Normality vs Molarity — When Each Is Used and How to Convert
A bottle in the laboratory says "0.1 N NaOH". The one beside it says "0.1 M NaOH". Are they the same solution? Here, yes — but change NaOH to H₂SO₄ and the two labels describe solutions that differ by a factor of two. Understanding why is the whole of this topic.
The two definitions side by side
Normality, N = gram-equivalents of solute ÷ litres of solution
Both are "amount per litre of solution"; they differ in what they count. Molarity counts particles. Normality counts reacting capacity — how many units of chemical work one formula unit can do. The bridge is a single small integer:
What the n-factor is, reaction by reaction
The n-factor is not a property you can read off a formula. It is decided by what the substance does in the reaction you are performing:
| Type of substance | n-factor is… | Examples |
|---|---|---|
| Acid | Number of H⁺ actually given up | HCl = 1, H₂SO₄ = 2, H₃PO₄ = 3 (when fully neutralised) |
| Base | Number of OH⁻ actually given up | NaOH = 1, Ca(OH)₂ = 2 |
| Salt | Total positive charge of the cations in one formula unit | NaCl = 1, Na₂CO₃ = 2 (as a base), Al₂(SO₄)₃ = 6 |
| Oxidising or reducing agent | Number of electrons gained or lost per formula unit | KMnO₄ = 5 in acidic medium, 3 in neutral or faintly alkaline medium |
Read that last row again — it is the point of the whole article. The same potassium permanganate solution has two different normalities depending on the medium, but only one molarity. Molarity is a property of the bottle; normality is a property of the bottle and the reaction. That is why IUPAC discourages normality in modern scientific writing, and equally why volumetric-analysis courses still teach it: in a titration it makes the arithmetic very short.
The one advantage of normality
At the equivalence point of any titration — acid–base, redox, precipitation — the equivalents of the two reactants are equal, so this relation always works. The molarity version needs the stoichiometry written in every time:
Both give the same answer. Normality hides the ratio inside the units; molarity keeps it visible.
Worked example 1 — the simplest conversion
Question: What is the normality of 0.500 M H₂SO₄ used for complete neutralisation?
Sulphuric acid releases 2 H⁺ per molecule when fully neutralised, so n-factor = 2.
N = M × n-factor = 0.500 × 2
N = 1.00 N
And for 0.500 M NaOH, n-factor = 1, so N = 0.500 N. Same molarity, different normality — the two bottles in the opening paragraph.
Worked example 2 — preparing a solution of stated normality
Question: What mass of anhydrous sodium carbonate, Na₂CO₃, is needed to make 250 mL of 0.100 N solution for use as a base?
Step 1 — molar mass. Na = 22.990, C = 12.011, O = 15.999
M(Na₂CO₃) = (2 × 22.990) + 12.011 + (3 × 15.999) = 45.980 + 12.011 + 47.997 =
105.988 g mol⁻¹
Step 2 — n-factor. Acting as a base, carbonate accepts 2 H⁺ per formula unit, so n-factor = 2.
Step 3 — equivalent weight. 105.988 ÷ 2 = 52.994 g eq⁻¹
Step 4 — equivalents required. 0.100 N × 0.250 L = 0.0250 equivalents
Step 5 — mass. 0.0250 × 52.994 = 1.3249
Mass required = 1.32 g
Cross-check by the molarity route: 0.100 N ÷ 2 = 0.0500 M.
moles = 0.0500 × 0.250 = 0.0125 mol.
mass = 0.0125 × 105.988 = 1.3249 g ✓ Same answer by a completely different path.
Worked example 3 — a titration, solved both ways
Question: 25.0 mL of NaOH solution is exactly neutralised by 20.0 mL of 0.100 N H₂SO₄. Find the concentration of the NaOH.
Route A — normality.
N₁ V₁ = N₂ V₂
N(NaOH) × 25.0 = 0.100 × 20.0
N(NaOH) = 2.00 / 25.0 = 0.0800 N
NaOH has n-factor 1, so its molarity equals its normality:
0.0800 M
Route B — molarity, stoichiometry written out.
