Solubility Equilibria and the Solubility Product Constant ($K_{sp}$)

Solubility Equilibria

Introduction to Solubility Equilibria

Solubility equilibria describe the equilibrium established between a solid ionic compound and its dissolved ions in solution. These reactions are heterogeneous because they involve both solid and aqueous phases.

• Solubility equilibria involve the dissolution or precipitation of ionic compounds.

• General rules for predicting solubility in water are qualitative, but solubility equilibria allow for quantitative predictions.

Solubility Rules (Review)

General Solubility Guidelines

Understanding which ionic compounds are soluble in water is essential for predicting precipitation reactions.

• All ionic solids have some degree of solubility in water.

• Group 1 metals and NH4+ compounds are always soluble.

• Group 2 metals are mostly soluble.

• Pb2+, Ag+, and Hg22+ compounds are usually insoluble.

• Reference tables (e.g., Table 4.1) are used to classify solubility.

Example: Predicting the products of a reaction:

• $\mathrm{Pb(NO_3)_2 (aq) + 2\ KI (aq) \rightarrow 2\ KNO_3 (aq) + PbI_2 (s)}$

Solubility Product Equilibrium Constant ($K_{sp}$)

Definition and Significance

The solubility product constant ($K_{sp}$) quantifies the equilibrium between a solid and its ions in a saturated solution. All salts have a limited solubility, and $K_{sp}$ values can range widely (e.g., $K_{sp} = 10^{-60}$ to $10^{36}$).

• A saturated solution contains the maximum amount of dissolved ions, with undissolved solid present.

• The extent of dissolution is expressed by the magnitude of $K_{sp}$.

Writing $K_{sp}$ Expressions

For a generic salt $A_xB_y$ (where A is the cation and B is the anion):

• Dissolution reaction: $A_xB_y(s) \rightleftharpoons xA^{m+}(aq) + yB^{n-}(aq)$

• $K_{sp} = [A^{m+}]^x [B^{n-}]^y$

Examples:

• $\mathrm{Ca_3(PO_4)_2(s) \rightleftharpoons 3\ Ca^{2+}(aq) + 2\ PO_4^{3-}(aq)}$ $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$

• $\mathrm{BaS(s) \rightleftharpoons Ba^{2+}(aq) + S^{2-}(aq)}$ $K_{sp} = [Ba^{2+}][S^{2-}]$

Solubility vs. $K_{sp}$

• Solubility is the amount of substance that dissolves to form a saturated solution (usually in g/L).

• Molar solubility is the number of moles of solute that dissolve per liter of saturated solution (mol/L).

• $K_{sp}$ is the equilibrium constant for the dissolution of a solid in water.

Example:

• $\mathrm{BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)}$

Factors Affecting Solubility

• pH of the solution: Solubility can increase or decrease depending on the acidity/basicity of the solution.

• Common ion effect: Presence of a common ion decreases solubility.

• Temperature: Solubility generally increases with temperature for most salts.

While solubility can change with these factors, $K_{sp}$ is constant for a given solute at a specific temperature.

Converting Between Solubility and $K_{sp}$

• To find $K_{sp}$ from solubility, use the stoichiometry of the dissolution reaction to determine ion concentrations, then substitute into the $K_{sp}$ expression.

• To find solubility from $K_{sp}$, set up an ICE (Initial, Change, Equilibrium) table and solve for the concentration of dissolved ions.

Example: For $\mathrm{PbBr_2}$ with a molar solubility of $1.0 \times 10^{-2}$ M:

• Dissolution: $\mathrm{PbBr_2(s) \rightleftharpoons Pb^{2+}(aq) + 2\ Br^-(aq)}$

• At equilibrium: $[Pb^{2+}] = 1.0 \times 10^{-2}$ M, $[Br^-] = 2.0 \times 10^{-2}$ M

• $K_{sp} = [Pb^{2+}][Br^-]^2 = (1.0 \times 10^{-2})(2.0 \times 10^{-2})^2 = 4.0 \times 10^{-6}$

ICE Table Method for Solubility Calculations

Using ICE Tables

ICE tables help track the changes in concentration as a solid dissolves and equilibrium is established.

• Set up initial concentrations (usually 0 for ions).

• Change: Add $+x$ or $+2x$ depending on stoichiometry.

• Equilibrium: Express concentrations in terms of $x$.

• Substitute into $K_{sp}$ and solve for $x$ (the molar solubility).

Example: For $\mathrm{AgCl}$, $K_{sp} = 1.33 \times 10^{-5}$:

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

• Let $x$ = molar solubility. $[Ag^+] = x$, $[Cl^-] = x$

• $K_{sp} = x^2 \Rightarrow x = \sqrt{K_{sp}} = \sqrt{1.33 \times 10^{-5}} = 3.65 \times 10^{-3}$ M

Common Ion Effect and Le Chatelier's Principle

Common Ion Effect

The common ion effect is an application of Le Chatelier’s Principle. The presence of a common ion reduces the solubility of a salt.

• Adding a soluble salt that shares an ion with the sparingly soluble salt shifts the equilibrium toward the solid, decreasing solubility.

Example: Adding NaCl to a solution of AgCl:

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

• $\mathrm{NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq)}$

• Increased $[Cl^-]$ shifts equilibrium left, decreasing $[Ag^+]$.

Mathematical Treatment of the Common Ion Effect

When a common ion is present, the equilibrium calculation must account for its initial concentration.

• For $\mathrm{AgCl}$ with $[Cl^-]_{initial} = 0.1$ M and $K_{sp} = 1.8 \times 10^{-10}$:

• $K_{sp} = [Ag^+][Cl^-] = x(0.1) = 1.8 \times 10^{-10}$

• $x = [Ag^+] = 1.8 \times 10^{-9}$ M

Comparing Solubilities Using $K_{sp}$

Simple 1:1 Salts

For salts with a 1:1 stoichiometry, the larger the $K_{sp}$, the greater the solubility.

• Example: Comparing AgCl, AgBr, and AgI:

Compound	Dissolution Reaction	$K_{sp}$	Molar Solubility (M)
AgCl	$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$	$1.8 \times 10^{-10}$	$1.3 \times 10^{-5}$
AgBr	$AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$	$5.0 \times 10^{-13}$	$7.1 \times 10^{-7}$
AgI	$AgI(s) \rightleftharpoons Ag^+(aq) + I^-(aq)$	$8.3 \times 10^{-17}$	$9.1 \times 10^{-9}$

AgCl is the most soluble, AgI is the least soluble.

Salts with More Complex Stoichiometry

For salts with different stoichiometry, compare molar solubility by solving the $K_{sp}$ expression for $x$.

• Example: $Ca_3(PO_4)_2$ with $K_{sp} = 1.0 \times 10^{-25}$:

• Dissolution: $Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)$

• $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 = (3x)^3 (2x)^2 = 108x^5$

• Solve for $x$ (molar solubility): $x = (K_{sp}/108)^{1/5}$

Summary Table: Factors Affecting Solubility

Factor	Effect on Solubility
pH	Can increase or decrease solubility depending on the salt
Common Ion	Decreases solubility
Temperature	Usually increases solubility for most salts

Key Concepts to Master

• Understanding equilibrium reactions and the meaning of $K$ values

• Writing $K_{sp}$ for heterogeneous reactions

• Predicting reaction shifts with concentration, pressure, or temperature changes (Le Chatelier’s Principle)

• Using ICE tables for solubility calculations

• Calculating $K_{sp}$ from molar solubility and vice versa

• Applying the common ion effect in equilibrium calculations