Equivalent Weight — The Concept Students Get Wrong
Equivalent weight is not a difficult formula. It is one division. What trips students up is the number you divide by — the n-factor — because it is not a property of the compound alone. It depends on what the compound is doing in that particular reaction. KMnO₄ has three different equivalent weights depending on the medium. Once that idea lands, the whole topic becomes routine.
The formula
And the two relations built on it:
N = Molarity × n-factor · At the end point: N₁V₁ = N₂V₂
How to find the n-factor — four cases
| Species | n-factor is… | Example |
|---|---|---|
| Acid | Number of replaceable H⁺ ions (basicity) | H₂SO₄ → 2, HCl → 1, H₃PO₄ → 3 |
| Base | Number of replaceable OH⁻ ions (acidity) | NaOH → 1, Ca(OH)₂ → 2 |
| Salt | Total positive charge of the cations in the formula | Na₂CO₃ → 2, AlCl₃ → 3 |
| Redox agent | Electrons gained or lost per formula unit | KMnO₄ in acid → 5, K₂Cr₂O₇ in acid → 6 |
| Element | Its valency | Al → 3, so E = 26.982 ÷ 3 = 8.99 |
For the redox case you cannot guess. You must write the half reaction and count electrons.
Why KMnO₄ has three equivalent weights
M(KMnO₄) = 39.098 + 54.938 + 4 × 15.999 = 158.032 g mol⁻¹. Now look at what manganese actually does:
| Medium | Half reaction | n | E (g/eq) |
|---|---|---|---|
| Acidic | MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | 5 | 158.032 ÷ 5 = 31.61 |
| Neutral / faintly alkaline | MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ | 3 | 158.032 ÷ 3 = 52.68 |
| Strongly alkaline | MnO₄⁻ + e⁻ → MnO₄²⁻ | 1 | 158.032 ÷ 1 = 158.03 |
Same compound, same molar mass, three answers. This is the single most tested idea in the whole topic, and the reason "the equivalent weight of KMnO₄" is an incomplete question unless the medium is stated.
Worked example 1 — a dibasic acid
Find the equivalent weight of H₂SO₄.
M = 2 × 1.008 + 32.06 + 4 × 15.999 = 2.016 + 32.06 + 63.996 = 98.072 g mol⁻¹
Both hydrogens are replaceable, so n-factor = 2.
E = 98.072 ÷ 2 = 49.04 g/eq
Consequence: a 1 M solution of H₂SO₄ is 2 N. Molarity and normality are not interchangeable words.
Worked example 2 — preparing a standard solution
6.3 g of oxalic acid dihydrate, H₂C₂O₄·2H₂O, is dissolved to make 250 mL of solution. Find its normality.
M = (2 × 1.008) + (2 × 12.011) + (4 × 15.999) + 2 × 18.015
= 2.016 + 24.022 + 63.996 + 36.030 = 126.064 g mol⁻¹
Oxalic acid is dibasic, so n-factor = 2 and E = 126.064 ÷ 2 = 63.03 g/eq
Equivalents = 6.3 ÷ 63.03 = 0.09995 ≈ 0.100 eq
N = 0.100 ÷ 0.250 L = 0.40 N
The waters of crystallisation are part of the molar mass. Leaving them out gives E = 45.02 and a normality of 0.56 N — a 40 % error, and the classic reason a titration "does not match the book value".
Worked example 3 — an acid–base titration
25.0 mL of 0.10 N Na₂CO₃ needs 20.0 mL of HCl to reach the methyl-orange end point. Find the normality of the HCl.
N₁V₁ = N₂V₂ → 0.10 × 25.0 = N₂ × 20.0
N₂ = 2.50 ÷ 20.0 = 0.125 N
Because HCl is monobasic, 0.125 N is also 0.125 M. For Na₂CO₃ (n-factor 2) the 0.10 N solution is only 0.05 M — the normality equation handled that difference automatically, which is exactly why volumetric analysis uses normality.
Worked example 4 — a redox titration
20.0 mL of 0.020 M KMnO₄ is used to titrate Fe²⁺ in acidic medium. How many moles of Fe²⁺ were present, and what mass of FeSO₄·7H₂O does that correspond to?
In acid, KMnO₄ has n-factor 5, so N = M × n = 0.020 × 5 = 0.100 N
Equivalents of KMnO₄ = 0.100 × 0.0200 L = 2.00 × 10⁻³ eq
At the end point, equivalents of Fe²⁺ = equivalents of KMnO₄ = 2.00 × 10⁻³ eq
Fe²⁺ → Fe³⁺ + e⁻, so its n-factor is 1 and moles = equivalents = 2.00 × 10⁻³ mol
M(FeSO₄·7H₂O) = 55.845 + 32.06 + 63.996 + 7 × 18.015 = 278.01 g mol⁻¹
Mass = 2.00 × 10⁻³ × 278.01 = 0.556 g
The whole point of equivalents: you never had to balance the full 5Fe²⁺ + MnO₄⁻ equation. The n-factors did that bookkeeping for you.
Common mistakes
- Treating the n-factor as fixed. It belongs to the reaction, not the bottle. KMnO₄ is 31.61, 52.68 or 158.03 g/eq depending on the medium.
- Confusing normality with molarity. N = M × n-factor. They are equal only when n = 1.
- Ignoring water of crystallisation. Hydrated salts — oxalic acid dihydrate, FeSO₄·7H₂O, CuSO₄·5H₂O — carry that water in their molar mass.
- Using the charge on one ion for a salt. For Na₂CO₃ the n-factor is the total cationic charge, 2 × 1 = 2, not 1.
- Guessing the redox n-factor. Write the half reaction and count electrons; H₂O₂ is n = 2 whether it acts as oxidant or reductant, and that only becomes obvious from the half equations.
- Mixing millilitres and litres. Normality is per litre. In N₁V₁ = N₂V₂ the volumes may both be in mL because they cancel, but in equivalents = N × V the volume must be in litres.
Where it appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11–12 | Normality of a prepared solution; simple acid–base titration |
| JEE / NEET | n-factor of redox agents, back titration, equivalent concept in mole problems |
| IIT-JAM / CUET-PG | Practical volumetric analysis, double-indicator Na₂CO₃/NaHCO₃ titrations |
| GATE / CSIR-NET | Iodometry, permanganometry, hardness of water in equivalents |
Get the n-factor right, every time. The Equivalent Weight calculator takes the formula and the reaction type, works out the molar mass and the n-factor, and shows the division — so you can see which of the two you had wrong.
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