Class 12 Chemistry Notes Chapter 6 (Titrimetric Analysis (Redox Rxn)) – Lab Manual (English) Book

Detailed Notes with MCQs of Chapter 6 from your Lab Manual: Titrimetric Analysis involving Redox Reactions. This is a crucial topic, not just for your practical exams, but also because the principles frequently appear in various government competitive examinations. Pay close attention to the concepts and calculations.

Chapter 6: Titrimetric Analysis (Redox Reactions)

1. Introduction to Volumetric Analysis & Titration:

• Volumetric Analysis: A quantitative chemical analysis method that involves determining the volume of a solution of accurately known concentration (standard solution) required to react quantitatively with a measured volume of the solution whose concentration is to be determined (analyte).
• Titration: The process of adding the standard solution (titrant) from a burette to the analyte solution (titrand) in a flask until the reaction is just complete.
• Redox Titration: A type of titration based on an oxidation-reduction reaction between the titrant and the analyte.

2. Fundamental Concepts in Redox Reactions:

• Oxidation: Loss of electrons, increase in oxidation number.
• Reduction: Gain of electrons, decrease in oxidation number.
• Oxidizing Agent (Oxidant): A substance that accepts electrons and gets reduced itself (e.g., KMnO₄, K₂Cr₂O₇).
• Reducing Agent (Reductant): A substance that donates electrons and gets oxidized itself (e.g., Oxalic acid, Mohr's salt/FAS).
• Oxidation Number: A hypothetical charge assigned to an atom in a molecule or ion, assuming bonds are ionic. Crucial for balancing redox reactions and calculating equivalent weights.

3. Key Terms in Titration:

• Standard Solution: A solution whose concentration is accurately known.
  • Primary Standard: A highly pure compound that is stable, non-hygroscopic, has a high molecular weight, and can be accurately weighed to prepare a standard solution directly (e.g., Oxalic acid crystals (H₂C₂O₄·2H₂O), Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O)).
  • Secondary Standard: A substance whose solution concentration cannot be determined directly by weighing; it needs to be standardized against a primary standard (e.g., KMnO₄, NaOH). KMnO₄ is not primary because it's difficult to obtain in a perfectly pure state and can decompose in sunlight or in the presence of MnO₂.
• Titrant: The solution added from the burette (usually the standard solution).
• Titrand/Analyte: The solution taken in the titration flask, whose concentration is to be determined.
• Equivalence Point: The theoretical point in a titration where the amount of titrant added is chemically equivalent to the amount of analyte present according to the stoichiometry of the reaction.
• Endpoint: The point in a titration where a physical change (e.g., color change) occurs, indicating that the reaction is complete. Ideally, the endpoint should coincide with the equivalence point.
• Indicator: A substance that undergoes a distinct visual change (usually color) at or near the equivalence point.
  • Self-Indicator: A reactant or product that has a distinct color change itself, eliminating the need for an external indicator. KMnO₄ is a classic example; it acts as its own indicator. The permanganate ion (MnO₄⁻) is intensely purple, while the product Mn²⁺ (in acidic medium) is virtually colorless. The first persistent faint pink color signals the endpoint.

4. Titrations Involving Potassium Permanganate (KMnO₄):

KMnO₄ is a strong oxidizing agent, commonly used in redox titrations. Its behavior depends heavily on the reaction medium (acidic, alkaline, or neutral). In your syllabus, the focus is primarily on titrations in acidic medium (using dilute H₂SO₄).

• Why Dilute H₂SO₄?
  • Provides the necessary H⁺ ions for the reduction of MnO₄⁻ to Mn²⁺.
  • It does not interfere with the redox reaction (unlike HCl, where Cl⁻ can be oxidized by KMnO₄, or HNO₃, which is itself an oxidizing agent).
• Reduction Half-Reaction (Acidic Medium):
  MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
  (Purple) (Colorless)
• Equivalent Weight of KMnO₄ (Acidic Medium):
  Equivalent Weight = Molar Mass / Number of electrons gained per mole
  Equivalent Weight = Molar Mass (158 g/mol) / 5 = 31.6 g/equivalent

