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).
- Cations from strong bases:
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 ).