Normality Equation for Mixing Acid and Base, Formulas, Solved Examples

ANAND CLASSES Study Material and Notes to learn the Normality equation for mixing acid and base with formulas, step-by-step solved examples, and methods to find resultant normality and nature of the solution.

🔷 What is Normality (N)?

📌 Definition:

Normality (N) is a measure of concentration that tells you how many gram equivalents of a solute are present per liter of solution. $$\text{Normality (N)} = \frac{\text{Gram Equivalents of Solute}}{\text{Volume of Solution in Liters}}$$

✅ Gram Equivalent:

A gram equivalent depends on the type of reaction:

• Acid-Base Reactions: Based on H⁺ or OH⁻ ions.
• Redox Reactions: Based on electrons transferred.
• Precipitation Reactions: Based on ions reacting.

💡 Example of Normality:

• 1 mole of H₂SO₄ can donate 2 H⁺ ions, so its normality is 2 N (if 1 M concentration).
• 1 mole of NaOH donates 1 OH⁻, so its normality is 1 N (if 1 M concentration).

🔷 Derivation and Meaning of the Equation

🧮 Equation:

$$N_1V_1 + N_2V_2 = N_3V_3$$

This is derived from the law of conservation of equivalents. It says:

The total number of equivalents in the initial solutions is equal to the total number of equivalents in the final solution (assuming no chemical reaction takes place, only mixing).

🔄 Equivalents in a solution:

$$\text{Equivalents} = \text{Normality} \times \text{Volume}$$

Thus: $$N_1V_1 + N_2V_2 = \text{Total equivalents after mixing} = N_3V_3$$

Where:

• $N_1, V_1$: Normality and Volume of Solution 1
• $N_2, V_2$: Normality and Volume of Solution 2
• $N_3, V_3$: Normality and Volume of the final solution after mixing

🔷 Assumptions and Conditions for Use

This equation is valid only under specific conditions:

✅ When:

• You are mixing solutions of the same solute or acid/base type (like HCl + HCl).
• There’s no chemical reaction changing the number of equivalents (just dilution or addition).

❌ Not to be used when:

• A chemical reaction changes the equivalence (like acid-base neutralization or redox reaction).
• The volume changes non-linearly due to interaction or contraction/expansion.

🔷 Detailed Numerical Example

🎯 Problem:

You mix:

• 200 mL of 2 N HCl
• 300 mL of 1 N HCl
  Find the normality of the resulting solution.

✏️ Step-by-step:

$$N_1 = 2 \, N,\quad V_1 = 200 \, mL$$

$$N_2 = 1 \, N,\quad V_2 = 300 \, mL$$

$$V_3 = V_1 + V_2 = 500 \, mL$$

Now apply: $$N_1V_1 + N_2V_2 = N_3V_3$$

$$(2)(200) + (1)(300) = N_3(500) $$

$$400 + 300 = 500N_3 $$

$$700 = 500N_3$$

$$ N_3 = \frac{700}{500} = 1.4 \, N$$

✅ Final normality after mixing = 1.4 N

🔷 Conceptual Understanding: Why Equivalents?

• When mixing two solutions, you’re just combining the total reactive units (equivalents).
• Since 1 N = 1 equivalent per liter, multiplying it by volume gives the total equivalents.
• By summing them up and dividing by the total volume, you get the final normality.

🔷 Do You Know?

🔹 If you dilute a solution (say by adding water), you’re increasing volume but not changing the total equivalents. So: $$N_1V_1 = N_2V_2$$

This is a special case of our main equation when you’re mixing with water or a diluting agent (N₂ = 0).

🔷 Real-Life Application

• Titrations: Used in standardizing solutions.
• Pharmaceutical industry: To prepare exact concentrations of drugs.
• Water treatment: For determining acid or base strength in a sample.

🧪 Mixing Acid and Base: Not Full Neutralization

When you mix an acid and a base, both with known normalities and volumes, but they’re not in stoichiometric (equal equivalents) amounts, some amount of acid or base will remain unreacted in the final solution.

This is common in practical chemistry labs when you’re titrating or buffering solutions.

🎯 Objective:

You’re mixing:

• An acid of normality $N_a$ and volume $V_a$
• A base of normality $N_b$ and volume $V_b$

You want to find:

1. The nature of the final solution (acidic or basic)
2. The resultant normality of the leftover unreacted species

🔍 Step-by-Step Method

1️⃣ Calculate total acid equivalents:

$$\text{Acid Equivalents} = N_a \times V_a$$

2️⃣ Calculate total base equivalents:

$$\text{Base Equivalents} = N_b \times V_b$$

Now compare:

• If acid equivalents > base equivalents, the solution is acidic
• If base equivalents > acid equivalents, the solution is basic
• If both are equal, it’s neutral (full neutralization)

3️⃣ Find the leftover equivalents:

$$\text{Leftover Equivalents} = |N_aV_a – N_bV_b|$$

4️⃣ Total final volume:

$$V_{\text{total}} = V_a + V_b$$

5️⃣ Resultant Normality:

$$N_{\text{final}} = \frac{\text{Leftover Equivalents}}{V_{\text{total}}}$$

The nature (acidic or basic) depends on which equivalents are in excess.

