N1 x V1 = N2 x V2 Normality Neutralisation Equation, Derivation, Solved Examples

ANAND CLASSES Study material and notes to learn the Normality and Neutralisation Equation with clear derivation, detailed explanation, and step-by-step solved examples. Perfect for JEE, NEET, and CBSE Class 11 Chemistry revision.

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🧪 What is Normality (N)?

Normality is a way to express concentration of a solution. $$\text{Normality (N)} = \frac{\text{Number of gram equivalents of solute}}{\text{Volume of solution in liters}}$$

• Gram Equivalent: The amount of a substance that can donate or accept 1 mole of protons (H⁺), electrons, or ions.
• Normality changes with the reaction type because the gram equivalent depends on the reaction.

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🧮 What is Gram Equivalent?

1. For Acid-Base Reactions:

$$\text{Equivalent of acid} = \frac{\text{Mass of acid (g)}}{\text{Equivalent weight of acid}} $$

$$\text{Equivalent weight of acid} = \frac{\text{Molar mass}}{\text{Basicity}} $$

$\quad (\text{Basicity = number of replaceable H⁺})$

2. For Bases:

$$\text{Equivalent weight of base} = \frac{\text{Molar mass}}{\text{Acidity}} $$

$\quad (\text{Acidity = number of replaceable OH⁻})$

3. For Redox Reactions:

$$\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} $$

$\quad (\text{n-factor = electrons lost or gained})$

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🧩 What is the Equation $N_1V_1 = N_2V_2\:$?

This formula is based on the Law of Equivalents:

When two substances react completely, the number of equivalents of one = number of equivalents of the other.

This is not based on moles, but on equivalents — a more universal quantity that factors in the chemical role the substance plays (e.g., how many H⁺ ions it donates or how many electrons it transfers).

So,

$$\text{Equivalents} = \text{Normality} \times \text{Volume (in L)}$$

Let substance A and B react:

• A is in solution 1, and B is in solution 2
• When they react completely: $$\text{Equivalents of A} = \text{Equivalents of B}$$

So: $$N_1V_1 = N_2V_2$$

This is the Normality Equation.

This equation arises from the idea that:

At the equivalence point in a reaction, the number of equivalents of one reactant = number of equivalents of the other.

$$\text{Equivalents of acid} = \text{Equivalents of base}$$

And since: $$\text{Equivalents} = \text{Normality} \times \text{Volume (L)}$$

$$N_1V_1 = N_2V_2$$

• $N_1, V_1 $: normality and volume of first solution (e.g., acid)
• $N_2, V_2$: normality and volume of second solution (e.g., base)

✅ It works only when the reaction is complete and balanced, usually at the equivalence point.

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🔷 Why Use Equivalents Instead of Moles?

• Moles just count particles.
• Equivalents consider reactive capacity.
• For example, 1 mole of H₂SO₄ gives 2 H⁺ ions (so 2 equivalents). But 1 mole of HCl gives only 1 H⁺ ion (1 equivalent).
• So: $$\text{1 mol H}_2\text{SO}_4 = 2 \text{ equivalents} $$ and $$\text{1 mol HCl} = 1 \text{ equivalent}$$

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🧪 Detailed Example 1: Acid-Base Titration

🌟 Case.1 : For Monoprotic Reaction:

$$\text{HCl (aq) + NaOH (aq)} \rightarrow \text{NaCl (aq) + H₂O (l)}$$

• 1 mole of HCl reacts with 1 mole of NaOH.
• So, n-factor is 1 for both.

Let’s say:

• You have 25 mL of HCl of unknown strength.
• You titrate it with 0.1 N NaOH.
• It takes 30 mL of NaOH to neutralize the acid.

Use the formula: $$N_1V_1 = N_2V_2$$

$$N_1 \times 25 = 0.1 \times 30$$

$$N_1 = \frac{3}{25} = 0.12 \, \text{N}$$

🌟 Case.2 : For Polyprotic Acid

Let’s use sulfuric acid $H_2SO_4$​, which gives 2 H⁺ ions (n-factor = 2).

Reaction:

$$H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O$$

Let:

• $N_1​=?$ (for $H_2SO_4$​)
• $V_1 = 20 \, \text{mL}$
• $N_2 = 0.1 \, \text{N (NaOH)}$
• $V_2 = 40 \, \text{mL}$

Apply: $$N_1 \times 20 = 0.1 \times 40 $$ $$N_1 = \frac{4}{20} = 0.2 \, \text{N}$$

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🔄 Example 2: Redox Reaction

Reaction: $$\text{Fe}^{2+} + \text{MnO}_4^- \rightarrow \text{Fe}^{3+} + \text{Mn}^{2+}$$

• In acidic medium, 5 Fe²⁺ ions are oxidized by 1 MnO₄⁻ ion.
• So, n-factor of MnO₄⁻ = 5, n-factor of Fe²⁺ = 1

Let’s say:

• 25 mL of $\text{FeSO}_4$ solution reacts with 20 mL of 0.02 N $\text{KMnO}_4$

Using: $$N_1V_1 = N_2V_2$$

$$N_1 \times 25 = 0.02 \times 20 \Rightarrow N_1 = \frac{0.4}{25} = 0.016 \, \text{N}$$

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🎯 Key Differences: Normality vs Molarity

Property	Molarity (M)	Normality (N)
Definition	Moles of solute / L of solution	Equivalents of solute / L of solution
Reaction-specific	❌ No	✅ Yes (depends on reaction type)
n-factor needed?	❌ No	✅ Yes
Use	General solutions	Titrations, redox, acid-base, precipitation

$$\text{Normality} = \text{Molarity} \times \text{n-factor}$$

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📌 6. Do You Know? (Quick Facts)

• Normality is not fixed for a substance—it depends on the type of reaction.
• Always balance the reaction to find n-factor correctly.
• In acid-base reactions, normality = molarity × basicity/acidity.
• $N_1V_1 = N_2V_2$ helps find unknown normality or volume during titration.

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📝 Practice Problems

Q1.

25 mL of H₂SO₄ solution completely reacts with 30 mL of 0.1 N NaOH. What is the normality of the acid?

(H₂SO₄ is diprotic ⇒ n-factor = 2)

Solution:

$$N_1 \times 25 = 0.1 \times 30 \Rightarrow N_1 = \frac{3}{25} = 0.12 \, \text{N}$$

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