Salt Hydrolysis: Calculating Kh, Degree h and Solution pH
Salt hydrolysis is a central topic in chemical equilibrium for secondary and entrance exams. When a salt dissolves in water, it dissociates into cations and anions. Depending on whether these ions originate from weak or strong electrolytes, they may react with water to generate excess hydronium or hydroxide ions. Understanding how to calculate the hydrolysis constant Kh, degree of hydrolysis h, and solution pH allows students to solve examination numericals methodically.
Types of Salts and Hydrolysis Behavior
Salts are grouped into four categories depending on their parent acids and bases:
- Strong Acid + Strong Base (e.g., NaCl): Neither ion hydrolyses. The solution remains neutral with pH = 7.00 at 25 °C.
- Weak Acid + Strong Base (e.g., CH3COONa): The conjugate base anion hydrolyses with water, releasing OH⁻ ions. The solution is basic (pH > 7).
- Strong Acid + Weak Base (e.g., NH4Cl): The conjugate acid cation hydrolyses with water, producing H3O⁺ ions. The solution is acidic (pH < 7).
- Weak Acid + Weak Base (e.g., CH3COONH4): Both cation and anion hydrolyse. The pH depends on the relative values of Ka and Kb, independent of salt concentration.
Fundamental Hydrolysis Formulas
Weak Acid – Strong Base Salt (Anionic Hydrolysis):
Hydrolysis equilibrium: A⁻(aq) + H2O(l) ⇌ HA(aq) + OH⁻(aq)
Kh = Kw / Ka
h = √(Kh / c) = √(Kw / (Ka · c)) [valid when h < 0.05]
[OH⁻] = c · h = √((Kw · c) / Ka)
pH = 7 + 0.5·pKa + 0.5·log10(c) [at 25 °C where pKw = 14]
Strong Acid – Weak Base Salt (Cationic Hydrolysis):
Hydrolysis equilibrium: B⁺(aq) + H2O(l) ⇌ BOH(aq) + H⁺(aq)
Kh = Kw / Kb
h = √(Kh / c) = √(Kw / (Kb · c)) [valid when h < 0.05]
[H⁺] = c · h = √((Kw · c) / Kb)
pH = 7 - 0.5·pKb - 0.5·log10(c) [at 25 °C]
Variables and units: Kw = ionic product of water (1.00 × 10⁻¹⁴ mol² L⁻² at 25 °C); Ka = acid dissociation constant (mol L⁻¹); Kb = base dissociation constant (mol L⁻¹); c = initial salt concentration (mol L⁻¹ or M); h = degree of hydrolysis (dimensionless fraction); Kh = hydrolysis constant (mol L⁻¹ or dimensionless).
Worked Examples with Step-by-Step Solutions
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: Class 11-12 Boards, JEE Main, NEET-UG, CUET-UG
- Topic: Chemical Equilibrium — Salt Hydrolysis
- Difficulty: School to entrance level
Question: Calculate the hydrolysis constant Kh, the degree of hydrolysis h, and the pH of a 0.10 M aqueous sodium acetate (CH3COONa) solution at 25 °C. Given: Ka(CH3COOH) = 1.80 × 10⁻⁵ and Kw = 1.00 × 10⁻¹⁴.
Step-by-step solution:
Step 1: Identify the salt type. Sodium acetate comes from a weak acid (CH3COOH) and a strong base (NaOH). The acetate ion undergoes anionic hydrolysis:
CH3COO⁻ + H2O ⇌ CH3COOH + OH⁻
Step 2: Calculate the hydrolysis constant Kh.
Kh = Kw / Ka = (1.00 × 10⁻¹⁴) / (1.80 × 10⁻⁵) = 5.56 × 10⁻¹⁰
Step 3: Calculate the degree of hydrolysis h.
For concentration c = 0.10 M, using the approximation (1 - h) ≈ 1:
h = √(Kh / c) = √((5.556 × 10⁻¹⁰) / 0.10) = √(5.556 × 10⁻⁹) = 7.45 × 10⁻⁵
Since h = 0.00745% (< 5%), the assumption holds valid.
Step 4: Compute [OH⁻] and pOH.
