Mohr's Salt Titration: Molarity and Purity Calculations
Mohr's salt titration is a core volumetric analysis topic in school chemistry practicals and competitive entrance examinations. It provides an accurate method to standardise potassium permanganate (KMnO4) solutions and determine the purity of iron salts. Students frequently encounter quantitative problems requiring them to calculate molarity, strength, and percentage purity. Mastering the underlying redox stoichiometry prevents errors in both practical and theory exams.
What Is Mohr's Salt and Why Is It Used?
Mohr's salt is a pale green double salt chemically named ferrous ammonium sulphate hexahydrate, with the formula FeSO4·(NH4)2SO4·6H2O. In redox titrations, it serves as a primary standard reducing agent. Unlike simple ferrous sulphate (FeSO4·7H2O), Mohr's salt resists oxidation by air because ammonium ions stabilize the crystal lattice. When preparing the solution, dilute sulphuric acid (H2SO4) is added to prevent hydrolysis of ferrous (Fe2+) ions into insoluble basic ferric salts.
Mohr's salt, FeSO4·(NH4)2SO4·6H2O: Fe (55.85) + 2 S (64.12) + 14 O (224.00) + 2 N (28.02) + 20 H (20.16) = 392.15 g/mol (commonly rounded to 392.0 g/mol; both values are standard).
Potassium permanganate, KMnO4: K (39.10) + Mn (54.94) + 4 O (64.00) = 158.04 g/mol (commonly rounded to 158.0 g/mol).
Redox Chemistry and the Equivalence Principle
In acidic medium, potassium permanganate acts as a strong oxidising agent. The manganese in MnO4- drops from +7 to +2 oxidation state by gaining 5 electrons. Simultaneously, Fe2+ from Mohr's salt loses 1 electron to form Fe3+.
The reduction half-reaction is:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O (n-factor of KMnO4 = 5)
The oxidation half-reaction is:
Fe2+ → Fe3+ + e- (n-factor of Fe2+ = 1)
The balanced ionic equation is:
MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Mole relation: (M1 × V1) / 1 = (M2 × V2) / 5
Equivalence relation: N1 × V1 = N2 × V2, where N1 = 5 × M1 and N2 = 1 × M2
Strength (g/L) = Molarity (mol/L) × Molar Mass (g/mol)
Percentage Purity (%) = [Mass of pure salt (g) / Mass of impure sample (g)] × 100
Here, M1 is the molarity of KMnO4, V1 is its volume in mL, M2 is the molarity of Mohr's salt, and V2 is its volume in mL. KMnO4 acts as a self-indicator: the first permanent pale pink drop marks the endpoint.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 12, ISC Class 12, State Boards, JEE Main, NEET, CUET
- Topic: Redox Titrations and Volumetric Analysis
- Difficulty: School / Competitive Entrance
Worked Examples with Step-by-Step Arithmetic
Step 1: Calculate molarity of the Mohr's salt solution (M2).
Moles of Mohr's salt = 3.920 g / 392.0 g/mol = 0.01000 mol.
Volume of solution = 250.0 mL = 0.2500 L.
M2 = 0.01000 mol / 0.2500 L = 0.04000 M.
Step 2: Calculate molarity of KMnO4 (M1) using stoichiometry.
1 mole of MnO4- reacts with 5 moles of Fe2+.
(M1 × V1) / 1 = (M2 × V2) / 5
(M1 × 20.00 mL) / 1 = (0.04000 M × 25.00 mL) / 5
M1 × 20.00 = 0.2000 ⇒ M1 = 0.01000 M.
Step 3: Verification by normality.
N2 = 0.04000 M × 1 = 0.04000 N.
N1 = (0.04000 N × 25.00 mL) / 20.00 mL = 0.05000 N.
M1 = N1 / 5 = 0.05000 / 5 = 0.01000 M. Both methods agree.
Step 4: Calculate strength in g/L.
Strength = 0.01000 mol/L × 158.0 g/mol = 1.580 g/L.
Answer: Molarity of Mohr's salt = 0.0400 M; Molarity of KMnO4 = 0.0100 M; Strength of KMnO4 = 1.58 g/LStep 1: Calculate moles of KMnO4 used.
Moles of KMnO4 = 0.02000 mol/L × 0.01800 L = 0.0003600 mol.
Step 2: Determine moles of Fe2+ in the 25.00 mL portion.
Moles of Fe2+ = 5 × 0.0003600 mol = 0.001800 mol.
Step 3: Scale up to total 250.0 mL solution.
Total moles of Fe2+ = 0.001800 mol × (250.0 / 25.00) = 0.01800 mol.
Mass of pure Mohr's salt = 0.01800 mol × 392.14 g/mol = 7.0585 g (or 7.0560 g using 392.0 g/mol).
Step 4: Compute percentage purity.
Purity = (7.0585 g / 8.000 g) × 100 = 88.23% (or 88.20% using 392.0 g/mol).
Step 5: Verification via milliequivalents.
meq of KMnO4 = 5 × 0.02000 N × 18.00 mL = 1.800 meq.
meq of Fe2+ in 250 mL = 1.800 × 10 = 18.00 meq = 0.01800 eq.
Pure mass = 0.01800 eq × 392.14 g/eq = 7.0585 g. Purity = (7.0585 / 8.000) × 100 = 88.23%.
Answer: Moles of Fe2+ in aliquot = 0.001800 mol; Mass of pure salt = 7.059 g; Percentage purity = 88.23%Common Practical and Calculation Mistakes
- Inverting the n-factor ratio: Writing M1V1 / 5 = M2V2 instead of 5 × M1V1 = M2V2 is a frequent error. 1 mole of KMnO4 oxidises 5 moles of Mohr's salt.
- Adding hydrochloric acid: Never acidify with HCl. KMnO4 oxidises Cl- to Cl2 gas, causing an artificially high titre. Nitric acid (HNO3) oxidises Fe2+ prematurely. Dilute H2SO4 must be used.
- Heating the solution: Unlike oxalic acid, Mohr's salt titrations are done at room temperature. Heating promotes aerial oxidation of Fe2+ to Fe3+.
- Omitting hydration water: Forgetting 6H2O gives ~284 g/mol instead of 392 g/mol, spoiling all purity calculations.
- Forgetting the aliquot factor: Failing to multiply moles in the 25.0 mL sample by 10 to get the full 250.0 mL content causes a tenfold error.
Exam Relevance and Testing Patterns
| Examination | Assessment Mode | Core Concepts Tested |
|---|---|---|
| CBSE & State Boards (Class 12) | Practical & Theory Papers | Standard solution preparation, calculating strength and molarity of KMnO4, viva on indicators and acid choice. |
| ISC Class 12 | Practical Exam | Volumetric titration tables, concordant readings, stoichiometry, percentage purity of hydrated double salts. |
| JEE Main & Advanced | Objective & Numerical Questions | Redox n-factor identification, gram-equivalent law, multi-step purity calculations, acid-base-redox interference. |
| NEET | Single-Choice Questions | Direct formula applications (N1V1 = N2V2), molarity to strength conversions, oxidation state changes. |
| CUET (UG) | Multiple-Choice Questions | Balanced ionic redox equations, identification of primary standards, stoichiometric mole ratios in solution. |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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