Solubility Equilibria and the Solubility Product Constant (Ksp)

Chapter 15: Solubility Equilibria

Introduction to Solubility and Solubility Product Constant

Solubility equilibria describe the balance between a solid ionic compound and its ions in solution. The solubility product constant (Ksp) quantifies the equilibrium for the dissolution of slightly soluble salts. Understanding Ksp is essential for predicting solubility, calculating concentrations, and determining when precipitation occurs.

• Solubility Product Constant (Ksp): The equilibrium constant for the dissolution of a slightly soluble ionic compound.

• Molar Solubility (s, mol/L): The number of moles of solute that dissolve in 1 L of a saturated solution.

• Solubility (g/L): The number of grams of solute that dissolve in 1 L of a saturated solution.

• Saturated Solution: A solution in which the maximum amount of solute has dissolved at a given temperature.

Example: For silver chloride:

• Equation: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

• Ksp expression: $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$

Solubility Product Expressions for Ionic Compounds

Each slightly soluble ionic compound has a unique Ksp expression based on its dissociation stoichiometry. The general form is:

• For $\text{A}_x\text{B}_y(s) \rightleftharpoons x\text{A}^{y+}(aq) + y\text{B}^{x-}(aq)$

• $K_{sp} = [\text{A}^{y+}]^x[\text{B}^{x-}]^y$

Examples:

Compound	Dissociation Equation	Ksp Expression
MgF2	$\text{MgF}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{F}^-(aq)$	$K_{sp} = [\text{Mg}^{2+}][\text{F}^-]^2$
Ag2CO3	$\text{Ag}_2\text{CO}_3(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CO}_3^{2-}(aq)$	$K_{sp} = [\text{Ag}^+]^2[\text{CO}_3^{2-}]$
Ca3(PO4)2	$\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$	$K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$

Calculating Ksp from Solubility

Given the solubility of a compound, Ksp can be calculated by determining the equilibrium concentrations of ions and substituting into the Ksp expression.

• Step 1: Convert solubility from g/L to mol/L using molar mass.

• Step 2: Use stoichiometry to find ion concentrations at equilibrium.

• Step 3: Substitute values into the Ksp expression.

Example: For CaSO4 (solubility = 0.67 g/L, MW = 136.2 g/mol):

• $s = \frac{0.67\ \text{g/L}}{136.2\ \text{g/mol}} = 4.9 \times 10^{-3}\ \text{mol/L}$

• $\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

• At equilibrium: $[\text{Ca}^{2+}] = s$, $[\text{SO}_4^{2-}] = s$

• $K_{sp} = s \times s = s^2 = (4.9 \times 10^{-3})^2 = 2.4 \times 10^{-5}$

Example: For Ag2CrO4 (solubility = 0.044 g/L, MW = 332 g/mol):

• $s = \frac{0.044}{332} = 1.3 \times 10^{-4}\ \text{mol/L}$

• $\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq)$

• At equilibrium: $[\text{Ag}^+] = 2s$, $[\text{CrO}_4^{2-}] = s$

• $K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2s)^2(s) = 4s^3$

• Substitute $s$: $K_{sp} = 4(1.3 \times 10^{-4})^3 = 9.3 \times 10^{-12}$

Calculating Solubility from Ksp

Given Ksp, the molar solubility of a compound can be determined by solving the equilibrium expression for s.

• Step 1: Write the dissociation equation and Ksp expression.

• Step 2: Set up variables for ion concentrations in terms of s.

• Step 3: Solve for s.

• Step 4: Convert molar solubility to g/L if needed.

Example: For AgCl ($K_{sp} = 1.6 \times 10^{-10}$, MW = 143.35 g/mol):

• $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

• $K_{sp} = s^2$

• $s = \sqrt{K_{sp}} = \sqrt{1.6 \times 10^{-10}} = 1.3 \times 10^{-5}\ \text{mol/L}$

• Convert to g/L: $\text{sol}(g/L) = s \times \text{MW} = 1.3 \times 10^{-5} \times 143.35 = 1.9 \times 10^{-3}\ \text{g/L}$

Example: For CaF2 ($K_{sp} = 3.9 \times 10^{-11}$, MW = 78.08 g/mol):

• $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$

• $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s \times (2s)^2 = 4s^3$

• $s = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} = 2.1 \times 10^{-4}\ \text{mol/L}$

• Convert to g/L: $\text{sol}(g/L) = s \times \text{MW} = 2.1 \times 10^{-4} \times 78.08 = 1.7 \times 10^{-2}\ \text{g/L}$

Precipitation Reactions and the Reaction Quotient (Q)

Precipitation occurs when the product of ion concentrations exceeds the Ksp value. The reaction quotient (Q) is calculated using initial ion concentrations to predict whether a precipitate will form.

• Q < Ksp: Unsaturated solution; no precipitate forms.

• Q = Ksp: Saturated solution; no precipitate forms.

• Q > Ksp: Supersaturated solution; a precipitate forms.

Condition	Solution Type	Precipitate Formation
Q < Ksp	Unsaturated	No precipitate
Q = Ksp	Saturated	No precipitate
Q > Ksp	Supersaturated	Precipitate forms

Example: For AgCl, if $Q = [\text{Ag}^+][\text{Cl}^-]$ is greater than $K_{sp}$, AgCl will precipitate from solution.

Additional info:

• These notes cover the core concepts of solubility equilibria, including calculation methods and precipitation prediction, as outlined in General Chemistry Chapter 15.

• All equations are provided in LaTeX format for clarity and academic rigor.