Normality in Chemistry: Complete ISC Grade 11 Guide with Concepts, Formulas & Applications

 

Normality  

The Complete Guide for Grade 11 Chemistry

If you’ve just stepped into the world of volumetric analysis and titrations, normality might look like one of those “extra” concentration terms. But trust me — once you understand equivalent weight, normality becomes beautifully logical. Let’s break it down step-by-step.

Equivalent Weight

Definition:
The equivalent weight of an element or compound is the number of parts by weight of that element or compound that combine with:

• 1 part by weight of hydrogen
• 8 parts by weight of oxygen
• 35.5 parts by weight of chlorine

This definition connects chemical reactivity directly to mass.

Gram Equivalent Weight

Definition:

• For an element or radical:

$$\text{Gram equivalent weight}=\frac{\text{Atomic weight}}{\text{Valency}}$$

• ​For a compound:

$$\text{Gram equivalent weight}=\frac{\text{Molecular weight}}{\text{Combining power in a specific reaction}}$$

 Important: Equivalent weight depends on the reaction taking place.

Calculation of Equivalent Weight

For Elements

$$\text{Equivalent weight}=\frac{\text{Atomic weight}}{\text{Valency}}$$Example:

• Na: $\frac{23}{1}=$

• Mg: $\frac{24}{2}=$

For Acids

$$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Basicity}}$$

Basicity = Number of replaceable H⁺ ions.

Examples:

• HCl → $\frac{36.5}{1}=$

• H₂SO₄ → $\frac{98}{2}=$

• H₃PO₄ → $\frac{98}{3}=$

For Bases

$$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Acidity}}$$

Acidity = Number of replaceable OH⁻ ions.

Examples:

• NaOH → $\frac{40}{1}=$

• Ca(OH)₂ → $\frac{74}{2}=$

• Al(OH)₃ → $\frac{78}{3}=$

For Salts

$$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Total positive or negative charge}}$$Examples:

• Na₂CO₃ → $\frac{106}{2}=$

• K₂SO₄ → $\frac{174}{2}=$

For Oxidising Agents

$$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Number of electrons gained per molecule}}$$
$$𝑀𝑛{𝑂}_4^{-}+8{𝐻}^{+}+5{𝑒}^{-}\to 𝑀{𝑛}^{2+}$$

Here, 5 electrons are gained.

$$\text{Equivalent weight}=\frac{158}{5}=31.6$$

 In neutral or alkaline medium, the equivalent weight changes because the number of electrons gained changes.

For Reducing Agents

$$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Number of electrons lost}}$$

Example: FeSO₄

$$𝐹{𝑒}^{2+}\to 𝐹{𝑒}^{3+}+{𝑒}^{-}$$

Electrons lost = 1

$$\text{Equivalent weight}=\frac{152}{1}=152$$

Number of Gram Equivalents

$$\text{Number of gram equivalents}=\frac{\text{Weight of substance}}{\text{Gram equivalent weight}}$$

• 1 gram equivalent = equal to its gram equivalent weight

• 2 gram equivalents = twice its gram equivalent weight

Normality (N)

Now comes the main concept 

Definition

Normality is the number of gram equivalents of solute per litre of solution.

$$𝑁=\frac{\text{Gram equivalents}}{\text{Volume in litres}}$$

A 1 Normal (1N) solution contains 1 gram equivalent per litre of solution.

Relation Between Normality and Molarity

$$𝑁=𝑛\times 𝑀$$

Where:

• N = Normality
• M = Molarity
• n = Reactivity factor

For Acids:

n = Basicity

Example:    For H₂SO₄,

$$𝑁=2\times 𝑀$$

For Bases:

n = Acidity

For Redox Reactions:

n = Number of electrons exchanged

Law of Chemical Equivalence

This law states:

Number of gram equivalents of reagent 1 completely reacts with number of gram equivalents of reagent 2.

Mathematically:

$${𝑁}_1{𝑉}_1={𝑁}_2{𝑉}_2$$

This equation is the backbone of titration calculations.

Why Normality is Reaction-Specific

Unlike molarity, normality depends on the type of reaction.

Example:

• H₃PO₄ can act as mono-, di-, or tribasic acid depending on the reaction.
• KMnO₄ has different equivalent weights in acidic and alkaline media.

That’s why always identify the reaction type first.

Quick Summary for Revision

✔ Equivalent weight depends on the reaction

✔ Gram equivalent = weight ÷ equivalent weight

✔ Normality = gram equivalent per litre

✔ N = n × M

✔ N₁V₁ = N₂V₂ for titration