Unit-1. Volumetric Analysis

Introduction to Gravimetric and Volumetric Analysis

Quantitative chemical analysis can be performed by measuring either the mass of a product (gravimetric analysis) or the volume of a solution required for a complete reaction (volumetric analysis, commonly known as titration).

• Gravimetric analysis: The analyte is converted into a stable, weighable precipitate or product; its mass is measured and related to the amount of analyte using stoichiometry.
• Volumetric analysis (titration): A solution of known concentration (titrant) is added to the analyte solution until the reaction reaches the equivalence point; the volume of titrant used is measured.

Both methods rely on the concept of equivalent weight, which links the reacting capacity of a substance to a reference amount of hydrogen or oxygen.

Equivalent Weight

The equivalent weight (E) of a substance is the mass that combines with or displaces 1 g of hydrogen (or 8 g of oxygen) in a chemical reaction.

General Relationship

Equivalent weight (E) = Atomic or Molecular weight (M) / Valency (n)

Where valency is the number of electrons lost, gained, or shared per formula unit in the reaction.

Specific Cases

Example: For sulfuric acid (H₂SO₄, M = 98 g mol⁻¹), basicity = 2 → E = 98 / 2 = 49 g eq⁻¹.

Concentration of Solution and Units

Concentration expresses the amount of solute present in a given quantity of solution or solvent. The following units are commonly used in volumetric analysis.

• Percentage (w/w): % (w/w) = (mass of solute / mass of solution) × 100
• g L⁻¹: Concentration (g L⁻¹) = mass of solute (g) / volume of solution (L)
• Molarity (M): M = moles of solute / volume of solution (L) (mol L⁻¹)
• Molality (m): m = moles of solute / mass of solvent (kg) (mol kg⁻¹)
• Normality (N): N = equivalents of solute / volume of solution (L) (eq L⁻¹)
• Formality (F): Used for ionic compounds that do not exist as discrete molecules; F = formula weight of solute / volume of solution (L).
• ppm (parts per million): ppm = (mass of solute (mg) / mass of solution (kg)) × 10⁶ ≈ mg kg⁻¹.
• ppb (parts per billion): ppb = (mass of solute (µg) / mass of solution (kg)) × 10⁹ ≈ µg kg⁻¹.

Relationship between molarity and normality:

Normality (N) = Molarity (M) × n where n is the valency (number of equivalents per mole) of the solute in the reaction.

Primary and Secondary Standard Substances

In titrimetric analysis, standards are used to determine the exact concentration of a solution.

Primary Standards

Characteristics:

• High purity (≥ 99.9 %)
• Stable towards air, light, heat
• Non‑hygroscopic (does not absorb moisture)
• Reasonably soluble in the solvent used
• High equivalent weight (to minimize weighing errors)

Examples: Sodium carbonate (Na₂CO₃), potassium dichromate (K₂Cr₂O₇), oxalic acid dihydrate (H₂C₂O₄·2H₂O), potassium hydrogen phthalate (KHP).

Secondary Standards

Solutions whose concentration is determined by titration against a primary standard. They are prepared for convenience (e.g., NaOH, HCl, KMnO₄) and must be standardized before use.

Law of Equivalence and Normality Equation

At the equivalence point of a titration, the number of equivalents of the titrant equals the number of equivalents of the analyte.

Acid–base titration:

N₁V₁ = N₂V₂

• N₁, V₁ = normality and volume of the acid
• N₂, V₂ = normality and volume of the base

Redox titration:

Equivalents of oxidizing agent = Equivalents of reducing agent

Using normality: NₒₓVₒₓ = NᵣₑdVᵣₑd

Titration and Its Types

1. Acid‑Base Titration

Based on neutralization reactions. An indicator (e.g., phenolphthalein, methyl orange) changes colour at the endpoint, which approximates the equivalence point.

Example: Titration of HCl with NaOH using phenolphthalein (colourless → pink).

