OCR A Level Chemistry A: Qualitative analysis, 5.3.2 (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

The questions here practise section 5.3.2, Qualitative analysis, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover learning outcome 5.3.2(a): anions in part (i), ammonium and transition metal cations in part (ii), and the PAG4 skills. H432 is the full A level; Module 5 is assessed in components 01 and 03. Questions 8 to 10 combine several tests.

Read the study guide and revision notes first. See also the OCR chemistry hub, printable checklist and free diagnostics.

Questions

1. Aqueous ammonia is added dropwise, and then in excess, to a pale blue solution of a copper(II) salt.

(a) Describe what is seen. [2] (b) Write an equation for the reaction that takes place in excess ammonia, starting from Cu(OH)₂. [2]

2. A few drops of aqueous ammonia or of aqueous sodium hydroxide give the same precipitate with Cu²⁺(aq). In excess, only ammonia dissolves it. Explain both observations. [3]

3. Aqueous sodium hydroxide is added dropwise, and then in excess, to chromium(III) nitrate solution.

(a) Describe what is seen. [2] (b) Write ionic equations for the reaction with a few drops and the reaction in excess. [2]

4. Aqueous sodium hydroxide added to iron(II) chloride solution gives a green precipitate. After fifteen minutes in an open tube, its top is orange-brown.

These half-equations apply:

Fe(OH)₂(s) + OH⁻(aq) → Fe(OH)₃(s) + e⁻ O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)

(a) Identify the orange-brown solid. [1] (b) Use the half-equations to construct the overall equation. [2] (c) State the change in the oxidation number of iron, and name the oxidising agent. [2]

5. Solid X is ammonium iron(II) sulfate, (NH₄)₂Fe(SO₄)₂. Describe test-tube tests, with the expected observations, that would show the presence of each of its three ions. [6]

6. A student dissolves 1.00 g of chromium(III) chloride hexahydrate, CrCl₃·6H₂O, in water and adds 0.500 mol dm⁻³ NaOH(aq) from a burette. Use these Ar values: H = 1.0, O = 16.0, Cl = 35.5, Cr = 52.0.

(a) Show that the Mr of CrCl₃·6H₂O is 266.5. [1] (b) Calculate the volume of NaOH(aq) needed to just precipitate all the chromium as Cr(OH)₃. [2] (c) Calculate the total volume of NaOH(aq) needed to convert all the chromium into [Cr(OH)₆]³⁻. [2]

7. Four unlabelled bottles contain dilute solutions of copper(II) sulfate, iron(II) sulfate, manganese(II) sulfate and chromium(III) sulfate. The only reagent available is aqueous ammonia. Describe how the student can identify each solution, giving the observations for each. [6]

8. Compound G is a single salt. A student records these results.

Test	Observation
G dissolved in distilled water	pale green solution
Solution of G acidified with dilute HNO₃(aq)	no fizzing
Barium nitrate solution then added	stays clear
Silver nitrate solution added to a second acidified portion	cream precipitate
Dilute NH₃(aq), then concentrated NH₃(aq), added to that precipitate	no change in dilute; dissolves in concentrated
NaOH(aq) to excess, fresh portion, warmed; damp red litmus at the mouth	green precipitate, unchanged in excess, turning orange-brown at the surface on standing; litmus stays red

(a) Identify G. Explain how each test result supports your answer. [5] (b) Write ionic equations, with state symbols, for the formation of the cream precipitate and the green precipitate. [2] (c) Suggest why the student used barium nitrate rather than barium chloride. [1]

9. A student must confirm that a blue-green solid is copper(II) chloride. Her method:

• Dissolve the solid in tap water.
• Add dilute hydrochloric acid, then aqueous silver nitrate. A white precipitate forms, so chloride is present.
• To a fresh portion, add two drops of aqueous ammonia. A pale blue precipitate forms, so Cu²⁺ is present.

