Normality is calculated by dividing the number of Equivalent Weights of a solute by the volume of the solution in litres. The formula for calculating Normality is N = number of Equivalent weights of solute/volume of Solution in liters, where N is normality. Each substance has different equivalent weights. In this article, we are going to learn what normality is, how to calculate normality and some sample problems on the normality concept.

What is Normality?

In chemistry, normality is a measure of concentration that represents the number of equivalents of a solute dissolved in a liter of solution. It is often used in acid-base reactions and other reactions where stoichiometry involves the transfer of multiple protons, ions, or other chemical species. According to the standard definition, normality is defined as the quantity of gram or mole reciprocals of solute present in one liter of an answer. At the point when we say the same, it is the number of moles of receptive units in a compound.

Normality Formula

Normality = Number of gram reciprocals × [volume of arrangement in liters]^-1

Number of gram reciprocals = weight of solute × [Equivalent weight of solute]^-1

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

N = Molarity × Molar mass × [Equivalent mass]^-1

N = Molarity × Basicity = Molarity × Acidity

Normality is many times meant by the letter N. A portion of different units of normality are likewise communicated as eq L^-1 or meq L^-1. The last option is much of the time utilized in clinical detailing.

How to Calculate Normality?

To calculate the normality of a solution, you need to know the molarity of the solute and the number of equivalents per mole of solute. The formula to calculate normality is:

N=M × E_q

The steps to calculate normality are mentioned below:

Step 1: Determine the molarity (M) of the solute in the solution. This is typically find in units of moles per liter (mol /L) or molarity.

Step 2: Determine the number of equivalents (Eq) per mole of solute based on the chemical reaction or species involved. The number of equivalents depends on the specific chemical reaction and the species involved. For example, for monoprotic acids and bases, the number of equivalents is 1. For diprotic acids and bases, the number of equivalents is 2. For polyprotic acids and bases or other chemical species that can donate or accept multiple protons or ions, the number of equivalents is equal to the number of protons or ions involved in the reaction.

Step 3: Multiply the molarity (M) by the number of equivalents (Eq) to calculate the normality (N) of the solution using the formula:

Calculation of Normality in Titration

Titration is a technique used to determine the concentration of a substance in a solution by reacting it with a solution of known concentration (a titrant). The process involves gradually adding the titrant to the analyze solution (the solution of unknown concentration) until a chemical reaction between the two solutions is complete. The formula to calculate normality in titration is given below:

N₁ V₁ = N₂ V₂

Where,

N₁ = Normality of the Acidic arrangement.
V₁ = Volume of the Acidic arrangement.
N₂ = Normality of the essential arrangement.
V₂ = Volume of the essential arrangement.

Normality Equations

The normality equation to find normality of solution before and after changes in given as

Beginning Normality (N₁) × Initial Volume (V₁) = Normality of the Final Solution (N₂) × Final Volume (V₂)

Assume four distinct arrangements with a similar solute of normality and volume are mixed; hence, the resultant normality is given by,

NR = [N _a V _a + N _b V _b + N _c V _c + N _d V _d] × [V _a+ V _b+ V _c +V _d]^-1

Assuming four arrangements having different solute of molarity, volume and H+ particles (n _a, n _b, n _c, n_d) are blended then the resultant normality is given by;

NR = [n _a M _a V _a + n _b M _b V _b + n _c M _c V _c + n _d M _d V _d] × [V _a+ V _b+ V _c+ V _d]^-1.

Calculation of Normality in Titration

Titration is the process of a gradual addition of a solution of a known concentration and volume with another solution of unknown concentration until the reaction approaches its neutralization.

To find the normality of the acid and base titration,

where,

• N₁= Normality of the Acidic solution
• V₁ = Volume of the Acidic solution
• N₂ = Normality of the basic solution
• V₃ = Volume of the basic solution

Uses of Normality

• Normality is used in precipitation reactions to measure the number of ions which are likely to precipitate in a specific reaction.
• Normality is used to describe the concentration of acids and bases.
• It is used in redox reactions to determine the number of electrons that a reducing or oxidizing agent can donate or accept.

Limits in Using Normality

While normality is a useful concept in certain types of chemical analysis, it has several limitations which are mentioned below:

• Normality is dependent on the stoichiometry of the reaction.
• The number of equivalents per mole of solute can vary depending on the specific chemical reaction or species involved.
• Normality is primarily used in acid-base and redox reactions where the stoichiometry involves the transfer of protons, ions, or electrons. It may not be applicable or meaningful for other types of reactions, such as precipitation reactions or complexation reactions.
• Normality does not account for the presence of other species in solution that may interfere with the reaction or affect the accuracy of the measurement.
• Determining the exact number of equivalents per mole of solute can be challenging, especially for complex reactions or mixtures of species.

