Hydrolysis Notes

Hydrolysis

Objectives

• Define the term “Hydrolysis.”
• Illustrate the “Nature” of salts.
• Determine the relationship between Hydrolysis Constant to $K<em>a$ or $K</em>b$.

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:
    • $Cl^-, Br^-, I^-, NO<em>3^-$, and $ClO</em>4^-$
  • 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)
  • $CH<em>3COOH + NaOH \rightleftharpoons CH</em>3COONa + H_2O$
• The reaction for the dissociation of this salt is:
  • $CH<em>3COONa(s) \rightarrow CH</em>3COO^-(aq) + Na^+(aq)$
• The conjugate base of the weak acid can then act as a base and react with water as follows:
  • $CH<em>3COO^- + H</em>2O \rightleftharpoons CH_3COOH + OH^-$

Basic Salts - Hydrolysis Constant

• $CH<em>3COO^- + H</em>2O \rightleftharpoons CH_3COOH + OH^-$
• From this reaction, we can write an expression for the “Hydrolysis Constant.”
  • $K<em>h = \frac{[CH</em>3COOH][OH^-]}{[CH_3COO^-]}$
• This expression also represents the $K_b$ expression.
  • Thus, $K<em>b = K</em>h$
• Now $K<em>w = K</em>a \times K_b$
  • So: $K<em>b \equiv K</em>h = \frac{K<em>w}{K</em>a}$

Acidic Salts

• Acidic salts are formed from a reaction of a weak base and a strong acid.
  • E.g. $NH_4Cl$
  • $NH<em>3(aq) + HCl(aq) \rightarrow NH</em>4Cl(aq)$
• The reaction for the dissociation of this salt is:
  • $NH<em>4Cl(aq) \rightarrow NH</em>4^+(aq) + Cl^-(aq)$
• The conjugate acid of the weak base can then act as an acid and react with water as follows:
  • $NH<em>4^+(aq) + H</em>2O(l) \rightleftharpoons NH<em>3(aq) + H</em>3O^+(aq)$

Acidic Salts - Hydrolysis Constant

• $NH<em>4^+(aq) + H</em>2O(l) \rightleftharpoons NH<em>3(aq) + H</em>3O^+(aq)$
• From this reaction, we can write an expression for the “Hydrolysis Constant.”
  • $K<em>h = \frac{[NH</em>3][H<em>3O^+]}{[NH</em>4^+]}$
• This expression also represents the $K_a$ expression.
  • Thus, $K<em>a = K</em>h$
• Now $K<em>w = K</em>a \times K_b$
  • So: $K<em>a \equiv K</em>h = \frac{K<em>w}{K</em>b}$

pH of Salt Solutions

• Calculate the pH of a 0.5M solution of $NH<em>4Cl$. ($K</em>b = 1.8 \times 10^{-5}$)
• The reaction for the dissociation of this salt is:
  • $NH<em>4Cl(aq) \rightarrow NH</em>4^+(aq) + Cl^-(aq)$
• The $NH_4^+(aq)$ then reacts with water as follows:
  • $NH<em>4^+(aq) + H</em>2O(l) \rightleftharpoons NH<em>3(aq) + H</em>3O^+(aq)$
• $K<em>h = K</em>a = \frac{[NH<em>3][H</em>3O^+]}{[NH_4^+]}$
• $K<em>a \equiv K</em>h = \frac{K<em>w}{K</em>b}$
  • 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. $NH<em>4F, CH</em>3COONH<em>4, NH</em>4CN$

• When $K<em>b > K</em>a$: Solution is basic

• When $K<em>b < K</em>a$: Solution is acidic

• When $K<em>b = K</em>a$: Solution is neutral

Salts formed from weak acids and weak bases - Example

• To determine whether an $(NH<em>4)</em>2CO<em>3$ solution is acidic, basic, or neutral, let’s work out the values of $K</em>a$ for $NH<em>4^+$ and $K</em>b$ for $CO_3^{2-}$.
• Because $K<em>a < K</em>b$, the solution is basic (pH > 7 ).