Volumetric Analysis

Indicators and Beyond: Making Titrations Work

Acid Base Indicators

Why Indicators Change Colour

In an acid–base titration, an indicator is used to show when the reaction has reached its endpoint, especially when both solutions are colourless and you cannot see the change directly. Indicators work because of a key property found in many coloured organic compounds. These molecules often contain multiple double bonds and, in some cases, lone pairs of electrons that are not involved in bonding. This structure allows them to interact with hydrogen ions (H⁺) and form a different chemical species with a different colour.

This is why indicators change colour. In alkaline conditions, there are very few hydrogen ions available, so the indicator remains in one form. As acid is added and the solution becomes more acidic, hydrogen ions begin to interact with the indicator molecules. This causes a structural change, which leads to a visible colour change. Interestingly, some naturally occurring pigments in plants behave in a similar way, which is why simple indicators can even be made by extracting colour from plant materials.

Choosing the right indicator is not as simple as picking any substance that changes colour. There are two key factors that need to be considered.

The first is visibility. The colour change must be clear and easy to recognise. In practice, this is not always straightforward. Some indicators change between colours that are quite similar, such as blue to purple in litmus, or yellow to orange in methyl orange. These changes can be subtle and harder to detect, particularly under certain lighting conditions or for individuals with colour vision deficiencies. A good indicator should produce a distinct and obvious change so that the endpoint can be identified with confidence.

The second factor is the pH at which the indicator changes colour. Ideally, you might expect an indicator to change exactly at pH 7, which is the true neutral point. However, in reality, most indicators change over a range of pH values rather than at a single point. For example, methyl orange changes colour in a slightly acidic range, typically between pH 3.0 and 4.5. On the other hand, phenolphthalein changes colour in a slightly alkaline range, usually between pH 8.0 and 10.0.

This means that the choice of indicator depends on the type of titration being carried out. You need an indicator whose colour change occurs as close as possible to the equivalence point of the reaction. Selecting the correct indicator is therefore essential for obtaining accurate and reliable results in titration experiments.

Robert Boyle

The Scientist Who Helped Turn Colour Change into Evidence

Robert Boyle lived in the 1600s, long before students had pH meters, digital sensors or even a modern periodic table. Chemistry at the time was still partly mixed with old alchemical ideas, but Boyle helped push it towards proper experimental science.

One of his important contributions was showing that certain coloured plant extracts could be used to identify acids and alkalis. He noticed that some natural substances changed colour when added to different chemical solutions. This was an early version of what we now call an acid–base indicator.

That might sound simple, but it was a massive shift. Boyle helped chemists realise that colour change was not just visual drama; it could be used as evidence. The same idea sits behind indicators in titrations today. When phenolphthalein turns from colourless to pink, or methyl orange changes colour in acidic conditions, the indicator is acting as a chemical signal.

Boyle’s work helped chemistry move from guesswork to observation, testing and evidence. In titration, every colour change at the endpoint is part of that same tradition: using visible evidence to measure what cannot be seen directly.

Back Titrations

Titrations Beyond Acid and Alkali Reactions

Titrations are not limited to reactions between acids and alkalis. In fact, any reaction that produces a visible colour change can be used in volumetric analysis. This opens up a much wider range of applications than students often expect.

Some reagents are particularly useful because they are self-indicating. This means they naturally change colour during the reaction, so no additional indicator is needed. A key example is potassium manganate(VII). In solution, it has an intense purple colour, but as it reacts and is reduced to manganese(II), the solution becomes pale pink. This built-in colour change makes it very easy to identify the endpoint of the reaction.

In other reactions, however, the colour change is not as clear, and an indicator is needed to make the endpoint visible. For example, when potassium dichromate(VI) is reduced to chromium(III), the colour changes from orange to green. While this is technically a colour change, it can be quite difficult to detect accurately in practice. Adding an appropriate indicator can sharpen this transition, making the titration more reliable and easier to interpret.

Another important technique you need to understand is the back titration. This is used when a direct titration is not practical, often because the reaction is too slow or does not produce a clear endpoint.

Imagine you want to determine the atomic mass of an unknown element, X, which reacts with sulphuric acid according to the equation:

X + H₂SO₄ → X₂SO₄ + H₂

You start by reacting a known mass of X with an excess of sulphuric acid of known concentration and volume. Because the acid is in excess, you know that all of X will react completely. However, this means some acid will remain unreacted.

To find out how much acid actually reacted, you perform a second titration on the remaining sulphuric acid. This allows you to measure how much acid is left over after the reaction.

By comparing the initial amount of acid added with the amount remaining, you can calculate how much sulphuric acid reacted. From the balanced equation, you can then determine the number of moles of X that must have reacted, since the mole ratio between X and H₂SO₄ is fixed.

At this point, you have both the mass of X you started with and the number of moles of X that reacted. This is enough to calculate the atomic mass of X.

Back titrations might feel like an extra step, but they are extremely powerful. They allow you to analyse reactions that would otherwise be difficult to measure directly, and they are a good example of how chemists design methods around practical limitations.

