Iodine/Thiosulfate Redox Titrations Revision Notes for OCR A-Level Chemistry A

Iodine/Thiosulfate Redox Titrations

Introduction to iodine/thiosulfate titrations

Iodine/thiosulfate titrations are a powerful analytical technique used to determine the concentration of various oxidising agents. This method involves a two-stage process where thiosulfate ions ( $\text{S}_2\text{O}_3^{2-}$ ) are used to reduce aqueous iodine ( $\text{I}_2$ ).

The fundamental principle relies on the following redox reactions:

Oxidation of thiosulfate: $2\text{S}_2\text{O}_3^{2-}\text{(aq)} \rightarrow \text{S}_4\text{O}_6^{2-}\text{(aq)} + 2e^-$

Reduction of iodine: $\text{I}_2\text{(aq)} + 2e^- \rightarrow 2\text{I}^-\text{(aq)}$

Overall equation: $2\text{S}_2\text{O}_3^{2-}\text{(aq)} + \text{I}_2\text{(aq)} \rightarrow 2\text{I}^-\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)}$

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This technique is particularly useful because directly titrating iodine would be very limiting on its own. However, by incorporating an additional step involving an oxidising agent, we can analyse many different substances indirectly. This makes it one of the most versatile analytical methods in chemistry.

Applications of iodine/thiosulfate titrations

This versatile analytical method can be used to determine the concentration of several important substances:

Chlorate(I) content ( $\text{ClO}^-$ ) in household bleach
Copper(II) content ( $\text{Cu}^{2+}$ ) in copper(II) compounds
Copper content in copper alloys such as brass and bronze

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The key requirement is that the oxidising agent must be capable of oxidising iodide ions ( $\text{I}^-$ ) to iodine ( $\text{I}_2$ ). This principle allows the technique to be adapted for analyzing many different oxidising agents.

General procedure for iodine/thiosulfate titrations

Step 1: Preparing the sample

Begin by adding a standard solution of sodium thiosulfate ( $\text{Na}_2\text{S}_2\text{O}_3$ ) to a burette. This will be your titrant.

Step 2: Creating the iodine solution

Prepare a solution containing the oxidising agent you wish to analyse. Using a pipette, transfer a measured volume of this solution into a conical flask. Then add an excess of potassium iodide (KI).

The oxidising agent reacts with the iodide ions to produce iodine, which gives the solution a distinctive yellow-brown colour.

Step 3: Initial titration

Begin titrating the solution with sodium thiosulfate from the burette. During this stage, the iodine is gradually reduced back to iodide ions ( $\text{I}^-$ ), and the brown colour begins to fade. Continue adding the thiosulfate solution until the mixture becomes a pale straw colour.

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This gradual colour change can make identifying the exact end point quite difficult, which is why we use a starch indicator in the next step. The pale straw colour indicates that most of the iodine has been reduced, but a small amount remains.

Step 4: Adding starch indicator

When the solution reaches a pale straw colour, add a small amount of starch indicator. The solution will immediately turn deep blue-black. This colour forms because starch creates a complex with iodine molecules.

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The starch indicator must only be added towards the end of the titration. If added too early, the starch-iodine complex may not fully break down, leading to an inaccurate end point. Always wait until the solution is pale straw colour before adding starch.

Step 5: Completing the titration

Continue adding sodium thiosulfate solution dropwise. The blue-black colour will gradually fade as more iodine is reduced. The end point is reached when the blue-black colour completely disappears, leaving a colourless solution. At this point, all the iodine has been reduced to iodide ions.

Step 6: Obtaining concordant results

Repeat the titration to obtain concordant (consistent) results, typically within 0.10 cm³ of each other. Calculate the mean titre for accurate analysis.

Analysis of household bleach

Background chemistry

Household bleach contains chlorate(I) ions ( $\text{ClO}^-$ ), commonly known as hypochlorite. The active ingredient is typically labelled as sodium chlorate(I) (NaClO) or sodium hypochlorite on product labels, showing the percentage content.

In this analysis, we calculate the concentration of $\text{ClO}^-$ ions in bleach, usually expressed in mol dm⁻³.

