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Iodine titrations: direct and indirect analysis

Key idea: Iodimetry uses iodine to titrate a reductant; iodometry releases iodine from an oxidant first. Thiosulfate reacts 2:1 with iodine and removes the starch colour at an iodometric end-point.

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  • GCE A-Level H2 Chemistry 2027
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Iodine can be the titrant, or it can be produced by the sample and then titrated. Distinguishing these arrangements tells you what the endpoint means and which reaction ratios connect the titre to the analyte, the substance being measured.

The direct redox titration method still applies: find the known titrant amount, follow the balanced reaction, then relate the analyte amount to its sample volume.

Iodimetry: iodine is delivered from the burette

In direct iodimetry, a standard iodine solution is used to titrate a reducing substance. The sample consumes the added iodine. With the specified starch indicator method, the first persistent blue iodine–starch colour signals a small excess of iodine after the reducing substance has reacted.

Here iodine is being added. The endpoint is the appearance of its starch colour. The method determines when and how starch is used; the late-addition rule below belongs to titration of an initially iodine-rich solution with thiosulfate.

Recognise the direct iodimetry endpoint

Direct iodimetry: the first persistent starch colourDirect iodimetry only. Iodine is added to a reducing sample using the specified starch method. Local blue colour disappears after mixing before the endpoint; the first persistent blue colour after mixing signals excess iodine. Indirect iodometry has the reverse starch endpoint.Before endpointlocal colourAfter mixingcolour disappearsEndpoint after mixingfirst persistent blue
Direct iodimetry: the first persistent starch colour
  1. A small iodine addition makes a local blue starch colour.
  2. After mixing, the colour disappears while reducing substance remains.
  3. The first blue colour that remains after mixing marks a small excess of iodine.

This schematic illustrates direct iodimetry with the specified starch method, not indirect iodometry, whose starch colour disappears at the endpoint. Colours, drop size, mixing and the 4.8-second playback are illustrative, not experimental evidence or the required persistence time. Follow the observation interval in the laboratory method.

Iodometry: the sample produces iodine first

In indirect iodometry, an oxidising analyte reacts with excess iodide to produce iodine. The iodine formed is titrated with standard sodium thiosulfate solution:

I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

Thus one mole of iodine requires two moles of thiosulfate. Start from the measured thiosulfate amount and work backwards:

n(S₂O₃²⁻) → n(I₂) → n(analyte)

The second conversion depends on the reaction that formed the iodine. Do not assume its ratio from the thiosulfate equation alone.

Follow both reactions in a bleach analysis

For a hypochlorite sample under the stated analysis conditions:

ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O

One mole of hypochlorite forms one mole of iodine, which then consumes two moles of thiosulfate. This analysis assumes quantitative iodine formation and no other oxidant or reductant that would change the iodine amount.

Use the supervised method

Acidifying bleach can release hazardous chlorine. The calculation below uses supplied data; actual analysis requires the laboratory’s specified reagents, quantities and safety controls. Never improvise by mixing bleach and acid.

Guided practice 1

Work backwards through an iodometric titration

About 8 min

Problem

A 25.0 cm³ bleach sample reacts with excess acidified iodide to form iodine. The iodine requires 18.60 cm³ of 0.100 mol dm⁻³ thiosulfate. Given that 1 mol ClO⁻ forms 1 mol I₂, calculate [ClO⁻].

Try this before viewing the solution

Show the steps Optional working
Unit: mol
Unit: mol
Unit: mol
Unit: mol dm⁻³

Hints

Hint 1

start at the burette

First calculate moles of thiosulfate.
Hint 2

use both equations

Two moles of thiosulfate react with one mole of iodine; iodine and hypochlorite are 1:1.
Show solution step by step

  1. Use the titre

    Method

    Find the amount of thiosulfate delivered.

    Reason

    This is the directly measured reacting amount.

    Working

    Answer

    n(S₂O₃²⁻) = 0.100(18.60 × 10⁻³) = 1.860 × 10⁻³ mol

  2. Follow both ratios

    Method

    Convert thiosulfate to iodine, then iodine to hypochlorite.

    Reason

    The analysis is indirect.

