Ionic Equilibrium Part 5 - By Prof Shashank Deep#

Overview#

This advanced module covers complex concepts in ionic equilibrium that are essential for NEET Chemistry. Building upon the foundations from previous parts, we explore sophisticated equilibrium systems and their applications.

Related Topics#

Explore these related concepts to enhance your learning:

• NEET Syllabus Overview
• Practice Questions
• Crash Course Schedule

Advanced Acid-Base Equilibria#

Polyprotic Acids#

Acids that can donate more than one proton per molecule.

Examples:

• H₂SO₄: Sulfuric acid (diprotic)
• H₃PO₄: Phosphoric acid (triprotic)
• H₂CO₃: Carbonic acid (diprotic)

Stepwise Dissociation Constants:

• K₁: First dissociation constant
• K₂: Second dissociation constant
• K₃: Third dissociation constant

Relationship: K₁ > K₂ > K₃

Buffer Solutions#

Solutions that resist changes in pH upon addition of small amounts of acid or base.

Types of Buffers:

1. Acidic Buffer: Weak acid + its salt

   • Example: CH₃COOH + CH₃COONa

2. Basic Buffer: Weak base + its salt

   • Example: NH₄OH + NH₄Cl

Henderson-Hasselbalch Equation:

pH = pKa + log([A⁻]/[HA])

Buffer Capacity:

• Amount of acid or base the buffer can neutralize
• Depends on concentration and ratio of components

Solubility Equilibrium#

Solubility Product (Ksp): The product of molar concentrations of ions in a saturated solution.

Common Ion Effect: Decrease in solubility of a salt due to presence of a common ion.

Examples:

• AgCl in NaCl solution
• CaF₂ in CaCl₂ solution

Complex Ion Formation#

Formation Constants (Kf)#

Stability constants for complex ion formation.

Example:

[Ag(NH₃)₂]⁺: Kf = 1.6 × 10⁷
[Cu(NH₃)₄]²⁺: Kf = 1.1 × 10¹³

Applications#

• Qualitative analysis: Separation and identification of ions
• Electroplating: Metal deposition using complex ions
• Biological systems: Metal ion transport and storage

Advanced Problem Solving#

Problem 1: Buffer pH Calculation#

Question: Calculate the pH of a buffer solution containing 0.1 M CH₃COOH and 0.1 M CH₃COONa. (Ka = 1.8 × 10⁻⁵)

Solution:

pKa = -log(1.8 × 10⁻⁵) = 4.74
pH = 4.74 + log(0.1/0.1) = 4.74

Problem 2: Solubility with Common Ion#

Question: Calculate the solubility of AgCl in 0.1 M NaCl solution. (Ksp = 1.8 × 10⁻¹⁰)

Solution:

AgCl ⇌ Ag⁺ + Cl⁻
[Ag⁺] = s, [Cl⁻] = 0.1 + s ≈ 0.1
Ksp = [Ag⁺][Cl⁻] = s × 0.1
s = Ksp/0.1 = 1.8 × 10⁻⁹ M

Indicators and Titration#

Acid-Base Indicators#

Substances that change color at specific pH ranges.

Common Indicators:

• Phenolphthalein: pH 8.2 - 10.0 (colorless to pink)
• Methyl orange: pH 3.1 - 4.4 (red to yellow)
• Bromothymol blue: pH 6.0 - 7.6 (yellow to blue)

Titration Curves#

Strong Acid - Strong Base:

• pH jump at equivalence point
• Equivalence pH = 7

Weak Acid - Strong Base:

• Gradual pH change
• Equivalence pH > 7

Strong Acid - Weak Base:

• Equivalence pH < 7

Hydrolysis of Salts#

Types of Hydrolysis#

1. Salt of Strong Acid + Weak Base: Acidic solution

   • Example: NH₄Cl

2. Salt of Weak Acid + Strong Base: Basic solution

   • Example: CH₃COONa

3. Salt of Weak Acid + Weak Base: pH depends on Ka and Kb

   • Example: CH₃COONH₄

Hydrolysis Constant (Kh)#

Kh = Kw/Ka (for basic salts)
Kh = Kw/Kb (for acidic salts)

Numerical Problems#

Problem 3: pH of Salt Solution#

Question: Calculate the pH of 0.1 M NH₄Cl solution. (Kb(NH₃) = 1.8 × 10⁻⁵, Kw = 1 × 10⁻¹⁴)

Solution:

Ka(NH₄⁺) = Kw/Kb = 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
[NH₄⁺] = 0.1, [NH₃] = [H₃O⁺] = x

Ka = [NH₃][H₃O⁺]/[NH₄⁺] = x²/0.1
x = √(Ka × 0.1) = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶
pH = -log(7.5 × 10⁻⁶) = 5.12

Key Points for NEET#

1. Memorize important constants: Ka, Kb, Ksp values
2. Understand Henderson-Hasselbalch equation thoroughly
3. Practice buffer calculations regularly
4. Know common ion effect applications
5. Master titration curve patterns

Common Mistakes to Avoid#

1. Confusing Ka and Kb values
2. Ignoring dilution effects in calculations
3. Wrong selection of indicators for titrations
4. Forgetting about autoprotolysis of water
5. Incorrect assumption about complete dissociation

Quick Reference Formulas#

pH and pOH#

pH = -log[H⁺]
pOH = -log[OH⁻]
pH + pOH = 14

Equilibrium Constants#

Ka × Kb = Kw
pH = pKa + log([A⁻]/[HA])
pOH = pKb + log([BH⁺]/[B])

Solubility Product#

Ksp = [Mⁿ⁺][Xᵐ⁻] for MXₘ
s = √(Ksp) for MX salts

Practice Questions#

1. Calculate the pH of 0.01 M HCl solution
2. Determine the solubility of BaSO₄ in pure water (Ksp = 1.1 × 10⁻¹⁰)
3. Find the pH of buffer containing 0.2 M NH₃ and 0.1 M NH₄Cl
4. Calculate the degree of hydrolysis of 0.1 M CH₃COONa

Related Resources#

• Practice Problems
• Video Lectures
• Previous Year Questions

---

This advanced content is part of Prof Shashank Deep’s comprehensive chemistry series designed to help NEET aspirants master ionic equilibrium concepts.