n-Factor and Equivalent Concept Problems
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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%.