*Normality* is a unit of concentration in acid-base chemistry that is usually expressed in equivalents per liter. An equivalent is the number of equivalent weights (not the mass) of a substance. Equivalent weight, in turn, is the molar mass of a substance divided by the number of hydrogen (H+) or hydroxide (OH-) ions with which one molecule of the substance reacts in solution.

For example, calcium carbonate, which has the formula CaCO_{3}, has a molar mass of 100.1 g. You can determine this from any periodic table of the elements. Ca has a molar mass of 40.1, C a molar mass of 12, and O a molar mass of 16, making the total molar mass of calcium carbonate equal to 40.1 + 12 + 3(16) = 100.1. Because a calcium ion has a positive charge of 2, and exists as Ca^{2+}, each molecule of CaCO_{3} can potentially react with two OH- ions. Thus the equivalent weight of CaCO_{3} is 100.1 ÷ 2 = 50.05 g/Eq.

The upshot of this is that a 1 L solution containing, for example, 200.2 g of CaCO_{3} (i.e., 2 mol) would have a molarity of 2 M, but would have a normality of 2 N, because the equivalent weight of CaCO_{3} is only half its molecular mass, meaning that 1 mol = 2 Eq.

This principle applies to other compounds as well, for example, sodium hydroxide (NaOH). To calculate the normality of a solution of NaOH:

## Step 1: Determine the Number of Moles of NaOH in the Sample

Assume for this problem that you have 0.5 L of a 2.5 M solution of NaOH. This means you have 1.25 mol of NaOH total.

## Step 2: Look Up the Molar Mass of NaOH

From the periodic table, the molar mass of Na = 23.0, that of ) = 16.0, and that of H = 1.0. 23 + 16 + 1 = 40 g.

## Step 3: Determine the Number of Equivalents

You have 1.25 mol of a substance with a molar mass of 40.0 g.

(1.25 mol)(40 g/mol) = 50 g

Since the valence of NaOH is 1, for this compound, 1 mol = 1 eq. This means that for NaOH solutions, normality and molarity are the same, unlike the case with CaCO_{3}.

Thus the normality of your NaOH solution = 2.5 N.