Learn Chemistry Normality Formula With Example for JEE Main 2025

Understanding the concept of Normality is crucial for excelling in JEE Main 2025. Normality is a key unit of concentration used in various chemistry calculations, especially in titrations and reaction stoichiometry. This article aims to provide a clear understanding of Normality, including its formula and practical examples, to help students grasp this essential concept for their exams. With step-by-step explanations and solved examples, students can build their confidence in solving problems involving Normality.

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What is Normality?

Ever heard of a 1N banana smoothie? Or wondered what it means when a solution is "normal"? Worry not, young chemist, "normality" in JEE Main chemistry isn't about measuring your fruit intake. It's a powerful tool used to understand the concentration of solutions, particularly when acids and bases are involved.

Imagine pouring a pack of Tang powder into water. The orange powder (solute) dissolves, spreading its tangy goodness (dissolved particles) throughout the water (solvent). But how much "tanginess" (concentration) do we have? Molarity, another unit we've met, uses moles per liter to answer this. But for acid-base reactions, there's a twist. Both acids and bases have special "reactive units" called equivalents. 

Definition of Normality:

We define normality as the number of grams equivalent to solute that is present in a one-liter solution. So, the unit of normality is gram/liter.

We denote normality with the letter ‘N’.

We can write the Normality Formula as: 

What is an Equivalent? 

It's the number of hydrogen (H+) ions an acid can donate or the number of hydroxide (OH-) ions a base can accept in a single reaction. Let’s understand what is equivalent and its significance as well.

Let’s say we have an equation:

              NaCl              +     H2SO4          →     Na2SO4                +      HCl

(Sodium Chloride) (Sulphuric Acid)   (Sodium Sulfate)    (Hydrochloric Acid)

Equivalence

You can see that this equation is not balanced. Now, let’s balance this equation:

                                 2NaCl  + H2SO4 → Na2SO4 + 2HCl

No of moles (n):   (2 moles)   (1 mole)     (1 mole)  (2 moles)

So, 2 moles of NaCl react with 1 mole of H2SO4 to give 1 mole of Na2SO4 and 2 moles of HCl.

This means, without balancing the equation, we can’t determine the quantity of reactant (or moles) that undergoes a reaction to form a product.

Normality Formula

• $\text{Normality} = \text{Number of gram equivalents} \times [\text{Volume of solution in litres}]^{-1}$

• $\text{Number of gram equivalents} = \text{Weight of solute} \times [\text{Equivalent weight of solute}]^{-1}$

• N = Weight of Solute (gram) × [Equivalent weight × Volume (L)]

• $N = \text{Molarity} \times \text{Molar mass} \times [\text{Equivalent mass}]^{-1}$

• N = Molarity × Basicity = Molarity × Acidity

Some of the other units of normality are also expressed as eq $L^{-1}$ or $\text{meq} L^{-1}$. The latter is often used in medical reporting.

Number of Gram Equivalent

So, let’s get forward with understanding the concept of gram equivalence.

We know that no of moles = Mass/Molecular weight

Number of gram equivalent = Mass/Equivalent weight, 

and Equivalent weight = Molecular weight X

(X = valence factor, where valence factor for acids and bases is the number of H+ and OH- ions they release in the solution, respectively).

We’ll understand these two formulas with an example.

Let’s Find Out Gram Equivalent

For example, Find the number of gram equivalents present in 0.5 g of HCl.

HCl releases one H+ ion in the solution, so its valence factor = 1.

The molecular weight of HCl = 36.46 g.

So, equivalent weight =

Molecular weight X = 36.46/1 = 36.46 g, and

Number of gram equivalent

= Mass/Equivalent weight 

= 0.5/36.46 = 0.0137

So, we get the number of gram equivalent  = 0.0137

Let us take another example of 1.06 g of Na2CO3 to understand this concept clearly

We are given the mass of Na2CO3 = 1.06 g.

Firstly, Find the equivalent weight of Na2CO3. 

