What is normality in chemistry?

Normality (N) is a measure of concentration expressed as equivalents of solute per litre of solution. One equivalent is the amount of substance that donates or accepts one mole of protons (acid-base) or one mole of electrons (redox). The formula is N = equivalents / volume(L) = (mass / EW) / V, where EW is the equivalent weight.

How is equivalent weight different from molar mass?

Molar mass (M) is the mass of one mole of a compound. Equivalent weight (EW) is the mass that provides exactly one equivalent — one transferable H⁺, OH⁻, or one mole of electrons. EW = M / n, where n is the n-factor. For H₂SO₄ (M = 98.08 g/mol, n = 2), EW = 49.04 g/eq.

What is the n-factor and how do I find it?

The n-factor (also called the valency factor) counts the equivalents per formula unit. For acids it is the number of replaceable H⁺ ions (HCl = 1, H₂SO₄ = 2, H₃PO₄ = 3). For bases it is the number of OH⁻ groups (NaOH = 1, Ca(OH)₂ = 2). For redox agents it is the change in oxidation state per formula unit — KMnO₄ in acid has n = 5 because Mn goes from +7 to +2.

How do I convert normality to molarity?

Use N = M × n, so M = N / n. A 2 N solution of H₂SO₄ (n = 2) has molarity 1 mol/L. A 1 N solution of NaOH (n = 1) is also 1 mol/L. The relationship depends entirely on the n-factor of the specific substance and reaction type.

What does N₁V₁ = N₂V₂ mean for dilutions?

This is the dilution law: the number of equivalents is conserved when you add solvent. If you dilute 10 mL of 6 N HCl with water to 60 mL, the final normality is (6 × 10) / 60 = 1 N. The same law applies at the equivalence point of a titration — equivalents of acid equal equivalents of base.

How do I use this for a titration calculation?

Switch to Titration mode and enter the normality and volume of your standard titrant (N₁, V₁) plus the volume of analyte (V₂). The calculator solves for the unknown normality N₂ = (N₁ × V₁) / V₂. For example, if 25 mL of 0.1 N NaOH neutralises 30 mL of HCl, the acid normality is (0.1 × 25) / 30 = 0.0833 N.

Normality Calculator

Normality is one of the most powerful concentration units in analytical chemistry because it accounts for the reactive capacity of a substance — not just how many molecules are present, but how many protons or electrons each molecule can donate or accept. This calculator handles every core normality calculation: computing N from a weighed sample, finding equivalent weights, interconverting normality and molarity, solving dilution problems using the N₁V₁ = N₂V₂ law, and determining unknown concentrations at a titration equivalence point.

Core formula and constants

The central equation is:

N = equivalents / V = (mass / EW) / V

where EW (equivalent weight) = M / n, M is the molar mass in g/mol, and n is the n-factor (the number of equivalents per mole). The n-factor is reaction-specific:

Acids: number of replaceable H⁺ ions (HCl = 1; H₂SO₄ = 2; H₃PO₄ = 3)
Bases: number of OH⁻ groups released (NaOH = 1; Ca(OH)₂ = 2)
Redox oxidising agents: change in oxidation state per formula unit (KMnO₄ in acidic solution: Mn⁷⁺→Mn²⁺, n = 5; K₂Cr₂O₇: 2 × Cr⁶⁺→Cr³⁺, n = 6)
Redox reducing agents: electrons lost per formula unit (FeSO₄: Fe²⁺→Fe³⁺, n = 1)

The dilution law N₁V₁ = N₂V₂ states that equivalents are conserved on dilution and holds at the equivalence point of any acid-base or redox titration.

How the calculator works

Select a calculation mode from the dropdown. For normality-from-mass problems, choose a preset substance (12 common acids, bases and redox agents with verified molar masses and n-factors) or enter custom values. Type in your known quantities and the result panel updates instantly, showing the formula, substituted values, and units at every step. The built-in reference table lists equivalent weights for all 12 presets so you can cross-check manually.

Worked example — preparing 0.5 L of 0.2 N H₂SO₄

H₂SO₄: M = 98.079 g/mol, n = 2 (both H⁺ ions are replaceable), EW = 98.079 / 2 = 49.04 g/eq.

To make 0.5 L of 0.2 N solution you need:

equivalents = N × V = 0.2 × 0.5 = 0.1 eq
mass = eq × EW = 0.1 × 49.04 = 4.904 g

Dissolve 4.904 g of H₂SO₄ in water and make up to 500 mL — you have exactly 0.2 N.

The same sample weighs the same as for 0.1 mol/L (0.1 M), confirming that for H₂SO₄: N = 2 × M, so 0.2 N = 0.1 M.

Formula note

All five modes use exact SI constants. There is no approximation. Results are displayed to six significant figures, switching to scientific notation for values outside the range 0.001 – 9 999 999 to keep the output readable whether you are working with micromolar reagents or concentrated laboratory stock solutions.