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Experiment: preparation of Mohr’s salt and potash alum

Written by TopicLamp Editorial Team Checked by TopicLamp Review Team Updated 10 min read

Beyond NCERT In the JEE Main and NEET syllabus; the Class 12 book has no section for it.

Quick answer

Mohr’s salt and potash alum are double salts crystallised from approximately equimolar amounts of their component sulphates. Dilute sulphuric acid suppresses hydrolysis, slow cooling promotes good crystals and prolonged heating of the iron solution increases oxidation risk. Their fixed crystal compositions do not mean that every dissolved salt crystallises quantitatively.

What is the double salt experiment?

Preparing Mohr’s salt and potash alum

The two double salts in this experiment can be crystallised from approximately equimolar amounts of their component sulphates. A double salt has a fixed stoichiometric composition; not every double salt is a 1 : 1 combination. The NCERT Laboratory Manual prepares two: ferrous ammonium sulphate (Mohr’s salt) and potash alum (Lab Manual, Experiment 9.1, page 100). Like complexes, double salts are combinations of stable compounds:

Both double salts as well as complexes are formed by the combination of two or more stable compounds in stoichiometric ratio.

NCERT Chemistry Part I (Class 12), Section 5.1, page 120

The NCERT Class 12 textbook explains double salts (NCERT Section 5.1, page 120) but does not describe their preparation. The JEE Main 2027 and NEET (UG) 2026 syllabuses list Mohr’s salt and potash alum among the preparations of Unit 20. The method below follows the NCERT Laboratory Manual (Chemistry, Class 12), Unit 9, Experiment 9.1 (pages 100–102).

In plain words: dissolve the two salts together in a little warm water, keep the solution slightly acidic, let it cool slowly, and crystals containing both salts in a fixed ratio grow. In water they fall apart again into simple ions, which is what makes them double salts and not complexes.

Double salts by crystallisation

  • The equations (Lab Manual page 100):
    • K2SO4 + Al2(SO4)3 + 24H2O → K2SO4·Al2(SO4)3·24H2O, or 2KAl(SO4)2·12H2O (potash alum)
    • FeSO4 + (NH4)2SO4 + 6H2O → FeSO4·(NH4)2SO4·6H2O (Mohr’s salt)
  • Equimolar amounts. For these two preparations, the double salt forms when a mixture with approximately equimolar proportions of the two salts is crystallised (Lab Manual page 100), because the crystal holds them in a 1 : 1 ratio.
  • Lab Manual slip (pages 100–101): the alum amounts. 2.4 g of K2SO4 (174.3 g mol−1) is 0.0138 mol, but 6.6 g of Al2(SO4)3, written anhydrous in its equation (342.2 g mol−1), is 0.0193 mol. They would be equimolar only for a molar mass of 479 g mol−1, and the Lab Manual names no hydrate. With the anhydrous salt, potassium sulphate limits the yield (reasoned).
  • A little dilute sulphuric acid. Fe2+ and Al3+ ions undergo hydrolysis, so dilute sulphuric acid is added when their sulphates are dissolved, to suppress it (Lab Manual page 100).
  • Lab Manual slip (pages 100–101): the acid volume. The theory says 2–3 mL of dilute sulphuric acid, but the procedures add about 0.4 mL (potash alum, in 10 mL of water) and 0.5 mL (Mohr’s salt, in 5 mL). Follow the procedures.
  • Double salt or complex. Double salts such as Mohr’s salt and potash alum dissociate into simple ions completely when dissolved in water, while a complex ion such as [Fe(CN)6]4− of K4[Fe(CN)6] does not dissociate into Fe2+ and CN− ions (NCERT page 120); strictly, it releases no appreciable free Fe2+ or CN− under ordinary test conditions (reasoned).

Apparatus (Lab Manual page 100): a 50 mL beaker, a 50 mL conical flask, a trough, a glass rod, a tripod stand, a funnel and wire gauze; potassium sulphate, aluminium sulphate, ferrous sulphate, ammonium sulphate, dilute sulphuric acid and ethanol.

