Equivalent weight

A titration flask holds two solutions of an acid and a base, and the moment they exactly neutralise each other the colour of the indicator flips. Behind that flip lies a single powerful idea: chemists can measure "how much" of a substance reacts not just in grams or moles, but in a currency that automatically accounts for how many hydrogen ions, hydroxide ions, or electrons each formula unit carries. That currency is the equivalent weight, and the concentration built on it is normality. ISC keeps equivalent weight and normality as examinable arithmetic within Class 11's "Some Basic Concepts of Chemistry," precisely because it turns messy titration calculations into one clean relation, $N_1V_1 = N_2V_2$ , that works no matter how lopsided the balanced equation looks. This article shows how the n-factor is read off a formula, how equivalent weight and normality are computed from it, and how the volumetric relation is used to solve a real titration problem.

The core insight

Dividing molar mass by the n-factor gives the gram equivalent weight, so equivalents (unlike moles) always react one-to-one, which is why $N_1V_1 = N_2V_2$ works for any acid-base or redox titration regardless of stoichiometric coefficients.

The core idea

The gram equivalent weight of a substance is its gram molecular (or atomic) mass divided by a small whole number called the n-factor:

E = \frac{\text{gram molecular mass (or atomic mass)}}{n\text{-factor}}

where: $E$ is the gram equivalent weight and the n-factor is the number of "reacting units" carried by one formula unit — replaceable $\text{H}^+$ , replaceable $\text{OH}^-$ , transferable positive charge, or electrons exchanged. The n-factor depends on what kind of substance is being weighed:

Element — the n-factor is its valency. Aluminium (valency 3) has n-factor 3.
Acid — the n-factor is its basicity, the number of replaceable $\text{H}^+$ per molecule: $\text{HCl}$ is 1, $\text{H}_2\text{SO}_4$ is 2, $\text{H}_3\text{PO}_4$ is 3.
Base — the n-factor is its acidity, the number of replaceable $\text{OH}^-$ per formula unit: $\text{NaOH}$ is 1, $\text{Ca(OH)}_2$ is 2.
Salt — the n-factor is the total positive (or negative) charge replaced per formula unit, i.e. (number of metal atoms) $\times$ (valency of metal). In $\text{Al}_2(\text{SO}_4)_3$ there are 2 aluminium atoms of valency 3, so the n-factor is $2\times3=6$ .
Oxidising or reducing agent — the n-factor equals the number of electrons gained or lost per formula unit, taken here as a stated number rather than derived from oxidation-number bookkeeping (that derivation belongs to the later Redox Reactions chapter).

Once $E$ is known, normality measures concentration in equivalents per litre:

N = \frac{\text{gram equivalents of solute}}{V_{\text{solution}}\ (\text{L})} = \frac{w/E}{V_{\text{solution}}\ (\text{L})}

where: $w$ is the mass of solute in grams, $E$ is its gram equivalent weight, and $V_{\text{solution}}$ is the solution volume in litres. Normality connects to the more familiar molarity through the n-factor itself:

N = M \times n\text{-factor}

where: $M$ is molarity (mol/L). When the n-factor is 1 — a monobasic acid or monoacidic base — normality and molarity are numerically identical; whenever the n-factor exceeds 1, normality is the larger number. The figure below shows why equivalents, not moles, are the natural currency for a titration: one mole of $\text{H}_2\text{SO}_4$ supplies two moles of $\text{H}^+$ , so it takes twice as many moles of $\text{NaOH}$ to neutralise it — but the number of equivalents on each side is always exactly equal at the endpoint.

Figure

A balance scale showing one mole of sulphuric acid on one pan balanced against two moles of sodium hydroxide on the other pan, with a caption noting that the equivalents on each side are equal even though the moles are not

Derivation ASolving a titration with N1V1 = N2V2

The goal is to show, starting only from the definition of normality as equivalents per litre, why setting the equivalents of acid equal to the equivalents of base at the point of exact neutralisation gives the compact relation $N_1V_1 = N_2V_2$ , and why this relation is safe to use even when the acid and base have different basicity or acidity, unlike the molarity version $M_1V_1 = M_2V_2$ which only works when both reagents happen to share the same n-factor.

Step 1. Write the number of equivalents delivered by each solution. Normality is equivalents per litre, so a volume $V_1$ (in litres) of a solution of normality $N_1$ contains $N_1 V_1$ gram equivalents of solute, and likewise the second solution contains $N_2 V_2$ gram equivalents.

Step 2. Apply the definition of the equivalence point. At the moment two solutions react exactly — the endpoint of a titration — the acid has supplied exactly as many equivalents of $\text{H}^+$ as the base has supplied of $\text{OH}^-$ (or, for a redox titration, the oxidiser has taken exactly as many electrons as the reducer has given). That is precisely the statement that the equivalents on the two sides are equal.

