Mohr's salt FeSO4·(NH4)2SO4·6H2O: dissolve equimolar FeSO4 + (NH4)2SO4 in dilute H2SO4, evaporate, cool — pale-green double salt crystallises. Potash alum K2SO4·Al2(SO4)3·24H2O: dissolve equimolar K2SO4 + Al2(SO4)3, evaporate, cool — octahedral colourless crystals. Both are double salts (ionise to constituent ions).
-- NCERT Chemistry Lab Manual Class 12, Unit 9, p. 100Mohr Salt Potash Alum
8 MCQs2 revision cards9-step worked example
Source: NCERT Practical and Analytical ChemistryPYQ coverage: NEET 2024, 2026Official key: NTA-verifiedLast updated: 25 Sep 2026
Mohr Salt Potash Alum, explained for NEET
The trap that costs marks here: aspirants confuse the indicator choice for titration of Mohr's salt with KMnO₄. Mohr's salt titration is a redox titration — KMnO₄ is self-indicating (decolourises until end-point, then first persistent pink). Picking phenolphthalein or methyl orange is a category error; those are acid-base indicators with pH-range logic that does not apply to permanganimetry.
What is Mohr's salt? Ferrous ammonium sulphate, FeSO₄·(NH₄)₂SO₄·6H₂O — a double salt (NCERT Chemistry Lab Manual Class 12, Unit 9, pages 100–101). It is preferred over plain FeSO₄ because the ammonium ion stabilises Fe²⁺ against aerial oxidation. In NEET practical chemistry, it appears as the primary standard for standardising KMnO₄.
What is potash alum? K₂SO₄·Al₂(SO₄)₃·24H₂O — prepared by mixing hot concentrated solutions of K₂SO₄ and Al₂(SO₄)₃ in stoichiometric proportions, then crystallising on slow cooling. Its NEET relevance is limited to identification (colourless octahedral crystals, acidic solution, gives white gelatinous Al(OH)₃ with NaOH).
Bridge to NEET: Questions test (a) why Mohr's salt is preferred over FeSO₄, (b) indicator choice for permanganimetric titration, (c) the normality equation N₁V₁ = N₂V₂ applied to redox end-points.
Watch-out: The end-point of KMnO₄ vs Mohr's salt is the first persistent faint pink — not deep purple. Overshooting (adding excess KMnO₄ past the first colour hold) invalidates the titre value. Slow drop-wise addition near end-point is the procedural safeguard.
Can you answer these Mohr Salt Potash Alum MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Mohr's salt has the formula:
Show answer and why every option is right or wrong
Answer: B. Mohr's salt is ferrous ammonium sulphate hexahydrate, FeSO₄·(NH₄)₂SO₄·6H₂O — a double salt of Fe²⁺ with ammonium sulphate (NCERT Chemistry Lab Manual Class 12, Unit 9, pages 100–101).
Why A is wrong: A describes ferrous sulphate heptahydrate (green vitriol), not the double salt.
Why C is wrong: C has Fe³⁺ (ferric) and 24 waters — this resembles ferric ammonium alum, not Mohr's salt.
Why D is wrong: D substitutes K₂SO₄ for (NH₄)₂SO₄ — this would be a potassium iron double salt, not Mohr's salt.
MCQ 2Easy RecallPractice
Mohr's salt is preferred over ferrous sulphate as a primary standard because:
Show answer and why every option is right or wrong
Answer: D. The presence of (NH₄)₂SO₄ stabilises Fe²⁺ against aerial oxidation, ensuring the titre value remains reproducible over time (NCERT Class 12 Chemistry, Chapter 4).
Why A is wrong: Higher molecular weight alone does not determine suitability as a primary standard — stability and purity are the criteria.
Why B is wrong: Reaction speed with KMnO₄ is comparable; the advantage is storage stability, not kinetics.
Why C is wrong: Solubility is adequate for both salts; it is not the deciding factor for primary-standard selection.
MCQ 3Easy RecallPractice
Potash alum has the formula:
Show answer and why every option is right or wrong
Answer: D. Potash alum is the double sulphate of potassium and aluminium with 24 waters of crystallisation: K₂SO₄·Al₂(SO₄)₃·24H₂O (NCERT Class 12 Chemistry, Chapter 4).