Mohr’s salt from iron(II) sulphate and ammonium sulphate

Procedure (Lab Manual page 101):

  1. Dissolve 3.5 g of ferrous sulphate and 1.7 g of ammonium sulphate, weighed separately, in 5 mL of distilled water in a 50 mL conical flask by heating.
  2. Add about 0.5 mL of dilute sulphuric acid and concentrate the solution by heating till the crystallisation point is reached.
  3. Let it cool slowly to room temperature: light green crystals of ferrous ammonium sulphate separate out.
  4. Decant the mother liquor and wash the crystals with very small amounts of a 1 : 1 mixture of cold water and alcohol.
  5. Filter, wash with alcohol, dry between the folds of filter paper and record the yield.
  • Do not heat for long. Prolonged heating may oxidise ferrous ions to ferric ions and change the stoichiometry of the crystals (Lab Manual page 101).
  • Its use. Mohr’s salt is the standard reducing agent in the permanganate titration.

Potash alum from potassium sulphate and aluminium sulphate

Procedure (Lab Manual pages 100–101):

  1. Heat 10 mL of distilled water in a 50 mL beaker to about 40 °C, dissolve 6.6 g of aluminium sulphate in it and add about 0.4 mL of dilute sulphuric acid.
  2. Add 2.4 g of powdered potassium sulphate and heat with constant stirring until it dissolves completely.
  3. Let the solution cool slowly to room temperature: white crystals of potash alum separate out.
  4. Decant the mother liquor and wash the crystals by gently shaking with a 1 : 1 mixture of cold water and alcohol.
  5. Filter, dry between the folds of filter paper and note the yield.
  • An acidic solution. Al3+ hydrolyses in water (Lab Manual page 100), which is why a solution of potash alum turns blue litmus red (reasoned; the Lab Manual asks this, page 102).

Precautions (Lab Manual page 101): cool the solution slowly to get good crystals and avoid rapid cooling; do not disturb the solution while it cools; do not heat the ferrous ammonium sulphate solution for long.

Solved examples

Example 1: Are the Mohr’s salt amounts equimolar?

Taking the ferrous sulphate as FeSO4·7H2O (the Lab Manual does not name the hydrate), check that 3.5 g of it and 1.7 g of (NH4)2SO4 are close to equimolar. (Fe = 55.85, S = 32.07, O = 16.00, N = 14.01, H = 1.008.)

Answer: M(FeSO4·7H2O) = 278.0 g mol−1, so 3.5 g is 0.0126 mol. M((NH4)2SO4) = 132.2 g mol−1, so 1.7 g is 0.0129 mol. The two agree to within 2.5%: equimolar for practical purposes (reasoned). With anhydrous FeSO4 (151.9 g mol−1), 3.5 g would be 0.0230 mol, almost twice the ammonium sulphate, so the heptahydrate, rather than the anhydrous salt, fits the Lab Manual’s “equimolar” (reasoned).

Example 2: Theoretical and percentage yield

With the amounts of Example 1, what mass of Mohr’s salt can form at most? If 4.2 g is obtained, what is the percentage yield?

Answer: FeSO4·7H2O is the smaller amount, 0.0126 mol, and each mole gives one mole of FeSO4·(NH4)2SO4·6H2O (M = 392.2 g mol−1): 0.0126 × 392.2 = 4.94 g. Percentage yield = 4.2/4.94 × 100 = 85% (reasoned).

Common mistakes to avoid

Common mistake

Leaving out the acid

Dilute sulphuric acid suppresses hydrolysis (Lab Manual page 100), reducing the risk of basic-salt or hydroxide contamination; it does not guarantee complete protection from iron(II) oxidation.

Common mistake

Cooling fast

Rapid cooling or disturbing the solution gives small, poor crystals; let it cool slowly and undisturbed.

Quick revision summary

  • These double salts crystallise from approximately equimolar component sulphates; in water they release their constituent hydrated ions.
  • Mohr’s salt: FeSO·(NH)SO·6HO, light green; potash alum: KAl(SO)·12HO (NCERT page 120), written KSO·Al(SO)·24HO in the Lab Manual, white.
  • Dilute HSO suppresses hydrolysis of Fe and Al; cool slowly; avoid prolonged heating of the iron(II) solution.

Formula sheet

  • FeSO + (NH)SO + 6HO → FeSO·(NH)SO·6HO
  • KSO + Al(SO) + 24HO → KSO·Al(SO)·24HO
  • Percentage yield = actual mass ÷ theoretical mass × 100

Questions students ask

Why are Mohr’s salt crystals light green but potash alum crystals white?