Chemical reactions involved

First reaction - producing iodine:

The chlorate(I) ions from the bleach react with iodide ions ( $\text{I}^-$ ) and hydrogen ions ( $\text{H}^+$ ) to form iodine:

$\text{ClO}^-\text{(aq)} + 2\text{I}^-\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)} + \text{H}_2\text{O}\text{(l)}$

Second reaction - titrating the iodine:

The iodine produced is then titrated with sodium thiosulfate solution:

$2\text{S}_2\text{O}_3^{2-}\text{(aq)} + \text{I}_2\text{(aq)} \rightarrow 2\text{I}^-\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)}$

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Key mole ratio:

From these equations, we can determine that:

1 mol $\text{ClO}^-$ produces 1 mol $\text{I}_2$
1 mol $\text{I}_2$ reacts with 2 mol $\text{S}_2\text{O}_3^{2-}$

Therefore: 1 mol ClO⁻ is equivalent to 2 mol S₂O₃²⁻

This 1:2 ratio is critical for all bleach calculations.

Practical procedure

Using a pipette, add 10.0 cm³ of bleach to a 250 cm³ volumetric flask and add water to prepare 250.0 cm³ of solution. This dilutes the bleach to a workable concentration.
Using a pipette, measure 25.0 cm³ of this diluted solution into a conical flask. Then add 10 cm³ of 1 mol dm⁻³ potassium iodide (KI) followed by sufficient 1 mol dm⁻³ hydrochloric acid (HCl) to acidify the solution. The HCl provides the $\text{H}^+$ ions needed for the reaction.
Using a burette, titrate this solution with a standard 0.0500 mol dm⁻³ solution of sodium thiosulfate ( $\text{Na}_2\text{S}_2\text{O}_3$ ).
Repeat the titration to obtain concordant results.
Analyse your results to determine the concentration of chlorate(I) ions in the bleach.

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Worked Example: Calculating the concentration of ClO⁻ in bleach

Let's assume a mean titre of 24.20 cm³ of sodium thiosulfate was required.

Step 1: Calculate the amount of S₂O₃²⁻ that reacted

Using the formula: $n = c \times \frac{V}{1000}$

$n(\text{S}_2\text{O}_3^{2-}) = 0.0500 \times \frac{24.20}{1000} = 1.21 \times 10^{-3} \text{ mol}$

Step 2: Determine the amounts of I₂ and ClO⁻ that reacted

Using the mole ratio: 1 mol $\text{ClO}^-$ is equivalent to 2 mol $\text{S}_2\text{O}_3^{2-}$

$n(\text{ClO}^-) = \frac{1}{2} \times n(\text{S}_2\text{O}_3^{2-})$

$n(\text{ClO}^-) = \frac{1}{2} \times 1.21 \times 10^{-3} = 6.05 \times 10^{-4} \text{ mol}$

Step 3: Determine the amount of ClO⁻ in the 250.0 cm³ solution

The 25.0 cm³ used in the titration contained $6.05 \times 10^{-4}$ mol $\text{ClO}^-$

Therefore, the 250.0 cm³ solution contains:

$n(\text{ClO}^-) = 6.05 \times 10^{-4} \times 10 = 6.05 \times 10^{-3} \text{ mol}$

Step 4: Determine the concentration of ClO⁻ in the bleach

The 250.0 cm³ solution was prepared using 10 cm³ of bleach, so:

10 cm³ bleach contains $6.05 \times 10^{-3}$ mol $\text{ClO}^-$ ions

1 dm³ bleach contains: $6.05 \times 10^{-3} \times 100 = 0.605$ mol NaClO

Therefore, the concentration of $\text{ClO}^-$ ions in the bleach is:

$\text{concentration} = 6.05 \times 10^{-3} \times 100 = 0.605 \text{ mol dm}^{-3}$

Analysis of copper in alloys

Background chemistry

Iodine/thiosulfate titrations can also be used to determine the copper content in copper(II) salts or copper alloys. For copper(II) salts, the compound is simply dissolved in water to produce $\text{Cu}^{2+}$ ions. For insoluble copper(II) compounds, they can be reacted with acid to form soluble $\text{Cu}^{2+}$ ions.

For copper alloys such as brass or bronze, the alloy is first reacted and dissolved in concentrated nitric acid. The resulting solution is then neutralised to form $\text{Cu}^{2+}$ ions in solution.

$\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)}$

Chemical reactions involved

First reaction - producing iodine:

When excess potassium iodide (KI) is added, the $\text{Cu}^{2+}$ ions react with iodide ions to form a white precipitate of copper(I) iodide (CuI) and release iodine. The mixture appears brown in colour:

$2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}$

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Oxidation states:

Reduction: Cu²⁺ (+2) → CuI (+1)
Oxidation: I⁻ (-1) → I₂ (0)

This is a redox reaction where copper is reduced and iodide is oxidised simultaneously.