Also, Check

• How to calculate the Molarity of a Solution? 
• Expressing Concentration of Solutions
• Mass Percent Formula

Sample Questions

Question 1: In the following equation find the normality when it is 2.0 M H₃PO₄.

H₃AsO₄ + 2NaOH → Na₂HAsO₄ + 2H₂O

Solution:

Assuming we take a gander at the given response we can distinguish that the main two the H⁺ particles of H₃AsO₄ respond with NaOH to shape the item. Consequently, the two particles are 2 reciprocals. To find the normality, we will apply the given recipe.

N = Molarity (M) × number of Moles

N = 2.0 × 2

In this way, the normality of the arrangement = 4.0.

Question 2: Find the normality of the arrangement got by dissolving 0.121 g of the salt sodium carbonate (Na₂CO₃) in 250 ml of water.

Solution:

Given: 0.121 g Na₂CO₃ (molar mass =106 g/mol) in 250 mL or 0.25 L arrangement. Presently, Na₂CO₃ structures an essential salt arrangement and can take part in balance equation as follows:

Na₂CO₃ + 2H⁺(from corrosive) → H₂CO₃ + 2Na⁺

Presently, normality is characterized as the number of counterparts per liter of arrangement. Or then again. normality= no. of counterparts/Volume(in Liter). We realize that number of counterparts is, Number of moles × n-factor. Likewise, the n-factor for bases is characterized as the quantity of OH-delivered per particle (for Arrhenius type bases) OR it is the quantity of H+ acknowledged per atom (for Lowry Bronzed type bases). Here, Na₂CO₃ is a Lowry Bronzed base with n-factor 2. Presently,

Number of moles, n = Mass/molar mass = 0.121/(106) = 0.0011

Number of reciprocals = n × n-factor = 0.0011 × 2 = 0.0022

Consequently, Normality = No. of reciprocals/V (in liter) = 0.0022/0.25 = 0.0088 N

Question 3: What is the normality of the following?

• 0.1481 M NaOH
• 0.0321 M H₃PO₄

Solution:

• N = 0.1481 mol/L × (1 eq/1mol) = 0.1481 eq/L = 0.1481 N
• N = 0.0321 mol/L × (3 eq/1mol) = 0.0963 eq/L = 0.0963 N

Question 4: What will the normality of citrus extract be assuming 20.00 ml of the citrus extract arrangement is titrated with 18.12 ml of 0.1718 N KOH?

Solution:

N _a × V _a = N _b × V _b

N_a × (20.00 ml) = (0.1718N) × (18.12 ml)

Accordingly, the convergence of citrus extract = 0.1556 N.

Question 5: Calculate the normality of the base assuming 23.87 mL of the base is utilized in the normalization of 0.4258 g of KHP (eq. wt. = 204.23)?

Solution:

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

= 2.085 × 10^-3 eq base/0.02387 L = 0.0873 N

Normality of the base is = 0.0873 N.

Question 6: Ascertain the normality of corrosive in the event that 41.18 ml is utilized to titrate 0.1369 g Na₂CO₃?

Solution:

0.1369 g Na₂CO₃ × (1 mol/105.99 g) × (2 eq/1 mol) × (1 eq corrosive/1 eq base):

= 2.583 × 10^-3 eq corrosive/0.04118 L = 0.0627 N

Normality of the corrosive = 0.0627N.

Question 7: Calculate the normality of NaOH solution prepared by dissolving 0.4m NaOH to make a 250 ml solution.

Solution:

Normality (N) = number of Gram Equivalent of solute/Volume of Solution in liter

Number of Gram Eq. of the Solute = weight/Equivalent weight

Presently, Equivalent weight= Molar Mass n =23+16+11=40

Thus, N = No. of gram eq. mass/Vol(liter)

= Weight/Equivalent weight × 1000/V(in ml)

= 4/40 × 1000/250

= 0.4N

FAQs – How to Calculate Normality of a Solution

What is Normality (N)?

Normality is a measure of the concentration of solute in a solution expressed as the gram equivalent weight per liter of solution.

How is Normality calculated?

Normality is calculated using the formula: Normality (N)=gram equivalent weight of solute /liters of solution

To find the gram equivalent weight: Gram Equivalent Weight=molar mass /valence or n-factor.

What is the difference between Normality and Molarity?

• Molarity (M) measures the number of moles of solute per liter of solution.
• Normality (N) measures the number of equivalents per liter of solution. It depends on the nature of the solute’s participation in the reaction (e.g., for sulfuric acid in a neutralization reaction, normality is twice the molarity because it provides two moles of H⁺ per mole of H₂SO₄).

Can Normality change with the type of reaction?

Yes, normality depends on the reaction context. For example, in different reactions, the same compound can have different normalities depending on how many equivalents it reacts with.

Is Normality temperature dependent?

Normality itself is not directly dependent on temperature, but since it involves volume measurement (liters of solution), and volume can change with temperature, the concentration expressed as normality can indirectly vary with temperature.