Class 11 Chemistry Notes Chapter 3 (Chapter 3) – Lab Manual (English) Book

Detailed Notes with MCQs of Chapter 3 from your Chemistry Lab Manual, which deals with Quantitative Estimation. This is a crucial area, not just for your practical exams, but also because the principles involved frequently appear in various government competitive exams. We'll break down the key experiments and concepts.

Chapter 3: Quantitative Estimation - Key Concepts & Notes

Quantitative estimation involves determining the exact amount or concentration of a substance. The experiments in this chapter primarily focus on volumetric analysis (titration) and determining molar mass.

1. Volumetric Analysis (Titration)

• Definition: A laboratory method of quantitative chemical analysis used to determine the concentration of an identified analyte (the substance to be analyzed). A reagent, termed the titrant or standard solution, of known concentration and volume is reacted with a solution of the analyte or titrand to determine its concentration.
• Equivalence Point: The point in a titration where the amount of titrant added is chemically equivalent to the amount of analyte present in the sample.
• End Point: The point in a titration where a physical change (like colour change) occurs, indicating that the equivalence point has been reached (or is very close). This change is usually brought about by an indicator or by the titrant/analyte itself.
• Standard Solution: A solution whose concentration is accurately known.
  • Primary Standard: A highly pure compound used to prepare a standard solution directly by weighing. It should be stable, non-hygroscopic, have a high molecular weight (to minimize weighing errors), and be readily soluble. Examples: Oxalic acid crystals (H₂C₂O₄·2H₂O), Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O), Anhydrous Sodium Carbonate (Na₂CO₃).
  • Secondary Standard: A substance whose solution concentration is determined by titration against a primary standard solution. They are usually less pure or less stable than primary standards. Examples: KMnO₄, NaOH, HCl, H₂SO₄.
• Concentration Terms:
  • Molarity (M): Moles of solute per litre of solution (mol L⁻¹). Formula: M = (Mass of solute / Molar mass of solute) / Volume of solution (L)
  • Normality (N): Gram equivalents of solute per litre of solution (eq L⁻¹). Formula: N = (Mass of solute / Equivalent mass of solute) / Volume of solution (L)
  • Relationship: Normality (N) = Molarity (M) × n-factor
    • n-factor: For acids, basicity (no. of replaceable H⁺ ions); for bases, acidity (no. of replaceable OH⁻ ions); for redox reagents, the number of electrons lost or gained per molecule in the reaction.

Experiment 1: Determination of Molarity of KMnO₄ solution by titrating against standard Oxalic Acid solution.

• Principle: Redox titration. Potassium permanganate (KMnO₄) is a strong oxidizing agent, and oxalic acid (H₂C₂O₄) is a reducing agent. The titration is carried out in an acidic medium (dilute H₂SO₄).
• Reactions:
  • Reduction half-reaction (MnO₄⁻): MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) (Purple to Colourless)
  • Oxidation half-reaction (C₂O₄²⁻): C₂O₄²⁻(aq) → 2CO₂(g) + 2e⁻
  • Overall Ionic Equation: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
• Procedure Highlights:
  • Prepare a standard solution of Oxalic Acid (Primary Standard).
  • Titrate this against the given KMnO₄ solution (Secondary Standard).
  • Acid Medium: Dilute H₂SO₄ is added to provide the necessary H⁺ ions for the reduction of MnO₄⁻ to Mn²⁺.
    • Why not HCl? HCl is a reducing agent itself (Cl⁻ can be oxidized to Cl₂ by KMnO₄), which would consume extra KMnO₄, leading to inaccurate results.
    • Why not HNO₃? HNO₃ is an oxidizing agent itself and can interfere with the reaction.
  • Heating: The oxalic acid solution (with H₂SO₄) needs to be heated to about 60-70°C before titration because the reaction between MnO₄⁻ and C₂O₄²⁻ is slow at room temperature. The Mn²⁺ ions produced act as an auto-catalyst, speeding up the reaction once it starts.
  • Indicator: KMnO₄ acts as a self-indicator. MnO₄⁻ is intensely purple, while Mn²⁺ is almost colourless.
  • End Point: The first appearance of a permanent light pink colour in the solution (due to a slight excess of unreacted MnO₄⁻ ions).
• Calculations:
  • Use the molarity equation derived from stoichiometry:
    (M₁V₁) / n₁ (for KMnO₄) = (M₂V₂) / n₂ (for Oxalic Acid)
    Where: M₁ = Molarity of KMnO₄, V₁ = Volume of KMnO₄ used, n₁ = 5 (from balanced equation/electrons gained)
    M₂ = Molarity of Oxalic Acid, V₂ = Volume of Oxalic Acid taken, n₂ = 2 (from balanced equation/electrons lost)
  • Alternatively, use Normality: N₁V₁ (KMnO₄) = N₂V₂ (Oxalic Acid)
    • Equivalent mass of KMnO₄ (in acidic medium) = Molar Mass / 5 = 158 / 5 = 31.6 g/eq
    • Equivalent mass of Oxalic Acid (H₂C₂O₄·2H₂O) = Molar Mass / 2 = 126 / 2 = 63 g/eq

Experiment 2: Determination of Molarity of KMnO₄ solution by titrating against standard Ferrous Ammonium Sulphate (Mohr's Salt) solution.

