Equivalent weight and n-factor — calculate for acid, base, salt, and redox agents

Question

Calculate the equivalent weight and n-factor for the following: (a) $\text{H}_2\text{SO}_4$ in reaction with $\text{NaOH}$ (b) $\text{Ca(OH)}_2$ in neutralisation (c) $\text{KMnO}_4$ in acidic medium (d) $\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium

(JEE Main 2022, similar pattern)

Solution — Step by Step

\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}}

The n-factor depends on the context — it changes based on whether the substance acts as an acid, base, oxidising agent, or reducing agent.

$\text{H}_2\text{SO}_4$ donates 2 $\text{H}^+$ ions to $\text{NaOH}$ :

n = 2, \quad \text{Eq. wt.} = \frac{98}{2} = 49

Note: if $\text{H}_2\text{SO}_4$ reacts with $\text{NaOH}$ to form $\text{NaHSO}_4$ (only 1 $\text{H}^+$ replaced), then $n = 1$ and eq. wt. = 98. Always check the reaction.

$\text{Ca(OH)}_2$ provides 2 $\text{OH}^-$ ions:

n = 2, \quad \text{Eq. wt.} = \frac{74}{2} = 37

$\text{KMnO}_4$ in acidic medium:

$\text{Mn}$ goes from $+7$ to $+2$ (forms $\text{Mn}^{2+}$ ), a change of 5 per $\text{Mn}$ atom.

n = 5, \quad \text{Eq. wt.} = \frac{158}{5} = 31.6

$\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium:

Each $\text{Cr}$ goes from $+6$ to $+3$ , a change of 3. There are 2 Cr atoms per formula unit.

n = 2 \times 3 = 6, \quad \text{Eq. wt.} = \frac{294}{6} = 49

Why This Works

The equivalent weight concept ensures that one equivalent of any substance reacts exactly with one equivalent of another. The n-factor normalises the molar mass by how many “reactive units” each molecule provides.

For acid-base reactions, the reactive unit is $\text{H}^+$ or $\text{OH}^-$ . For redox reactions, the reactive unit is the electron transferred. When $\text{KMnO}_4$ gains 5 electrons per $\text{Mn}$ , each mole of $\text{KMnO}_4$ is worth 5 equivalents. Dividing the molar mass by 5 gives the mass per equivalent.

This is why titration calculations become simpler with equivalents: at the endpoint, milliequivalents of acid = milliequivalents of base (or oxidant = reductant).

Alternative Method — Using Milliequivalents Directly

For titration problems, skip molarity and use:

N_1 V_1 = N_2 V_2

where $N = \text{Normality} = \text{Molarity} \times n\text{-factor}$ .

JEE Main loves to change the medium for $\text{KMnO}_4$ . In acidic medium: $\text{Mn}^{7+} \to \text{Mn}^{2+}$ , $n = 5$ . In neutral/weakly basic medium: $\text{Mn}^{7+} \to \text{Mn}^{4+}$ ( $\text{MnO}_2$ ), $n = 3$ . In strongly basic medium: $\text{Mn}^{7+} \to \text{Mn}^{6+}$ ( $\text{MnO}_4^{2-}$ ), $n = 1$ . Three different n-factors for the same compound — the medium decides everything.

Common Mistake

The most dangerous error: students calculate n-factor for $\text{K}_2\text{Cr}_2\text{O}_7$ as 3 instead of 6. They forget there are two chromium atoms per formula unit. Each Cr changes by 3, so the total change per formula unit is $2 \times 3 = 6$ . Always count the number of atoms undergoing the oxidation state change and multiply.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Milliequivalents Directly

Common Mistake

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