Class 12 Chemistry Volumetric Analysis Notes

Unit 1

General and Physical Chemistry

Class 12 Chemistry

Volumetric Analysis

Class 12 Chemistry – Volumetric Analysis Notes PDF

On mobile, swipe inside the PDF to read all pages and pinch to zoom.

Introduction

Definition Volumetric analysis is a quantitative analytical method in which the amount or concentration of an unknown substance is determined by measuring the volume of a standard solution required to react completely with it.

Unit 1 develops the quantitative foundations required for titration: gravimetric and volumetric analysis, equivalent weight, concentration units, standard solutions, law of equivalence, normality calculations, acid–base titration, redox titration and related numerical problems.

1. Quantitative Chemical Analysis

Gravimetric Analysis

A quantitative method in which the amount of an analyte is determined from a carefully measured mass, usually after converting the analyte into a suitable solid form.

Volumetric Analysis

A quantitative method in which the amount of an analyte is calculated from the measured volume of a solution of known concentration that reacts completely with it.

FeatureGravimetric AnalysisVolumetric Analysis
Main measurementMassVolume
Common operationPrecipitation, filtration, drying and weighingTitration with a standard solution
Result based onMass relationshipStoichiometric volume/concentration relationship
Typical apparatusBalance, crucible, filter apparatusBurette, pipette, volumetric flask, conical flask

Fig. 1 – Gravimetric vs Volumetric Analysis

Quantitative Analysis Gravimetric measure mass Volumetric measure reacting volume

2. Equivalent Weight

Definition The equivalent weight of a substance is the mass that reacts with or supplies one equivalent of the reacting species under the specified chemical reaction.

2.1 Element

Equivalent weight of element = Atomic weight ÷ Valency
Example For calcium: atomic weight ≈ 40 and valency = 2.
Equivalent weight = 40 ÷ 2 = 20 g eq⁻¹

2.2 Acid

Equivalent weight of acid = Molar mass ÷ Basicity

Basicity is the number of replaceable H+ ions furnished per molecule in the reaction considered.

Example For H2SO4: molar mass = 98, basicity = 2.
Equivalent weight = 98 ÷ 2 = 49 g eq⁻¹

2.3 Base

Equivalent weight of base = Molar mass ÷ Acidity

For this purpose, acidity of a base represents the number of replaceable OH groups taking part in neutralization.

Example For Ca(OH)2: molar mass ≈ 74, acidity = 2.
Equivalent weight = 74 ÷ 2 = 37 g eq⁻¹

2.4 Salt

Equivalent weight of salt = Formula mass ÷ Total positive or negative ionic charge involved

2.5 Oxidizing and Reducing Agents

Equivalent weight = Molar mass ÷ n-factor

For redox substances, the n-factor is the number of electrons accepted or donated per formula unit in the specified reaction. It can depend on reaction conditions and medium.

Fig. 2 – Equivalent Weight Decision Chart

Equivalent Weight Element AW / valency Acid MM / basicity Base MM / acidity Salt FM / charge Redox agent MM / n-factor Always identify the reaction first when the n-factor can vary.

3. Concentration of Solutions

Concentration expresses how much solute is present in a specified amount of solution or solvent.

UnitDefinition / FormulaTypical Unit
Mass percentage (w/w) (Mass of solute ÷ Mass of solution) × 100 %
Mass/volume percentage (w/v) (Mass of solute in g ÷ Volume of solution in mL) × 100 %
Volume percentage (v/v) (Volume of solute ÷ Volume of solution) × 100 %
Strength Mass of solute per litre of solution g L⁻¹
Molarity (M) Moles of solute ÷ Volume of solution in litres mol L⁻¹
Molality (m) Moles of solute ÷ Mass of solvent in kilograms mol kg⁻¹
Normality (N) Gram-equivalents of solute ÷ Volume of solution in litres eq L⁻¹
Formality (F) Formula-weight units of solute ÷ Volume of solution in litres F
ppm Parts of solute per 10⁶ parts of solution ppm
ppb Parts of solute per 10⁹ parts of solution ppb

3.1 Molarity

M = Number of moles of solute ÷ Volume of solution in litre
M = (Mass × 1000) ÷ (Molar mass × Volume in mL)

3.2 Molality

m = Number of moles of solute ÷ Mass of solvent in kg
Difference Molarity depends on volume of solution and therefore varies slightly with temperature. Molality depends on mass of solvent and is independent of volume change due to temperature.

