Normality expresses a concentration of a solution. It is abbreviated by the letter N. Normality is defined as the gram equivalent weight per liter of solution.

## Understanding Equivalent Weight

In order to solve a normality calculation, equivalent weight must be understood. Think of equivalent weight as the reactive capacity of a chemical species, like electrons or ions. The letters **Eq** or **eq** commonly abbreviate equivalent weight.

The valence state of an element or compound or the number of hydrogen ions a molecule transfers describes the number of electrons or protons transferred in reactions. For example, the Al^{+3} ion has a valence of 3, and n (the number of equivalents) also equals 3.

## Understanding Equivalent Weight in Acid-Base Chemistry

The amount of hydrogen ions transferred in acid-base reactions will give the equivalent weight of that acid. For example, sulfuric acid, H_{2}SO_{4}, with two H^{+} ions has an n (number of equivalents) of 2, while hydrochloric acid, HCl, with one H^{+} ion has an equivalent weight of 1.

An acid supplies ions, and the base reacts with those ions. Equivalent weight is applied not only to hydrogen ions of the acid, but to the ions that form a base, too. For example, NaOH dissociates into Na^{+} and OH^{-}, where the OH^{-} has an equivalent weight of 1.

## Calculating Gram Equivalent Weight

Once it is understood how a chemical species, like ions or electrons, reacts in a chemical reaction, the gram equivalent weight may be calculated. Gram equivalent weight is simply equivalent weight expressed in units of mass. Gram equivalent weight is numerically equal to the calculated equivalent weight.

To find gram equivalent weight, use the formula **Eq = MW / n**

- Eq = Equivalent weight
- MW = Atomic or molecular weight in grams/mole, from periodic table
- n = number of equivalents

A couple examples are below:

*Example 1: H _{2}SO_{4}*

For every mole of sulfuric acid, there are two H^{+} ions, n = 2. Look at a periodic table and find the sum of the atomic masses of the S, O and H in your formula:

S = 32.07; O = 16.00; H = 1.01. Sum the molecular weight of H_{2}SO_{4}: 32.07 + 4(16.00) + 2(1.01) = 98.08 g/mol

Eq = 98.08 / 2 = 49.04 g/eq.

The gram equivalent weight of H_{2}SO_{4} is 49.04 g/eq. Only half as much sulfuric acid than, for example, HCl, is needed to react something with an acid.

*Example 2: NaOH*

There is only 1 OH^{-}, so the number of equivalents is 1. Look at a periodic table and find the sum of the atomic masses of the Na, O and H in your formula:

Na = 22.99; O = 16.00; H = 1.01. To sum, 22.99 + 16.00 + 1.01 = 40.00 g/mole

Eq = 40.00 / 1 = 40.00 g/eq

The gram equivalent weight of NaOH is 40.00 g/eq.

## Normality Equation

Once gram equivalent weight is understood, it is easier to understand the equation for normality:

**Normality (N)** = **m /V** **× 1** **/ Eq**

- m = mass of solute in grams
- V = total volume of solution in Liters
- Eq = equivalent weight

## Normality Calculation of NaOH

Example: How is a 1N solution of NaOH prepared?

**Normality (N)** = **m /V** **× 1** **/ Eq**

- N = 1
- m = unknown
- V = 1 liter
- Eq = 40.00g/eq (go back to the gram equivalent weight section if help is needed in remembering why this is so)

**1 N** = **m /1L*× 1* / 40.00 g/eq**

Using algebra and remembering that N is in eq/L:

**m** = **1 eq/L** **× 1 L × 40.00 g/eq** **;** therefore **m** = **40 g**

To make a 1N solution of NaOH, 40 grams of NaOH are dissolved in 1 L.

Likewise, for a 0.1 N solution of NaOH, divide by a factor of 10 and 4 grams of NaOH per liter is needed.