    Working

    Answer

    n(I₂) = (1.860 × 10⁻³)/2 = 9.30 × 10⁻⁴ mol

    Hence n(ClO⁻) = 9.30 × 10⁻⁴ mol.

  3. Use the sample volume

    Method

    Divide by 0.0250 dm³.

    Reason

    The calculated moles came from the 25.0 cm³ aliquot.

    Working

    Answer

    [ClO⁻] = 0.0372 mol dm⁻³.

Judge the starch endpoint

As thiosulfate consumes iodine, the iodine colour fades. In this method, add starch near the endpoint when the iodine solution is pale yellow, then add thiosulfate dropwise until the blue-black iodine–starch colour just disappears. At high iodine concentration, the iodine–starch association can retain iodine and make a sharp endpoint slow to judge.

Exam-style question 1

Explain the starch endpoint

3 marks

Examination question

In an iodometric titration, explain when starch should be added, describe the final endpoint and state why starch is not added at the start. [3 marks]

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Show solution step by step

  1. Choose the time

    1 mark

    Method

    Add starch when the iodine solution is pale yellow, close to the endpoint.

    Reason

    Only a small amount of iodine remains.

    Working

    Answer

    Add starch at pale yellow, then continue the titration dropwise.

  2. State the endpoint

    1 mark

    Method

    Add thiosulfate dropwise until the blue-black colour just disappears.

    Reason

    The remaining iodine has reacted.

    Working

    Answer

    Endpoint: the blue-black iodine–starch colour just disappears. Other species may still give the mixture a colour.

  3. Explain the timing

    1 mark

    Method

    Do not add starch at the start.

    Reason

    The intense iodine–starch association at high iodine concentration can make the endpoint slow and difficult to judge.

    Working

    Answer

    High iodine concentration can retain iodine in the complex and delay a sharp colour change.

Disappearance of the iodine–starch colour does not require the whole mixture to become colourless: other ions or a precipitate may remain. Follow the method’s stated observation for the actual sample.

Iodine lost before titration would require less thiosulfate and lead to an underestimate of the analyte amount. Use the specified timing and handling conditions to keep the iodine formed available for reaction.

Check your understanding 1 Find the overall ratio for a different analyte

Under suitable conditions, copper(II) ions react quantitatively with excess iodide:

2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂

The iodine from a 25.0 cm³ copper(II) sample requires 20.00 cm³ of 0.100 mol dm⁻³ thiosulfate. Assume both reactions are quantitative. Calculate the copper(II) concentration. Show both reaction ratios in the following question.

Show hint

One mole of iodine came from two moles of copper(II), and reacts with two moles of thiosulfate. Link these two equations before calculating the sample concentration.

Show the steps Optional working
Unit: mol
Unit: mol dm⁻³
Show answer
n(S₂O₃²⁻) = 0.100(0.02000) = 2.000 × 10⁻³ mol
n(I₂) = (2.000 × 10⁻³)/2 = 1.000 × 10⁻³ mol
n(Cu²⁺) = 2(1.000 × 10⁻³) = 2.000 × 10⁻³ mol
c(Cu²⁺) = (2.000 × 10⁻³)/0.0250 = 0.0800 mol dm⁻³

Check your understanding 2 Show the two ratios and the overall ratio

Show both reaction ratios for the 25.0 cm³ copper(II) aliquot requiring 20.00 cm³ of 0.100 mol dm⁻³ thiosulfate. Use the copper–iodide and iodine–thiosulfate equations above, and state the overall ratio.

Show answer
n(S₂O₃²⁻) = 0.100(0.02000) = 2.000 × 10⁻³ mol
n(I₂) = (2.000 × 10⁻³)/2 = 1.000 × 10⁻³ mol
n(Cu²⁺) = 2(1.000 × 10⁻³) = 2.000 × 10⁻³ mol
c(Cu²⁺) = (2.000 × 10⁻³)/0.0250 = 0.0800 mol dm⁻³

The overall copper(II):thiosulfate ratio is 1:1, although iodine:thiosulfate is 1:2. The bleach example’s overall 1:2 ratio therefore cannot be reused for this analyte.

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