Understanding Weak Acid and Base Equilibria: Essential Guide for AP Chemistry

While strong acids and bases completely dissociate in water, weak acids and bases only partially dissociate, establishing an equilibrium between their dissociated and undissociated forms. In this section, we’ll explore how to calculate the pH of weak acid and base solutions using equilibrium concepts, offering key insights into their behavior in aqueous solutions.

Strong vs. Weak Acids and Bases

Strong Acids and Bases

Strong acids and bases completely dissociate in water, meaning they exist entirely as ions in solution. For example, HCl (hydrochloric acid) dissociates fully as: HCl → H⁺ + Cl⁻

Weak Acids and Bases

Weak acids and bases, on the other hand, only partially dissociate in water, establishing a dynamic equilibrium. For instance, acetic acid (CH₃COOH) partially dissociates: CH₃COOH ⇌ CH₃COO⁻ + H⁺

Key Characteristics:

• Strong acids/bases have negligible reverse reactions, and their conjugates are extremely weak.
• Weak acids/bases exist in an equilibrium state, meaning their conjugates retain measurable acidic or basic properties.

Equilibrium Constants for Weak Acids and Bases

The extent to which a weak acid or base dissociates is represented by the equilibrium constant:

• Ka (acid dissociation constant) for weak acids
• Kb (base dissociation constant) for weak bases

Example:

For acetic acid, Ka = 1.8 × 10⁻⁵. A smaller Ka value indicates weaker dissociation, meaning most of the acid remains undissociated.

Calculating pH for Weak Acids

Example Problem:

Find the pH of a 2M solution of CH₃COOH (Ka = 1.8 × 10⁻⁵).

Step 1: Write the Dissociation Equation

CH₃COOH ⇌ CH₃COO⁻ + H⁺

Step 2: Set Up an ICE Table

Step 3: Write the Ka Expression

Ka = [CH₃COO⁻][H⁺] / [CH₃COOH]
1.8 × 10⁻⁵ = (x)(x) / (2 – x)

Since x is small, we approximate 2 – x ≈ 2: 1.8 × 10⁻⁵ = x² / 2
x² = 1.8 × 10⁻⁵ × 2
x ≈ 0.006 M

Step 4: Calculate pH

pH = -log[H⁺] = -log(0.006) ≈ 2.22

Calculating pH for Weak Bases

Example Problem:

Find the pH of a 1M solution of NH₃ (Kb = 1.8 × 10⁻⁵).

Step 1: Write the Base Dissociation Equation

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

Step 2: Set Up an ICE Table

Step 3: Write the Kb Expression

Kb = [NH₄⁺][OH⁻] / [NH₃]
1.8 × 10⁻⁵ = (x)(x) / (1 – x)

Approximate 1 – x ≈ 1: 1.8 × 10⁻⁵ = x²
x ≈ 0.0042 M

Step 4: Calculate pOH and pH

pOH = -log[OH⁻] = -log(0.0042) ≈ 2.38
pH = 14 – pOH = 14 – 2.38 = 11.62

Key Points to Remember

• Weak acids and bases partially dissociate and establish equilibrium.
• Use ICE tables to track changes in concentrations and calculate pH or pOH.
• For weak acids/bases, Ka and Kb values provide a measure of dissociation.