What is Normality in Chemistry?

Normality is a key concept in Chemistry used to measure the concentration of a solution. It represents the number of gram equivalents of a solute in one liter of solution. This unit is especially useful in acid-base titrations, redox reactions, and precipitation reactions, as it helps determine the exact reacting capacity of solutions. Understanding normality simplifies calculations in stoichiometry and enhances precision in chemical analysis.

This page provides a clear explanation of normality, its formula, and its practical applications. We will also highlight the difference between normality and molarity, helping students and professionals build a solid foundation for mastering chemical reactions. 

What is Noramlity?

Normality is a measure of concentration in chemistry, defined as the number of gram equivalents of solute per liter of solution. It is particularly useful in acid-base titrations, redox reactions, and precipitation processes. The formula for calculating normality (N) is:

Where the number of gram equivalents is determined by dividing the mass of the solute by its equivalent weight. The equivalent weight depends on the specific reaction and is calculated by dividing the molar mass of the solute by its valence factor—the number of moles of reactive units (such as H⁺ ions in acids or OH⁻ ions in bases) the solute provides per mole.

We can write the Normality Formula as: 

The Number of Gram Equivalents is Calculated as:

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

Normality Formulas

1. General Formula: \[N = \text{Number of gram equivalents} \times [\text{Volume of solution in litres}]^{-1}\]

2. Using Molarity: $N = \text{Molarity} \times \text{Molar mass} \times [\text{Equivalent mass}]^{-1}$

3. For Acids and Bases: N = Molarity × Basicity = Molarity × Acidity

How to Calculate Normality

Steps to Calculate Normality:

Step 1: Identify the equivalent weight of the solute (using molar mass and valence).

Step 2: Calculate the number of gram equivalents using the solute's weight and equivalent weight.

Step 3: Ensure the volume of the solution is expressed in liters.

Step 4: Apply the formula for normality.

Normality in Titration

For titration calculations, use the formula:

N1 V1 = N2 V2

Where:

• N1​, V1​: Normality and volume of the acidic solution.

• N2​, V2​: Normality and volume of the basic solution.

Relation Between Normality and Molarity

Normality and molarity are related as:

$N = M \times \text{Number of equivalents}$

For example, a 1 M solution of H₂SO₄ has a normality of 2 N because it donates 2 H⁺ ions per molecule.

Differences Between Normality and Molarity

Applications of Normality:

• Acid-Base Titrations: Helps determine the exact point of neutralisation by accounting for the number of reactive units.

• Redox Reactions: Assists in calculating the equivalent amounts of oxidising and reducing agents.

• Precipitation Reactions: Used to measure the concentration of ions that will form a precipitate.

Limitations:

Normality is reaction-specific; the same solution can have different normalities depending on the reaction considered. This specificity can lead to confusion, so it's essential to clearly define the context when using normality. In many cases, molarity or molality may be preferred for their consistency across different reactions.

Conclusion

Understanding normality is crucial for accurately preparing and analysing chemical solutions, particularly in titrations and reactions involving equivalents. While it offers a detailed view of a solute's reactive capacity, its reaction-specific nature makes it less versatile than molarity in general applications. For a deeper understanding, practice normality calculations and explore real-world examples in chemical analysis.

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 standardisation 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