🧪 Example 1: Acid in Excess

You mix:

• 100 mL of 1 N HCl (acid)
• 50 mL of 1 N NaOH (base)

Step 1: $$\text{Acid Equivalents} = 1 \times 100 = 100 $$

$$\text{Base Equivalents} = 1 \times 50 = 50$$

→ Excess Acid = 100 – 50 = 50 eq

Step 2: $$V_{\text{total}} = 100 + 50 = 150 \, \text{mL}$$

Step 3: $$N_{\text{final}} = \frac{50}{150} = 0.333 \, \text{N}$$

✅ Final Solution:

• Acidic
• Normality = 0.333 N

🧪 Example 2: Base in Excess

Mix:

• 100 mL of 1 N HCl
• 150 mL of 1 N NaOH

Acid Equivalents = 100
Base Equivalents = 150

→ Excess base = 50 eq
Total volume = 250 mL

$$N_{\text{final}} = \frac{50}{250} = 0.2 \, \text{N}$$

✅ Final Solution:

• Basic
• Normality = 0.2 N

🔁 General Formula

$$N_{\text{result}} = \frac{|N_aV_a – N_bV_b|}{V_a + V_b}$$

Where:

• Resultant solution is acidic if $N_aV_a > N_bV_b$
• Resultant solution is basic if $N_bV_b > N_aV_a$

🎯 Quick Summary Table

🔥 Extension: Acid-Base Mixture Case (Neutralization Reactions)

⚗️ What Happens in Acid-Base Reactions?

In acid-base neutralization, an acid donates H⁺ ions, and a base donates OH⁻ ions. These react to form water and salt, neutralizing each other.

✅ Reaction Example:

$$\text{HCl (acid)} + \text{NaOH (base)} \rightarrow \text{NaCl (salt)} + \text{H}_2\text{O}$$

This is a 1:1 reaction:

• 1 mole of HCl neutralizes 1 mole of NaOH
• Or 1 equivalent of HCl neutralizes 1 equivalent of NaOH

🔷 So, What Changes?

When an acid reacts with a base, the normality equation becomes: $$\text{Acid Equivalents} = \text{Base Equivalents}$$

Which means: $$N_aV_a = N_bV_b$$

Where:

• $N_a, V_a$: Normality and Volume of Acid
• $N_b, V_b$: Normality and Volume of Base

This is called the Neutralization Equation.

🧮 Derivation: Why This Works?

During neutralization:

• 1 equivalent of H⁺ = 1 equivalent of OH⁻
• The number of equivalents of acid equals those of base at neutralization point

So, $$\text{Acid Normality} \times \text{Volume of Acid} = \text{Base Normality} \times \text{Volume of Base}$$

Or: $$N_aV_a = N_bV_b$$

🧪 Example: Neutralization

🎯 Problem:

What volume of 0.5 N NaOH is required to neutralize 50 mL of 1 N HCl?

Solution:

$$N_aV_a = N_bV_b \Rightarrow (1)(50) = (0.5)(V_b) \Rightarrow V_b = \frac{50}{0.5} = 100 \, \text{mL}$$

✅ So, 100 mL of 0.5 N NaOH is needed to neutralize 50 mL of 1 N HCl.

🔍 What If the Reaction Isn’t 1:1?

In cases like: $$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

Here:

• 1 mole of H₂SO₄ = 2 moles of NaOH
• So:
  • 1 equivalent of H₂SO₄ = 2 equivalents of NaOH
  • Still balanced on the equivalents basis

So the equation: $$N_aV_a = N_bV_b$$

still applies as long as you use normalities that already account for the number of H⁺ or OH⁻.

🧠 Concept Check

🟩 Case 1: Just Mixing (no reaction)

$$N_1V_1 + N_2V_2 = N_3V_3$$

Used when two solutions of same type (like 2 acids or 2 bases) are mixed.

🟨 Case 2: Neutralization Reaction (acid + base)

$$N_aV_a = N_bV_b$$

Used when acid and base react to neutralize each other.

📝 Practice Question:

Q: How many mL of 0.2 N H₂SO₄ is needed to completely neutralize 50 mL of 0.4 N NaOH?

🧠 H₂SO₄ is a dibasic acid (2 H⁺ per molecule), but normality already includes that.

✅ Solution:

$$N_aV_a = N_bV_b \Rightarrow (0.2)(V_a) = (0.4)(50) $$

$$\Rightarrow V_a = \frac{20}{0.2} = 100 \, \text{mL}$$

✅ Answer: 100 mL of H₂SO₄