Equivalent Concepts & Normality

In advanced stoichiometry, chemists use a system of 'equivalents' that goes beyond simple moles to describe the reacting capacity of a substance. The concept of an equivalent accounts for how many reactive units — protons, electrons, or charge — a single formula unit of a substance can donate or accept in a specific reaction. This makes it possible to compare very different substances (acids, bases, oxidants, reductants, and salts) on the same scale of reactivity. The n-factor (also called the valence factor) is the central idea: it is the number of equivalents per mole of a substance. For acids, n equals the number of H⁺ ions furnished per molecule; for bases, it is the number of OH⁻ ions (or H⁺ ions accepted); for redox reagents, it is the change in oxidation state per formula unit. Equivalent weight is then simply the molar mass divided by the n-factor, giving the mass of substance that provides one equivalent of reactive species. Normality (N) is the concentration unit built on these ideas: it expresses the number of equivalents of solute per litre of solution. Because one equivalent of any acid exactly neutralises one equivalent of any base — and one equivalent of any oxidant exactly reacts with one equivalent of any reductant — normality allows the elegant relationship N₁V₁ = N₂V₂ to be used universally in titrations. This topic is particularly important for competitive examinations such as the JEE, where n-factor calculations appear frequently in both theoretical and numerical problems.

Key takeaways
  • n-factor for Acids and Bases
  • n-factor for Redox Reactions
  • n-factor for Salts (Disproportionation and Double Decomposition)
  • Equivalent Weight
  • Normality Calculations

n-factor for Acids and Bases

For an acid, the n-factor equals the number of replaceable (ionisable) H⁺ ions per molecule. HCl has n = 1; H₂SO₄ has n = 2; H₃PO₄ has n = 3 (but H₃PO₃ has n = 2 because one H is directly bonded to P and is non-ionisable). For a base, n equals the number of OH⁻ ions furnished per formula unit: NaOH → n = 1; Ca(OH)₂ → n = 2; Al(OH)₃ → n = 3. Context matters: if H₂SO₄ reacts with NaOH to form NaHSO₄, only one proton is transferred per molecule, so n = 1 in that reaction.

n-factor for Redox Reactions

For oxidising and reducing agents, n-factor = total change in oxidation state per formula unit. For KMnO₄ in acidic medium, Mn goes from +7 to +2 (change = 5), so n = 5. In neutral/alkaline medium, Mn goes from +7 to +4 (change = 3), so n = 3. For K₂Cr₂O₇, Cr goes from +6 to +3 (change = 3) per Cr atom, and there are 2 Cr atoms, giving n = 6. For oxalic acid (H₂C₂O₄) as a reducing agent, each C goes from +3 to +4 (change = 1), with 2 C atoms, giving n = 2.

n-factor for Salts (Disproportionation and Double Decomposition)

For salts that do not undergo redox, the n-factor equals the total positive valency of the cation(s) in one formula unit. For Na₂CO₃, Na has valency 1 and there are 2 Na atoms, giving n = 2. For FeSO₄, Fe²⁺ has valency 2, giving n = 2. For a double-decomposition reaction between two salts, the equivalents of both salts must match: equivalents of salt A = equivalents of salt B.

Equivalent Weight

Equivalent weight (E) is defined as the molar mass (M) divided by the n-factor: E = M / n. This is the mass in grams that provides one equivalent of the reactive species. For H₂SO₄ (M = 98 g/mol, n = 2): E = 49 g/equiv. For KMnO₄ in acid (M = 158 g/mol, n = 5): E = 31.6 g/equiv. For Ca(OH)₂ (M = 74 g/mol, n = 2): E = 37 g/equiv. The law of equivalents states that in any complete chemical reaction, the number of equivalents of all reactants and products are equal.

Normality Calculations

Normality N = (number of equivalents of solute) / (volume of solution in litres) = (mass of solute / equivalent weight) / (volume in litres). It also relates to molarity: N = n × M (where M here is molarity). For example, a 2 M H₂SO₄ solution is 4 N (since n = 2). The key titration relationship is N₁V₁ = N₂V₂, where subscript 1 refers to the acid (or oxidant) and subscript 2 refers to the base (or reductant). This equality holds regardless of the stoichiometric ratio, because equivalents always react 1:1 by definition.

