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.
Open the Equivalent Weight Calculator →Volumetric analysis is where school chemistry meets the laboratory, and it rewards careful practice. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.