2. Redox Titration

Involves oxidation‑reduction reactions. Common titrants:

• Potassium permanganate (KMnO₄) – self‑indicator (purple → colourless)
• Potassium dichromate (K₂Cr₂O₇) – requires external indicator (e.g., diphenylamine)
• Sodium thiosulfate (Na₂S₂O₃) – used with iodine (I₂/I⁻) system

The equivalence point is detected by colour change of the indicator or by potentiometric methods.

Numerical Problems

Worked examples illustrate the application of the concepts above.

Problem 1: Equivalent Weight Calculation

Calculate the equivalent weight of (a) phosphoric acid (H₃PO₄, M = 98 g mol⁻¹) and (b) potassium permanganate (KMnO₄, M = 158 g mol⁻¹) in acidic medium (MnO₄⁻ → Mn²⁺).

Solution:

• (a) Phosphoric acid is tribasic (basicity = 3).
  E = M / Basicity = 98 / 3 ≈ 32.67 g eq⁻¹
• (b) In acidic medium, Mn changes from +7 to +2 → change in oxidation number = 5.
  E = M / Δox = 158 / 5 = 31.6 g eq⁻¹

Problem 2: Normality from Molarity

A 0.2 M solution of sulfuric acid (H₂SO₄) is used in a titration. What is its normality?

Solution:
Sulfuric acid is dibasic (n = 2).
N = M × n = 0.2 × 2 = 0.4 N

Problem 3: Dilution and Normality

25 mL of 0.1 N NaOH is diluted to 100 mL. Calculate the normality of the diluted solution.

Solution:
Using N₁V₁ = N₂V₂:
0.1 N × 25 mL = N₂ × 100 mL → N₂ = (0.1 × 25) / 100 = 0.025 N

Problem 4: Acid‑Base Titration Calculation

20.0 mL of an HCl solution requires 25.0 mL of 0.10 N NaOH for complete neutralization. Find the molarity of the HCl solution.

Solution:
First find normality of HCl using N₁V₁ = N₂V₂:
N_HCl × 20.0 mL = 0.10 N × 25.0 mL → N_HCl = (0.10 × 25.0) / 20.0 = 0.125 N
Since HCl is monoprotic (n = 1), normality = molarity.
M_HCl = 0.125 M

Problem 5: Redox Titration (KMnO₄ vs. Fe²⁺)

In a titration, 15.0 mL of 0.02 N KMnO₄ solution oxidizes Fe²⁺ to Fe³⁺. Calculate the number of moles of Fe²⁺ present.

Solution:
Equivalents of KMnO₄ = N × V (L) = 0.02 eq L⁻¹ × 0.015 L = 3.0 × 10⁻⁴ eq.
In the reaction MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, each Fe²⁺ loses 1 electron → 1 eq per mole.
Thus, moles of Fe²⁺ = equivalents = 3.0 × 10⁻⁴ mol.

Problem 6: Concentration Conversion

A solution contains 5 g of NaCl dissolved in 500 mL of water. Express the concentration in (a) g L⁻¹, (b) molarity, and (c) ppm.

Solution:
(a) g L⁻¹ = 5 g / 0.5 L = 10 g L⁻¹
(b) Molar mass NaCl ≈ 58.44 g mol⁻¹.
Moles = 5 g / 58.44 g mol⁻¹ = 0.0856 mol.
Molarity = 0.0856 mol / 0.5 L = 0.171 M.
(c) Mass of solution ≈ ppm): Assume density of solution ≈ 1 g mL⁻¹ → mass of solution ≈ 500 g = 0.5 kg.
ppm = (mass solute in mg / mass solution in kg) = (5000 mg / 0.5 kg) = 10 000 ppm.

Summary

This chapter has laid the theoretical foundation for volumetric analysis: defining equivalent weight, deriving concentration units, distinguishing primary and secondary standards, applying the law of equivalence, and executing acid‑base and redox titrations. The numerical problems reinforce the ability to interconvert units, calculate normalities, and determine unknown concentrations from titration data.