• Mohr's Salt: FeSO₄·(NH₄)₂SO₄·6H₂O (Molar Mass ≈ 392 g/mol). It's a good primary standard as it's stable and not readily oxidized by air in solid form.
• Principle: Redox titration in acidic medium (dilute H₂SO₄). KMnO₄ oxidizes Fe²⁺ ions from Mohr's salt to Fe³⁺ ions.
• Reactions:
  • Reduction half-reaction (MnO₄⁻): MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
  • Oxidation half-reaction (Fe²⁺): Fe²⁺(aq) → Fe³⁺(aq) + e⁻
  • Overall Ionic Equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
• Procedure Highlights:
  • Prepare a standard solution of Mohr's Salt (Primary Standard). Add some dilute H₂SO₄ during preparation to prevent hydrolysis of Fe²⁺ ions.
  • Titrate this against the given KMnO₄ solution.
  • Acid Medium: Dilute H₂SO₄ is essential to provide H⁺ ions and prevent precipitation of MnO₂. It also prevents the hydrolysis of Fe³⁺ formed.
  • Heating: Not required. The reaction proceeds smoothly at room temperature.
  • Indicator: KMnO₄ acts as a self-indicator.
  • End Point: Appearance of a permanent light pink colour.
• Calculations:
  • Use the molarity equation:
    (M₁V₁) / n₁ (for KMnO₄) = (M₂V₂) / n₂ (for Mohr's Salt)
    Where: M₁ = Molarity of KMnO₄, V₁ = Volume of KMnO₄ used, n₁ = 1 (stoichiometric coefficient of MnO₄⁻) or 5 (electrons gained)
    M₂ = Molarity of Mohr's Salt, V₂ = Volume of Mohr's Salt taken, n₂ = 5 (stoichiometric coefficient of Fe²⁺) or 1 (electron lost per Fe²⁺)
    Important: When using n₁ and n₂ as stoichiometric coefficients from the balanced equation: (M₁V₁)/1 = (M₂V₂)/5. When using n₁ and n₂ as electrons transferred per formula unit (n-factor): (M₁V₁)/5 = (M₂V₂)/1. Both lead to the same result: M₁V₁ = M₂V₂ / 5 or 5M₁V₁ = M₂V₂.
  • Alternatively, use Normality: N₁V₁ (KMnO₄) = N₂V₂ (Mohr's Salt)
    • Equivalent mass of KMnO₄ = Molar Mass / 5 = 31.6 g/eq
    • Equivalent mass of Mohr's Salt = Molar Mass / 1 = 392 / 1 = 392 g/eq (since only one Fe²⁺ ion loses one electron per formula unit).

2. Determination of Molar Mass

Experiment 3: Determination of Molar Mass of a Metal by Hydrogen Displacement Method

• Principle: An accurately weighed sample of an active metal (like Mg, Zn, Al) is reacted with an excess of non-oxidizing acid (like dilute HCl or H₂SO₄). The volume of hydrogen gas evolved is measured under known conditions of temperature and pressure. Using the Ideal Gas Law and stoichiometry, the molar mass of the metal is calculated.
• Reaction Example: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
  • Stoichiometry: 1 mole of Mg produces 1 mole of H₂ gas.
• Procedure Highlights:
  • React a known mass (w) of metal with excess acid in a setup (like a gas burette or collection over water in an inverted measuring cylinder) that allows collection and measurement of the evolved H₂ gas.
  • Measure the volume (V) of H₂ gas collected.
  • Measure the laboratory temperature (T) in Kelvin (T = °C + 273.15).
  • Measure the atmospheric pressure (P_atm) using a barometer.
  • Aqueous Tension: If the gas is collected over water, the measured pressure includes the partial pressure of water vapour (aqueous tension, P_aq) at that temperature. The pressure of dry H₂ gas (P) is calculated as: P = P_atm - P_aq. (Values for P_aq at different temperatures are available in tables).
• Calculations:
  1. Calculate the pressure of dry H₂ gas (P) in appropriate units (e.g., atm, Pa, bar).
  2. Convert the measured volume (V) to Litres (L) or m³.
  3. Use the Ideal Gas Equation: PV = nRT
     • Where: n = moles of H₂ gas, R = Ideal Gas Constant (e.g., 0.0821 L atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹ or 8.314 Pa m³ K⁻¹ mol⁻¹). Choose the value of R consistent with the units of P, V, and T.
     • Calculate moles of H₂: n = PV / RT
  4. Use stoichiometry to find moles of metal. For Mg + 2HCl → MgCl₂ + H₂, moles of Mg = moles of H₂. For Zn + H₂SO₄ → ZnSO₄ + H₂, moles of Zn = moles of H₂. For 2Al + 6HCl → 2AlCl₃ + 3H₂, moles of Al = (2/3) × moles of H₂.
  5. Calculate Molar Mass (M) of the metal:
     M = Mass of metal taken (w) / Moles of metal calculated.

