Hydrolysis of Salts

Hydrolysis

• Hydrolysis is the reaction of an anion and/or cation of a salt with water.
• Salt hydrolysis usually affects the pH of a solution.

Nature of Salts

• Salts are ionic compounds formed from the reaction of an acid and a base.
• The strength of the acid/base affects the nature of the salt (neutral, acidic, or basic).

Neutral Salts

• Have a pH of 7 and equal concentrations of H^+ and OH^- ions.
• Formed from strong acids and strong bases (e.g., NaCl).
• Ions that don't react appreciably with water:
  • Cations from strong bases: Alkali metal cations of group 1A (Li+, Na+, K+), Alkaline earth cations of group 2A (Mg2+, Ca2+, Sr2+, Ba2+) except for Be2+
  • Anions from strong monoprotic acids: Cl-, Br-, I-, NO3-, and ClO4-

Basic Salts

• Formed from reactions of weak acids and strong bases (e.g., CH3COONa).
• Reaction: CH3COONa(s) \rightarrow CH3COO^-(aq) + Na^+(aq)
• Hydrolysis: CH3COO^- + H2O \rightleftharpoons CH_3COOH + OH^-
• Hydrolysis Constant: Kh = \frac{[CH3COOH][OH^-]}{[CH_3COO^-]}
• Kb \equiv Kh = \frac{Kw}{Ka}

Acidic Salts

• Formed from a reaction of a weak base and a strong acid (e.g., NH4Cl).
• Reaction: NH4Cl(aq) \rightarrow NH4^+(aq) + Cl^-(aq)
• Hydrolysis: NH4^+(aq) + H2O(l) \rightleftharpoons NH3(aq) + H3O^+(aq)
• Hydrolysis Constant: Kh = \frac{[NH3][H3O^+]}{[NH4^+]}
• Ka \equiv Kh = \frac{Kw}{Kb}

pH of Salt Solutions

• Example calculation for 0.5M NH4Cl:
  • Ka = \frac{Kw}{K_b} = \frac{1.0 × 10^{-14}}{1.8 × 10^{-5}} = 5.6 × 10^{-10}
  • Ka = \frac{[NH3][H3O^+]}{[NH4^+]} = \frac{x^2}{0.50-x} ≈ \frac{x^2}{0.50}
  • x = \sqrt{(5.6 × 10^{-10}) × (0.50)} = 1.67 × 10^{-5}M
  • pH = -log(1.67 × 10^{-5}) = 4.77

Salts from Weak Acids and Weak Bases

• The resulting solution can be acidic, basic, or neutral.
• Examples: NH4F, CH3COONH4, NH4CN
  • If Kb > Ka, the solution is basic.
  • If Kb < Ka, the solution is acidic.
  • If Kb = Ka, the solution is neutral.
• Example: (NH4)2CO3 solution is basic because Ka < Kb