Practice Question

Collection of Carbon Dioxide

Beachy Head, close to Eastbourne in East Sussex, is a well-known chalk cliff that dominates the scenery along this part of the south coast. The soft chalk is gradually eroded by the sea, so areas near the cliff edge are cordoned off when they become unstable and may eventually collapse into the sea. Continuous monitoring of the cliff is needed for public safety.

A soil sample taken from the area contains a large amount of chalk, which is mainly calcium carbonate, together with clay, other silicate minerals and decomposing organic matter. Assuming that chalk is the only substance in the sample that reacts with dilute acid, there are several possible methods for determining the percentage of chalk in the soil sample.

1 Collection of carbon dioxide

A 1.00 g sample of soil is reacted with excess hydrochloric acid and the carbon dioxide evolved is collected over water. After cooling to room temperature, 89 cm³ of carbon dioxide is obtained.

(a) Write a balanced equation for the reaction between calcium carbonate and hydrochloric acid.

(b) Calculate the amount in moles of carbon dioxide obtained.

(c) Deduce the number of moles of calcium carbonate that react.

(d) Calculate the mass of calcium carbonate that reacts.

(e) Calculate the percentage of chalk in the soil sample.

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Model Answer

Carbonate Stoichiometry

(a) Balanced equation

Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide and water.

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

The carbonate ion reacts with acid to produce carbon dioxide, so the gas collected can be used to calculate how much calcium carbonate was present.

(b) Amount in moles of carbon dioxide

At room temperature and pressure, 1 mol of gas occupies 24 000 cm³.

amount = volume ÷ molar gas volume

amount of CO₂ = 89 ÷ 24 000 = 3.71 × 10⁻³ mol

Therefore, the amount of carbon dioxide obtained is 3.71 × 10⁻³ mol.

(c) Moles of calcium carbonate that react

From the balanced equation:

1 mol CaCO₃ produces 1 mol CO₂

Therefore, the mole ratio between CaCO₃ and CO₂ is 1 : 1.

moles of CaCO₃ = 3.71 × 10⁻³ mol

(d) Mass of calcium carbonate that reacts

Relative formula mass of CaCO₃:

CaCO₃ = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹

mass = amount × molar mass

mass of CaCO₃ = 3.71 × 10⁻³ × 100.1 = 0.371 g

Therefore, the mass of calcium carbonate that reacts is 0.371 g.

(e) Percentage of chalk in the soil sample

The original soil sample has a mass of 1.00 g.

percentage of chalk = (mass of CaCO₃ ÷ mass of soil sample) × 100

percentage = (0.371 ÷ 1.00) × 100 = 37.1%

Therefore, the percentage of chalk in the soil sample is 37.1%.

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Practice Question

Back Titration

10.00 g of soil is added to 100.0 cm³ of hydrochloric acid with a concentration of 1.00 mol dm⁻³. Once all the chalk has reacted, the remaining solid residue is allowed to settle. Excess acid remains in the solution.

Using a pipette, 25.00 cm³ of this solution is carefully removed and titrated against 0.0500 mol dm⁻³ sodium carbonate solution. A volume of 23.10 cm³ is required to neutralise the acid.

(a) Write a balanced equation for the reaction between sodium carbonate and hydrochloric acid.

(b) Calculate the amount in moles of sodium carbonate in the 23.10 cm³ required by the titration.

(c) Use your answer to b to deduce the amount in moles of hydrochloric acid contained in 25.00 cm³ of the solution.

(d) Deduce the amount in moles in 100.0 cm³ of the hydrochloric acid after it reacts with the soil sample.

(e) Calculate the amount in moles of hydrochloric acid in the 100.0 cm³ before it reacts with the soil sample.

(f) Use your answers to d and e to deduce the amount in moles of hydrochloric acid that react with the soil sample.

(g) Write the equation for the reaction between calcium carbonate and hydrochloric acid.

(h) Use the equation in g and your answer to f to state the amount in moles of calcium carbonate in 10.00 g of soil.

(i) Deduce the mass of calcium carbonate in 10.00 g of soil.

(j) Calculate the percentage of chalk in the soil sample.

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Model Answer

Back Titration Calculation

(a) Balanced equation

Sodium carbonate reacts with hydrochloric acid to form sodium chloride, carbon dioxide and water.

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)

(b) Moles of sodium carbonate

Convert the titre volume into dm³:

23.10 cm³ = 0.02310 dm³

amount = concentration × volume

amount of Na₂CO₃ = 0.0500 × 0.02310 = 1.155 × 10⁻³ mol

(c) Moles of hydrochloric acid in 25.00 cm³

From the equation, 1 mol of Na₂CO₃ reacts with 2 mol of HCl.

amount of HCl = 2 × 1.155 × 10⁻³ = 2.310 × 10⁻³ mol

This is the amount of excess hydrochloric acid in the 25.00 cm³ sample removed by pipette.

(d) Moles of hydrochloric acid in 100.0 cm³ after reaction

The titrated sample was 25.00 cm³, while the full solution was 100.0 cm³.