Why A is wrong: A would be soda alum — Na⁺ replaces K⁺. Potash alum specifically has potassium.
Why B is wrong: B is ammonium alum (NH₄ alum), not potash alum — the univalent cation differs.
Why C is wrong: C is chrome alum (violet colour), not potash alum — the trivalent cation is Cr³⁺, not Al³⁺.
MCQ 4Direct ApplicationPractice
In the titration of Mohr's salt against KMnO₄, the indicator used is:
Show answer and why every option is right or wrong
Answer: A. KMnO₄ acts as its own indicator in acidic medium: it decolourises while Fe²⁺ remains, and the first persistent faint pink marks the end-point. No external indicator is needed (NCERT Class 12 Chemistry, Chapter 4).
Why B is wrong: Methyl orange is an acid-base indicator (pH 3.1–4.4); using it here confuses acid-base and redox titration categories (trap: indicator-selection error).
Why C is wrong: Phenolphthalein is an acid-base indicator (pH 8.2–10); it does not respond to the redox equivalence point in permanganimetry (trap: indicator-selection error).
Why D is wrong: Starch solution is the indicator for iodometric titrations (I₂ + starch → blue), not for permanganimetric titrations.
MCQ 5Direct ApplicationPractice
In an acid-base titration of a weak acid with a strong base, which indicator is appropriate?
Show answer and why every option is right or wrong
Answer: A. The equivalence point of a weak-acid + strong-base titration lies above pH 7 (typically pH 8–9). Phenolphthalein changes colour in the range 8.2–10, which brackets this equivalence point (NCERT Class 11 Chemistry, Chapter 7).
Why B is wrong: Methyl orange changes at pH 3.1–4.4 — far below the eq-pt pH of a weak-acid/strong-base titration. It would signal a false early end-point (trap: indicator pH-range mismatch).
Why C is wrong: Litmus has a broad, indistinct transition (pH 5–8) — unsuitable for precise volumetric end-point detection.
Why D is wrong: Methyl red (pH 4.4–6.2) is still below the equivalence pH of ~8–9 for this combination; the colour change would occur before equivalence.
MCQ 6Direct ApplicationPractice
25.0 mL of 0.1 N Mohr's salt solution requires how many mL of 0.02 N KMnO₄ to reach the end-point?
Show answer and why every option is right or wrong
Answer: C. N₁V₁ = N₂V₂ → 0.1 × 25.0 = 0.02 × V₂ → V₂ = 2.5 / 0.02 = 125.0 mL. The normality equation applies directly at the redox equivalence point (NCERT Class 11 Chemistry, Chapter 7).
Why A is wrong: 5.0 mL results from dividing 0.1 by 0.02 without multiplying by volume — arithmetic shortcut error.
Why B is wrong: 12.5 mL results from computing 25 × 0.1 / 0.2 (wrong denominator — using 0.2 instead of 0.02).
Why D is wrong: 50.0 mL results from computing 25 × 0.02 / 0.01 or similar ratio inversion.
MCQ 7Concept TrapPractice
A student performing KMnO₄ vs Mohr's salt titration notices the pink colour appears but fades on swirling. The correct action is:
Show answer and why every option is right or wrong
Answer: C. The transient pink that fades on swirling means unreacted Fe²⁺ remains and reduces the added MnO₄⁻. The true end-point is the first persistent faint pink (≥30 s). Continue drop-wise addition until persistence is achieved (NCERT Class 12 Chemistry, Chapter 4 — procedure notes).
Why A is wrong: Recording a transient colour as end-point gives a falsely low titre (trap: end-point overshoot/undershoot — premature recording).
Why B is wrong: KMnO₄ has not decomposed; the colour fading indicates remaining Fe²⁺ is consuming the MnO₄⁻. The reagent is intact.
Why D is wrong: Adding H₂SO₄ maintains acidity but does not create MnO₄⁻ colour — it cannot replace the titrant addition step.
