolved
• Successful separation!

General rule:

Compound with smaller K_sp precipitates first (when cations at equal concentration)