Solubility, Solubility Product, and Common Ion Effect: Study Notes

Solubility and Solubility Product (Ksp)

Introduction to Solubility

Solubility refers to the amount of a salt (an ionic compound) that dissolves in a solution to form a saturated solution. Not all ionic compounds are equally soluble in water, and their solubility is governed by specific rules and equilibrium concepts.

• Solubility: The maximum amount of solute that can dissolve in a given quantity of solvent at a specific temperature.

• Saturated solution: A solution in which no more solute can dissolve at the given conditions; excess solute remains undissolved.

General Solubility Rules

Certain patterns help predict whether an ionic compound will dissolve in water:

• Group 1 metals (e.g., Na+, K+) and NH4+ are always soluble.

• Group 2 metals are mostly soluble.

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

These rules are useful for predicting whether a precipitate will form in a reaction. For example:

• Example: Predict the products for lead(II) nitrate and potassium iodide:

Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)

The Solubility Product Equilibrium Constant (Ksp)

The solubility product constant (Ksp) quantifies the equilibrium between a solid ionic compound and its dissolved ions in a saturated solution. It is a specific type of equilibrium constant for sparingly soluble salts.

• Solubility Product (Ksp): The equilibrium constant for the dissolution of a solid salt into its ions.

• Represents a heterogeneous equilibrium (solid and aqueous phases).

For a general salt:

$A_mB_n(s) \rightleftharpoons mA^{x+}(aq) + nB^{y-}(aq)$

• A = cation, B = anion

• m = number of cations, n = number of anions

• x = cation charge, y = anion charge

The Ksp expression is:

$K_{sp} = [A^{x+}]^m [B^{y-}]^n$

• Only aqueous ions are included; solids are omitted from the expression.

Examples:

• For Ca3(PO4)2:

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

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

• For BaS:

• $BaS(s) \rightleftharpoons Ba^{2+}(aq) + S^{2-}(aq)$

  $K_{sp} = [Ba^{2+}][S^{2-}]$

Saturated Solutions: Understanding Solubility

In a saturated solution:

• There is more solid present than can be dissolved at that temperature.

• Constant dissolution (solid to ions) and precipitation (ions to solid) occur simultaneously.

• Both solid and dissolved ions are present at equilibrium.

Example:

$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$

• The equilibrium concentration of the compound (not the ions) is called its molar solubility.

Calculating Ksp and Molar Solubility

Finding Ksp from Molar Solubility

Given the molar solubility of a salt, you can calculate its Ksp:

1. Write out the solubility product reaction.

2. Express the molar solubility as the change in concentration for the whole compound; use stoichiometry to find equilibrium concentrations for each ion.

3. Write the Ksp expression and solve.

Example: The molar solubility of PbBr2 is $1.0 \times 10^{-2}$ mol/L. Calculate Ksp.

Finding Molar Solubility from Ksp

Given Ksp, you can find the molar solubility:

1. Write out the equilibrium reaction.

2. Let x represent the molar solubility; use stoichiometry to relate x to ion concentrations.

3. Substitute into the Ksp expression and solve for x.

Example: The Ksp of Ag2CO3 is $8.1 \times 10^{-12}$. What is its molar solubility?

Solubility Product Equilibria & the Common Ion Effect

The Common Ion Effect

The common ion effect occurs when a salt is dissolved in a solution that already contains one of its ions. The presence of a common ion reduces the solubility of the salt due to Le Châtelier's principle.

• If a common ion is added, the equilibrium shifts to favor the solid, decreasing the salt's solubility.

Example:

$HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$

If HCl is added to a solution containing AgCl, the increased [Cl-] causes more AgCl(s) to precipitate.

Ksp Calculations with a Common Ion

1. Write out the equilibrium reaction.

2. Make an ICE (Initial, Change, Equilibrium) table, using the common ion concentration as an initial value.

3. Solve for the unknown (usually the molar solubility).

Example: What is the molar solubility of CuBr(s) if the solid is added to a solution containing 0.45 M KBr? Ksp for CuBr is $6.27 \times 10^{-9}$.

Additional Concepts from Chapter 17

Key Topics to Know

• Common ion effect for weak acid-base equilibria

• Buffers: definition, buffer range, and buffer capacity

• How to recognize titration graphs and identify stages

• How to write dissociation reactions

• Use of the Henderson-Hasselbalch (H-H) equation in buffer calculations

• Titration calculations for:

  • A strong acid and strong base

  • A weak acid and strong base

• Finding Ksp from molar solubility and vice versa

Summary Table: Solubility Product Calculations

Given	Find	Steps
Molar solubility (s)	Ksp	1. Write dissolution equation 2. Express ion concentrations in terms of s 3. Substitute into Ksp expression 4. Solve for Ksp
Ksp	Molar solubility (s)	1. Write dissolution equation 2. Express ion concentrations in terms of s 3. Substitute Ksp value 4. Solve for s
Ksp and common ion concentration	Molar solubility (s)	1. Write dissolution equation 2. Set up ICE table with initial common ion 3. Substitute into Ksp expression 4. Solve for s