MCQ 8CalculationPractice
A Mohr's salt solution is prepared by dissolving 3.92 g in 100 mL. Given M(Mohr's salt) = 392 g/mol and that Fe²⁺ → Fe³⁺ involves a 1-electron change, what is the normality of the solution with respect to the KMnO₄ titration?
Show answer and why every option is right or wrong
Answer: B. Step 1: Molarity = 3.92 / 392 / 0.1 L = 0.1 M. Step 2: n-factor of Mohr's salt = 1 (Fe²⁺ loses 1 electron to become Fe³⁺). Normality = Molarity × n-factor = 0.1 × 1 = 0.1 N.
Why A is wrong: 0.01 N results from forgetting to divide by the volume: 3.92/392 = 0.01 is the number of moles, not a concentration.
Why C is wrong: 0.2 N results from incorrectly assigning n-factor = 2 (confusing Fe²⁺ → Fe³⁺ with a 2-electron change, perhaps mixing up with Cu²⁺ or Sn²⁺/Sn⁴⁺).
Why D is wrong: 1.0 N results from using volume as 10 mL instead of 100 mL, inflating the concentration tenfold.
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Mohr Salt Potash Alum: quick recall before you leave
How do you solve a Mohr Salt Potash Alum question? A worked example
- 1
Given
• Volume of Mohr's salt solution (analyte): V₁ = 25.00 mL• Normality of Mohr's salt: N₁ = 0.1 N (n-factor = 1 for Fe²⁺ → Fe³⁺)• Normality of KMnO₄ solution: N₂ = unknown• Titre volume (KMnO₄ used): V₂ = 20.00 mL• Medium: dilute H₂SO₄ (acidic), no separate indicator required
- 2
Required
Find the normality (N₂) of the KMnO₄ solution.
- 3
Concept
At the redox equivalence point, equivalents of reducing agent = equivalents of oxidising agent. This is expressed as N₁V₁ = N₂V₂.
- 4
Formula
N₁V₁ = N₂V₂
- 5
Substitution
0.1 × 25.00 = N₂ × 20.00
- 6
Calculation
N₂ = (0.1 × 25.00) / 20.00 = 2.500 / 20.00 = 0.125 N
Note on exact values: The volumes 25.00 mL and 20.00 mL are measured values (4 significant figures from a burette). The normality 0.1 N is given as a 1-sig-fig value, so the final answer is limited to 1 significant figure. However, in NEET practical contexts, 0.1 N is conventionally treated as exact (defined concentration), giving the answer as 0.125 N (3 sig figs from the volume ratio). - 7
Final answer
N₂ = 0.125 N
The n-factor of KMnO₄ in acidic medium is 5 (Mn⁷⁺ → Mn²⁺), so the corresponding molarity = 0.125 / 5 = 0.025 M. This cross-check confirms the answer is physically reasonable. - 8
Common trap
Overshooting the end-point: if the student records the titre as, say, 21.5 mL (added KMnO₄ past the first persistent pink), the calculated N₂ would be falsely low (0.116 N). The procedural safeguard is drop-wise addition near end-point and accepting the first 30-second-persistent faint pink.
- 9
Similar NEET-style question
"25.0 mL of a FeSO₄ solution of unknown concentration is titrated against 0.04 N KMnO₄. If 12.5 mL of KMnO₄ is consumed, what is the concentration of FeSO₄ in g/L?" (Apply N₁V₁ = N₂V₂ to find N₁, then convert normality to g/L using equivalent weight of FeSO₄.)
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What to remember before solving Mohr Salt Potash Alum questions
More in Practical and Analytical Chemistry: 6 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Mohr Salt Potash Alum questions from past NEET papers
2 questions from NEET 2024, 2026. Answers verified against NTA official keys.
NEET 2026
In potash alum, the ratio of K⁺ and SO₄²⁻ ions is
11:2
22:1
32:3
43:2
NTA Answer: Option 1(final)
NEET 2024Revised key
1dilute hydrochloric acid
2concentrated sulphuric acid
3dilute nitric acid
4dilute sulphuric acid
NTA Answer: Option 4(revised_final)
All 7 past-paper questions from Practical and Analytical Chemistry →
Sources
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