The Key Equation

N=Equivalents of soluteVolume of solution (L)=wE×V=nfactor×MolarityN = \frac{\text{Equivalents of solute}}{\text{Volume of solution (L)}} = \frac{w}{E \times V} = n_{\text{factor}} \times M_{\text{olarity}}
  • N — Normality (equiv/L (eq L⁻¹)): Number of equivalents of solute per litre of solution.
  • w — Mass of solute (grams (g)): The mass of the solute dissolved to make the solution.
  • E — Equivalent weight (g/equiv): Molar mass divided by n-factor; mass providing one equivalent of reactive species.
  • V — Volume of solution (litres (L)): Total volume of the prepared solution.
  • n_{\text{factor}} — n-factor (valence factor) (dimensionless): Number of equivalents per mole; depends on the specific reaction type.

Worked Example

Problem. A 25.0 mL sample of H₂SO₄ solution is titrated against 0.1 N NaOH solution. The end point is reached after adding 40.0 mL of NaOH. (a) Find the normality of the H₂SO₄ solution. (b) Find the molarity of the H₂SO₄ solution. (c) What mass of H₂SO₄ is present in 1 litre of this solution?

NH2SO4=0.16 N,M=0.08 mol L−1,w=7.84 g L−1N_{\text{H}_2\text{SO}_4} = 0.16\,\text{N},\quad M = 0.08\,\text{mol L}^{-1},\quad w = 7.84\,\text{g L}^{-1}

Answer. Normality of H₂SO₄ = 0.16 N; Molarity = 0.08 M; Mass per litre = 7.84 g

Figure
Diagram
n-factor Summary: Common Reagents in Acid-Base & RedoxAcid-Base n-factorsHCl→ n = 11 H⁺/moleculeH₂SO₄→ n = 22 H⁺/moleculeH₃PO₃→ n = 22 ionisable H⁺NaOH→ n = 11 OH⁻/unitCa(OH)₂→ n = 22 OH⁻/unitRedox n-factorsKMnO₄ (acid)n = 5+7→+2KMnO₄ (neut.)n = 3+7→+4K₂Cr₂O₇n = 62×(+6→+3)FeSO₄n = 1+2→+3H₂C₂O₄n = 22×(+3→+4)Law of Equivalents: At equivalence point in any titrationN₁V₁ = N₂V₂ ↔ Equivalents of Reactant 1 = Equivalents of Reactant 2Equivalent Weight FormulaE = M / n-factore.g. KMnO₄ (acid): E = 158/5 = 31.6 g/equive.g. H₂SO₄: E = 98/2 = 49 g/equiv
Origins · 1 minWilliam Hyde Wollaston and how this was discovered
WH
William Hyde Wollaston
1766–1828 · British

Wollaston was one of the first chemists to develop and promote the concept of equivalents in chemistry. In 1814, he constructed an early 'equivalents table' listing the relative masses of substances that react with each other, laying the groundwork for what would later become equivalent weights and normality. His systematic approach helped bridge the gap between Dalton's atomic theory and practical quantitative analysis. Wollaston discovered two elements — palladium (1803) and rhodium (1804) — within the same year, making him one of very few chemists to discover two elements simultaneously. He also invented the reflecting goniometer and made important contributions to optics.

  • Mole Concept and Avogadro's Number — Normality and equivalent weight are extensions of the mole concept; understanding moles is essential before working with equivalents.
  • Molarity and Solution Concentration — Normality = n-factor × Molarity; mastery of molarity calculations is a prerequisite for normality problems.
  • Oxidation States and Redox Reactions — Calculating the n-factor for redox reagents requires accurately assigning oxidation states to determine the electron change per formula unit.
  • Acid-Base Titrations — The N₁V₁ = N₂V₂ law is the fundamental tool used in titration calculations; normality makes all acid-base and redox titrations analytically equivalent.