# Normality Formula

• Last Updated : 20 May, 2022

In chemistry, one of the key phrases is the normality formula. As a result, we utilize it to calculate a solution’s concentration. Similarly, the equivalent concentration of a solution is denoted by the letter ‘N’. It is most commonly used to determine the number of reactive species present in a solution. In addition, during the titration reactions. It’s also occasionally utilized in acid-base chemistry settings. Let’s go through the Normality Formula notion.

### Normality

It’s the number of gram or mole equivalents of solute in a liter of solution. The letter N is used to represent normalcy; eq L-1 and meq L-1 are two other normalcy units. In medical writing, the latter is commonly employed. Normality is also commonly used in three scenarios,

• In acid-base chemistry, to determine the concentrations.
• Used to calculate the number of ions that are expected to precipitate in a given reaction in precipitation reactions.
• In redox reactions, it’s utilized to determine how many electrons a reducing or oxidizing chemical can donate or take.

Normality Formula

The following is the Normality Formula,

Normality (N) = Number of gram equivalents / Volume of the solution in liters

Where,

Number of gram equivalents = weight of solute / Equivalent weight of solute

Thus,

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

∴ N = (Molarity × Molar mass) / Equivalent mass

∴ N = Molarity × Basicity = Molarity × Acidity

Steps to Calculate Normality

In order to calculate normalcy, students might use a few strategies,

• The first piece of advice for students is to collect information on the equivalent weight of the reacting substance or solute. To learn about molecular weight and valence, consult your textbook or reference books.
• The number of gram equivalents of solute is calculated in the second stage.
• Remember to determine the volume in liters.
• Finally, the formula is used to calculate normalcy, with the values replaced.

Equations of Normality Formula

The normalcy equation, which can be used to estimate the volume of a solution needed to make a solution of different normality, is as follows,

Initial Normality (N1) × Initial Volume (V1) = Final Normality (N2) × Final Volume (V2)

If four distinct solutions with the same normalcy and volume are combined, the resultant normality is;

NR = (NaVa + NbVb + NcVc + NdVd) / (Va+Vb+Vc+Vd)

When four solutions (na, nb, nc, nd) with varied solute molarity, volume, and H+ ions are mixed, the resultant normalcy is given by,

NR = (naMaVa + nbMbVb + ncMcVc + ndMdVd) / (Va+Vb+Vc+Vd)

### Sample Questions

Question 1: What is the difference between Molarity and Normality?

Answer:

The number of moles of solute per liter of solution is known as morality, whereas the number of grammes equivalent of solute per liter of solution is known as normality.
In addition, whereas Molarity is a measurement of the moles in respect to the total volume of the solution, Normality is a measurement of the gramme equivalent in relation to the total volume of the solution.

Question 2: Calculate the normality of a 310 mL NaOH solution made by dissolving 0.4 gram of NaOH.

Answer:

Normality (N) = Number of gram equivalents / Volume of the solution in liters

Number of gram equivalents = weight of solute / Equivalent weight of solute

∴ Equivalent weight of solute = 23+16+1 = 40

Since,

Normality (N) = weight of solute / Volume of the solution in liters × Equivalent weight of solute

∴ N = (4/40) × (1000/310)

∴ N = 0.1 × 3.2258

∴ N = 0.3225 N

Question 3: When 19.0 ml of the citric acid solution is titrated with 30.09 mL of 0.1811 N KOH, what is the citric acid concentration?

Answer:

Na × Va = Nb × Vb

∴ Na × 19.0 = 0.1811 × 30.09

∴ Na = (0.1811 × 30.09) / 19.0

∴ Na = 5.4492 / 19.0

Na = 0.2868 N

Question 4: When 0.321 g sodium carbonate is mixed in a 250 mL solution, determine its normality.

Answer:

Chemical Formula of Sodium carbonate is Na2CO3

N = Na2CO3 × (1 mol/105.99 g) × (2 eq/1 mol)

∴ N = 0.1886/0.2500

∴ N = 0.07544 N

Question 5: When the concentration of H3PO4 is 2.2 M, calculate and discover the normality.

H3AsO4 + 2NaOH → Na2HAsO4 + 2H2O

Answer:

Only two of the H+ ions of H3AsO4 react with NaOH to generate the product, according to the described reaction. As a result, there are two equivalents of the two ions. The above formula will be used to find the normality.

N = Molarity × Number of equivalents

∴ N = 2.2 × 2

∴ N = 4.4

Question 6: What is the normality of the following 1L aqueous solution with 55 gram NaOH dissolved in it?

Answer:

Equivalent weight of solute = 40g

Since, Number of gram equivalents = weight of solute / Equivalent weight of solute

∴ Number of gram equivalents = 55/40

∴ Number of gram equivalents = 1.375 eq

Normality (N) = Number of gram equivalents / Volume of the solution in liters

∴ N = 1.375/1

∴ N = 1.375 eq/L

Question 7: Write any 2 uses of Normality.

Answer:

Uses of Normality,

• In precipitation reactions, normality is used to determine how many ions are expected to precipitate.
• It determines how many electrons a reducing or oxidizing substance can contribute or absorb in redox processes.
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