Topic 16 · page 15 of 19

Core Practical 13b: sodium thiosulfate and iodine

Use a known amount of iodine, made from potassium iodate(V) and excess iodide, to find the exact concentration of a sodium thiosulfate solution. Starch is added near the end, and the end point is blue-black to colourless.

13 min read16.17(ii)16.15

Read first: Redox titration reactions and ratios

Method 16.17(ii)

Aim: to find the concentration of a sodium thiosulfate solution by titrating it against iodine made from a standard solution of potassium iodate(V).

The iodine is made in the flask:

IOX3X−(aq)+5 IX−(aq)+6 HX+(aq)→3 IX2(aq)+3 HX2O(l) \begin{aligned} &\ce{IO3-(aq) + 5I-(aq) + 6H+(aq)} \\ &\ce{-> 3I2(aq) + 3H2O(l)} \end{aligned}

and then reacts with thiosulfate:

IX2(aq)+2 SX2OX3X2−(aq)→2 IX−(aq)+SX4OX6X2−(aq) \small \ce{I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq)}

PracticalStandardising sodium thiosulfate

  1. Pipette 25.0 cm³ of standard potassium iodate(V) solution into a conical flask.
  2. Add excess potassium iodide and dilute sulfuric acid. Iodine forms and the solution is brown.
  3. Titrate with sodium thiosulfate from the burette until the colour is pale yellow.
  4. Add about 1 cm³ of starch solution. The mixture turns blue-black.
  5. Add thiosulfate dropwise until the blue-black colour just disappears.
  6. Repeat until you have concordant titres, within 0.10 cm³ of each other.
Four conical flasks: brown with iodine, pale yellow as thiosulfate is added, blue-black after starch is added, and colourless at the end point.
The colour sequence for thiosulfate with iodine. The colours are drawn to show the sequence, not photographed.

Explaining the method 16.17(ii)

The overall ratio is IO₃⁻ : I₂ : S₂O₃²⁻ = 1 : 3 : 6. Iodate(V) is a primary standard: a pure solid that can be weighed accurately, so the iodine it makes is accurately known.

Other oxidising agents can make the iodine instead. In the brass experiment on page 16, copper(II) ions do, but the titration with thiosulfate and starch is the same.

Using the results 16.17(ii)16.15

Worked exampleConcentration of sodium thiosulfate

25.0 cm³ of 0.0100 mol dm⁻³ potassium iodate(V) is treated with excess KI and acid. The iodine needs a mean titre of 24.80 cm³ of sodium thiosulfate.

  1. n(IO₃⁻) = 0.0100 × 25.0 ÷ 1000 = 2.50 × 10⁻⁴ mol
  2. Ratio IO₃⁻ : S₂O₃²⁻ = 1 : 6, so n(S₂O₃²⁻) = 6 × 2.50 × 10⁻⁴ = 1.50 × 10⁻³ mol
  3. [S₂O₃²⁻] = 1.50 × 10⁻³ ÷ 0.02480 = 0.0605 mol dm⁻³

Check yourself

  1. Why is starch not added at the start of the titration?
  2. State the colour change at the end point.
  3. Why is excess potassium iodide added to the iodate(V)?
  4. State the oxidation number change of iodine in IO₃⁻ → I₂, and say whether IO₃⁻ is the oxidising or reducing agent.
  5. 25.0 cm³ of 0.0150 mol dm⁻³ KIO₃ is treated with excess KI and acid. The iodine needs 27.30 cm³ of sodium thiosulfate. Calculate [S₂O₃²⁻] to 3 significant figures.
  6. A student uses the ratio 1 : 2 in question 5. By what factor is their answer wrong?

Answers cover these until you have tried the questions

  1. A lot of iodine present would give a lot of starch–iodine complex, which releases iodine slowly, so the end point would not be sharp. Starch is added when the solution is pale yellow.
  2. Blue-black to colourless.
  3. The equation needs 5 I⁻ for each IO₃⁻. An excess makes sure all of the iodate(V) reacts, so the amount of iodine formed is fixed by the iodate(V).
  4. Iodine goes from +5 to 0, a decrease, so IO₃⁻ is reduced: it is the oxidising agent.
  5. n(IO₃⁻) = 0.0150 × 25.0 ÷ 1000 = 3.75 × 10⁻⁴ mol. n(S₂O₃²⁻) = 6 × 3.75 × 10⁻⁴ = 2.25 × 10⁻³ mol. [S₂O₃²⁻] = 2.25 × 10⁻³ ÷ 0.02730 = 0.0824 mol dm⁻³.
  6. With 1 : 2 they would get 2 × 3.75 × 10⁻⁴ = 7.50 × 10⁻⁴ mol, so the answer is 3 times too small (1 : 2 against 1 : 6).