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AQA GCSE Chemistry revision notes

Chemical analysis

Section 4.8
14 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 8462 section 4.8

Checked against AQA 8462 section 4.8. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

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In the exam: No formula or equation sheet · calculator allowed in every paper

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4.8.1.1

Pure substances

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In chemistry, a pure substance contains one element or one compound and is not mixed with another substance.
  • Test purity by measuring a melting point or boiling point and comparing it with reliable reference data.
  • For example, a sample that melts sharply at the accepted melting point is consistent with a pure substance; a mixture usually changes the value and melts over a range.
  • A common error is to use the everyday meaning of pure, such as natural or unadulterated, instead of the chemical meaning.
Worked example

Reference data give the melting point of substance P as 64C64\,^\circ\mathrm{C}. Batch A melts sharply at 64C64\,^\circ\mathrm{C}, while batch B melts from 5858 to 61C61\,^\circ\mathrm{C}. Which batch is more likely to be pure? Explain.

  1. 1.Compare both results with the reference value. Batch A changes state at one temperature matching 64C64\,^\circ\mathrm{C}. Batch B changes state across a range below the reference value, which is evidence of a mixture.

Answer: Batch A; it melts at the accepted temperature and has a sharp melting point, whereas batch B melts over a different range.

Common mistakes

  • Don't fall into the trap of using the everyday meaning of pure, such as natural or unadulterated, instead of the chemical meaning.
  • Don't fall into the trap of calling a sample pure because it looks uniform, without comparing a measured melting or boiling point with reference data.

Exam tip

When data are supplied, compare both the measured value and whether melting occurs sharply or across a range.

Tier 1 · Easy

ORIGINAL

A sealed sample contains only solid sulfur. Decide whether it is a pure substance in the chemical sense and justify your decision.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A sample contains exactly two compounds and no elements. A student says it is chemically pure because it contains only compounds. Evaluate this statement.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A drink is advertised as 'pure fruit juice'. Analysis shows water, sugars, acids and flavour compounds. Explain why the label may be reasonable in everyday language but the drink is not pure in chemistry.

[4 marks]

Total for this question: 4

Your progress and exam materials
4.8.1.2

Formulations

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A formulation is a mixture designed to be a useful product, with every component included for a particular purpose.
  • Make a formulation by measuring and mixing its components carefully so the final product has the required properties.
  • For example, a paint may contain a pigment for colour, a solvent to control flow and a binder that leaves a solid coating.
  • A common error is to call every mixture a formulation; a formulation must have a designed use and controlled composition.
Worked example

A 250g250\,\mathrm{g} fertiliser formulation is 18%18\% nitrogen compound, 12%12\% potassium compound and the remainder filler. Calculate the mass of filler and explain one purpose of measuring the components accurately.

  1. 1.The named components total 18+12=30%18+12=30\%, so filler is 70%70\%. Its mass is 0.70×250=175g0.70\times250=175\,\mathrm{g}. Careful measurement keeps each batch at the designed composition, so it performs consistently.

Answer: Mass of filler =175g=175\,\mathrm{g}; accurate measurement gives the product its required composition or properties.

Common mistakes

  • Don't fall into the trap of calling every mixture a formulation; a formulation must have a designed use and controlled composition.
  • Don't fall into the trap of listing ingredients without linking each component or proportion to the product's required properties.

Exam tip

For an ‘identify a formulation’ question, state the designed purpose and the carefully controlled composition.

Tier 1 · Easy

ORIGINAL

A cleaning spray is made from measured amounts of solvent, detergent, fragrance and dye, each with a stated purpose. State why the spray is a formulation.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A paint contains carefully measured pigment, solvent and binder. Explain why the paint is a formulation and give the purpose of each named component.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Two batches of a medical cream contain the same chemicals. In batch X the components were weighed precisely; in batch Y their proportions varied. Evaluate which batch better fits the definition of a formulation.

[4 marks]

Total for this question: 4

4.8.1.3

Chromatography

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Chromatography separates a mixture because its substances distribute differently between a stationary phase and a mobile phase; a mixture may give several spots, while a pure compound gives one spot in every solvent.
  • For paper chromatography, draw a pencil origin line, add small sample spots, keep the solvent below the line, allow the solvent to rise, then mark the solvent front.
  • Calculate $R_f=\frac{\text{distance moved by the centre of the spot}}{\text{distance moved by the solvent front}}$; for example, 3.6/6.0=0.603.6/6.0=0.60.
  • A common error is to measure from the paper edge or spot boundary instead of from the origin to the centre of the spot.
  • Separation also depends on solubility in the mobile phase and attraction to the stationary phase, so changing the solvent can change the pattern.
A paper chromatogram showing the origin, separated spots and marked solvent front.
Worked example

A chromatogram has a solvent front 8.0cm8.0\,\text{cm} above the origin. A spot centre is 5.6cm5.6\,\text{cm} above the origin. Calculate its RfR_f value.

