Winkler Method: Dissolved Oxygen Titration Calculations
Dissolved oxygen (DO) indicates water purity and aquatic life support. In Class 11 and 12 chemistry, JEE Main, NEET, and CUET, dissolved oxygen numericals appear under redox titrations and environmental chemistry. The classic analytical technique is the Winkler titration method. Understanding its chemical sequence, mole ratios, and conversion factors allows students to solve multi-step problems rapidly and accurately.
Chemical Principle and Reaction Steps
Dissolved molecular oxygen (O2) cannot be titrated directly because it reacts sluggishly in dilute aqueous solution. The Winkler method overcomes this through a three-stage fixation and displacement process:
Step 1: Fixation of Oxygen
Manganese(II) sulfate (MnSO4) and alkaline-iodide reagent (NaOH or KOH with KI) are added to the sample bottle without trapping air bubbles. White manganese(II) hydroxide forms:
Mn2+ + 2OH− → Mn(OH)2 (white solid)
Dissolved oxygen oxidizes this precipitate into brown manganese(IV) oxyhydroxide:
2Mn(OH)2 + O2 → 2MnO(OH)2 (brown solid)
In either notation, 1 mole of O2 accepts 4 electrons to oxidize manganese ions.
Step 2: Acidification and Iodine Release
Concentrated sulfuric acid (H2SO4) dissolves the brown precipitate. In acidic medium, manganese oxidizes iodide ions (I−) to molecular iodine (I2):
MnO(OH)2 + 2I− + 4H+ → Mn2+ + I2 + 3H2O
Because 1 mole of O2 produces 2 moles of MnO(OH)2, it liberates exactly 2 moles of I2.
Step 3: Titration with Sodium Thiosulfate
The liberated iodine is titrated against standardized sodium thiosulfate (Na2S2O3) with starch indicator:
I2 + 2S2O32− → 2I− + S4O62−
Combining the reactions gives the central equivalence: 1 mol O2 ≡ 2 mol I2 ≡ 4 mol S2O32−. Exactly 4 moles of thiosulfate correspond to 1 mole of dissolved oxygen.
Key Formula and Equivalent Weight
Because each O2 molecule gains 4 electrons during reduction, its n-factor is 4. With a molar mass of 32.00 g/mol, the equivalent weight of oxygen is 32.00 / 4 = 8.00 g/equiv (8000 mg/equiv). For sodium thiosulfate, oxidation of S2O32− to S4O62− involves 1 electron per molecule, so its n-factor is 1 and normality equals molarity.
Or using normality: DO (mg/L) = [Vthio (mL) × Nthio (eq/L) × 8000] / Vsample (mL)
Where: Vthio = volume of Na2S2O3 used (mL); Mthio = molarity of Na2S2O3; Nthio = normality of Na2S2O3; 8000 = equivalent mass of O2 in mg; Vsample = volume of water sample titrated (mL). In dilute aqueous solutions, 1 mg/L equals 1 ppm.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: Class 11 and 12 Chemistry, JEE Main, NEET, CUET
- Topic: Redox Titrations and Water Analysis
- Difficulty: Moderate
Worked Numerical Examples
Question: A 100.0 mL sample of river water was analysed for dissolved oxygen using the Winkler method. The liberated iodine required 12.50 mL of 0.0100 M Na2S2O3 to reach the starch end point. Calculate the dissolved oxygen content of the water in mg/L (ppm).
Solution:
Step 1: Calculate moles of thiosulfate consumed.
Moles of S2O32− = 0.0100 mol/L × 0.01250 L = 1.250 × 10−4 mol.
Step 2: Relate moles of thiosulfate to moles of O2.
From reaction stoichiometry, 4 moles of S2O32− react per 1 mole of O2:
Moles of O2 = (1.250 × 10−4 mol) / 4 = 3.125 × 10−5 mol.
Step 3: Convert moles of O2 to mass in milligrams.
Mass of O2 = 3.125 × 10−5 mol × 32.00 g/mol = 1.000 × 10−3 g = 1.000 mg.
Step 4: Compute concentration in mg/L.
Volume of water sample = 100.0 mL = 0.1000 L.
Dissolved oxygen = 1.000 mg / 0.1000 L = 10.00 mg/L.
Double check by second route:
Using normality and equivalent weight:
Equivalents of thiosulfate = 0.0100 eq/L × 0.01250 L = 1.250 × 10−4 eq.
Mass of O2 = 1.250 × 10−4 eq × 8.00 g/eq = 1.000 × 10−3 g = 1.000 mg.
DO = 1.000 mg / 0.1000 L = 10.00 mg/L (10.00 ppm). Both routes match.
Answer: 10.00 mg/L
Question: A 250.0 mL lake water sample was treated by the Winkler method. The liberated iodine required 8.40 mL of 0.0250 M Na2S2O3 solution to decolorize the starch-iodine complex. What is the dissolved oxygen level in ppm?
Solution:
Step 1: Calculate millimoles of thiosulfate used.
Millimoles of S2O32− = 0.0250 mmol/mL × 8.40 mL = 0.2100 mmol.
Step 2: Find millimoles of O2.
Millimoles of O2 = 0.2100 mmol / 4 = 0.0525 mmol.
Step 3: Calculate mass of O2 in milligrams.
Mass of O2 = 0.0525 mmol × 32.00 mg/mmol = 1.680 mg.
Step 4: Find concentration in ppm.
Volume of water = 250.0 mL = 0.2500 L.
DO = 1.680 mg / 0.2500 L = 6.720 mg/L = 6.72 ppm.
Double check by second route:
Using the master formula directly:
DO = [8.40 × 0.0250 × 8000] / 250.0 = 1680 / 250.0 = 6.72 ppm. Both methods agree.
Answer: 6.72 ppm
Common Mistakes to Avoid
- Assuming 1:1 or 1:2 Mole Ratio: Believing that 1 mole O2 equals 1 or 2 moles S2O32−. Remember: 1 mole O2 yields 2 moles I2, requiring 4 moles S2O32−.
- Using 16 g/equiv for Oxygen: Molecular oxygen O2 has molar mass 32.00 g/mol and gains 4 electrons. Its equivalent weight is 32.00 / 4 = 8.00 g/equiv, not 16.00.
- Adding Starch Too Early: Adding starch when iodine is concentrated forms an irreversibly bound blue complex. Add starch only after the yellow colour fades to pale straw.
- Trapping Air Bubbles: Introducing air bubbles during sampling dissolves atmospheric oxygen, yielding artificially high values.
- Confusing ppm with Percentages: In dilute aqueous solutions of unit density, 1 mg/L equals 1 ppm. A result of 6.72 mg/L represents 6.72 ppm, not 0.0672% or 6.72%.
Exam Relevance and Question Patterns
| Examination | Question Focus | Key Skills Tested |
|---|---|---|
| CBSE & State Boards | Practical questions, equations, short calculations | Writing balanced redox equations, indicator action, n-factor |
| JEE Main | Numerical questions on dissolved oxygen and water chemistry | Multi-step redox stoichiometry, thiosulfate titration, mg/L conversions |
| JEE Advanced | Multi-concept questions and reagent displacement | Bottle volume displacement, interfering species, equivalent concepts |
| NEET & CUET | Objective problems linking dissolved oxygen to water quality | Applying DO = (V × M × 8000) / Vsample, environmental chemistry concepts |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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