Normality Calculator
ChemistryCalculate the normality of a solution from its molarity and n-factor (equivalents per mole). Get results in equivalents/litre with step-by-step working for acids, bases, and salts.
Normality (N)
What is a Normality?
The Normality Calculator computes the normality of a solution from its molarity and n-factor (equivalents per mole of solute). Normality (symbol N, unit eq/L or N) is a concentration measure that quantifies the reactive capacity of a solution rather than its raw molar concentration. While molarity counts total moles of solute per litre, normality counts how many equivalents — reactive units — are present per litre.
An equivalent is defined as the amount of solute that reacts with or provides one mole of reactive species in a given type of reaction. For acids, one equivalent provides one mole of H⁺ ions; for bases, one equivalent provides one mole of OH⁻ ions; for oxidising and reducing agents in redox reactions, one equivalent involves one mole of electron transfer. The n-factor converts between moles and equivalents: equivalents = moles × n-factor, and normality = molarity × n-factor.
For monoprotic acids and monobasic bases (n = 1), normality equals molarity — 1 M HCl = 1 N HCl. For diprotic acids like H₂SO₄ (n = 2), normality is double the molarity — 1 M H₂SO₄ = 2 N. For KMnO₄ in acidic medium (n = 5), 1 M KMnO₄ = 5 N. This is why normality is so useful in titrations: at the equivalence point, equivalents of titrant always equal equivalents of analyte, giving the clean relationship N₁V₁ = N₂V₂, regardless of the acids' proton counts.
Normality is heavily used in Indian chemistry education. Class 12 practicals, CBSE volumetric analysis experiments, and undergraduate analytical chemistry labs all express solution concentrations in normality for acid-base and redox titrations. Use the Molarity Calculator when preparing solutions by mass; use this Normality Calculator to convert that molarity to the normality needed for titration calculations.
How to use this Normality calculator
- Prepare or measure the molarity — determine the molarity of your solution (mol/L). Use the Molarity Calculator if starting from mass and volume.
- Enter Molarity — type the molarity of the solution into the Molarity field in mol/L. For example, for a 0.5 M H₂SO₄ solution, enter 0.5.
- Determine and enter the n-Factor — type the number of equivalents per mole of your solute into the n-Factor (Equivalents per Mole) field. For H₂SO₄ (diprotic acid, donates 2 H⁺) enter 2; for NaOH (monobasic) enter 1; for KMnO₄ in acidic medium (gains 5 electrons) enter 5.
- Read Normality (N) — the highlighted result shows the normality in equivalents per litre. This is the value to use in N₁V₁ = N₂V₂ for titration calculations.
- Confirm Equivalents per Litre — verify the eq/L output matches what your analytical method specifies, as some protocols express normality in this notation.
- Apply in titration — use the normality to calculate equivalents used: equivalents = normality × volume (L). At the equivalence point, N₁V₁ = N₂V₂ gives the concentration of the unknown.
Formula & Methodology
Normality formula: > N = M × n Where: - N = normality (equivalents per litre, eq/L) - M = molarity (mol/L) - n = n-factor (dimensionless; number of equivalents per mole of solute) Titration equivalence relation: > N₁V₁ = N₂V₂ Where subscripts 1 and 2 refer to the titrant and analyte respectively. n-Factor reference table: | Reagent | Reaction type | n-Factor | |---|---|---| | HCl | Acid-base | 1 | | H₂SO₄ | Acid-base | 2 | | H₃PO₄ | Acid-base (complete) | 3 | | NaOH | Acid-base | 1 | | Ca(OH)₂ | Acid-base | 2 | | Na₂CO₃ | Acid-base | 2 | | KMnO₄ (acidic) | Redox | 5 | | K₂Cr₂O₇ (acidic) | Redox | 6 | | FeSO₄ | Redox | 1 | Worked example — H₂SO₄ standardisation: A 0.49 M H₂SO₄ solution is prepared. What is its normality? - n-factor of H₂SO₄ = 2 (donates 2 H⁺ per molecule) - N = 0.49 × 2 = 0.98 N If this solution is used to titrate 25 mL of NaOH (n = 1) and the titration uses 20 mL of H₂SO₄: - N₁V₁ = N₂V₂ → 0.98 × 20 = N(NaOH) × 25 - N(NaOH) = 19.6 ÷ 25 = 0.784 N = 0.784 M (since n = 1 for NaOH)
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