The acid is 0.100 N with n-factor 2, so M(H₂SO₄) = 0.100 / 2 = 0.0500 M
moles of H₂SO₄ = 0.0500 × 0.0200 L = 1.00 × 10⁻³ mol
Balanced equation: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O, so 1 mol acid needs 2 mol base
moles of NaOH = 2 × 1.00 × 10⁻³ = 2.00 × 10⁻³ mol
M(NaOH) = 2.00 × 10⁻³ / 0.0250 L = 0.0800 M ✓
Identical answers. Route A is faster; Route B still works when you have forgotten what an n-factor is. Note that volumes may stay in millilitres in Route A because they appear on both sides — but in Route B they must be in litres, since there the volume is not part of a ratio.
Worked example 4 — the redox trap
Question: A bottle contains 0.0200 M KMnO₄. State its normality (a) in acidic medium and (b) in neutral medium, and give the equivalent weight in each case.
Molar mass. K = 39.098, Mn = 54.938, O = 15.999
M(KMnO₄) = 39.098 + 54.938 + (4 × 15.999) = 39.098 + 54.938 + 63.996 = 158.032 g mol⁻¹
(a) Acidic medium. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O — five electrons, so
n-factor = 5.
N = 0.0200 × 5 = 0.100 N
Equivalent weight = 158.032 ÷ 5 = 31.61 g eq⁻¹
(b) Neutral or faintly alkaline medium. MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ —
three electrons, so n-factor = 3.
N = 0.0200 × 3 = 0.0600 N
Equivalent weight = 158.032 ÷ 3 = 52.68 g eq⁻¹
Same bottle, same 0.0200 M, two different normalities — because normality answers "how much work can it do?" and that depends on the job. This is the question examiners use to separate students who memorised a number from students who understood the definition.
Quick reference
| Substance | Reaction context | n-factor | N when M = 0.100 |
|---|---|---|---|
| HCl | Neutralisation | 1 | 0.100 N |
| H₂SO₄ | Complete neutralisation | 2 | 0.200 N |
| H₃PO₄ | Complete neutralisation (all 3 H⁺) | 3 | 0.300 N |
| NaOH | Neutralisation | 1 | 0.100 N |
| Ca(OH)₂ | Neutralisation | 2 | 0.200 N |
| Na₂CO₃ | As a base, to CO₂ and water | 2 | 0.200 N |
| KMnO₄ | Acidic medium | 5 | 0.500 N |
| K₂Cr₂O₇ | Acidic medium | 6 | 0.600 N |
Because every n-factor is 1 or more, normality is never smaller than molarity for the same solution. If your working gives N < M, you divided instead of multiplying.
Common mistakes that cost marks
- Reading the n-factor off the formula alone. H₃PO₄ has n-factor 3 only when all three protons react. In a titration stopped at the first endpoint it is 1.
- Using KMnO₄ n-factor 5 everywhere. It is 5 in acidic medium and 3 in neutral or faintly alkaline medium.
- Putting molarity values into N₁V₁ = N₂V₂. That relation is true for equivalents. With molarities you must include the mole ratio from the balanced equation.
- Dividing to convert M to N. N = M × n-factor. You divide the other way: M = N ÷ n-factor.
- Using molar mass instead of equivalent weight. To weigh out a solution of known normality you need molar mass ÷ n-factor.
- Volume in millilitres where it is not a ratio. In N₁V₁ = N₂V₂ both sides may be in mL. In n = M × V, V must be in litres.
- Assuming N and M are interchangeable because they matched once. They match only when the n-factor is 1.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11–12 | Concentration terms; titration calculations in practicals |
| BSc volumetric analysis | Standardising solutions, equivalent weights, redox titrations |
| IIT-JAM / CUET-PG | n-factor reasoning, especially for permanganate and dichromate |
| GATE / CSIR-NET | Multi-step analytical problems where the medium changes the n-factor |
Convert between M, N and mass in one place. The concentration calculator handles molarity, normality and the mass to weigh out, so an n-factor slip does not silently double or halve every later step of a titration.
Open the Concentration (Molarity / Normality) Calculator →Volumetric analysis is one of the most scoring parts of a chemistry practical, because the method is fixed and repeatable. ABC Chemistry covers it in the Class 11–12 batches at the Gurugram centre and online across India: abcchemistry.in.