5. Specific Titrations:

• A. Titration of KMnO₄ vs. Oxalic Acid (H₂C₂O₄·2H₂O):

  • Oxidation Half-Reaction: C₂O₄²⁻ → 2CO₂ + 2e⁻
  • Overall Balanced Ionic Equation: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
  • Stoichiometry (Mole Ratio): 2 moles KMnO₄ react with 5 moles Oxalic Acid.
  • n-factor for KMnO₄: 5 (electrons gained)
  • n-factor for Oxalic Acid (H₂C₂O₄): 2 (electrons lost per molecule)
  • Equivalent Weight of Oxalic Acid (H₂C₂O₄·2H₂O): Molar Mass (126 g/mol) / 2 = 63 g/equivalent
  • Important Condition: The reaction is slow at room temperature, so the oxalic acid solution must be heated to 60-70°C before titration. The Mn²⁺ formed acts as an auto-catalyst, speeding up the reaction after it begins.
  • Indicator: KMnO₄ acts as a self-indicator.
  • Endpoint: Appearance of the first persistent pale pink color.

• B. Titration of KMnO₄ vs. Mohr's Salt (Ferrous Ammonium Sulphate, FAS - FeSO₄·(NH₄)₂SO₄·6H₂O):

  • Mohr's salt provides Fe²⁺ ions, which are oxidized to Fe³⁺.
  • Oxidation Half-Reaction: Fe²⁺ → Fe³⁺ + e⁻
  • Overall Balanced Ionic Equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
  • Stoichiometry (Mole Ratio): 1 mole KMnO₄ reacts with 5 moles Mohr's Salt (or 5 moles Fe²⁺).
  • n-factor for KMnO₄: 5 (electrons gained)
  • n-factor for Mohr's Salt (FAS): 1 (only Fe²⁺ is oxidized, losing 1 electron per formula unit)
  • Equivalent Weight of Mohr's Salt (FeSO₄·(NH₄)₂SO₄·6H₂O): Molar Mass (392 g/mol) / 1 = 392 g/equivalent
  • Important Condition: This titration is carried out at room temperature. No heating is required. Dilute H₂SO₄ must be added to the Mohr's salt solution before titration to prevent hydrolysis of Fe²⁺ and provide the acidic medium.
  • Indicator: KMnO₄ acts as a self-indicator.
  • Endpoint: Appearance of the first persistent pale pink color.

6. Calculations:

The core principle at the equivalence point is that the number of equivalents of the oxidizing agent equals the number of equivalents of the reducing agent.

• Using Normality (N):
  N₁V₁ = N₂V₂
  Where:
  N₁ = Normality of solution 1 (e.g., KMnO₄)
  V₁ = Volume of solution 1 used
  N₂ = Normality of solution 2 (e.g., Oxalic acid or FAS)
  V₂ = Volume of solution 2 taken
• Using Molarity (M):
  (M₁V₁) / n₁ = (M₂V₂) / n₂
  Where:
  M₁ = Molarity of solution 1
  V₁ = Volume of solution 1 used
  n₁ = n-factor (stoichiometric coefficient in the balanced equation if considering the other reactant, OR more reliably, the number of electrons transferred per mole) of reactant 1
  M₂ = Molarity of solution 2
  V₂ = Volume of solution 2 taken
  n₂ = n-factor of reactant 2
• Relationship: Normality (N) = Molarity (M) × n-factor
• Calculating Strength: Strength (g/L) = Molarity (mol/L) × Molar Mass (g/mol)
  Strength (g/L) = Normality (eq/L) × Equivalent Weight (g/eq)

7. Essential Precautions:

• Rinse the burette with the titrant (KMnO₄ solution).
• Rinse the pipette with the analyte solution (Oxalic acid or Mohr's salt).
• Remove the air bubble from the burette tip before starting.
• Read the upper meniscus for colored solutions like KMnO₄.
• Add dilute H₂SO₄ to the analyte before titration.
• Heat the oxalic acid solution to 60-70°C; do not heat Mohr's salt solution.
• Add KMnO₄ dropwise, especially near the endpoint, with constant swirling.
• Note the reading when the first persistent pale pink color appears throughout the solution.

