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🧪 Titration

Normality Calculator

Find normality from molarity and the n-factor (N = M × n), or from mass, equivalent weight and volume. Built for acids, bases and redox titrations.

N = M × n
From mass & volume
Equivalent weight
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Normality — Quick answer

Normality counts reactive equivalents per litre. It is molarity times the n-factor — the number of H⁺, OH⁻ or electrons per molecule.

N = M × n  ·  N = mass / (equiv. weight × VL)
equiv. weight = molar mass / n

Worked example: 0.5 M H₂SO₄ donates 2 H⁺ (n = 2). N = 0.5 × 2 = 1.0 N.

n-factor of common reagents

Reagentn-factor1 M =
HCl, NaOH11 N
H₂SO₄, Ca(OH)₂22 N
H₃PO₄33 N

Used for: titrations, acid/base standards, redox, analytical chemistry.

🧪 Normality Calculator

Choose a method: from molarity & n-factor, or from mass, molar mass, n-factor & volume.

Normality
Molarity
Equivalent weight
Equivalents

⚠️ The n-factor depends on the reaction, not just the formula — e.g. a reducing agent's electron transfer can vary. IUPAC discourages normality, but it remains common in titrations (N₁V₁ = N₂V₂).

Normality measures concentration in reactive equivalents per litre rather than moles per litre. The point is that some molecules pack more punch: one sulfuric acid molecule donates two protons, so a 1 M acid is "twice as reactive" — 2 N. The bridge is the n-factor, the number of reactive units (H⁺, OH⁻ or transferred electrons) per molecule, giving the simple rule N = M × n. It is the natural unit for titrations, where equivalents react one-for-one and the endpoint is just N₁V₁ = N₂V₂.

Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: equivalent-concentration definitions.

The normality equations

From molarity
N = M × n  (n = equivalents per molecule)
From mass
N = mass / (equivalent weight × Vlitres)
Equivalent weight
equivalent weight = molar mass / n

The n-factor is the heart of it: 1 for monoprotic acids and bases (HCl, NaOH), 2 for diprotic species (H₂SO₄, Ca(OH)₂), 3 for triprotic H₃PO₄, and the number of electrons transferred for redox reagents. Equivalent weight is just the molar mass shared out among those equivalents. Both routes give the same answer — molarity times n, or mass per equivalent weight per litre — so use whichever data you have.

Worked example — standardising sulfuric acid

Scenario: A 0.5 M sulfuric acid solution; H₂SO₄ donates 2 H⁺, so n = 2 (molar mass 98 g/mol).

Normality
N = M × n = 0.5 × 2 = 1.0 N
By mass (100 mL)
eq. wt = 98/2 = 49 g · 4.9 g in 0.1 L → N = 4.9/(49×0.1) = 1.0 N

The acid is 1.0 N — twice its molarity, because each molecule supplies two protons. Both methods agree: 0.5 M × 2, or 4.9 g of acid (one-tenth of an equivalent weight per 100 mL) gives 1.0 N. In a titration this 1.0 N acid would neutralise a 1.0 N base in equal volumes, since equivalents always react 1:1 — the convenience that keeps normality in use for analytical work.

Frequently Asked Questions

How do you calculate normality?

N = M × n (n-factor). 0.5 M H₂SO₄ (n=2) = 1.0 N. Or N = mass/(equiv. weight × V_L).

Normality vs molarity?

Molarity = mol/L; normality = equivalents/L. Equal when n=1 (HCl, NaOH); differ otherwise. N = M × n.

What is the n-factor?

Reactive units per molecule: H⁺ for acids, OH⁻ for bases, electrons for redox. HCl 1, H₂SO₄ 2, H₃PO₄ 3.

What is equivalent weight?

molar mass ÷ n. H₂SO₄: 98/2 = 49 g/equivalent. Mass ÷ eq. weight = equivalents.

When is normality used?

Acid–base and redox titrations: equivalents react 1:1, so N₁V₁ = N₂V₂. IUPAC discourages it elsewhere.

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