Hydrolysis Of Salts

1.0What is Hydrolysis of Salts?

Hydrolysis refers to the reaction of the cation, the anion, or salt ions with water, resulting in an acidic or basic solution. It is the reverse of neutralisation. The extent of hydrolysis depends on the strength of the acid and base from which the salt is derived.

Hydrolysis Constant (K_h)

Reaction: Consider the general hydrolysis reaction for a salt (BA):

$BA+H_{2}O\rightleftharpoons HA+BOH$

Where HA  is the acid and BOH is the base.

• Applying the law of chemical equilibrium:

$\frac{[\text{HA}][\text{BOH}]}{[\text{BA}][{\text{H}}_2\text{O}]}=K$

• Since water (H_2O) is present in large excess, its concentration is considered constant:

$\frac{[\text{HA}][\text{BOH}]}{[\text{BA}]}=K\cdot [H_{2}O]=K_h$

Here, $K_h$ ​is called the hydrolysis constant.

Degree of Hydrolysis (h)

• The degree of hydrolysis is the fraction or percentage of the total salt hydrolysed at equilibrium:

$h=\frac{\text{Number of moles of salt hydrolyzed}}{\text{Total number of moles of salt taken}}$

• For example, if 90% of the salt is hydrolysed, h=0.90

Hydrolysis behaviour varies based on the strength of the acid and base that form the salt. Understanding the equilibrium expressions, hydrolysis constant (K_h​), degree of hydrolysis (h), and their relationship to pH is essential for predicting the nature of the solution.

2.0Types of Salts and Hydrolysis Behavior

Salt of Weak Acid and Strong Base

• Example: Sodium acetate (CH_3COONa)
• Hydrolysis constant: $K_h=\frac{K_w}{K_a}$
• Degree of hydrolysis: $h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{K_w}{K_a\cdot C}}$
• pH:$\text{pH}=\frac{1}{2}\left[\log K_w+\log K_a-\log C\right]$

Cations (e.g., Na⁺, K⁺, Li⁺): These cations come from strong bases and do not hydrolyse. They remain inert in solution and do not affect the pH.

Anions (e.g., F⁻, CN⁻, CH₃COO⁻, CO₃²⁻​): These anions are the conjugate bases of weak acids. They are more substantial bases and hydrolyse in water to produce OH− ions, making the solution basic.

The solution is basic (pH>7) due to the production of hydroxide ions (OH−).

Salt of Strong Acid and Weak Base

• Example: Ammonium chloride (NH₄Cl)
• Hydrolysis constant: $K_h=\frac{K_w}{K_b}$
• Degree of hydrolysis: $h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{K_w}{K_b\cdot C}}$
• $\text{pH}=\frac{1}{2}\left[\log K_w-\log K_b+\log C\right]$
• The solution becomes acidic due to the production of H₃O⁺, and the overall reaction can be written as:
  NH₄Cl + H₂O → NH₃ + H₃O⁺ + Cl⁻

Effect on pH: The production of H₃O⁺ ions lowers the pH of the solution, making it acidic (pH<7).

Cations from Weak Bases:
Examples: NH₄⁺​, Zn²⁺, Fe³⁺, Cu²⁺, Al³⁺
These cations hydrolyse in water to produce H⁺ ions (or H₃O⁺).

Anions from Strong Acids:
Examples: Cl⁻, Br⁻, NO⁻₃​, SO²⁻₄​, ClO⁻₄
​These anions do not hydrolyse because they are conjugate bases of strong acids.

Nature of the Solution:
The solution is acidic because only the cation hydrolyses, increasing H⁺ concentration.

Salt of Weak Acid and Weak Base

Example: Ammonium acetate (NH₄CH₃COO)

• Hydrolysis constant: $K_h=\frac{K_w}{K_a\cdot K_b}$
• Degree of hydrolysis: $h=\sqrt{K_h}=\sqrt{\frac{K_w}{K_a\cdot K_b}}$
• $\text{pH}=\frac{1}{2}\left[\log K_a+\log K_w-\log K_b\right]$

When a salt is formed from a weak acid and a weak base (e.g., NH₄CH₃COO), the cation and the anion undergo hydrolysis.

The pH of the solution depends on the relative strengths of the acid K_a​) and the base (K_b​):

• If K_a > K_b​: The solution is acidic (pH<7).
• If K_a< K_b​: The solution is basic (pH>7).
• If K_a= K_b​: The solution is neutral (pH=7).

Example Reaction: For ammonium acetate (NH₄CH₃COO):

• NH₄⁺+H₂O⇌NH₃+H₃O⁺
• CH3COO⁻+H₂O⇌CH₃COOH+OH⁻

Salt of Strong Acid and Strong Base

Cations from Strong Bases:

• Examples: Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺
• These cations are derived from strong bases like LiOH, NaOH, KOH, etc.
• Why no hydrolysis?
• • These cations have low polarising power because they are large and have a single positive charge.
  • They do not react with water to form any acidic or basic species.
  • Therefore, they do not affect the pH of the solution.
• These salts ionise entirely, and their aqueous solutions are neutral (pH=7).

Anions from Strong Acids:

• Examples: Cl⁻, Br⁻, I⁻, NO₃⁻​, SO₄²⁻​, ClO⁻₄
• These anions are derived from strong acids like HCl, HBr, HI, HNO₃, H₂SO₄​, and HClO₄​.
• Why no hydrolysis?
• • These anions are very weak bases because they are the conjugate bases of strong acids, which dissociate completely in water.
  • They do not react with water to produce hydroxide ions (OH⁻).

