Normality Calculator

How it works

N = M × n · equivalent weight = molar mass ÷ n · N = mass ÷ (equivalent weight × V) · mass = N × V × equivalent weight · stock M = % w/w × density × 1000 ÷ molar mass · N₁V₁ = N₂V₂

Normality expresses concentration as equivalents of a reactive species per litre. For an acid, n is the number of H⁺ ions it gives up in the reaction; for a base, the OH⁻ (or H⁺ accepted); for a redox reagent, the electrons transferred per formula unit. Multiply molarity by that n-factor: 1 M H₂SO₄ is 2 N because each molecule supplies two protons. Equivalent weight is the molar mass divided by n, so the mass you need for a target normality is N × V × equivalent weight. Concentrated acids and bases come as liquids labelled with % w/w and density; their molarity is % × density × 1000 ÷ molar mass, and from that you get the volume of stock to dilute. In a titration the equivalents of titrant and sample are equal at the end point, so the unknown normality is N₁V₁ ÷ V₂. Choose a common reagent to fill in n, the molar mass and the typical stock strength, or pick Other and type your own. Normality depends on the reaction, so check n for the end point or redox change you are using.

Worked example

Sulfuric acid releases 2 H⁺ per molecule, so a 0.5 M solution is 0.5 × 2 = 1 N. To make 1 L of 0.1 N sodium carbonate (105.99 g/mol, n = 2): equivalent weight = 52.995 g/eq and mass = 0.1 × 1 × 52.995 = 5.30 g. Concentrated HCl at 37% and 1.19 g/mL is 0.37 × 1.19 × 1000 ÷ 36.46 = 12.08 N, so 1 L of 0.1 N needs 8.28 mL of it. In a titration, 25.0 mL of 0.100 N NaOH neutralises 20.0 mL of an acid, so the acid is 0.100 × 25.0 ÷ 20.0 = 0.125 N.

Sources: IUPAC Gold Book — equivalent · PubChem — compound molar masses · Equivalent concentration — Wikipedia

Frequently asked questions