Since Na2CO3 is a salt, so the number of positive charges on the cation gives X = 2

Molecular weight = 106 g

So, Equivalent Weight = 

Molecular weightX

= 106/2 = 53 g, 

and number of gram equivalent is:

Mass/Equivalent weight

= 1.06/53 = 0.02

In a chemical equation, the number of grams equivalent of both reactions always remains the same.

Normality Chemistry

There are three types of Normality

1. Seminormal - The solutions whose normality is ½ or N/2.

2. Binormal - The solutions having normality as 2 or 2 N.

3. Decinormal - Normality is 1/10 or N/10.

4. CentiNormal - Normality is 1/100 or N/100.

Normality Calculation Formula

Let’s take an example of how to calculate normality:

If 13 g of N2O4 is present in 500 ml of solution. Find normality.

We are given a mass of N2O4 = 0.65 g, and volume = 500 ml = 0.5 l.

We know that normality, N = no of gram equivalent/volume of solution in liters

Let’s find out equivalent weight to find out the number of gram equivalent for N2O4

Molecular weight of N2O4 = (2 x 14) + (4 x 16) = 28 + 64 = 92 g.

Since the number of negative charges on oxygen = 4. So, X = 4

So, equivalent weight = Molecular weight X = 92/4 = 13 g, 

and Number of gram equivalent =

Mass/Equivalent weight

= 13/13 = 1

Now, let’s calculate the normality, by the formula given below:

N =  No of gram equivalent / Volume of solution in liters = 1/0.5  

N = 2 gram / liter.

Here, the Normality is N = 2, which means the solution of N2O4 is BiNormal.

Normality Equation

For deriving the normality equation, let’s understand normality in mixtures 

Let’s consider two ideal solutions having their normalities as Na and Nb, and the volume as Va and Vb respectively as shown below:

Where Na = The normality of the acidic solution,

Va = Volume of the acidic solution, 

Nb = Normality of the basic solution, and

Vb = Volume of the basic solution.

On combining these two solutions, we get a mixture whose volume is Va + Vb and the normality as N.

So, we got the Normality Formula for the mixture as:  

N = (Na Va + Nb Vb)/ (Va + Vb)

Let’s consider three Cases:

Case 1: The concentration of the acidic solution > concentration of the basic solution (the release of H+ ions > OH- ions)

So,  Na Va >  Nb Vb

Case 2: The concentration of the basic solution < acidic solution, then

Na Va <  Nb Vb (release of OH- ions > H+ ions)

Case 3: When concentration is equal, then

Na Va = Nb Vb is the normality equation.

In this case, there is no release of both OH- ions and H+ ions.

This means the number of gram equivalent of H+ ions = the number of gram equivalent of OH- ions. Such a type of solution is neutral and this process is called neutralization.

How is Normality Different from Molarity?

Molarity tells you the number of moles of solute (the dissolved substance) per liter of solution. But what if the solute has more than one "reactive unit"? For example, sulfuric acid (H₂SO₄) has two replaceable hydrogens, meaning it can potentially react with two other molecules. Molarity wouldn't capture this double punch, that's where normality comes in.

Normality takes into account the number of equivalent weights of the solute per liter. An equivalent weight is like dividing the molar mass by the number of reactive units in the molecule. So, for sulfuric acid, its normality would be twice its molarity, because it has two reactive hydrogens.

Uses of Normality

Normality is used mostly in three common situations:

• In determining the concentrations in acid-base chemistry. For instance, normality is used to indicate hydronium ions (H3O+) or hydroxide ions (OH–) concentrations in a solution.

• Normality is used in precipitation reactions to measure the number of ions which are likely to precipitate in a specific reaction.

• It is used in redox reactions to determine the number of electrons that a reducing or an oxidising agent can donate or accept.

Why Use Normality?

Normality shines in situations involving acid-base reactions and titrations. Here's why:

• Acid-Base Titrations: Imagine a battle between an acid army (with its hydrogen soldiers) and a base army (with hydroxide soldiers). Normality tells you how many "soldier units" each side has, making it easier to calculate the amount of acid needed to neutralize the base (and vice versa).