Evaluate her method and conclusions, and describe the improvements she should make. [7]

10. Violet crystals, R, are a double salt with two cations and one anion. Results of tests on a solution of R:

Test	Observation
Solution of R acidified with dilute HNO₃(aq)	no fizzing
Barium nitrate solution then added	white precipitate
Silver nitrate solution added to a second acidified portion	stays clear
NaOH(aq) to excess, warmed; damp red litmus at the mouth	grey-green precipitate, dissolving in excess to a dark green solution; litmus stays red
NH₃(aq) to a large excess, fresh portion	grey-green precipitate, slowly dissolving to a purple solution

(a) Identify the transition metal ion and the anion in R. Use the results to explain your answer. [4] (b) State the evidence that the second cation is not NH₄⁺. [1] (c) Give an equation showing how the grey-green solid becomes the purple solution. [1] (d) The second cation is a Group 1 ion. Suggest why the tests in 5.3.2 do not identify it. [2]

Answers

1. (a) Pale blue precipitate with a little ammonia. [1] It dissolves in excess to give a deep blue solution. [1] (b) Correct complex, [Cu(NH₃)₄(H₂O)₂]²⁺. [1] Balanced equation: Cu(OH)₂(s) + 4NH₃(aq) + 2H₂O(l) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2OH⁻(aq) [1] Examiner insight: Check charge as well as atoms: the two OH⁻ released balance the 2+ on the complex, so both sides total zero.

2. Ammonia reacts with water in an equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. [1] This supplies enough OH⁻ to precipitate Cu(OH)₂, the same solid that NaOH gives. [1] In excess, NH₃ acts as a ligand, forming the soluble complex [Cu(NH₃)₄(H₂O)₂]²⁺; excess NaOH only adds OH⁻, which Cu(OH)₂ does not react with. [1] Examiner insight: “Ammonia is a base” is not enough for the first point; show where the hydroxide ions come from with the equilibrium.

3. (a) Grey-green precipitate. [1] Dissolves in excess to give a dark green solution. [1] (b) Cr³⁺(aq) + 3OH⁻(aq) → Cr(OH)₃(s) [1] Cr(OH)₃(s) + 3OH⁻(aq) → [Cr(OH)₆]³⁻(aq) [1] Examiner insight: The complex is 3− (+3 from chromium, six × −1 from hydroxide); writing 3+ unbalances the equation.

4. (a) Iron(III) hydroxide, Fe(OH)₃. [1] (b) Multiply the iron half-equation by 4 so the electrons cancel. [1] Add and cancel the four OH⁻ on each side: 4Fe(OH)₂(s) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s) [1] (c) Iron goes from +2 to +3. [1] The oxidising agent is oxygen, O₂, from the air. [1] Examiner insight: Cancel species that appear on both sides; no electrons should remain in the overall equation.

5. Fe²⁺: add aqueous sodium hydroxide (or aqueous ammonia) to a solution of X. [1] A green precipitate forms, insoluble in excess, browning at the surface on standing. [1] NH₄⁺: warm X with aqueous sodium hydroxide, holding moist red litmus paper near the open end. [1] The litmus turns blue (ammonia gas). [1] SO₄²⁻: acidify a solution of X with dilute nitric acid and add aqueous barium nitrate (or barium chloride). [1] A white precipitate forms. [1] Examiner insight: For each ion give both the reagent and the result; either alone is only half a test.

6. (a) Mr = 52.0 + 3(35.5) + 6(2 × 1.0 + 16.0) = 52.0 + 106.5 + 108.0 = 266.5 [1] (b) n(Cr³⁺) = 1.00 ÷ 266.5 = 3.752 × 10⁻³ mol [1] n(OH⁻) = 3 × 3.752 × 10⁻³ = 1.126 × 10⁻² mol, so volume = 1.126 × 10⁻² ÷ 0.500 = 2.251 × 10⁻² dm³ = 22.5 cm³ [1] (c) Cr(OH)₃ + 3OH⁻ → [Cr(OH)₆]³⁻ needs a further 3 mol OH⁻ per mol Cr, another 22.5 cm³. [1] Total = 45.0 cm³ [1] Examiner insight: Carry unrounded values between steps and round only the final answer, to three significant figures.

7. To a fresh portion of each, add ammonia dropwise, then in large excess, shaking; leave to stand. [1] Copper(II): pale blue precipitate, dissolving in excess to a deep blue solution. [1] Chromium(III): grey-green precipitate, dissolving in excess to a purple solution. [1] Iron(II): green precipitate, insoluble in excess. [1] Manganese(II): light brown precipitate, insoluble in excess. [1] Iron(II) and manganese(II) differ in precipitate colour; on standing the iron(II) one turns orange-brown at the surface. [1] Examiner insight: When two ions behave alike in excess, state which observation separates them.

8. (a) No effervescence with acid, so no carbonate. [1] No precipitate with Ba²⁺ in acid, so no sulfate. [1] Cream precipitate with Ag⁺ that dissolves only in concentrated ammonia, so bromide. [1] Green precipitate with NaOH, insoluble in excess and turning orange-brown in air, so Fe²⁺. [1] Litmus stays red, so no ammonium; G is iron(II) bromide, FeBr₂. [1] (b) Ag⁺(aq) + Br⁻(aq) → AgBr(s) [1] Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s) [1] (c) Barium chloride adds Cl⁻, which would interfere with a later halide test. [1] Examiner insight: Work through every row of the table; a correct name without the reasoning for each test leaves most of a “use the results” answer unsupported.

9. Tap water can contain chloride ions. [1] She should use distilled water. [1] Hydrochloric acid adds Cl⁻, so a white precipitate was certain whatever the solid. [1] She should acidify with dilute nitric acid. [1] She could confirm AgCl: it dissolves in dilute aqueous ammonia. [1] Two drops of ammonia show only the first stage; excess ammonia should dissolve the precipitate to a deep blue solution, confirming Cu²⁺. [1] Judgement: her chloride result is invalid; her copper result is supported but not fully confirmed. [1] Examiner insight: “Evaluate” needs a judgement as well as a list of faults; end by stating which conclusions stand and which do not.

10. (a) The grey-green precipitate dissolves in excess NaOH; only Cr(OH)₃ does this, so the ion is Cr³⁺. [1] It also dissolves in excess NH₃ to give a purple solution, [Cr(NH₃)₆]³⁺, which confirms Cr³⁺. [1] White precipitate with Ba²⁺ after acidifying, so the anion is SO₄²⁻. [1] No effervescence and no precipitate with Ag⁺ rule out carbonate and halides. [1] (b) On warming with NaOH(aq), no gas turned damp red litmus blue, so no ammonia was given off. [1] (c) Cr(OH)₃(s) + 6NH₃(aq) → [Cr(NH₃)₆]³⁺(aq) + 3OH⁻(aq) [1] (d) Group 1 hydroxides are soluble, so there is no precipitate with NaOH(aq) or NH₃(aq). [1] The 5.3.2 cation tests cover only NH₄⁺ and five transition metal ions, so none gives a positive result. [1] Examiner insight: In (d), only claim what the evidence supports: the tests show which ions are absent, not which Group 1 ion is present.

Where marks are usually lost

• Writing “deep blue precipitate” for copper with excess ammonia; it is a solution.
• Claiming Fe(OH)₂, Fe(OH)₃ or Mn(OH)₂ dissolves in excess of either reagent.
• Missing the change on standing for Fe²⁺ and Mn²⁺ precipitates.
• Getting the charge on [Cr(OH)₆]³⁻ wrong, or leaving charges off complex ions altogether.
• Using hydrochloric acid, or barium chloride, before a halide test.
• Giving a reagent without an observation in “describe a test” answers.

Next steps

• 5.3.2 revision notes
• 5.3.2 study guide
• Redox for oxidation numbers and half-equations
• OCR A Level chemistry hub
• Printable OCR chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 5.3.2 Qualitative analysis, learning outcome (a)(i) and (a)(ii), with additional guidance (PAG4, HSW4) and the linked statements 3.1.4(a) and 5.3.1(j). Published by OCR.