  1. 1.Use Rf=distance moved by spotdistance moved by solventR_f=\dfrac{\text{distance moved by spot}}{\text{distance moved by solvent}}.
  2. 2.Substitute distances measured from the origin: Rf=5.6÷8.0R_f=5.6\div8.0.
  3. 3.Evaluate the dimensionless ratio: Rf=0.70R_f=0.70.

Answer: Rf=0.70R_f=0.70

Common mistakes

  • Don't fall into the trap of measuring from the paper edge or spot boundary instead of from the origin to the centre of the spot.
  • Don't fall into the trap of allowing the solvent to cover the origin line, so the samples dissolve directly into the solvent reservoir.

Exam tip

For an RfR_f calculation, show both distances measured from the origin and give a ratio between 00 and 11.

Tier 1 · Easy

ORIGINAL

On a chromatogram, a pigment centre is 4.2cm4.2\,\mathrm{cm} above the origin and the solvent front is 7.0cm7.0\,\mathrm{cm} above the origin. Calculate the pigment's RfR_f value.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Describe how to use paper chromatography to find out whether a purple pen ink contains more than one soluble dye. Include two setup details that prevent misleading results.

[5 marks]

Total for this question: 5

Tier 3 · Hard

ORIGINAL

A solvent front moves 8.0cm8.0\,\mathrm{cm}. An unknown gives spots at 2.4cm2.4\,\mathrm{cm} and 5.6cm5.6\,\mathrm{cm}. Reference RfR_f values in this solvent are: J 0.300.30, K 0.550.55, L 0.700.70. Identify the substances present and state why this evidence does not prove the sample contains only those substances.

[5 marks]

Total for this question: 5

4.8.2.1

Test for hydrogen

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Hydrogen is identified by the characteristic pop produced when it burns rapidly. Hold a burning splint at the open end of the test tube containing the collected gas.
  • For example, a gas that gives a pop with a lit splint has a positive hydrogen-test result.
  • A common error is to insert a glowing splint; that is the test for oxygen, not hydrogen.
  • Only a small collected sample should be tested because hydrogen burns rapidly in oxygen.
  • The splint stays outside the test tube opening.
Worked example

Give the procedure and positive observation for confirming that gas collected from a metal-acid reaction is hydrogen.

  1. 1.Keep the flame at the open end of the container rather than pushing it deep inside. A pop sound is the required positive observation.

Answer: Place a burning splint at the tube opening; hydrogen burns with a pop.

Common mistakes

  • Don't fall into the trap of inserting a glowing splint; that is the test for oxygen, not hydrogen.
  • Don't fall into the trap of reporting only that the gas burns, rather than the characteristic squeaky pop at the mouth of the tube.

Exam tip

State ‘burning splint’ and ‘squeaky pop’ together; the observation earns the identification.

Tier 1 · Easy

ORIGINAL

A colourless gas makes a pop when tested at the mouth of its tube with a lit splint. Identify the gas.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A student tests a tube before it has filled reasonably fully with the collected gas. Explain why this could give an unreliable negative result in the hydrogen test.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A student tests a gas using a glowing splint and sees no change, then concludes that hydrogen is absent. Evaluate the conclusion and describe the correct confirmation test.

[3 marks]

Total for this question: 3

4.8.2.2

Test for oxygen

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Oxygen supports combustion and is identified when a glowing splint relights. Insert a glowing wooden splint into the test tube containing the gas.
  • For example, a splint with no flame that bursts back into flame gives a positive oxygen result.
  • A common error is to use a burning splint and look for a pop, which tests for hydrogen.
  • The test uses oxygen's ability to support combustion and should be performed on a collected gas sample.
  • Relighting is the required positive observation.
Worked example

Describe a test that distinguishes oxygen from a gas that does not support combustion. State the positive result.

  1. 1.First blow out the flame so the splint is glowing. Put it into the gas sample. Relighting is the diagnostic result for oxygen; merely remaining warm is not sufficient.

Answer: Insert a glowing splint; it relights in oxygen.

Common mistakes

  • Don't fall into the trap of using a burning splint and looking for a pop, which tests for hydrogen.
  • Don't fall into the trap of describing the splint as burning before insertion, rather than glowing with no visible flame.

Exam tip

The exact positive observation is that a glowing splint relights, not merely that it glows more brightly.

Tier 1 · Easy

ORIGINAL

A glowing splint returns to flame inside a jar of colourless gas. Name the gas.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A gas causes a glowing splint to burst back into flame. Identify the gas and state the property demonstrated by this test.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Three observations are reported for one gas: a lit splint gives no pop, limewater stays clear, and a glowing splint relights. Use all three results to identify the gas and explain which observation is decisive.

[4 marks]

Total for this question: 4

4.8.2.3

Test for carbon dioxide

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Carbon dioxide is identified because it forms a fine white precipitate in limewater, making the liquid look milky or cloudy.
  • Bubble the gas through, or shake it with, aqueous calcium hydroxide solution.
  • For example, clear limewater becoming cloudy is a positive result for carbon dioxide under this test.
  • A common error is to report that limewater becomes colourless; the required observation is that it turns milky or cloudy.
  • The cloudiness is caused by insoluble calcium carbonate, so ‘white precipitate forms’ is also a valid observation.
Worked example

State the reagent and the expected visible change when testing a gas sample for carbon dioxide.

  1. 1.Pass the gas through limewater or shake the two together. Name both the reagent and the formation of cloudiness for a complete answer.

Answer: Use limewater, aqueous calcium hydroxide; it changes from clear to milky or cloudy.

Common mistakes

  • Don't fall into the trap of reporting that limewater becomes colourless; the required observation is that it turns milky or cloudy.
  • Don't fall into the trap of naming calcium hydroxide but omitting the observation that a white precipitate makes the limewater cloudy.

Exam tip

Give reagent and observation as a pair: limewater changes from colourless to milky or cloudy.

Tier 1 · Easy

ORIGINAL

An unknown gas is shaken with clear limewater, which becomes cloudy. Identify the gas.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Carbon dioxide makes clear limewater turn cloudy. Explain what causes the cloudiness and name the solid formed.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A learner writes: 'Carbon dioxide is confirmed because it extinguishes a flame.' Improve this claim to give the specification test, its reagent and its positive observation.

[3 marks]

Total for this question: 3

4.8.2.4

Test for chlorine

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Chlorine gas bleaches damp litmus paper white. Expose damp litmus paper to the gas while using appropriate small-scale safety precautions because chlorine is toxic.
  • For example, litmus losing all its colour and becoming white is the positive chlorine result.
  • A common error is to use dry litmus paper; moisture is required for the specified test.
  • A positive result is bleaching rather than a permanent acid-base indicator colour, so the final white appearance is essential.
  • The gas must not be inhaled.
Worked example

Describe how litmus paper is used to test for chlorine and state the observation that confirms a positive result.

  1. 1.Moisten the litmus paper before exposing it to the gas. Record complete loss of colour to white, not simply a colour change between red and blue.

Answer: Use damp litmus paper; chlorine bleaches it white.

Common mistakes

  • Don't fall into the trap of using dry litmus paper; moisture is required for the specified test.
  • Don't fall into the trap of reporting only an initial colour change and omitting that the paper is ultimately bleached white.

Exam tip

Include both test conditions and result: damp litmus paper is bleached white by chlorine.

Tier 1 · Easy

ORIGINAL

Damp litmus paper loses its colour and becomes white in an unknown gas. Identify the gas.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A student says a gas must be chlorine because damp litmus paper turned white. Another student says any acidic gas would do this. Explain which student is correct.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Two students test the same gas. One uses dry blue litmus and sees no change; the other uses damp litmus and it turns white. Explain which result should be used and what conclusion follows.

[3 marks]

Total for this question: 3

4.8.3.1

Flame tests (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Flame tests identify some metal ions: Li+ crimson, Na+ yellow, K+ lilac, Ca2+ orange-red and Cu2+ green.
  • Place a small amount of a clean sample into a non-luminous flame and compare the observed colour with the known colours.
  • For example, a lilac flame indicates potassium ions, whereas an orange-red flame indicates calcium ions.
  • A common error is to treat a flame test as reliable for every mixture; a strong colour, especially sodium yellow, can mask another ion.
  • Contamination must be controlled by using clean apparatus because traces of sodium can produce an intense yellow flame.
Worked example

Samples M and N give a green flame and an orange-red flame respectively. Identify the metal ion in each sample.

  1. 1.Use the required colour associations. Green corresponds to copper compounds, and orange-red corresponds to calcium compounds.

Answer: M contains copper(II) ions, Cu2+. N contains calcium ions, Ca2+.

Common mistakes

  • Don't fall into the trap of treating a flame test as reliable for every mixture; a strong colour, especially sodium yellow, can mask another ion.
  • Don't fall into the trap of confusing lithium's crimson flame with calcium's orange-red flame when matching an unknown.

Exam tip

Record the observed flame colour first, then match it to one named ion from the specified colour list.

Tier 1 · Easy

ORIGINAL

A compound produces a crimson flame. Identify the metal ion present.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A pure potassium salt unexpectedly gives an intense yellow flame instead of a clear lilac flame. Suggest a likely cause and one improvement to the procedure.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A mixture known to contain two metal ions gives only an intense yellow flame. State one ion supported by the observation and explain why the second ion cannot be identified confidently from this test alone.

[3 marks]

Total for this question: 3

4.8.3.2

Metal hydroxides (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • With sodium hydroxide, Cu2+ gives a blue precipitate, Fe2+ green and Fe3+ brown.
  • Add sodium hydroxide solution dropwise, record any precipitate, then add excess when distinguishing aluminium from other white precipitates.
  • Al3+, Ca2+ and Mg2+ form white hydroxide precipitates, but only aluminium hydroxide dissolves in excess sodium hydroxide.
  • A common error is to claim every white precipitate dissolves in excess; calcium and magnesium hydroxides remain.
  • The coloured precipitates identify copper(II), iron(II) and iron(III) directly, whereas the three white precipitates require careful discrimination.
Worked example

Solutions A and B each form a white precipitate when a little sodium hydroxide is added. The precipitate from A dissolves after excess sodium hydroxide is added, but B's remains. What can be concluded about A, and why is B not fully identified?

  1. 1.All three candidate ions can first give a white precipitate. Only aluminium hydroxide dissolves in excess sodium hydroxide, so A is aluminium. An insoluble white precipitate is consistent with either calcium or magnesium, leaving B ambiguous.

Answer: A contains aluminium ions; B could contain calcium or magnesium ions, so the observations do not distinguish those two.

Common mistakes

  • Don't fall into the trap of claiming every white precipitate dissolves in excess; calcium and magnesium hydroxides remain.
  • Don't fall into the trap of identifying a white precipitate as aluminium without checking whether it dissolves in excess sodium hydroxide.

Exam tip

For a white precipitate, state the excess-sodium-hydroxide result before distinguishing aluminium from calcium or magnesium.

Tier 1 · Easy

ORIGINAL

Adding sodium hydroxide solution to an unknown ionic solution forms a brown precipitate. Identify the metal ion.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Sodium hydroxide solution gives a green precipitate with sample P and a brown precipitate with sample Q. Identify the metal ion in each sample.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Three solutions are tested with sodium hydroxide. W gives a blue precipitate, X gives a green precipitate, and Y gives a white precipitate that dissolves in excess sodium hydroxide. Identify the metal ion in W, X and Y.

[3 marks]

Total for this question: 3

4.8.3.3

Carbonates (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Carbonate ions are tested by adding a dilute acid; a carbonate reacts to release carbon dioxide gas.
  • Pass the gas produced into limewater to confirm that it is carbon dioxide.
  • For example, effervescence followed by limewater turning cloudy is a positive sequence for carbonate ions.
  • A common error is to identify a carbonate from fizzing alone; the gas should be confirmed with limewater.
  • The confirmatory gas test matters because effervescence alone can be produced by reactions involving gases other than carbon dioxide.
Worked example

Describe a complete chemical test for carbonate ions in an unknown powder, including how the gaseous product is identified.

  1. 1.Add a dilute acid to the powder and collect the gas released. Bubble it through aqueous calcium hydroxide. Fizzing plus cloudy limewater provides the two-stage positive result.

Answer: Add dilute acid, then pass the gas into limewater; effervescence occurs and the limewater turns milky or cloudy.

Common mistakes

  • Don't fall into the trap of identifying a carbonate from fizzing alone; the gas should be confirmed with limewater.
  • Don't fall into the trap of naming hydrogen as the gas from the acid-carbonate reaction instead of carbon dioxide.

Exam tip

A complete carbonate test gives both stages: effervescence with dilute acid, then limewater turning cloudy.

Tier 1 · Easy

ORIGINAL

A solid fizzes when dilute acid is added, and the gas makes limewater cloudy. Name the ion detected in the solid.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

An unknown powder and a known carbonate are tested in parallel with dilute acid and limewater. Neither test makes the limewater cloudy. Explain why the unknown result cannot be used to rule out carbonate ions, and state what should be done next.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Powders C and D both fizz when dilute acid is added. The gas from C turns limewater cloudy, but the gas from D leaves limewater clear. Explain which powder has tested positive for carbonate ions and why fizzing alone is insufficient evidence.

[4 marks]

Total for this question: 4

4.8.3.4

Halides (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Halide ions are tested by acidifying the solution with dilute nitric acid and then adding silver nitrate solution.
  • Use clean samples and record the precipitate colour: chloride white, bromide cream and iodide yellow.
  • For example, a cream silver-halide precipitate identifies bromide ions under the specified test conditions.
  • A common error is to swap the cream and yellow results: silver bromide is cream, while silver iodide is yellow.
  • Acidification removes interfering ions, while nitric acid is chosen because nitrate ions do not form a precipitate with silver ions.
Worked example

Give the reagents, in order, used to test an aqueous sample for halide ions, and state the result for chloride ions.

  1. 1.First acidify the sample using dilute nitric acid. Then add aqueous silver nitrate. If chloride is present, insoluble silver chloride forms as a white precipitate.

Answer: Add dilute nitric acid, then silver nitrate solution; chloride ions give a white precipitate.

Common mistakes

  • Don't fall into the trap of swapping the cream and yellow results: silver bromide is cream, while silver iodide is yellow.
  • Don't fall into the trap of using hydrochloric acid to acidify, which introduces chloride ions and can create a misleading white precipitate.

Exam tip

Write the reagents in order—dilute nitric acid, then silver nitrate—before giving the precipitate colour.

Tier 1 · Easy

ORIGINAL

After dilute nitric acid and silver nitrate are added to a solution, a yellow precipitate forms. Identify the halide ion.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A student acidifies an unknown with dilute hydrochloric acid before adding silver nitrate solution. Explain why this can give a false positive for chloride ions and name the correct acid.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Samples R, S and T form white, cream and yellow precipitates respectively after the correct halide test. Identify all three ions and state the two reagents that must have been added, in order.

[5 marks]

Total for this question: 5

4.8.3.5

Sulfates (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Sulfate ions form a white precipitate when barium chloride solution is added under acidic conditions.
  • Acidify the sample with dilute hydrochloric acid, then add barium chloride solution and observe any precipitate.
  • For example, a white barium sulfate precipitate is the positive result for sulfate ions.
  • A common error is to report only that the mixture turns white without naming the formation of a precipitate.
  • The acidic conditions remove interfering carbonate ions before barium ions are added to test for sulfate.
Worked example

Describe the reagent sequence and positive observation for testing an unknown solution for sulfate ions.

  1. 1.Acidify the unknown with dilute hydrochloric acid before adding aqueous barium chloride. Record the formation of a white solid rather than merely a pale solution.

Answer: Add dilute hydrochloric acid, then barium chloride solution; a white precipitate forms if sulfate ions are present.

Common mistakes

  • Don't fall into the trap of reporting only that the mixture turns white without naming the formation of a precipitate.
  • Don't fall into the trap of using sulfuric acid to acidify, which adds sulfate ions and can produce a false positive.

Exam tip

State ‘dilute hydrochloric acid, then barium chloride’ and identify the white solid as a precipitate.

Tier 1 · Easy

ORIGINAL

An acidified solution forms a white precipitate when barium chloride solution is added. Identify the ion being tested.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A solution contains carbonate ions but no sulfate ions. Explain why adding barium chloride directly could be misleading and how acidifying first prevents this problem.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A student adds barium chloride directly to an unknown solution, sees a white precipitate and reports sulfate ions. Evaluate the procedure and give the complete test needed before accepting the identification.

[4 marks]

Total for this question: 4

4.8.3.6

Instrumental methods (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Instrumental methods detect and identify elements or compounds using measured signals rather than only visible chemical-test observations.
  • Choose an instrumental method when rapid analysis, high sensitivity or accurate measurement is important.
  • For example, a sensitive instrument can detect a component at a concentration too low to give a clear precipitate or colour change.
  • A common error is to list 'easy' as a specification advantage; the required comparison is that instrumental methods are accurate, sensitive and rapid.
  • Instrumental methods generate signals that can be compared with reference or calibration data to identify or measure substances.
Worked example

A laboratory must screen 180180 water samples in one day for a contaminant present at very low concentration. Give two reasons for choosing an instrumental method.

  1. 1.Link each circumstance to an advantage. The large number of samples requires speed, while the very small amount of contaminant requires sensitivity.

Answer: It is rapid enough for many samples and sensitive enough to detect a low concentration.

Common mistakes

  • Don't fall into the trap of listing 'easy' as a specification advantage; the required comparison is that instrumental methods are accurate, sensitive and rapid.
  • Don't fall into the trap of claiming instrumental analysis is always cheaper, even though the specification advantages are speed, sensitivity and accuracy.

Exam tip

When comparing methods, link ‘sensitive’ to small amounts and ‘accurate’ to reliable quantitative measurement.

Tier 1 · Easy

ORIGINAL

State one advantage of an instrumental method over a chemical test based on a visible colour change.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

An instrument reads 5.02mgdm35.02\,\mathrm{mg\,dm^{-3}} for a certified 5.00mgdm35.00\,\mathrm{mg\,dm^{-3}} standard and can detect 0.001mgdm30.001\,\mathrm{mg\,dm^{-3}}. State which result demonstrates accuracy and which demonstrates sensitivity.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

For a certified 10.0mgdm310.0\,\mathrm{mg\,dm^{-3}} standard, method A reports 12mgdm312\,\mathrm{mg\,dm^{-3}} after 2525 minutes and misses very dilute samples. Method B reports 10.1mgdm310.1\,\mathrm{mg\,dm^{-3}} after 4040 seconds and detects every sample. Compare the methods using the three specification advantages of instrumental analysis.

[4 marks]

Total for this question: 4

4.8.3.7

Flame emission spectroscopy (chemistry only)

Notes
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Explanation

  • In flame emission spectroscopy, a solution sample enters a flame and the emitted light passes through a spectroscope.
  • Identify metal ions by comparing the positions of lines in the sample spectrum with a reference set recorded in the same form.
  • Line intensity can be compared with calibration data to measure concentration; for example, an intensity halfway between two standards gives an intermediate concentration when the calibration is linear.
  • A common error is to use line brightness to identify the ion; line position identifies the ion, while intensity is used for concentration.
  • A sample can contain several metal ions because each element contributes its characteristic set of emission lines.
Worked example

A sodium calibration is linear through the origin. An intensity of 2424 units corresponds to 3.0mgdm33.0\,\mathrm{mg\,dm^{-3}}. A sample gives 4040 units under identical conditions. Calculate its sodium-ion concentration.

  1. 1.For a linear calibration through the origin, concentration is proportional to intensity. Calculate 3.0×4024=5.0mgdm33.0\times\frac{40}{24}=5.0\,\mathrm{mg\,dm^{-3}}.

Answer: 5.0mgdm35.0\,\mathrm{mg\,dm^{-3}}

Common mistakes

  • Don't fall into the trap of using line brightness to identify the ion; line position identifies the ion, while intensity is used for concentration.
  • Don't fall into the trap of comparing a sample with reference spectra recorded on different axes or under unsuitable conditions.

Exam tip

Use line position for identity and a calibration relationship between intensity and concentration for quantity.

Tier 1 · Easy

ORIGINAL

Reference lines occur at 589nm589\,\mathrm{nm} for sodium and 671nm671\,\mathrm{nm} for lithium. A sample has one line at 671nm671\,\mathrm{nm}. Identify the metal ion.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A flame emission spectrum contains lines at the reference positions for lithium and sodium. The sodium line is brighter. State what can be concluded about the ions present and why the brighter line alone does not prove that sodium has the higher concentration.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A solution produces emission lines at 589nm589\,\mathrm{nm} and 766nm766\,\mathrm{nm}. References assign these to sodium and potassium. At 766nm766\,\mathrm{nm}, standards of 2.02.0 and 6.0mgdm36.0\,\mathrm{mg\,dm^{-3}} give intensities 1515 and 4747 units; the sample gives 3131 units. Identify both ions and estimate the potassium-ion concentration, assuming a linear response.

[5 marks]

Total for this question: 5

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