Molar Solubility and Ksp Calculations

Core Concept

Molar solubility (s) is the number of moles of a salt that dissolve per liter of solution to reach saturation. It is not the same as Ksp: Ksp is a constant, while s depends on stoichiometry (and on the presence of common ions). The two are linked through the dissolution equation — for a 1:1 salt, Ksp = s²; for a 1:2 or 2:1 salt, Ksp = 4s³. Solving for s from Ksp (or back) is a simple ICE-table exercise, and the common-ion effect lowers s whenever one of the salt's ions is already present.

Key Ideas

• Molar solubility s = mol of salt dissolved per liter of saturated solution.
• Ksp ↔ s via stoichiometry: the coefficient of each ion becomes a multiplier of s.
• 1:1 salt (AgCl, BaSO₄): Ksp = s² → s = √Ksp.
• 1:2 or 2:1 salt (PbCl₂, Ag₂CrO₄): Ksp = 4s³ → s = (Ksp/4)^(1/3).
• Common-ion effect: an ion already present lowers s compared to pure water.
• Larger Ksp does not always mean larger s across different stoichiometries — always convert to s to compare.

Equations and Variables

• AgCl(s) ⇌ Ag⁺ + Cl⁻: Ksp = s·s = s²
• PbCl₂(s) ⇌ Pb²⁺ + 2 Cl⁻: Ksp = s·(2s)² = 4s³
• Ag₂CrO₄(s) ⇌ 2 Ag⁺ + CrO₄²⁻: Ksp = (2s)²·s = 4s³
• Ca₃(PO₄)₂(s) ⇌ 3 Ca²⁺ + 2 PO₄³⁻: Ksp = (3s)³·(2s)² = 108s⁵

How It Works

1. Write the balanced dissolution equation.
2. Let s be the molar solubility; assign equilibrium ion concentrations as multiples of s (e.g., 2s for an ion with coefficient 2).
3. Substitute into the Ksp expression and solve for s.
4. To compare solubilities of different salts, convert Ksp → s; a direct Ksp comparison is only valid for identical stoichiometries.
5. For a common-ion solution, the shared ion starts at a nonzero concentration, and s is smaller (solve the resulting algebra rather than using the pure-water shortcut).

Worked Example

Calculate the molar solubility of (a) AgCl (Ksp = 1.8 × 10⁻¹⁰) and (b) PbCl₂ (Ksp = 1.7 × 10⁻⁵) in pure water.

(a) AgCl(s) ⇌ Ag⁺ + Cl⁻, so Ksp = s².

s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M

(b) PbCl₂(s) ⇌ Pb²⁺ + 2 Cl⁻, so Ksp = [Pb²⁺][Cl⁻]² = s(2s)² = 4s³.

s = (Ksp/4)^(1/3) = (1.7 × 10⁻⁵ / 4)^(1/3) = (4.25 × 10⁻⁶)^(1/3) = 1.6 × 10⁻² M

Note that PbCl₂ (Ksp = 1.7 × 10⁻⁵) is more soluble than AgCl (Ksp = 1.8 × 10⁻¹⁰) despite the Ksp comparison being across different stoichiometries — the conversion to s makes the ranking unambiguous.

How it works

1. Write the balanced dissolution equation.
2. Let s be the molar solubility; assign equilibrium ion concentrations as multiples of s (e.g., 2s for an ion with coefficient 2).
3. Substitute into the Ksp expression and solve for s.
4. To compare solubilities of different salts, convert Ksp → s; a direct Ksp comparison is only valid for identical stoichiometries.
5. For a common-ion solution, the shared ion starts at a nonzero concentration, and s is smaller (solve the resulting algebra rather than using the pure-water shortcut).

Common confusions

• "Ksp = s² always." — Only for 1:1 salts; use 4s³ for AB₂/A₂B, and 108s⁵ for A₃B₂-type salts.
• "Larger Ksp always means more soluble." — False across different stoichiometries; Ag₂CrO₄ (Ksp = 1.1 × 10⁻¹²) is more soluble than AgCl (Ksp = 1.8 × 10⁻¹⁰) because of the 4s³ vs s² relationship.
• "Forgetting to square the coefficient." — In PbCl₂, [Cl⁻] = 2s, so its Ksp term is (2s)², not 2s.
• "Common ion increases solubility." — It decreases solubility (Le Châtelier: adding a product shifts dissolution left).
• "s is fixed for a salt." — s depends on the medium; it is smaller in a common-ion solution than in pure water.

Quick review

• s = molar solubility (mol/L at saturation).
• 1:1 → Ksp = s²; AB₂/A₂B → Ksp = 4s³.
• Convert Ksp → s to compare different salts.
• Common ion lowers s.
• ICE table relates s to each ion concentration via coefficients.