Table of Contents

## Definition of Normality

Normality is a measure of concentration equal to the gram equivalent weight per liter of solution. Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute’s role in the reaction determines the solution’s normality.

Normality is also known as the equivalent concentration of a solution.

As per the standard definition, it is described as the number of gram or mole equivalents of solute present in one liter of a solution. When we say equivalent, it is the number of moles of reactive units in a compound.

## Formula of Normality

Normality = Number of gram equivalents × [volume of solution in liters]^{-1}

Number of gram equivalents = weight of solute × [Equivalent weight of solute]^{-1}

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

= Molarity × Molar mass × [Equivalent mass]^{-1}

N = Molarity × Basicity = Molarity × Acidity

Normality is often denoted by the letter N. Some of the other units of normaliity are also expressed as eq L^{-1} or meq L^{-1}.

The latter is often used in medical reporting per the standard definition, and N is described as the Number of gram or mole equivalents of solute present in one liter of a solution. When we say equivalent, it is the number of moles of reactive units in a compound.

## Units of Normality

The capital letter N is used to indicate concentration in terms of normalit,y. It can also be expressed as eq/L (equivalent per liter) or meq/L (milliequivalent per liter of 0.001 N).

## Normality Equations

The equation of normality that helps to estimate the volume of a solution required to prepare a solution of different N is given by

Initial Normality (N_{1}) × Initial Volume (V_{1}) = Normality of the Final Solution (N_{2}) × Final Volume (V_{2})

Suppose four different solutions with the same solute of normality and volume are mixed; therefore, 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}

If four solutions having different solutes of molarity, volume, and H^{+} ions (n_{a}, n_{b}, n_{c}, n_{d}) are mixed 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}

## Uses of Normality

Normality is used mostly in three common situations:

- In determining the concentrations in acid-base chemistry. For instance, normaliity indicates hydronium ions (H
_{3}O^{+}) or hydroxide ions (OH^{–}) concentrations in a solution. - Normality is used in precipitation reactions to measure the number of ions likely to precipitate in a specific reaction.
- It is used in redox reactions to determine the number of electrons that a reducing or an oxidizing agent can donate or accept.

## Limitations in Using Normality

Many chemists use normality in acid-base chemistry to avoid the calculations’ mole ratios or get more accurate results. While normalitiy is used commonly 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 is a better option 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 contain different normalities for different reactions.

## Examples

Here are some solved examples of normality

### Example-1

Find the normality of 0.1 M H_{2}SO_{4} (sulfuric acid) for the reaction:

H_{2}SO_{4} + 2NaOH → Na_{2}SO_{4} + 2H_{2}O

According to the equation, 2 moles of H+ ions (2 equivalents) from sulfuric acid react with sodium hydroxide (NaOH) to form sodium sulfate (Na_{2}SO_{4}) and water. Using the equation:

N = molarity x equivalents N = 0.1 x 2 N = 0.2 N

Don’t be confused by the number of moles of sodium hydroxide and water in the equation. Since you’ve been given the molarity of the acid, you don’t need the additional information. All you need to figure out is how many moles of hydrogen ions are participating in the reaction.

Since sulfuric acid is a strong acid, you know it completely dissociates into its ions.

### Example-2

In the following reaction calculate and find the normality when it is 1.0 M H_{3}PO_{4}

H_{3}AsO_{4} + 2NaOH → Na_{2} HAsO_{4} + 2H_{2}O

**Solution:** If we look at the given reaction, we can identify that only two of the H^{+} ions of H_{3}AsO_{4} react with NaOH to form the product.

Therefore, the two ions are 2 equivalents. In order to find the normaliity, we will apply the given formula.

N = Molarity (M) × number of equivalents

Therefore, normaliti of the solution = 2.0. N = 1.0 × 2 (replacing the values)

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