Acid-Base Equilibria and Solubility: Practice Exam Study Notes (Chapters 16-17)

Acid-Base Equilibria

Bronsted-Lowry Acids and Bases

The Bronsted-Lowry theory defines acids as proton donors and bases as proton acceptors. Acid-base reactions involve the transfer of a proton (H+) from the acid to the base.

• Conjugate Acid-Base Pair: When an acid donates a proton, it forms its conjugate base; when a base accepts a proton, it forms its conjugate acid.

• Example: $\mathrm{PO_4^{3-} + H_2O \rightleftharpoons HPO_4^{2-} + OH^-}$ Here, $\mathrm{PO_4^{3-}}$ acts as a base, accepting a proton from water.

Lewis Acids and Bases

Lewis acids are electron pair acceptors, while Lewis bases are electron pair donors.

• Examples: $\mathrm{Al^{3+}}$ is a Lewis acid because it can accept electron pairs.

Acid and Base Strength

The strength of an acid or base is determined by its degree of ionization in water.

• Strong Acids/Bases: Completely ionize in solution (e.g., HCl, NaOH).

• Weak Acids/Bases: Partially ionize (e.g., CH3COOH, NH3).

pH and pOH Calculations

pH is a measure of the hydrogen ion concentration in a solution. It is calculated as:

• $\mathrm{pH = -\log[H^+]}$

• $\mathrm{pOH = -\log[OH^-]}$

• $\mathrm{pH + pOH = 14}$ (at 25°C)

• Example: For $[H^+] = 1.0 \times 10^{-3}$ M, $\mathrm{pH} = 3$.

Buffer Solutions

Buffers are solutions that resist changes in pH upon addition of small amounts of acid or base. They typically consist of a weak acid and its conjugate base, or a weak base and its conjugate acid.

• Common Buffer Pair: CH3COOH and CH3COO-

• Buffer Capacity: The amount of acid or base a buffer can neutralize before the pH begins to change appreciably.

• Henderson-Hasselbalch Equation: $\mathrm{pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)}$

Indicators

Indicators are substances that change color depending on the pH of the solution. They are used to determine the endpoint of titrations.

• Methyl Orange: Changes color in the pH range 3.1–4.4.

Titrations

Titration is a technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration.

• Equivalence Point: The point at which the amount of titrant added is stoichiometrically equivalent to the amount of analyte in solution.

• Example: Titrating HCl with NaOH.

Solubility and Solubility Product (Ksp)

Solubility Equilibria

Solubility product constant ($K_{sp}$) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound.

• General Form: For $\mathrm{AB_2}$: $\mathrm{AB_2 (s) \rightleftharpoons A^{2+} (aq) + 2B^- (aq)}$

• $K_{sp} = [A^{2+}][B^-]^2$

• Molar Solubility: The number of moles of solute that dissolve per liter of solution.

• Example: For $\mathrm{Ag_3PO_4}$, $K_{sp} = [Ag^+]^3[PO_4^{3-}]$

Calculating Molar Solubility

Molar solubility can be calculated from $K_{sp}$ using stoichiometry.

• Example: For $\mathrm{CaF_2}$, $K_{sp} = [Ca^{2+}][F^-]^2$

• If $s$ is the molar solubility: $[Ca^{2+}] = s$, $[F^-] = 2s$ $K_{sp} = s \times (2s)^2 = 4s^3$

Common Ion Effect

The presence of a common ion decreases the solubility of a salt.

• Example: Adding NaCl to a solution of AgCl decreases AgCl solubility due to increased [Cl-].

Mathematical Expressions for Solubility

• For $\mathrm{Sr(OH)_2}$: $K_{sp} = [Sr^{2+}][OH^-]^2$ If $s$ is the molar solubility: $[Sr^{2+}] = s$, $[OH^-] = 2s$ $K_{sp} = s \times (2s)^2 = 4s^3$ $s = \sqrt[3]{\frac{K_{sp}}{4}}$

Sample Solubility Table

Compound	Ksp	Molar Solubility (s)
Ag3PO4	1.6 × 10-18	1.57 × 10-6 M
CaF2	1.2 × 10-10	2.9 × 10-4 M
BaSO4	1.1 × 10-10	1.1 × 10-5 M
Fe(OH)3	1.6 × 10-39	1.6 × 10-13 M

Summary Table: Key Concepts

Concept	Definition	Example
Buffer	Resists pH change	CH3COOH/CH3COO-
Strong Acid	Completely ionizes	HCl
Weak Acid	Partially ionizes	CH3COOH
Solubility Product (Ksp)	Equilibrium constant for dissolution	AgCl: $K_{sp} = [Ag^+][Cl^-]$

Additional info:

• Some questions referenced methyl orange and buffer calculations; these are standard topics in acid-base equilibria and solution chemistry.

• Solubility product calculations and common ion effect are key for understanding precipitation and selective dissolution.