Identification of Ions and Gases Worked ExamplesIGCSE Chemistry 0620

(a) A metal reacting with a dilute acid gives hydrogen. Test: hold a lighted splint at the mouth of the tube — it pops.

(b) Manganese(IV) oxide catalyses the decomposition of hydrogen peroxide into water and oxygen. Test: a glowing splint relights.

(c) At the positive electrode (anode) chloride ions are oxidised to chlorine. Test: damp litmus paper is bleached (turns white).

(d) An ammonium salt warmed with an alkali gives ammonia. Test: damp red litmus paper turns blue.

A red flame shows lithium ions. A yellow precipitate with acidified silver nitrate shows iodide ions. The answer is C.

A: calcium gives an orange-red flame, not red. B: bromide gives a cream precipitate. D: sodium gives a yellow flame — the colour in the question belongs to the precipitate, not the flame.

A white precipitate that dissolves in excess sodium hydroxide narrows it to aluminium or zinc. Only the aluminium precipitate is insoluble in excess ammonia. The answer is A.

D (zinc): its precipitate dissolves in excess of both alkalis. B (calcium): the precipitate with sodium hydroxide does not dissolve in excess, and ammonia gives no (or very slight) precipitate. C (chromium(III)): the precipitate is green.

(a) Flame 2 — the blue (non-luminous) flame with the air hole open. The yellow flame 1 is itself coloured, so it would hide the colour produced by the metal ion.

(b) Dip a clean flame test wire into the solid (moistened so that some sticks to it), then hold the sample in the edge of the blue flame and note the colour.

(c) A blue-green flame shows copper(II) ions, $\text{Cu}^{2+}$.

Confirm: dissolve the solid in water and add aqueous sodium hydroxide, a few drops and then in excess. A light blue precipitate forms, which is insoluble in excess.

(a) A red-brown precipitate, insoluble in excess of both alkalis, is the result for iron(III) ions, $\text{Fe}^{3+}$.

(b) Test 4: no precipitate, so R contains no chloride, bromide or iodide ions. Test 5: the manganate(VII) is not decolourised, so R contains no sulfite ions.

A white precipitate with acidified barium nitrate (test 3) shows sulfate ions, $\text{SO}_4^{\,2-}$.

(c) R contains iron(III) and sulfate ions: iron(III) sulfate.

(a) A few drops: white precipitate. Excess: the precipitate dissolves, giving a colourless solution.

(b) Zinc behaves the same way with ammonia: white precipitate, which dissolves in excess to give a colourless solution.

(c) Zinc bromide contains no sulfate ions, so there is no change (no precipitate).

(d) Bromide ions give a cream precipitate with acidified silver nitrate.

(a) Ammonia — the only gas in the tests that turns damp red litmus paper blue.

(b) Ammonia given off on warming with sodium hydroxide alone (no aluminium) shows ammonium ions, $\text{NH}_4^{\,+}$.

(c) The nitrate test also gives ammonia. If any ammonia from the ammonium ions were still being released, it would turn the litmus blue in stage 2 even if there were no nitrate. Removing it first means any ammonia in stage 2 must come from nitrate ions being reduced by the aluminium.

Stage 2 gives ammonia, so nitrate ions, $\text{NO}_3^{\,-}$, are present. (d) The fertiliser is ammonium nitrate.

(a)(i) Iron(III) hydroxide is the red-brown precipitate. Three hydroxide ions balance the $3+$ charge.

(ii) Silver iodide is the yellow precipitate.

(iii) Barium sulfate is the white precipitate.

Check each equation: the atoms balance and the total charge on the left is zero, matching the neutral precipitate on the right.

(b) Hydrochloric acid contains chloride ions, which would form a white precipitate of silver chloride with the silver nitrate — you could not tell whether the chloride came from the sample or from the acid.

(a) The solution changes from purple to colourless. This is the test for sulfur dioxide.

(b) Sulfur dioxide from a solid and dilute acid suggests sulfite ions, $\text{SO}_3^{\,2-}$.

(c) A carbonate would give carbon dioxide, which does not decolourise acidified potassium manganate(VII) (it turns limewater milky instead).

(d) Test 1: add a small volume of acidified aqueous potassium manganate(VII) — it stays purple, so there is no sulfite.

Test 2: acidify with dilute nitric acid, then add aqueous barium nitrate — a white precipitate forms, showing sulfate.

A reagent can only distinguish two substances if it gives different results with them.

B: with excess aqueous ammonia, the zinc precipitate dissolves (colourless solution) but the aluminium precipitate does not. The results differ, so the answer is B.

A: both give a white precipitate that dissolves in excess — identical. C: both are sulfates, so both give a white precipitate. D: neither contains a halide, so both show no change.

(a) At P: a light blue precipitate in the blue solution. At R: the precipitate has dissolved, leaving a dark blue solution (no precipitate).

(b) The precipitate is copper(II) hydroxide:

(c) Moles of copper(II) ions, converting cm³ to dm³:

One mole of $\text{Cu(OH)}_2$ forms per mole of $\text{Cu}^{2+}$. $M_r = 64 + 2(16 + 1) = 98$.

The graph's maximum, at Q, is $0.49$ g — the same value.

(d) The mass would rise to the same maximum ($0.49$ g) but then stay level instead of falling to zero, because copper(II) hydroxide is insoluble in excess aqueous sodium hydroxide.

(a) A colourless liquid (water) condensing on heating shows Z is hydrated — it contains water of crystallisation.

(b) A green precipitate, insoluble in excess, turning brown near the surface on standing shows iron(II) ions, $\text{Fe}^{2+}$.

The litmus turning blue on warming with sodium hydroxide — with no aluminium added — shows ammonia is given off, so ammonium ions, $\text{NH}_4^{\,+}$, are present.

(c) A white precipitate with acidified barium nitrate shows sulfate ions. No precipitate with acidified silver nitrate rules out chloride, bromide and iodide.

(d) Ammonium ions react with hydroxide ions on warming:

Z is hydrated ammonium iron(II) sulfate.