Iodometric Titrations: Thiosulfate and Iodine Calculations
Iodometric titration is an indirect method. The substance you want to measure oxidises iodide (I−) to iodine (I2), and the iodine formed is then titrated with standard sodium thiosulfate. It is used for copper, bleaching powder, dissolved oxygen, and many other oxidising agents. In IIT-JAM, GATE, CSIR-NET and CUET-PG, the whole skill is writing the mole chain correctly: analyte → I2 → S2O32−.
The reactions and the key ratio
So: 1 mol I2 ≡ 2 mol S2O32−
n(analyte) = n(S2O32−) / (thiosulfate moles per mole of analyte)
Thiosulfate is first standardised, then used. The ratios that matter:
| Analyte | Reaction with iodide | mol S2O32− per mol analyte |
|---|---|---|
| KIO3 (standard) | IO3− + 5 I− + 6 H+ → 3 I2 + 3 H2O | 6 |
| Cu2+ | 2 Cu2+ + 4 I− → 2 CuI + I2 | 1 |
| Cl2 (available chlorine) | Cl2 + 2 I− → 2 Cl− + I2 | 2 |
| O2 (Winkler method) | O2 → 2 MnO(OH)2 → 2 I2 | 4 |
Excess iodide is added so that all the analyte reacts and the iodine formed stays dissolved as I3−. Starch is added near the end point, when the colour is pale yellow. Starch added too early holds iodine strongly and makes the end point sluggish. The blue colour disappears at the end point.
Worked example 1: standardising thiosulfate with KIO3
0.1070 g of KIO3 (M = 214.00 g/mol) is dissolved, excess KI and acid are added, and the iodine formed needs 30.00 mL of sodium thiosulfate solution. Find its molarity.
n(KIO3) = 0.1070 / 214.00 = 5.000 × 10−4 mol.
Each IO3− gives 3 I2, and each I2 needs 2 S2O32−, so n(S2O32−) = 6 × 5.000 × 10−4 = 3.000 × 10−3 mol.
M = 3.000 × 10−3 / 0.03000 L = 0.1000 M.
Worked example 2: copper in brass
0.5000 g of brass is dissolved so that all copper is Cu2+. Excess KI is added and the liberated iodine requires 22.40 mL of 0.1000 M thiosulfate. Find the percentage of copper (Cu = 63.546).
n(S2O32−) = 0.1000 × 0.02240 = 2.240 × 10−3 mol.
Two Cu2+ give one I2, which needs two S2O32−. So 1 mol Cu2+ ≡ 1 mol S2O32−, and n(Cu) = 2.240 × 10−3 mol.
Mass of Cu = 2.240 × 10−3 × 63.546 = 0.14234 g. Percentage = 0.14234 / 0.5000 × 100 = 28.47%.
Worked example 3: available chlorine in bleaching powder
1.000 g of bleaching powder is made up to 250 mL. A 25.00 mL portion is treated with excess KI and acid, and needs 9.80 mL of 0.1000 M thiosulfate. Find the percentage of available chlorine (Cl2 = 70.90 g/mol).
n(S2O32−) = 0.1000 × 0.00980 = 9.80 × 10−4 mol. n(Cl2) = 9.80 × 10−4 / 2 = 4.90 × 10−4 mol in the 25.00 mL portion.
Whole 250 mL: 4.90 × 10−4 × (250 / 25.00) = 4.90 × 10−3 mol. Mass = 4.90 × 10−3 × 70.90 = 0.3474 g.
Available chlorine = 0.3474 / 1.000 × 100 = 34.7%.
Worked example 4: dissolved oxygen by the Winkler method
A 200.0 mL water sample is fixed with Mn(II) and alkali, acidified with iodide, and the iodine formed needs 16.20 mL of 0.01000 M thiosulfate. Find the dissolved oxygen in mg/L (O2 = 31.998 g/mol).
n(S2O32−) = 0.01000 × 0.01620 = 1.620 × 10−4 mol. One O2 ≡ 4 S2O32−, so n(O2) = 4.050 × 10−5 mol.
Mass = 4.050 × 10−5 × 31.998 = 1.296 × 10−3 g = 1.296 mg in 0.2000 L. Concentration = 1.296 / 0.2000 = 6.48 mg/L.
Cross-check on the chain: each O2 takes 4 electrons (to become 2 O2−), and each S2O32− gives up 1 electron, so 4 thiosulfate per O2 is consistent.
Common mistakes
- Using 1:1 for IO3− to thiosulfate. The correct ratio is 1:6, because of the 3 I2 formed.
- Using 1:2 for copper (taking the thiosulfate ratio for I2). Per Cu2+ the ratio is 1:1.
- Forgetting the aliquot factor (250/25.00) when only part of the solution is titrated.
- Adding starch at the start, when iodine is still dark brown.
- Writing S4O62− as SO42−. In this titration thiosulfate is only mildly oxidised, to tetrathionate.
- Mixing up direct and indirect. The sample is not titrated directly; iodine is the stepping stone.
Exam relevance
| Question type | Where to look |
|---|---|
| Standardising thiosulfate | Use the 1:6 ratio with KIO3 (or the ratio for K2Cr2O7, which is also given in the question) |
| Percentage of an analyte in a sample | mol thiosulfate → mol analyte → mass → % |
| Aliquot of a flask | Multiply by the flask volume / pipette volume |
| Why excess KI | Complete reaction and I3− formation |
| Why starch late | Avoid iodine trapped in starch |
Check the syllabus given in the current official notification of your exam. The numbers on this page are practice values.
Once you have the moles of thiosulfate, you can turn moles and volume into molarity with the Concentration calculator.
Open the Concentration Calculator →Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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