• Stoichiometry: Normality helps you balance chemical equations involving acids and bases, ensuring you have the right number of "reactive units" on both sides.

Limitations in Using Normality

Many chemists use normality in acid-base Chemistry to avoid the mole ratios in the calculations or simply to get more accurate results. While normality is commonly used in precipitation and redox reactions, there are some limitations to it. These limitations are as follows:

• It is not a proper unit of concentration in situations apart from the ones that are mentioned above. It is an ambiguous measure, and molarity or molality are better options for units.

• Normality requires a defined equivalence factor.

• It is not a specified value for a particular chemical solution. The value can significantly change depending on the chemical reaction. To elucidate further, one solution can actually contain different normalities for different reactions.

Problems on Normality

1. In this reaction, when 1.0 M $H_3PO_4$ reacts, find the normality.

$H_3AsO_4 + 2NaOH \to Na_2HAsO_4 + 2H_2O$

Solution: Looking at the reaction, only two $H^+$ ions of $H_3AsO_4$ react, making it 2 equivalents. 

Using the formula N = Molarity (M) × Number of equivalents

N = 1.0 × 2

Hence, normality of the solution = 2.0

2. Calculate the normality of a solution with 0.321 g sodium carbonate $(Na_2CO_3)$ dissolved in 250 mL water.

Solution: Given, 0.321g $Na_2CO_3$ (Molar mass = 106 g/mol) in 250 mL or 0.25 L solution. For $Na_2CO_3$, the n-factor is 2.

Number of moles, n = Mass/Molar mass = 0.321/ (106) =  0.003

Number of equivalents = n x n-factor = 0.003 x 2 = 0.006

Hence, normality = No. of equivalents/ V (in litre) = 0.006/0.25 = 0.024 N

3. Find the normality of 0.1381 M NaOH and 0.0521 M $H_3PO_4$.

Solution: 

a. N = 0.1381 mol/L × (1 eq/1mol) = 0.1381 eq/L = 0.1381 N

b. N = 0.0521 mol/L × (3 eq/1mol) = 0.156 eq/L = 0.156 N

4. Calculate the concentration of citric acid if titrated with 28.12 mL of 0.1718 N KOH.

Solution: 

$N_a \times V_a = N_b \times V_b$

$N_a \times (25.00 mL) = (0.1718N) (28.12 mL)$

Hence, concentration of citric acid = 0.1932 N

5. Determine the normality of the base used in the standardization of 0.4258 g of KHP (eq. wt = 204.23).

Solution: 0.4258 g KHP × (1 eq/204.23g) × (1 eq base/1eq acid)

Normality $= \dfrac{2.085 \times 10^{-3} \text{eq base}}{0.03187 L} = 0.6542 N$

Normality of the base = 0.6542 N

Significance of Normality in JEE Main 2025

• In JEE Main Chemistry, questions related to normality are typically found within the "Some Basic Concepts of Chemistry" and "Solutions" chapters. 

• While these topics are fundamental, the exact frequency of normality-based questions can vary each year.

• Historically, these chapters have contributed approximately 1 to 2 questions per exam session.

Preparation Tips for Normality for JEE Main 2025

1. Grasp the concepts of equivalents, molarity, and the relationship between them.

2. Memorise N = Weight of Solute Equivalent Weight × Volume (L)N = $\frac{\text{Weight of Solute}}{\text{Equivalent Weight} \times \text{Volume (L)}}$ N = Equivalent Weight × Volume (L) Weight of Solute​.

3. Solve examples of titrations and reaction stoichiometry using the Normality formula.

Conclusion

By mastering the basics of normality, you can tackle acid-base problems and titrations with ease. Remember, practice makes perfect, so don't hesitate to solve practice problems and clarify doubts with your teachers. With dedication and a clear understanding of this concept, you'll be well on your way to cracking the normality section of your JEE Main exam!

Keep in mind that while normality is a valuable tool, it's not used as frequently as molarity in modern chemistry. However, understanding both concepts will demonstrate your depth of knowledge and analytical skills, making you a stand-out candidate.

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