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Qualitative analysis questions test more than memory. You need to know what each reagent does, why tests are done in a particular order, and how to combine several observations into one conclusion. These twenty questions build from single tests up to full unknowns. Try each one before reading the answer.
If you need a refresher first, work through identifying ions: a qualitative analysis flowchart and flame test colours.
Part A: Single tests
1. A solid gives a lilac flame. Which ion is present, and why is the result sometimes hard to see?
Answer: Potassium, K⁺. Traces of sodium give an intense yellow flame that masks the pale lilac. Viewing the flame through blue cobalt glass absorbs the yellow light so the lilac can be seen.
2. Sodium hydroxide solution is added to a solution. A blue precipitate forms. Name the ion and write the ionic equation.
Answer: Copper(II), Cu²⁺. Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
3. A solution gives a white precipitate with a few drops of NaOH, which dissolves when more NaOH is added. List the possible cations.
Answer: Al³⁺, Zn²⁺ or Pb²⁺. Their hydroxides are amphoteric and dissolve in excess hydroxide to form soluble complex ions such as [Al(OH)₄]⁻ and [Zn(OH)₄]²⁻. See amphoteric substances.
4. Why must a solution be acidified with dilute nitric acid, rather than hydrochloric acid, before silver nitrate is added?
Answer: The acid removes carbonate ions, which would otherwise form a precipitate of silver carbonate. Hydrochloric acid can’t be used because it adds chloride ions, which would always give a white AgCl precipitate, a false positive for chloride.
5. A gas turns limewater milky. Identify the gas and write the equation for the limewater reaction.
Answer: Carbon dioxide. Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l) The milkiness is a fine suspension of calcium carbonate.
6. A solution is acidified with HCl and barium chloride is added. A white precipitate forms. What does this show?
Answer: Sulfate ions are present: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). The HCl ensures the precipitate is not barium carbonate or barium sulfite, which dissolve in acid.
7. How can you distinguish between a white precipitate of AgCl and a cream precipitate of AgBr, if the colours look similar?
Answer: Add dilute ammonia solution. AgCl dissolves; AgBr does not (it dissolves only in concentrated ammonia). AgI is insoluble even in concentrated ammonia.
Part B: Explaining observations
8. A pale green solution gives a green precipitate with NaOH. After a few minutes, the top of the precipitate turns orange-brown. Explain.
Answer: The solution contains Fe²⁺. The green precipitate is iron(II) hydroxide, Fe(OH)₂. At the surface, it’s oxidised by oxygen from the air to orange-brown iron(III) hydroxide, Fe(OH)₃. This is a redox reaction: iron is oxidised from +2 to +3. See oxidation and reduction.
9. A student tests a solution for nitrate by warming with NaOH and aluminium powder. Damp red litmus turns blue. Why can’t the student conclude that nitrate is present without a further check?
Answer: Ammonium ions would also release ammonia when warmed with NaOH, turning the litmus blue. The student must first warm the solution with NaOH alone. If no ammonia is given off, ammonium is absent, and ammonia produced after adding aluminium must come from nitrate.
10. Why is distilled water, rather than tap water, used to dissolve samples?
Answer: Tap water contains dissolved ions such as Cl⁻, SO₄²⁻, Ca²⁺ and Mg²⁺. These could give positive results (for example a white precipitate with silver nitrate) that don’t come from the sample.
11. Zinc and aluminium ions both give white precipitates that dissolve in excess NaOH. Describe a test to tell them apart.
Answer: Add aqueous ammonia drop by drop, then in excess. Both form white precipitates at first. The zinc hydroxide precipitate dissolves in excess ammonia, forming the complex ion [Zn(NH₃)₄]²⁺. Aluminium hydroxide does not dissolve.
12. Why is the carbonate test usually done before the sulfate and halide tests?
Answer: Carbonate ions react with Ba²⁺ and Ag⁺ to form white or pale precipitates (BaCO₃, Ag₂CO₃) that could be mistaken for positive sulfate or halide results. Knowing whether carbonate is present, and acidifying to destroy it, prevents these false positives.
Part C: Identifying compounds
13. A white solid gives an orange-red flame. Its solution fizzes with dilute HCl and the gas turns limewater milky. Name the compound.
Answer: Calcium carbonate, CaCO₃. (Strictly, a solid that fizzes with acid is dissolved by it, so the flame test is often done on the solid directly. Calcium carbonate is insoluble in water, which fits.)
14. A colourless solution gives no flame colour and no precipitate with NaOH. When warmed with NaOH, it gives a gas that turns damp red litmus blue. With nitric acid then silver nitrate, it gives a white precipitate that dissolves in dilute ammonia. Name the compound.
Answer: Ammonium chloride, NH₄Cl. The gas is ammonia (from NH₄⁺), and the white precipitate soluble in dilute ammonia is AgCl (from Cl⁻).
15. A blue-green flame is seen. The solution gives a blue precipitate with NaOH. With nitric acid and silver nitrate there’s no precipitate, but with HCl and barium chloride a white precipitate forms. Name the compound.
Answer: Copper(II) sulfate, CuSO₄. Blue-green flame and blue hydroxide → Cu²⁺. White precipitate with barium chloride → SO₄²⁻.
16. An orange-brown solution gives an orange-brown precipitate with NaOH. After acidifying with nitric acid, silver nitrate gives a white precipitate. Name the compound and write ionic equations for both precipitates.
Answer: Iron(III) chloride, FeCl₃. Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
17. A white solid gives an apple-green flame. Its solution gives a white precipitate with dilute sulfuric acid. Silver nitrate (after nitric acid) gives a yellow precipitate insoluble in concentrated ammonia. Name the compound.
Answer: Barium iodide, BaI₂. Apple-green flame → Ba²⁺, confirmed by the white BaSO₄ precipitate with sulfuric acid. Yellow precipitate insoluble in concentrated ammonia → AgI, so I⁻.
Part D: Harder problems
18. A white solid X gives a yellow-orange flame. Its solution gives no precipitate with NaOH, and warming with NaOH gives no ammonia. Adding NaOH and aluminium powder then warming gives a gas that turns damp red litmus blue. Identify X.
Answer: Sodium nitrate, NaNO₃. Yellow flame → Na⁺. No ammonia with NaOH alone → no NH₄⁺. Ammonia after adding aluminium → NO₃⁻ reduced to NH₃.
19. A student has four unlabelled solutions: sodium chloride, sodium carbonate, sodium sulfate and sodium iodide. They are allowed only dilute nitric acid, barium nitrate and silver nitrate. Plan a sequence of tests to identify all four.
Answer:
- Add dilute nitric acid to a sample of each. The one that fizzes is sodium carbonate.
- To fresh samples of the remaining three, add nitric acid then barium nitrate. The one giving a white precipitate is sodium sulfate.
- To the remaining two, add nitric acid then silver nitrate. White precipitate → sodium chloride; yellow precipitate → sodium iodide.
Barium nitrate is used rather than barium chloride so that no chloride is added, keeping the samples usable for the silver nitrate test if needed.
20. A green solid Y dissolves in water. With NaOH it forms a green precipitate, insoluble in excess, that does not change colour on standing in air. With nitric acid and silver nitrate there’s no precipitate. With HCl and barium chloride there’s a white precipitate. Suggest an identity for Y and explain why iron(II) sulfate is less likely.
Answer: Nickel(II) sulfate, NiSO₄. Ni²⁺ gives a pale green hydroxide precipitate, Ni(OH)₂, that is stable in air. Iron(II) hydroxide, although also green, is oxidised in air to orange-brown iron(III) hydroxide, starting at the surface within minutes. The absence of this colour change points away from Fe²⁺.
How to score well on these questions
- Name the reagent fully: “dilute nitric acid, then silver nitrate solution”, not just “silver nitrate”.
- State the observation exactly: “white precipitate”, not just “goes white”; “fizzing” and then “limewater turns milky”, not “carbon dioxide forms”.
- Say what happens with excess reagent where it matters (NaOH and ammonia tests).
- Give ionic equations with state symbols when asked; spectator ions are left out.
- Explain the order of tests in planning questions.
For a wider problem-solving approach that includes instruments, see how chemists identify an unknown compound. For the difference between identifying and measuring, see qualitative vs quantitative analysis.
Key takeaways
- Each observation eliminates possibilities; combine them to reach one answer.
- The order of anion tests (carbonate, sulfate, halide, nitrate) and the choice of acid prevent false positives.
- Behaviour in excess NaOH or ammonia separates the white hydroxides.
- Colour changes on standing (Fe(OH)₂ turning brown) are evidence too.
- Planning questions reward a logical sequence, with each test using a fresh sample.
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