Multiple Choice Questions (MCQs):

1. In the titration of KMnO₄ against Oxalic acid in acidic medium, the n-factor (valency factor) for KMnO₄ is:
   (a) 1
   (b) 3
   (c) 5
   (d) 7

2. Which acid is generally used to provide the acidic medium in KMnO₄ titrations?
   (a) HCl
   (b) HNO₃
   (c) H₂SO₄
   (d) CH₃COOH

3. Mohr's salt is considered a:
   (a) Secondary Standard
   (b) Primary Standard
   (c) Self-Indicator
   (d) Oxidizing Agent

4. During the titration of KMnO₄ vs Mohr's salt, the oxidation state of Iron changes from:
   (a) +2 to +3
   (b) +3 to +2
   (c) 0 to +2
   (d) +2 to 0

5. Why is the solution of oxalic acid heated before titrating with KMnO₄?
   (a) To increase its volume
   (b) To ensure complete ionization
   (c) To increase the rate of reaction
   (d) To prevent precipitation of MnO₂

6. What is the color change at the endpoint when titrating oxalic acid or Mohr's salt with KMnO₄?
   (a) Pink to colorless
   (b) Colorless to permanent pale pink
   (c) Yellow to green
   (d) Orange to green

7. The equivalent weight of crystalline oxalic acid (H₂C₂O₄·2H₂O, Molar Mass = 126 g/mol) is:
   (a) 126 g/eq
   (b) 63 g/eq
   (c) 42 g/eq
   (d) 31.5 g/eq

8. If 20 mL of 0.1 M KMnO₄ solution reacts completely with 25 mL of FAS solution in acidic medium, what is the Molarity of the FAS solution?
   (a) 0.1 M
   (b) 0.2 M
   (c) 0.4 M
   (d) 0.5 M

9. KMnO₄ acts as a self-indicator in redox titrations because:
   (a) It is intensely colored and its product (Mn²⁺) is colorless.
   (b) It reacts with an external indicator.
   (c) It changes pH drastically at the endpoint.
   (d) It forms a precipitate at the endpoint.

10. A common precaution while reading the burette filled with KMnO₄ solution is:
    (a) Read the lower meniscus.
    (b) Read the upper meniscus.
    (c) Ignore the meniscus and read the middle level.
    (d) Add an indicator before reading.

Answer Key for MCQs:

1. (c)
2. (c)
3. (b)
4. (a)
5. (c)
6. (b)
7. (b)
8. (c) [Calculation: (M₁V₁)/n₁ = (M₂V₂)/n₂ => (0.1 * 20)/5 = (M₂ * 25)/1 => M₂ = (0.1 * 20 * 1) / (5 * 25) = 2 / 125 = 0.016 M. Correction: Let's recheck the stoichiometry. 1 MnO₄⁻ reacts with 5 Fe²⁺. So n₁=5 for KMnO₄ and n₂=1 for FAS. (M₁V₁)/n₁ = (M₂V₂)/n₂ => (0.1 * 20)/5 = (M₂ * 25)/1 => 0.4 = 25 * M₂ => M₂ = 0.4 / 25 = 4 / 250 = 0.016 M. Wait, the common way to write the formula is M₁V₁/coeff₁ = M₂V₂/coeff₂ or N₁V₁ = N₂V₂. Let's use the coefficients from the balanced equation: MnO₄⁻ + 5Fe²⁺ + ... -> ... Coefficient for KMnO₄ is 1, coefficient for Fe²⁺ (FAS) is 5. So, M(KMnO₄)V(KMnO₄) / 1 = M(FAS)V(FAS) / 5 => (0.1 * 20) / 1 = (M(FAS) * 25) / 5 => 2 = M(FAS) * 5 => M(FAS) = 2/5 = 0.4 M. Yes, this is correct.]
9. (a)
10. (b)

Study these notes thoroughly. Focus on understanding the underlying principles, the stoichiometry of the reactions, and the calculation methods. Good luck with your preparation!