Step 3. Equate the two expressions from Step 1 using the equality from Step 2. Since equivalents of solution 1 equal equivalents of solution 2 at the endpoint, $N_1 V_1 = N_2 V_2$ .

Step 4. See why this beats molarity. Because the n-factor is already folded into the definition of $E$ (and hence of $N$ ), one equivalent of acid always neutralises exactly one equivalent of base, regardless of how many $\text{H}^+$ or $\text{OH}^-$ each molecule carries. $M_1V_1 = M_2V_2$ would only hold if both substances happened to have the same n-factor; $N_1V_1 = N_2V_2$ needs no such coincidence, which is why normality is the standard tool for volumetric analysis.

Worked example

$25.0\ \text{mL}$ of a sodium hydroxide solution of unknown normality is exactly neutralised by $30.0\ \text{mL}$ of $0.20\ \text{N}$ hydrochloric acid. Find (a) the normality of the $\text{NaOH}$ solution, and (b) its molarity.

Step 1. Apply the volumetric relation. Let $N_1, V_1$ be for $\text{NaOH}$ and $N_2, V_2$ for $\text{HCl}$ , so $N_1 \times 25.0 = 0.20 \times 30.0$ .

Step 2. Solve for $N_1$ . $N_1 = \dfrac{0.20 \times 30.0}{25.0} = \dfrac{6.0}{25.0} = 0.24\ N$ .

Step 3. Convert to molarity. $\text{NaOH}$ has acidity (n-factor) $=1$ , so $M = N/n\text{-factor} = 0.24/1 = 0.24\ M$ .

Step 4. Sanity-check. For a monoprotic acid or monoacidic base, n-factor $=1$ makes normality and molarity numerically identical — exactly what Step 3 shows.

Answer: $N_{\text{NaOH}} = 0.24\ N$ and $M_{\text{NaOH}} = 0.24\ \text{mol L}^{-1}$ .

Competitive stretch

The same $30.0\ \text{mL}$ of $0.20\ \text{N}$ $\text{HCl}$ is instead titrated against a $\text{Ca(OH)}_2$ solution of unknown normality, requiring $20.0\ \text{mL}$ to reach the endpoint. Its normality is $N_1 = (0.20 \times 30.0)/20.0 = 0.30\ N$ . Since $\text{Ca(OH)}_2$ has acidity (n-factor) $2$ , its molarity is $M = N/2 = 0.30/2 = 0.15\ \text{mol L}^{-1}$ — half the normality, because each formula unit supplies two $\text{OH}^-$ ions. This is the case where $M_1V_1=M_2V_2$ would have silently failed: the acid and base have different n-factors, so only the normality relation gives the right answer directly.

Common traps

Confusing molar mass with equivalent weight — equivalent weight is molar mass divided by the n-factor, and the two coincide only when the n-factor is 1.
Forgetting to multiply by the number of metal atoms when finding a salt's n-factor, e.g. treating $\text{Al}_2(\text{SO}_4)_3$ as n-factor 3 instead of $2\times3=6$ .
Using $M_1V_1 = M_2V_2$ for a titration between reagents of different basicity/acidity, such as $\text{H}_2\text{SO}_4$ against $\text{NaOH}$ — only $N_1V_1=N_2V_2$ is guaranteed to work.
Swapping basicity and acidity — basicity counts replaceable $\text{H}^+$ in an acid, acidity counts replaceable $\text{OH}^-$ in a base; the names look reversed but are standard usage.

Ask the tutor: want a few more n-factor puzzles for tricky salts, or a walkthrough of setting up a titration with a diprotic acid and a triacidic base?

Why does normality change with the reaction but molarity does not?

Molarity just counts moles of solute per litre, which is fixed once the solution is made. Normality counts equivalents, and the n-factor used to convert moles to equivalents depends on the role the substance plays in a particular reaction — for example, a polyprotic acid can act as monobasic in a partial reaction or fully polybasic in another. In the ISC syllabus the n-factor is usually fixed by the substance's formula (basicity, acidity, or salt charge), so in most problems normality behaves like a fixed multiple of molarity, but you should still identify the n-factor context each question gives you.

Why is N1V1 = N2V2 true even when the balanced equation has different coefficients on each side?

Because normality already bakes the n-factor into its definition, one equivalent of any acid always reacts with exactly one equivalent of any base, one-to-one, no matter how many molecules of each are written in the balanced equation. The messy stoichiometric coefficients you would otherwise need are silently absorbed into the n-factor used to compute each substance's equivalent weight, which is exactly why titration calculations are done in normality rather than molarity.

Is equivalent weight the same as atomic or molecular weight?

Only when the n-factor is exactly 1. Equivalent weight is always molar (or atomic) mass divided by the n-factor, so for aluminium (valency 3, atomic mass 27) the equivalent weight is $27/3=9$ , not 27. The two quantities coincide only for substances like $\text{HCl}$ or $\text{NaOH}$ whose n-factor happens to be 1 — do not assume this holds in general.

Equivalent weight

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads