A student tests three unknown solutions to identify the halide ions present. They add a few drops of silver nitrate solution to each unknown solution and observe the colour of the precipitate formed. Solution A gives a white precipitate, Solution B gives a cream precipitate, and Solution C gives a yellow precipitate. Explain how these observations allow the student to identify which halide ion is present in each solution.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Silver nitrate solution reacts with halide ions to form silver halide precipitates. Silver chloride is white, silver bromide is cream, and silver iodide is yellow.
Model answer (4 marks)
Solution A gives a white precipitate, so it contains chloride ions (Cl⁻). Solution B gives a cream precipitate, so it contains bromide ions (Br⁻). Solution C gives a yellow precipitate, so it contains iodide ions (I⁻). The different colours arise because each halide ion reacts with Ag⁺ to form a different silver halide (AgCl, AgBr, AgI) which have different solubilities and colours.
Examiner tips
- Use the colour–halide correlation directly; state the ion for each solution. Include the reason for colour differences. Keep answer concise and use correct terminology.
Mark scheme (4 marks)
- Solution A contains chloride ions (Cl⁻)
- Solution B contains bromide ions (Br⁻)
- Solution C contains iodide ions (I⁻)
- The different coloured precipitates are formed because each halide ion reacts with silver ions to form a different, insoluble silver halide compound
Key terms in this question
Related
- All OCR GCSE Chemistry A: Gateway Science (J248) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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