Unit-1. Volumetric Analysis

Chemistry – Class 12

Unit-1. Volumetric Analysis

No MCQ questions available for this chapter.

🔹 Unit 1: Volumetric Analysis (8 Teaching Hours)


🧪 1. Introduction

a. Gravimetric Analysis

It is a quantitative analysis method where the amount of substance is determined by measuring mass.
Example: Determination of sulfate as BaSO₄.

b. Volumetric Analysis

It is a method of quantitative chemical analysis used to determine the concentration of a solute in a solution by measuring the volume of a standard solution required to react completely.
Example: Titration of HCl with NaOH.


⚖️ 2. Equivalent Weight

  • Definition: The equivalent weight of a substance is the amount that reacts with 1 mole of H⁺ ions or 1 mole of electrons in redox reactions.

a. Relationship between Equivalent Weight (E), Atomic Weight (A), and Valency (V):

E=AVE = \frac{A}{V}


📋 3. Equivalent Weight of Compounds

Compound TypeFormula
AcidEq. wt.=Molecular wt.Basicity\text{Eq. wt.} = \frac{\text{Molecular wt.}}{\text{Basicity}}
BaseEq. wt.=Molecular wt.Acidity\text{Eq. wt.} = \frac{\text{Molecular wt.}}{\text{Acidity}}
SaltDepends on ions replaced during reaction
Oxidizing AgentEq. wt.=Molecular wt.No. of electrons accepted\text{Eq. wt.} = \frac{\text{Molecular wt.}}{\text{No. of electrons accepted}}
Reducing AgentEq. wt.=Molecular wt.No. of electrons donated\text{Eq. wt.} = \frac{\text{Molecular wt.}}{\text{No. of electrons donated}}

🧂 4. Concentration of Solutions

UnitDescription
% (w/v)Grams of solute per 100 mL solution
g/LGrams of solute per liter of solution
Molarity (M)Moles of solute per liter of solution
Molality (m)Moles of solute per kg of solvent
Normality (N)Equivalents of solute per liter of solution
Formality (F)Formula units per liter (used for ionic compounds)
ppmParts per million (mg/L)
ppbParts per billion (μg/L)

🧪 5. Primary and Secondary Standard Substances

TypeDescriptionExamples
Primary StandardPure, stable, non-hygroscopic, high molar mass, used to prepare standard solutionsK₂Cr₂O₇, Na₂CO₃, Oxalic acid
Secondary StandardSolutions standardized against primary standardsKMnO₄, HCl, NaOH

⚖️ 6. Law of Equivalence

  • Statement: In a chemical reaction, equivalents of one substance = equivalents of another.

  • Normality Equation:

N1V1=N2V2N_1V_1 = N_2V_2

Where:
N1N_1 = normality of acid, V1V_1 = volume of acid
N2N_2 = normality of base, V2V_2 = volume of base


🧪 7. Titration

a. Definition

Titration is a method to determine the unknown concentration of a solution using a solution of known concentration.

b. Types of Titration

TypeDescription
Acid-Base TitrationInvolves neutralization reaction between acid and base
Redox TitrationInvolves electron transfer reaction between oxidizing and reducing agents

c. Indicators Used

Titration TypeIndicator
Strong acid + strong basePhenolphthalein or Methyl orange
Weak acid + strong basePhenolphthalein
Redox titrationSelf-indicator (KMnO₄) or external (starch for I₂)

🧮 8. Related Numerical Problems

Example 1:

Calculate the normality of a solution prepared by dissolving 4.9 g of H₂SO₄ in 500 mL of solution.
Molecular wt of H₂SO₄ = 98, Basicity = 2

Eq. wt.=982=49Normality (N)=mass in g×1000Eq. wt.×volume in mL=4.9×100049×500=0.2N\text{Eq. wt.} = \frac{98}{2} = 49 \\ \text{Normality (N)} = \frac{\text{mass in g} \times 1000}{\text{Eq. wt.} \times \text{volume in mL}} = \frac{4.9 \times 1000}{49 \times 500} = 0.2 N

Example 2:

25 mL of 0.1 N HCl is completely neutralized by NaOH. Calculate the volume of 0.05 N NaOH used.

N1V1=N2V20.1×25=0.05×V2V2=2.50.05=50 mLN_1V_1 = N_2V_2 \\ 0.1 \times 25 = 0.05 \times V_2 \Rightarrow V_2 = \frac{2.5}{0.05} = 50\ \text{mL}


📷 Suggested Diagrams (Insert manually)

  • Volumetric apparatus (Burette, Pipette, Conical Flask)

  • Acid-base titration setup

  • Titration curve for strong acid–strong base

  • Color change of indicators