Normality is a unit of the concentration of a chemical solution defined as the gram equivalent weight of solute per liter of solution. Normality is also called equivalent concentration. It is indicated by the symbol “N” or “eq/L” (equivalents per liter). To find the gram equivalent weight, you need to know how many hydrogen ions (H+ or H3O+), hydroxide ions (OH–), or electrons (e–) are transferred in a reaction or you need to know the valence of the chemical species.
The International Union of Pure and Applied Chemistry discourages the use of this unit, but you may encounter it in chemistry classes or the lab, particularly with acid-base titrations and redox reactions. Here is a look at the different ways to calculate normality of solution, along with examples.
Steps for Solving Normality Problems
- Get information to determine the number of equivalents formed or the equivalent weight of the solute or reactants. Usually, you need to know the valence, molecular weight, and whether or not a substance fully dissociates or dissolves.
- Calculate the gram equivalent of the solute.
- Remember the volume of the solution is in liters.
There are a few formulas used to calculate normality. Which one you use depends on the situation:
N = M x n
Here, M is molarity in moles per liter and n is the number of equivalents produced. The number of equivalents is an integer for acid-base reactions, but could be a fraction in a redox reaction.
N = Number of gram equivalents / volume of solution in liters
N = Weight of solute in grams / [volume in liters x equivalent weight]
N = Molarity x Acidity
N = Molarity x Basicity
N1 V1 = N2 V2
In a titration:
- N1 = Normality of the acidic solution
- V1 = Volume of the acidic solution
- N2 = Normality of the basic solution
- V23 = Volume of the basic solution
Alternatively, you can use this equation to make solutions with different volumes:
Initial Normality (N1) × Initial Volume (V1) = Normality of the Final Solution (N2) × Final Volume (V2)
Calculate Normality From Molarity
It’s easy to calculate normality from molarity for an acid or base solution if you know the number of hydrogen (acid) or hydroxide (base) ions produced. Often, you don’t need to break out the calculator.
For example, a 2 M hydrochloric acid (HCl) solution is also a 2 N HCl solution because each hydrochloric acid molecule forms one mole of hydrogen ions. Similarly, a 2 M sulfuric acid H2SO4) solution is a 4 N H2SO4 solution because each sulfuric acid molecule produces two moles of hydrogen ions. A 2 M phosphoric acid solution (H3PO4) is a 6 N H3PO4 solution because phosphoric acid produces 3 moles of hydrogen ions. Switching to bases, a 0.05 M NaOH solution is also a 0.05 N NaOH solution because sodium hydroxide produces one mole of hydroxide ions.
Sometimes even simple problems require a calculator. For example, let’s find the normality of 0.0521 M H3PO4.
N = M x n
N = (0.0521 mol/L)(3 eq/1mol)
N = 0.156 eq/L = 0.156 N
Keep in mind, normality depends on the chemical species. So, if you have one liter of a 1 N H2SO4 solution it will give you 1 N of hydrogen ions (H+) in an acid-base reaction, but only 0.5 N sulfate ions (SO4–) in a precipitation reaction.
Normality also depends on the chemical reaction. For example, let’s find the normality of 0.1 M H2SO4 (sulfuric acid) for the reaction:
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
According to the equation, 2 moles of H+ ions (2 equivalents) from sulfuric acid react with sodium hydroxide (NaOH) to form sodium sulfate (Na2SO4) and water. Using the equation:
N = molarity x equivalents
N = 0.1 x 2
N = 0.2 N
Even though you’re given extra information (number of moles of sodium hydroxide and water), they don’t affect the answer to this problem. Normality depends on the number of hydrogen ions participating in the reaction. Since sulfuric acid is a strong acid, you know it completely dissociates into its ions.
Sometimes not all of the hydrogen ions in a reactant participate in the reaction. For example, let’s find the normality of 1.0 M H3AsO4 in this reaction:
H3AsO4 + 2 NaOH → Na2HAsO4 + 2 H2O
If you look at the reaction, you see only two of the hydrogen ions in H3AsO4 react with NaOH to form the product. So, there are 2 equivalents and not 3 like you might expect. You can find normality using the equation:
N = Molarity x number of equivalents
N = 1.0 x 2
N = 2.0 N
Example: Normality of a Salt Solution
Find the normality of 0.321 g sodium carbonate in a 250 mL solution.
First, you need to know the formula for sodium carbonate to calculate its molecular weight and so you can see what ions it forms when it dissolves. Sodium carbonate is Na2CO3 and its molecular weight is 105.99 g/mol. When it dissolves, it forms two sodium ions and one carbonate ion. Set up the problem so the units cancel out to give an answer in equivalents per liter:
N = (mass in grams x equivalents) / (volume in liters x molecular weight)
Re-writing to make unit cancelling easy to see:
N = (0.321 g) x (1 mol/105.99 g) x (2 eq/1 mol) / 0.250 L
N = 0.0755 eq/L = 0.0755 N
Example: Acid-Base Titration
Find the normal concentration of citric acid when 25.00 mL citric acid solution is titrated with 28.12 mL of 0.1718 N KOH solution.
To solve this problem, use the formula:
Na × Va = Nb × Vb
Na × (25.00 mL) = (0.1718 N) (28.12 mL)
Na = (0.1718 N) (28.12 mL)/(25.00 mL)
Na = 0.1932 N
Limitations of Using Normality
There are considerations to remember when using normality:
- Normality always requires an equivalence factor.
- Normality depends on temperature. As long as you do all lab work at the same temperature (i.e., room temperature), it’s stable, but if you boil or refrigerate a solution, all bets are off. If you expect dramatic temperature changes, use a different unit, like molarity or mass percent.
- Normality depends on the substance and chemical reaction being studied. For example, if you calculate the normality of an acid with respect to a certain base, it may be different if you change the base.
- IUPAC (1997). “Equivalent entity”. Compendium of Chemical Terminology (The Gold Book) (2nd ed.). doi: 10.1351/goldbook
- IUPAC. The Use of the Equivalence Concept.