Lecture Notes: Ammonia and Base Chemistry

Ammonia as a Weak Base

• Ammonia (NH3) is a weak base.
• In reaction with water, acts as a Brønsted–Lowry base:
  • Water donates a proton, acting as a Brønsted–Lowry acid.
  • Ammonia accepts the proton, forming ammonium ion (NH4+).
  • The reaction results in hydroxide ions (OH-).

Brønsted–Lowry Acid-Base Theory

• Proton donation and acceptance define acids and bases:
  • Ammonia (NH3) is the base; ammonium (NH4+) is the conjugate acid.
  • Water (H2O) is the acid; hydroxide (OH-) is the conjugate base.

Generic Base Reaction

• Generic base (B) with water:
  • Forms BH+ and OH-.
  • Equilibrium expression uses base ionization constant (Kb).

Base Ionization Constant (Kb)

• Kb measures base strength:
  • Higher Kb = stronger base.
  • Equilibrium expression: [ K_b = \frac{[BH^+][OH^-]}{[B]} ]

Comparison of Weak Bases: Ammonia vs Aniline

• Ammonia (NH3):
  • Kb = 1.8 x 10^-5
• Aniline (C6H5NH2):
  • Kb = 4.3 x 10^-10
  • Ammonia is the stronger base.

pKb Calculation

• pKb = -log(Kb)
• Example:
  • Ammonia: pKb = 4.74
  • Aniline: pKb = 9.37

Calculating pH of Ammonia Solution

• Task: Find pH of 0.500 M ammonia solution.
• Reaction: NH3 + H2O → NH4+ + OH-

Equilibrium Concentrations

• Initial: [NH3] = 0.500 M, [NH4+] = 0, [OH-] = 0
• Change: [NH3] decreases by X, [NH4+] and [OH-] increase by X.
• Equilibrium: [NH3] = 0.500 - X, [NH4+] = X, [OH-] = X

Calculating Equilibrium

• Use Kb value for NH3: 1.8 x 10^-5
• Assumption: X << 0.500, simplifies math.
• Solve: [ X^2 = 9.0 \times 10^{-6} ]
• X = 0.0030 M (concentration of OH-).

Determining pH

• Calculate pOH: pOH = -log[OH-] = 2.52

• Use relationship: pH + pOH = 14

• pH = 14 - 2.52 = 11.48

• Conclusion: pH of the ammonia solution is 11.48.