Mohr’s salt is light green because its Fe2+ ions, present as [Fe(H2O)6]2+ (beyond the NCERT text; Wikipedia, “Ammonium iron(II) sulfate”), absorb visible light in d–d transitions; NCERT Class 12 Chemistry, Table 4.8 (page 103) lists Fe2+ as green in water. Pure potash alum has no partially filled d shell to support d–d absorption. Its clear crystals are colourless; a mass of small crystals or powder can look white because it scatters light. Absence of d–d transitions alone is not a general proof of whiteness (reasoned). A yellow or brown tinge points to Fe3+, yellow in that table, from oxidation on prolonged heating (Lab Manual page 101; reasoned).

How does potash alum purify water?

Potash alum clears muddy water because, once dissolved, it neutralises the negative charge on fine clay and silt particles, so that they stop repelling one another and clump into flocs that settle or can be filtered off (beyond the NCERT text; Wikipedia, “Potassium alum”). Hydrolysed aluminium species and aluminium hydroxide flocs help neutralise or capture the suspended particles; the process depends on pH and alkalinity (reasoned), and the NCERT Laboratory Manual (Chemistry, Class 12) explains why the charge matters: charged colloidal particles stay suspended and do not coagulate (Unit 1, pages 1–2). Alum only clears the water, which must still be filtered and disinfected before drinking (Wikipedia, “Potassium alum”).

What is the difference between alum and potash alum?

Alum is a family of double sulphates, MIMIII(SO4)2·12H2O, with a monovalent ion such as K+ or NH4+ and a trivalent one such as Al3+ or Cr3+; potash alum, KAl(SO4)2·12H2O, phitkari in Hindi, is the one usually called just alum (beyond the NCERT text; Wikipedia, “Alum” and “Potassium alum”). Chrome alum, KCr(SO4)2·12H2O, is dark violet. Different alums can crystallise together as mixed crystals, so a little chrome alum tints potash alum crystals violet (Wikipedia, “Alum”), the behaviour of isomorphous salts (reasoned).

Is the preparation of Mohr’s salt and potash alum in the JEE Main 2027 and NEET (UG) 2026 syllabus?

Yes, the preparation of Mohr’s salt and potash alum is in Unit 20, Principles Related to Practical Chemistry, of both syllabuses: JEE Main 2027 lists “The chemistry involved in the preparation of the following: Inorganic compounds, Mohr’s salt, potash alum”, and NEET (UG) 2026 prints the same with a semicolon after “Inorganic compounds”. Both also list the “Mohr’s salt vs KMnO4” titration. The NCERT Class 12 textbook explains double salts (NCERT page 120) but not their preparation, which the NCERT Laboratory Manual gives in Unit 9 (pages 100–102).

Practice questions

Try each question first, then open the answer to check your working.

  1. Q1. Can concentrated sulphuric acid be used in place of dilute sulphuric acid in preparing ferrous ammonium sulphate? Explain. (Lab Manual, Discussion Question (ii), page 101.)

    Show answerHide answer

    No. The dilute acid is used to suppress hydrolysis (Lab Manual page 100). NCERT notes that heating with concentrated sulphuric acid oxidises some iron(II) ions to iron(III) ions (NCERT Class 11 Chemistry, Unit 8, page 285). Fe3+ would spoil the stoichiometry of the crystals, as prolonged heating does (Lab Manual page 101; reasoned).

  2. Q2. What is the difference between K4[Fe(CN)6] and FeSO4·(NH4)2SO4·6H2O? (Lab Manual, Discussion Question (iii), page 101.)

    Show answerHide answer

    Mohr’s salt is a double salt: in water it gives Fe2+, NH4+ and SO42− ions. K4[Fe(CN)6] is a complex: the [Fe(CN)6]4− ion does not dissociate into Fe2+ and CN− ions (NCERT page 120), so it gives no test for Fe2+ (reasoned).

More practice questions

Sources and review

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  • : OpenAI Codex read the complete published lesson, FAQs and practice; corrected factual errors and overclaims, verified sources and recomputed answers and isomer counts. New unpublished draft; simulations unchanged.

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