Hydrolysis Notes

Hydrolysis

Objectives

  • Define the term “Hydrolysis.”
  • Illustrate the “Nature” of salts.
  • Determine the relationship between Hydrolysis Constant to Ka or Kb.

What is 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

  • When an acid and a base react, an ionic compound called a salt is formed.
  • The strength of the acid/base affects the nature of the salt.
  • Nature of Salts:
    • Neutral
    • Acidic
    • Basic

Neutral Salts

  • Salts that have a pH of 7 and an equal concentration of H^+ and OH^- ions.
  • These salts can be formed from:
    • Strong acids and strong bases.
    • E.g. NaCl

Neutral Salts - Ions that do not react with water

  • The following ions do not react appreciably with water to produce either H_3O^+ or OH^- ions:
    • Cations from strong bases:
      • Alkali metal cations of group 1A (Li^+, Na^+, K^+)
      • Alkaline earth cations of group 2A (Mg^{2+}, Ca^{2+}, Sr^{2+}, Ba^{2+}) except for Be^{2+}
    • Anions from strong monoprotic acids:
      • Cl^-, Br^-, I^-, NO3^-, and ClO4^-
    • Salts that contain only these ions give neutral solutions in pure water (pH 7).

Basic Salts

  • Basic salts are formed from reactions of a weak acid and a strong base.
    • E.g. CH_3COONa (sodium ethanoate)
    • CH3COOH + NaOH \rightleftharpoons CH3COONa + H_2O
  • The reaction for the dissociation of this salt is:
    • CH3COONa(s) \rightarrow CH3COO^-(aq) + Na^+(aq)
  • The conjugate base of the weak acid can then act as a base and react with water as follows:
    • CH3COO^- + H2O \rightleftharpoons CH_3COOH + OH^-

Basic Salts - Hydrolysis Constant

  • CH3COO^- + H2O \rightleftharpoons CH_3COOH + OH^-
  • From this reaction, we can write an expression for the “Hydrolysis Constant.”
    • Kh = \frac{[CH3COOH][OH^-]}{[CH_3COO^-]}
  • This expression also represents the K_b expression.
    • Thus, Kb = Kh
  • Now Kw = Ka \times K_b
    • So: Kb \equiv Kh = \frac{Kw}{Ka}

Acidic Salts

  • Acidic salts are formed from a reaction of a weak base and a strong acid.
    • E.g. NH_4Cl
    • NH3(aq) + HCl(aq) \rightarrow NH4Cl(aq)
  • The reaction for the dissociation of this salt is:
    • NH4Cl(aq) \rightarrow NH4^+(aq) + Cl^-(aq)
  • The conjugate acid of the weak base can then act as an acid and react with water as follows:
    • NH4^+(aq) + H2O(l) \rightleftharpoons NH3(aq) + H3O^+(aq)

Acidic Salts - Hydrolysis Constant

  • NH4^+(aq) + H2O(l) \rightleftharpoons NH3(aq) + H3O^+(aq)
  • From this reaction, we can write an expression for the “Hydrolysis Constant.”
    • Kh = \frac{[NH3][H3O^+]}{[NH4^+]}
  • This expression also represents the K_a expression.
    • Thus, Ka = Kh
  • Now Kw = Ka \times K_b
    • So: Ka \equiv Kh = \frac{Kw}{Kb}

pH of Salt Solutions

  • Calculate the pH of a 0.5M solution of NH4Cl. (Kb = 1.8 \times 10^{-5})
  • The reaction for the dissociation of this salt is:
    • NH4Cl(aq) \rightarrow NH4^+(aq) + Cl^-(aq)
  • The NH_4^+(aq) then reacts with water as follows:
    • NH4^+(aq) + H2O(l) \rightleftharpoons NH3(aq) + H3O^+(aq)
  • Kh = Ka = \frac{[NH3][H3O^+]}{[NH_4^+]}
  • Ka \equiv Kh = \frac{Kw}{Kb}
    • So K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}

pH of Salt Solutions - Calculation

  • Now K_a = \frac{(x)(x)}{(0.50-x)}
  • We assume that x is so small, thus 0.50-x = 0.50
  • 5.6 \times 10^{-10} = \frac{x^2}{(0.50)}
  • x = \sqrt{(5.6 \times 10^{-10}) \times (0.50)} = 1.67 \times 10^{-5}M
  • So pH = -log(1.67 \times 10^{-5}) = 4.77
  • pH is below 7 which proves the acidic nature of the salt

Salts formed from weak acids and weak bases

  • If the reaction has a “weak acid” and a “weak base,” it can form either acidic, basic, or neutral salt solution

    • E.g. NH4F, CH3COONH4, NH4CN
  • When Kb > Ka: Solution is basic

  • When Kb < Ka: Solution is acidic

  • When Kb = Ka: Solution is neutral

Salts formed from weak acids and weak bases - Example

  • To determine whether an (NH4)2CO3 solution is acidic, basic, or neutral, let’s work out the values of Ka for NH4^+ and Kb for CO_3^{2-}.
  • Because Ka < Kb, the solution is basic (pH > 7 ).