Lecture Summary: Calculating pH of Weak Acid and Base Solutions

Overview

This lecture discusses how to calculate the pH of weak acid and weak base solutions, with a focus on concepts like percent ionization. It also reviews basics of strong acids/bases and equilibrium expressions.

Basics of Strong and Weak Acids/Bases

Strong Acids

• Complete dissociation in water (e.g., HCl).
• Use single arrow in reactions.
• pH Calculation:
  • pH = -log[H₃O⁺]
  • POH via -log[OH⁻]
  • pH + POH = 14 at 25°C

Weak Acids and Bases

• Partial dissociation, reversible reactions (use double arrow).
• Equilibrium concept:
  • Weak acids use Ka:
    • Ka = [products]/[reactants]
    • Water (liquid) is excluded.
  • Weak bases use Kb:
    • Kb = [products]/[reactants]

Relationship Between Ka and Kb

• Ka × Kb = Kw = 1 × 10⁻¹⁴ (at 25°C)

Acid Strength

• Stronger acids have higher Ka values.
• pKa relates to acid strength inversely (lower pKa = stronger acid).
• Formulas:
  • pKa = -log(Ka)
  • pKb = -log(Kb)
  • pKa + pKb = 14 at 25°C

Percent Ionization

• Formula:
  • Percent ionization = (x / initial concentration) × 100%
  • x is the concentration of ion formed (H₃O⁺ for acids, OH⁻ for bases).

Example Problems

1. pH of 0.75 M Acetic Acid

• Steps:
  1. Write dissociation reaction.
  2. Use ICE table for equilibrium concentrations.
  3. Calculate using Ka = x² / (initial - x).
  4. Approximate x if Ka is small.
  5. Solve for [H₃O⁺] and then pH using -log[H₃O⁺].

2. pH of 0.25 M Ammonia Solution

• Steps:
  1. Write base dissociation reaction.
  2. Use ICE table.
  3. Calculate using Kb = x² / (initial - x).
  4. Approximate if Kb is small.
  5. Solve for [OH⁻], calculate pOH and then pH.

3. pH of 0.4 M Ammonium Chloride Solution

• Concept:
  • Recognize NH₄⁺ as a weak acid.
  • Use Ka (calculate from Kb).
  • Follow similar steps as above.

4. pH of 1.5 M Sodium Fluoride Solution

• Concept:
  • Recognize F⁻ as a weak base.
  • Use Kb (calculate from given Ka of HF).
  • Follow similar steps as above.

Additional Notes

• Quadratic Formula may be needed if approximation isn’t valid:
  • ax² + bx + c = 0
  • x = [-b ± sqrt(b² - 4ac)] / 2a