Normality Formula and Normality Calculator
Normality Formula and Normality Calculator
Calculate normality, the mass to weigh or the solution volume, with a table of equivalents for common acids, bases and oxidants.
What is normality?
Normality (N) is the number of equivalents of solute per litre of solution. An equivalent expresses reacting capacity: how many H+ an acid releases, how many OH- a base releases, or how many electrons are transferred in a redox reaction. Normality is used in titration because one equivalent of acid always neutralises exactly one equivalent of base.
Normality formula
Equivalents of common reagents
| Substance | Reaction | e (eq/mol) | 1 M equals |
|---|---|---|---|
| HCl, HNO3 | acid-base | 1 | 1 N |
| H2SO4 | acid-base | 2 | 2 N |
| H3PO4 | acid-base (full neutralisation) | 3 | 3 N |
| NaOH, KOH | acid-base | 1 | 1 N |
| Ca(OH)2, Ba(OH)2 | acid-base | 2 | 2 N |
| KMnO4 | redox, acidic | 5 | 5 N |
| K2Cr2O7 | redox, acidic | 6 | 6 N |
| Na2S2O3 | redox (iodometry) | 1 | 1 N |
The value of e depends on the reaction. H3PO4 titrated only to its first equivalence point has e = 1, and KMnO4 in neutral solution has e = 3.
Worked examples
Example 1: normality of sulfuric acid
4.904 g of H2SO4 (98.08 g/mol) is made up to 500 mL. What is the normality?
- n = 4.904 ÷ 98.08 = 0.05 mol, so M = 0.05 ÷ 0.5 L = 0.1 M
- H2SO4 releases 2 H+, so e = 2
- N = 0.1 × 2 = 0.2 N
Answer: 0.2 N.
Example 2: preparing 0.1 N NaOH
How many grams of NaOH (40.00 g/mol) for 250 mL of 0.1 N solution?
- e = 1, so EW = 40.00 g/eq
- mass = N × V × EW = 0.1 × 0.25 × 40.00 = 1.000 g
Answer: 1.000 g of NaOH.
Example 3: 0.1 N KMnO4 for permanganometry
How many grams of KMnO4 (158.03 g/mol) for 1 L of 0.1 N in acidic solution?
- MnO4- to Mn2+ gains 5 electrons, so e = 5
- EW = 158.03 ÷ 5 = 31.61 g/eq
- mass = 0.1 × 1 × 31.61 = 3.161 g
Answer: 3.161 g of KMnO4 (0.02 M).
Frequently asked questions
What is the difference between normality and molarity?
Molarity counts moles per litre, normality counts equivalents per litre: N = M × e. For HCl (e = 1) they are equal; for H2SO4 (e = 2) 0.1 M = 0.2 N.
What is equivalent weight?
The mass of substance equal to one equivalent: molar mass divided by e. For H2SO4, 98.08 ÷ 2 = 49.04 g/eq.
Does the dilution formula work with normality?
Yes, N1 × V1 = N2 × V2. See the dilution formula.
Why is normality used less today?
IUPAC prefers molarity because e depends on the reaction, but normality is still common in standard methods and on titrant labels.
Prepare solutions with analytical-grade reagents
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