Key Takeaways for Exams:

• Know the definitions: Molarity, Normality, Primary/Secondary Standard, Titration, End Point, Equivalence Point, n-factor, Equivalent Mass.
• Understand the principles of redox titrations involving KMnO₄.
• Know the specific reactions, role of H₂SO₄, reasons for heating (or not), indicator action, and endpoint colours.
• Be able to calculate equivalent masses and n-factors for KMnO₄, Oxalic Acid, and Mohr's Salt in acidic medium.
• Be proficient in using the Molarity equation (M₁V₁/n₁ = M₂V₂/n₂) and Normality equation (N₁V₁ = N₂V₂).
• Understand the principle of determining molar mass by hydrogen displacement.
• Know how to apply the Ideal Gas Law (PV=nRT) and correct for aqueous tension.
• Relate gas volume to moles of metal using reaction stoichiometry.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts we just discussed:

1. Which of the following is commonly used as a primary standard in volumetric analysis?
   (a) KMnO₄
   (b) NaOH
   (c) HCl
   (d) Anhydrous Na₂CO₃

2. In the titration of KMnO₄ against oxalic acid in acidic medium, KMnO₄ acts as:
   (a) Primary Standard
   (b) Indicator
   (c) Reducing Agent
   (d) Catalyst

3. Why is dilute H₂SO₄ added during the titration of Mohr's salt with KMnO₄?
   (a) To oxidize Fe²⁺ to Fe³⁺
   (b) To reduce MnO₄⁻ to Mn²⁺
   (c) To provide an acidic medium and prevent precipitation of MnO₂
   (d) To act as a catalyst

4. During the titration of oxalic acid with KMnO₄, the solution is heated to 60-70°C because:
   (a) It increases the solubility of oxalic acid
   (b) The reaction is slow at room temperature
   (c) It prevents the decomposition of KMnO₄
   (d) It sharpens the endpoint colour change

5. What is the n-factor (number of electrons gained per molecule) for KMnO₄ when it reacts in an acidic medium to form Mn²⁺?
   (a) 1
   (b) 3
   (c) 5
   (d) 7

6. Hydrochloric acid (HCl) is not used to acidify the medium in KMnO₄ titrations because:
   (a) HCl is a weak acid
   (b) HCl is volatile
   (c) Cl⁻ ions can be oxidized by KMnO₄
   (d) HCl forms an insoluble precipitate with KMnO₄

7. When hydrogen gas is collected over water, the measured pressure is:
   (a) Equal to the pressure of dry hydrogen gas
   (b) Less than the pressure of dry hydrogen gas
   (c) Equal to the sum of the pressure of dry hydrogen gas and aqueous tension
   (d) Equal to the aqueous tension only

8. In the reaction Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g), if 0.1 moles of H₂ gas are produced at STP, what mass of Zinc reacted? (Atomic mass of Zn = 65.4 g/mol)
   (a) 3.27 g
   (b) 6.54 g
   (c) 13.08 g
   (d) 0.654 g

9. The endpoint in the titration of KMnO₄ against ferrous ammonium sulphate is the appearance of:
   (a) Colourless solution
   (b) Permanent light pink colour
   (c) Brown precipitate
   (d) Green colour

10. The equivalent mass of oxalic acid dihydrate (H₂C₂O₄·2H₂O, Molar Mass = 126 g/mol) as a reducing agent is:
    (a) 126 g/eq
    (b) 63 g/eq
    (c) 42 g/eq
    (d) 31.5 g/eq

Answer Key for MCQs:

1. (d)
2. (b)
3. (c)
4. (b)
5. (c)
6. (c)
7. (c)
8. (b) [Stoichiometry is 1:1, so 0.1 mol Zn reacted. Mass = moles × molar mass = 0.1 × 65.4 = 6.54 g]
9. (b)
10. (b) [Oxalic acid (C₂O₄²⁻ → 2CO₂ + 2e⁻) loses 2 electrons per molecule, so n-factor = 2. Eq. Mass = Molar Mass / 2 = 126 / 2 = 63]

Study these notes carefully, focusing on the underlying principles and calculation methods. Good luck with your preparation!