Second reaction - titrating the iodine:

The iodine in the brown mixture is then titrated with a standard solution of sodium thiosulfate:

$2\text{S}_2\text{O}_3^{2-}\text{(aq)} + \text{I}_2\text{(aq)} \rightarrow 2\text{I}^-\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)}$

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Key mole ratio:

From the equation, 2 mol $\text{Cu}^{2+}$ produces 1 mol iodine, which reacts with 2 mol $\text{S}_2\text{O}_3^{2-}$

Therefore: 1 mol Cu²⁺ is equivalent to 1 mol S₂O₃²⁻

This is a simpler 1:1 ratio compared to the bleach analysis, making calculations more straightforward.

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Worked Example: Analysis of brass

Brass is an alloy of copper and zinc used in many ornaments and brass musical instruments. We can use an iodine/thiosulfate titration to find the percentage composition of copper and zinc in brass.

Given information:

A 0.500 g sample of brass is reacted with concentrated nitric acid to form a solution containing $\text{Cu}^{2+}$ and $\text{Zn}^{2+}$ ions
The solution is neutralised
Excess KI is added, causing $\text{Cu}^{2+}$ ions to react with $\text{I}^-$ ions, forming $\text{I}_2$
The iodine is titrated with 0.200 mol dm⁻³ sodium thiosulfate
25.20 cm³ of thiosulfate solution is required to reach the end point

Step 1: Calculate the amount of S₂O₃²⁻ that reacted

$n(\text{S}_2\text{O}_3^{2-}) = c \times \frac{V}{1000} = 0.200 \times \frac{25.20}{1000} = 5.04 \times 10^{-3} \text{ mol}$

Step 2: Determine the amounts of I₂ and Cu²⁺ that reacted

Using the 1:1 mole ratio: 1 mol $\text{Cu}^{2+}$ is equivalent to 1 mol $\text{S}_2\text{O}_3^{2-}$

$n(\text{Cu}^{2+}) = n(\text{S}_2\text{O}_3^{2-}) = 5.04 \times 10^{-3} \text{ mol}$

Step 3: Work out the percentages of Cu and Zn

Mass of Cu²⁺ in the sample:

$5.04 \times 10^{-3} \text{ mol} \times 63.5 \text{ g mol}^{-1} = 0.320 \text{ g}$

Mass of Zn in the sample:

$0.500 - 0.320 = 0.180 \text{ g}$

Percentage composition of Cu:

$\frac{0.320}{0.500} \times 100 = 64.0\%$

Percentage composition of Zn:

$\frac{0.180}{0.500} \times 100 = 36.0\%$

Therefore, the brass sample is approximately 64% copper and 36% zinc by mass.

Key exam tips

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Understanding mole ratios:

The most critical aspect of these calculations is identifying the correct mole ratio between the analyte and thiosulfate. Remember:

For $\text{ClO}^-$ : 1 mol $\text{ClO}^-$ ≡ 2 mol $\text{S}_2\text{O}_3^{2-}$
For $\text{Cu}^{2+}$ : 1 mol $\text{Cu}^{2+}$ ≡ 1 mol $\text{S}_2\text{O}_3^{2-}$

Using mean titres:

Always use the mean titre value from concordant results and the concentration of the standard sodium thiosulfate solution in your calculations.

Scaling up calculations:

When a sample is diluted (as in the bleach example), remember to scale up your answer to find the amount in the original solution.

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Common mistakes to avoid:

Adding starch indicator too early in the titration
Forgetting to account for dilution factors
Using incorrect mole ratios
Not using concordant results

Being aware of these common pitfalls will help you avoid errors in both practical work and calculations.

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Key Points to Remember:

Iodine/thiosulfate titrations involve oxidising iodide ions ( $\text{I}^-$ ) to iodine ( $\text{I}_2$ ) using an oxidising agent, then titrating the iodine with sodium thiosulfate ( $\text{Na}_2\text{S}_2\text{O}_3$ )
Starch indicator is added only when the solution becomes pale straw colour (near the end point) to give a blue-black colour, which disappears at the end point
For bleach analysis: The mole ratio is 1 mol ClO⁻ ≡ 2 mol S₂O₃²⁻, so divide the moles of thiosulfate by 2
For copper analysis: The mole ratio is 1 mol Cu²⁺ ≡ 1 mol S₂O₃²⁻, giving a simpler 1:1 relationship
This technique can determine the concentration of any oxidising agent capable of oxidising $\text{I}^-$ to $\text{I}_2$ , making it highly versatile for different analytical applications

Iodine/Thiosulfate Redox Titrations (OCR A-Level Chemistry A): Revision Notes