3.3 Normality

N = Number of gram equivalents ÷ Volume of solution in litre
N = (Mass × 1000) ÷ (Equivalent weight × Volume in mL)
N = M × n-factor

3.4 ppm and ppb

ppm = (Amount of solute ÷ Amount of solution) × 10⁶
ppb = (Amount of solute ÷ Amount of solution) × 10⁹

Fig. 3 – Concentration Units at a Glance

Concentration of Solution Percentage w/w, w/v, v/v Molarity mol / L solution Molality mol / kg solvent Normality eq / L solution Formality formula wt / L Trace units ppm / ppb

4. Primary and Secondary Standard Substances

4.1 Primary Standard

A primary standard substance is sufficiently pure and stable that a solution of accurately known concentration can be prepared directly by weighing the substance and dissolving it to a definite volume.

Desirable Properties

  • Very high purity.
  • Stable in air.
  • Not appreciably hygroscopic or volatile.
  • Relatively high equivalent or molar mass.
  • Readily soluble in the chosen solvent.
  • Reacts rapidly and stoichiometrically.

4.2 Secondary Standard

A secondary standard solution is a solution whose exact concentration is determined by standardization against a suitable primary standard.
FeaturePrimary StandardSecondary Standard
Purity/stabilityVery high and reliableMay change on storage or preparation
PreparationCan be prepared directly to known concentrationMust be standardized
RoleReference substanceWorking titrant or solution
Examples often used in teachingOxalic acid, sodium carbonate, potassium hydrogen phthalate depending on applicationNaOH, HCl, KMnO₄ solutions often require standardization

Fig. 4 – Standardization Concept

Primary standard accurately known Titration / standardization Secondary solution exact concentration known

5. Law of Equivalence and Normality Equation

Law of Equivalence At the equivalence point, chemically equivalent quantities of reacting substances have reacted with each other according to the stoichiometric reaction.
Number of equivalents of A = Number of equivalents of B

Since number of equivalents in a solution is proportional to normality × volume:

N₁V₁ = N₂V₂

When both volumes are expressed in the same unit, this relationship is especially useful for acid–base and redox titration calculations.

Example 25.0 mL of 0.100 N acid exactly neutralizes a base solution. If 20.0 mL of base is required:

N₁V₁ = N₂V₂
0.100 × 25.0 = N₂ × 20.0
N₂ = 0.125 N

Fig. 5 – Law of Equivalence

Solution A N₁ × V₁ equivalents proportional = Solution B N₂ × V₂ equivalents proportional At equivalence: N₁V₁ = N₂V₂

6. Titration

Definition A titration is an analytical procedure in which a solution of known concentration is gradually added to a measured amount of another solution until the reaction reaches the required stoichiometric point.

Important Terms

Titrant

The standard solution delivered, usually from a burette.

Analyte

The solution whose concentration or amount is being determined.

Equivalence Point

The theoretical point at which stoichiometrically equivalent amounts have reacted.

End Point

The experimentally observed signal used to stop the titration, often a colour change.

Fig. 6 – Standard Titration Apparatus

Burette containing titrant Stopcock Conical flask containing analyte Titrant is added gradually while the flask is mixed.

Basic Procedure

  1. Rinse and fill the burette with the titrant.
  2. Measure a known volume of analyte using a pipette.
  3. Transfer the analyte to a conical flask.
  4. Add a suitable indicator when required by the method.
  5. Add titrant gradually while swirling the flask.
  6. Near the end point, add titrant dropwise.
  7. Record initial and final burette readings.
  8. Repeat until concordant titres are obtained.

7. Acid–Base Titration

Acid–base titration is based on a neutralization reaction between an acid and a base.

Acid + Base → Salt + Water

For calculations using normality at equivalence:

NacidVacid = NbaseVbase
Worked Example 20.0 mL of HCl is neutralized by 25.0 mL of 0.080 N NaOH.

NHCl × 20.0 = 0.080 × 25.0
NHCl = 2.00 ÷ 20.0
NHCl = 0.100 N

Fig. 7 – Acid–Base Titration Concept

Acid solution H⁺ equivalents + Base solution OH⁻ equivalents Neutralization products salt + water At equivalence, reacting acid and base equivalents are equal.

8. Redox Titration

A redox titration is a titration based on an oxidation–reduction reaction in which electrons are transferred between oxidizing and reducing agents.

Equivalent weight and normality in redox reactions depend on the n-factor, which is related to the number of electrons transferred in the balanced reaction.

N = M × n-factor
N₁V₁ = N₂V₂

Fig. 8 – Redox Titration: Electron-Transfer Concept

Reducing agent loses electrons electron transfer Oxidizing agent gains electrons Oxidation and reduction always occur together.
Important For substances such as permanganate, dichromate or other redox reagents, the n-factor can depend on the chemical medium. Use the balanced reaction specified in the question.