3.0Buffer Solution

A buffer solution is a special type of solution that resists significant changes in its pH when diluted or when small amounts of a strong acid or strong base are added. This stability is crucial in many chemical and biological processes. A buffer solution is typically composed of:

1. A weak acid and its salt with a strong base (Acidic Buffer)
2. A weak base and its salt with a strong acid (Basic Buffer)

Mechanism: A buffer neutralises added acids or bases through equilibrium reactions.

Acidic Buffer 

(Example: Acetic Acid + Sodium Acetate)

An example of an acidic buffer is a solution containing acetic acid (CH₃COOH) and its salt, sodium acetate (CH₃COONa).

• Dissociation of Acetic Acid (Weak Acid)
  CH₃COOH⇌H⁺+ CH₃COO⁻
  Since acetic acid is a weak acid, it partially ionises in solution.
• Dissociation of Sodium Acetate (Salt of Strong Base and Weak Acid)
  CH₃COONa→Na⁺+CH₃COO⁻
  Sodium acetate completely dissociates, increasing the concentration of acetate ions (CH₃COO⁻) in solution.

Effect of Adding a Strong Acid (H⁺)

When a strong acid (e.g., HCl) is added, the extra H⁺ ions are neutralised by acetate ions:
CH₃COO⁻ + H⁺ → CH₃COOH
Since CH₃COOH is a weak acid, it remains primarily unionised, preventing a sharp drop in pH.

Effect of Adding a Strong Base (OH⁻)

When a strong base (e.g., NaOH) is added, the extra OH⁻ ions react with acetic acid to form water:
CH₃COOH + OH⁻→ CH3COO ⁻+ H₂O

This prevents a significant rise in pH.

Thus, the buffer resists drastic pH changes by adjusting the equilibrium between acetic acid and acetate ions.

Basic Buffer 

(Example: Ammonium Hydroxide + Ammonium Chloride)

A basic buffer consists of a weak base and its salt with a strong acid, such as ammonium hydroxide (NH₄OH) and ammonium chloride (NH₄Cl).

• Dissociation of Weak Base: NH₄OH⇌NH⁴⁺+OH⁻
• Dissociation of Salt (NH₄Cl): NH₄Cl→NH⁴⁺+Cl⁻

The NH₄⁺ ions help neutralise added OH⁻ ions, while the weak base (NH₄OH) neutralises added H⁺ ions, keeping the pH stable.

Significance Of Buffer Solutions

1. Biological Systems: Buffers maintain pH in blood and cellular fluids. The human blood buffer (carbonic acid-bicarbonate system) maintains a pH of around 7.4.
2. Chemical Reactions: Many reactions, such as enzyme activity in biological processes, require a constant pH for optimal results.
3. Industrial Applications: Buffers are used in fermentation, pharmaceuticals, and food industries to maintain stability.
   

4.0Solved Examples

1. Calculate for 0.01 N solution of sodium acetate -

(i)   Hydrolysis constant   (ii) Degree of hydrolysis             (iii) pH

Solution:

Given:

1. K_h=5.26×10⁻¹ (assume this is the ionisation constant or equilibrium constant for water in the context).
2. h =2.29×10⁻⁴ M (possibly the hydroxide ion concentration contributed by some species).
3. Additional [OH⁻] from a strong base, NaOH:

                         C_h=0.01×h=2.29×10⁻⁶ M.

Steps:

1. Total  [OH⁻]: Combine contributions from h and NaOH. If the dominant contribution is h=2.29×10⁻⁴ M, then total [OH⁻]=h+C_h​. However, C_h​ is negligible compared to h, so: [OH⁻]≈h=2.29×10⁻⁴ M.
2. pOH: pOH=−log⁡[OH⁻].

Substituting [OH⁻]=2.29 ×10⁻⁴:
pOH =−log⁡(2.29)−log⁡(10⁻⁴) = 0.36 + 4 = 5.64.

3. pH: Using the relationship pH+pOH=14:
   pH =14−pOH =14−5.6 = 8.36

2. Consider a 1.0 M solution of sodium bisulfite (NaHSO₃) dissolved in water.

a) Write the hydrolysis reactions that occur when NaHSO₃ dissolves in water.
b) Determine whether the NaHSO₃ solution will be acidic, basic, or neutral. Justify your answer with appropriate calculations.

Solution

a) Hydrolysis Reactions

Sodium bisulfite (NaHSO₃) is an acidic salt. In water, it dissociates into its ions, and the bisulfite ion (HSO₃⁻) undergoes hydrolysis:

Dissociation: NaHSO₃(aq) → Na⁺(aq) + HSO₃⁻(aq)
Hydrolysis of HSO₃⁻:
Acidic Hydrolysis: HSO₃⁻(aq) + H₂O(l) ⇌ H₂SO₃(aq) + OH⁻(aq)

Basic Hydrolysis: HSO₃⁻(aq) + H₂O(l) ⇌ SO₃²⁻(aq) + H₃O⁺(aq)

b) Acidity/Basicity of the Solution

We need to compare the equilibrium constants for the acidic and basic hydrolysis reactions to determine whether the solution is acidic or basic.

• Kₐ₂ (for H₂SO₃): 1.02 x 10⁻⁷
• Kₐ₁ (for HSO₃⁻): 6.3 x 10⁻⁸

Since Kₐ₁ (basic hydrolysis) is greater than Kₐ₂ (acidic hydrolysis), the basic hydrolysis reaction will dominate. This means that more H₃O⁺ ions will be produced, producing an acidic solution.

The 1.0 M solution of sodium bisulfite (NaHSO₃) will be acidic due to the greater extent of the basic hydrolysis of the bisulfite ion (HSO₃⁻).