Scale up by a factor of 4:

2.310 × 10⁻³ × 4 = 9.240 × 10⁻³ mol

Therefore, the amount of excess HCl remaining after reaction with the soil is 9.240 × 10⁻³ mol.

(e) Initial moles of hydrochloric acid

Convert the original acid volume into dm³:

100.0 cm³ = 0.1000 dm³

amount of HCl = 1.00 × 0.1000 = 0.100 mol

(f) Moles of hydrochloric acid that reacted with the soil

The amount of acid that reacted is the initial amount minus the excess amount left after reaction.

0.1000 − 0.009240 = 0.09076 mol

Therefore, 0.09076 mol of hydrochloric acid reacted with the chalk in the soil sample.

(g) Calcium carbonate and hydrochloric acid equation

Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide and water.

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

(h) Moles of calcium carbonate in 10.00 g of soil

From the equation, 1 mol of CaCO₃ reacts with 2 mol of HCl.

amount of CaCO₃ = 0.09076 ÷ 2 = 0.04538 mol

(i) Mass of calcium carbonate

Relative formula mass of CaCO₃:

40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹

mass = amount × molar mass

mass of CaCO₃ = 0.04538 × 100.1 = 4.54 g

(j) Percentage of chalk in the soil sample

The original soil sample had a mass of 10.00 g.

percentage of chalk = (4.54 ÷ 10.00) × 100 = 45.4%

Therefore, the percentage of chalk in the soil sample is 45.4%.

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Practice Question

Comparing Experimental Methods

Consider the experimental methods used in Questions 1 and 2.

Suggest reasons for the difference in the answers for the percentage of chalk in the soil sample obtained by the two methods.

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Model Answer

Possible Sources of Difference

The two methods may give different percentages of chalk because each method has different experimental uncertainties and possible sources of error.

In Question 1, carbon dioxide is collected over water. Some carbon dioxide may dissolve in the water, so the measured volume of CO₂ could be too low. This would make the calculated amount of calcium carbonate too low.

Some carbon dioxide may also escape before it is collected, especially when the acid is first added and the reaction is rapid. This would also reduce the measured gas volume.

The gas volume may also be affected if the apparatus is not airtight, if the gas is not fully cooled to room temperature, or if the gas is collected while still mixed with water vapour.

In Question 2, the back titration depends on accurate measurement of acid concentration, pipette volume, titre volume and complete neutralisation. Any error in the titre will affect the calculated amount of excess hydrochloric acid.

The soil sample may also be uneven, so the 1.00 g sample used in Question 1 and the 10.00 g sample used in Question 2 may not contain exactly the same proportion of chalk.

Another possible reason is that other substances in the soil could react with hydrochloric acid. If this happens, the back titration method may overestimate the amount of calcium carbonate, because it assumes that only chalk reacts with the acid.

Overall, the difference is likely due to a combination of gas loss or CO₂ dissolving in the gas collection method, and titration or sampling uncertainties in the back titration method.

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Practice Question

Accuracy of Alternative Methods

Two other possible methods for determining the percentage of chalk in a sample of the soil are:

• A weighed sample of the soil is heated strongly and the volume of carbon dioxide evolved is measured.

• A weighed sample of the soil is reacted with dilute hydrochloric acid until all the chalk dissolves. The solution is filtered and the mass of the unreacted residue is found.

Consider each of these methods and suggest, with reasons, whether they would be likely to produce an accurate result.

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Model Answer

Evaluating the Methods

Method 1: heating the soil and measuring carbon dioxide

This method is unlikely to give a very accurate result because the carbon dioxide produced may not come only from chalk.

Calcium carbonate decomposes on strong heating:

CaCO₃(s) → CaO(s) + CO₂(g)

However, the soil also contains decomposing organic matter. When heated strongly, this organic material may also produce carbon dioxide.

This would make the volume of carbon dioxide too high, so the calculated percentage of chalk would be overestimated.

There may also be practical errors because some carbon dioxide could escape before collection, or the calcium carbonate may not fully decompose if the heating is insufficient.

Method 2: reacting with hydrochloric acid and weighing the residue

This method may give a reasonable estimate, but it is still not guaranteed to be fully accurate.

The method assumes that all the chalk reacts and dissolves, while all the other materials remain as an unreacted solid residue.

Calcium carbonate reacts with hydrochloric acid:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

If some chalk remains trapped in the residue or does not fully react, the residue mass would be too high and the calculated chalk percentage would be too low.

If some non-chalk material dissolves in the acid, the residue mass would be too low and the calculated chalk percentage would be too high.

The residue must also be washed, dried and weighed carefully. If it is not fully dried, its mass will be too high, causing the chalk percentage to be underestimated.

Overall judgement

The heating method is likely to be less accurate because carbon dioxide can be produced from substances other than calcium carbonate, especially organic matter.

The residue method could be more reliable, but only if the chalk reacts completely and the residue is carefully filtered, washed and dried before weighing.

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Author

Luke Edwards-Stuart

Chemistry teacher, curriculum specialist and educational leader. Luke runs the free student resource website a-levelchemistry.co.uk, supporting students with high-quality Chemistry content.

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