[OH⁻] = c · h = 0.10 × 7.454 × 10⁻⁵ = 7.454 × 10⁻⁶ M
pOH = -log10(7.454 × 10⁻⁶) = 6 - 0.8724 = 5.128
Step 5: Determine solution pH.
pH = 14.000 - 5.128 = 8.87
Step 6: Double-check by second route using the direct formula.
pKa = -log10(1.80 × 10⁻⁵) = 4.745; log10(c) = log10(0.10) = -1.000
pH = 7 + 0.5·pKa + 0.5·log10(c) = 7 + (0.5 × 4.745) + (0.5 × (-1.000)) = 7 + 2.3725 - 0.5000 = 8.87
Both calculation paths confirm the exact same value.
Answer: Kh = 5.56 × 10⁻¹⁰, h = 7.45 × 10⁻⁵, pH = 8.87
Question: Find the hydrolysis constant Kh, degree of hydrolysis h, and pH of a 0.050 M ammonium chloride (NH4Cl) solution at 25 °C. Given: Kb(NH3) = 1.80 × 10⁻⁵ and Kw = 1.00 × 10⁻¹⁴.
Step-by-step solution:
Step 1: Identify the salt type. Ammonium chloride is formed from a strong acid (HCl) and a weak base (NH3). The ammonium cation hydrolyses:
NH4⁺ + H2O ⇌ NH3 + H3O⁺
Step 2: Calculate the hydrolysis constant Kh.
Kh = Kw / Kb = (1.00 × 10⁻¹⁴) / (1.80 × 10⁻⁵) = 5.56 × 10⁻¹⁰
Step 3: Determine the degree of hydrolysis h.
With c = 0.050 M and assuming (1 - h) ≈ 1:
h = √(Kh / c) = √((5.556 × 10⁻¹⁰) / 0.050) = √(1.1112 × 10⁻⁸) = 1.054 × 10⁻⁴
Here h = 0.0105%, fully verifying the approximation.
Step 4: Determine [H⁺] and pH.
[H⁺] = c · h = 0.050 × 1.054 × 10⁻⁴ = 5.270 × 10⁻⁶ M
pH = -log10(5.270 × 10⁻⁶) = 6 - 0.7218 = 5.28
Step 5: Double-check by second route using the direct formula.
pKb = -log10(1.80 × 10⁻⁵) = 4.745; log10(0.050) = -1.3010
pH = 7 - 0.5·pKb - 0.5·log10(c) = 7 - (0.5 × 4.745) - (0.5 × (-1.3010)) = 7 - 2.3725 + 0.6505 = 5.28
Both routes give an identical pH of 5.28.
Answer: Kh = 5.56 × 10⁻¹⁰, h = 1.05 × 10⁻⁴, pH = 5.28
Common Mistakes in Hydrolysis Calculations
- Swapping Ka and Kb: The denominator in Kh always contains the dissociation constant of the weak parent species. For acetate salts use Ka; for ammonium salts use Kb.
- Using the simplified square-root formula when h is large: When h exceeds 0.05 (5%), you cannot assume (1 - h) ≈ 1. You must solve the quadratic equation c·h² + Kh·h - Kh = 0.
- Formula sign errors: Remember that basic solutions require a plus sign (pH = 7 + 0.5·pKa + 0.5·log10(c)), while acidic solutions require a minus sign (pH = 7 - 0.5·pKb - 0.5·log10(c)).
- Ignoring temperature effects on Kw: The value Kw = 1.00 × 10⁻¹⁴ holds at 25 °C. At different temperatures, neutral pH shifts and Kw must be adjusted accordingly.
Exam Relevance and Question Patterns
Hydrolysis problems appear routinely in Class 11 examinations as well as entrance tests. The following overview indicates how this topic is evaluated:
| Examination | Question Format | Typical Problem Focus |
|---|---|---|
| CBSE and State Boards | Derivation and direct numericals | Deriving expressions for Kh and h; calculating pH for acetate and ammonium salts. |
| JEE Main | Multiple-choice and numerical entry | Calculating pH after incomplete acid-base neutralisation or dilution of salt solutions. |
| NEET-UG | Direct objective questions | Rapid evaluation of salt acidity or basicity, and formula-based pH determinations. |
| CUET-UG | Conceptual and numerical MCQs | Relating Kh to parent dissociation constants and comparing degrees of hydrolysis. |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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