9. Important Numerical Patterns

9.1 Equivalent Weight of an Acid

Question: Calculate the equivalent weight of H3PO4 when all three acidic hydrogens are neutralized.

Molar mass = 98
Basicity = 3
Equivalent weight = 98 ÷ 3
≈ 32.67 g eq⁻¹

9.2 Normality from Mass

Question: 4.9 g of H2SO4 is dissolved to make 500 mL of solution. Find normality for complete neutralization.

Equivalent weight of H2SO4 = 49 g eq⁻¹
Number of equivalents = 4.9 ÷ 49 = 0.1 eq
Volume = 0.500 L
N = 0.1 ÷ 0.500
N = 0.20 N

9.3 Molarity from Mass

Question: 5.85 g NaCl is dissolved to make 500 mL of solution. Calculate molarity.

Molar mass NaCl ≈ 58.5 g mol⁻¹
Moles = 5.85 ÷ 58.5 = 0.100 mol
Volume = 0.500 L
M = 0.100 ÷ 0.500
M = 0.200 M

9.4 Titration Using N₁V₁ = N₂V₂

Question: 25.0 mL of an acid requires 20.0 mL of 0.150 N base. Find acid normality.

Nacid × 25.0 = 0.150 × 20.0
Nacid = 3.00 ÷ 25.0
Nacid = 0.120 N

9.5 Molarity to Normality

Question: Find the normality of 0.25 M H2SO4 for complete neutralization.

n-factor = 2
N = M × n-factor
N = 0.25 × 2
N = 0.50 N

Fig. 9 – Numerical Problem Solving Flow

Identify given quantities Determine n-factor / Eq. wt. Choose correct formula Substitute units carefully Check answer

10. Common Mistakes to Avoid

  • Confusing molarity with molality.
  • Using molar mass where equivalent weight is required.
  • Forgetting that n-factor can depend on the reaction.
  • Mixing mL and L without conversion in molarity or normality formulas.
  • Using an unbalanced redox equation to determine electron change.
  • Confusing end point with the exact theoretical equivalence point.
  • Reading a burette scale in the wrong direction.
  • Using a single rough titre instead of concordant titration readings.

11. Quick Revision & Exam Points

Important Questions
  • Define gravimetric analysis and volumetric analysis.
  • Define equivalent weight and derive its relation with atomic weight and valency.
  • Calculate equivalent weight of acids, bases, salts and redox agents.
  • Define molarity, molality, normality, formality, ppm and ppb.
  • Differentiate molarity and molality.
  • State the relationship between molarity and normality.
  • Define primary and secondary standard substances.
  • State and explain the law of equivalence.
  • Derive or apply the normality equation N₁V₁ = N₂V₂.
  • Define titration, titrant, analyte, equivalence point and end point.
  • Explain acid–base titration.
  • Explain redox titration.
  • Solve numerical problems involving concentration, equivalent weight and titration.

One-Minute Revision

  • Volumetric analysis uses reacting solution volume.
  • Element Eq. wt. = atomic weight / valency.
  • Acid Eq. wt. = molar mass / basicity.
  • Base Eq. wt. = molar mass / acidity.
  • Redox Eq. wt. = molar mass / n-factor.
  • M = moles / litre of solution.
  • m = moles / kg of solvent.
  • N = equivalents / litre of solution.
  • N = M × n-factor.
  • Primary standard can prepare an accurately known solution directly.
  • Secondary standard must be standardized.
  • At equivalence: N₁V₁ = N₂V₂.
  • Acid–base titration is based on neutralization.
  • Redox titration is based on electron transfer.
Diagram Practice Practice the quantitative-analysis classification, equivalent-weight chart, concentration-units chart, standardization flow, law of equivalence, titration apparatus, acid–base titration and redox titration diagrams.

Source handling: The original Nepal eNotes Volumetric Analysis PDF is embedded at the top using the exact Google Drive file linked by the Nepal eNotes chapter page. The typed notes follow the verified NEB Grade 12 Chemistry Unit 1 scope and are written as a searchable, responsive study companion. Because the Drive viewer does not expose the handwritten page text as readable document text here, the typed section is not presented as a word-for-word transcription.

Discussion

Share a helpful question, idea, or explanation with other students.

Leave a Comment

Write a clear question, answer, or helpful explanation.
Your email will not be published.

Download Our Offline App

Study class-wise notes even when internet is not available. Get the app from Play Store.

Nepal eNotes offline app preview
Get it on Google Play