# n-Factor and Equivalent Concept Problems

> n-factor problems for JEE Chemistry: equivalents per mole, KMnO4 in acid neutral and base, normality-molarity conversion, double indicator titrations, purity.

- Canonical URL: https://prepelephant.com/topics/jee/chemistry/n-factor-problems
- Exam / course: JEE · Subject: Chemistry
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "n-Factor and Equivalent Concept Problems", PrepElephant, https://prepelephant.com/topics/jee/chemistry/n-factor-problems

## Direct answer

Equivalents are the common currency of titration, and the n-factor is the exchange rate: the number of equivalents one mole delivers — protons exchanged for acids and bases, electrons transferred per formula unit for oxidants and reductants. Equivalent weight = molar mass/n, normality = n × molarity, and every titration ends when equivalents are equal: N1V1 = N2V2. The JEE workhorse is potassium permanganate with n = 5 in acid (MnO4^- → Mn^2+), n = 3 in neutral or faintly alkaline medium (→ MnO2), n = 1 in strongly alkaline (→ MnO4^2-) — one salt, three currencies, and the question's medium decides which to use.

## What you must remember

- **Acid-base n:** H2SO4 = 2; H3PO4 = 3 total, but only 1 to the methyl-orange end point (→ NaH2PO4) and 2 to phenolphthalein (→ Na2HPO4) — n follows the end point, not the formula alone.
- **Redox roster:** K2Cr2O7 = 6; Na2S2O3 = 1 (two thiosulphates give two electrons as S4O6^2- forms); Mohr salt = 1 (Fe^2+); oxalic acid = 2 in both acid-base and redox roles.
- **KMnO4 medium trio:** 5 acidic, 3 neutral, 1 strongly alkaline — the most examined single fact in the chapter.
- **Hydrate honesty:** equivalent and molar masses include water of crystallisation — oxalic acid dihydrate at 126 g/mol, Mohr salt at 392 g/mol.
- **Double indicator analysis:** in NaOH + Na2CO3 mixtures, phenolphthalein volume V1 stops when carbonate is half-neutralised, methyl orange volume V2 at total; carbonate equivalents = 2(V2 − V1)N, hydroxide = (2V1 − V2)N.
- **Purity formula:** percentage purity = (normality × volume in mL × equivalent weight)/(1000 × sample mass) × 100.

## Working a titration and a mixture

First the direct titration: 25.0 mL of dilute H2SO4 needs 20.0 mL of 0.10 N NaOH. Then N(acid) × 25.0 = 0.10 × 20.0, so N = 0.080, and since sulphuric acid's n is 2, molarity = 0.080/2 = 0.040 M. The normality equation did the stoichiometry silently — that is its entire appeal.

Now the mixture. A 1.0 g sample of NaOH and Na2CO3 in water consumes 20.0 mL of 0.1 N HCl to phenolphthalein and 35.0 mL total to methyl orange. The extra 15.0 mL after the first end point is the second proton of the carbonate: carbonate equivalents = 15.0 × 0.1 × 2/1000 = 0.003, and at 53 g per equivalent (106/2), Na2CO3 = 0.003 × 53 = 0.159 g. Hydroxide consumed the rest: (2 × 20.0 − 35.0) = 5.0 mL × 0.1 N = 0.0005 equivalents × 40 = 0.020 g. Percentages: 15.9% carbonate, 2.0% hydroxide, remainder water or impurity. The logic is bookkeeping, not chemistry — which is precisely why it scores.

## Where students slip

Mixing currencies loses the most marks: N1V1 = N2V2 demands normalities on both sides, and molarities on both sides only if you also carry the balanced stoichiometry — candidates divide a normality by a molarity and halve or double the true answer. The KMnO4 medium switch is the second trap: 0.1 M permanganate is 0.5 N in acid but 0.3 N in neutral solution, and forgetting the medium flips every downstream number. Third, the H3PO4 end-point dependence: to phenolphthalein only two protons count. Finally, watch hydrates — solving with anhydrous oxalic acid (90 g/mol) instead of the dihydrate (126 g/mol) produces answers exactly 40% off, a margin no rounding survives.

## Frequently asked questions

### What is the n-factor of KMnO4 in acidic medium?

Five, because MnO4^- + 8H+ + 5e^- → Mn^2+ + 4H2O — five electrons per formula unit, so 0.1 M KMnO4 is 0.5 N in acid.

### Why does H3PO4 have different n-factors with different indicators?

Phosphoric acid neutralises stepwise: to the methyl-orange end point one proton reacts (n = 1), to phenolphthalein two (n = 2) — the indicator fixes how far the reaction runs.

### How do normality and molarity relate?

Normality = n-factor × molarity, since each mole supplies n equivalents; the relation flips to molarity = normality/n when you convert back.

### What does the extra methyl-orange volume measure in a double indicator titration?

The second half of the carbonate neutralisation — 2(V2 − V1) gives the carbonate's acid consumption, from which its mass follows directly.

### Why must hydrated salts be weighed with their water of crystallisation?

Titration counts molecules as they exist in the crystal; oxalic acid dihydrate has 126 g per mole of redox-active acid, and using 90 g inflates every result by 40%.