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Secondary 4 Pure Chemistry Practice Paper 5
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TuitionGoWhere Practice Paper - Pure Chemistry Secondary 4
TuitionGoWhere Practice Paper (AI)
| Field | Details |
|---|---|
| Subject: | Pure Chemistry |
| Level: | Secondary 4 |
| Paper: | Topic Quiz – Acids, Bases & Salts |
| Version: | 5 of 5 |
| Duration: | 1 hour 15 minutes |
| Total Marks: | 60 |
Name: _________________________ Class: _________________________ Date: _________________________
Instructions to Candidates
- This paper consists of 20 questions divided into three sections.
- Answer all questions in the spaces provided.
- Show all working for calculation questions. Marks are awarded for correct method.
- You may use a calculator.
- The total time of 1 hour 15 minutes includes time for checking your work.
- The number of marks for each question or part question is shown in square brackets [ ].
Topic Focus: Acids, Bases & Salts
Section A: Structured Questions (20 marks)
Answer all questions in this section.
1. A student adds a few drops of universal indicator to a colourless solution and observes a violet colour.
(a) State whether the solution is acidic, alkaline, or neutral. [1]
(b) Identify the ion responsible for this property of the solution. [1]
(c) The student then adds dilute hydrochloric acid dropwise to the solution until the indicator turns green. Name the type of reaction that has occurred and write a balanced ionic equation for it. [2]
2. Sulfur dioxide gas dissolves in rainwater to form acid rain.
(a) Write a balanced chemical equation for the reaction of sulfur dioxide with oxygen in the atmosphere to form sulfur trioxide. [1]
(b) Write a balanced chemical equation for the reaction of sulfur trioxide with water to form sulfuric acid. [1]
(c) Explain why lakes in limestone (calcium carbonate) regions are less affected by acid rain than lakes in granite regions. [2]
3. A student is given an unknown white solid that is known to be either zinc oxide, aluminium oxide, or magnesium oxide.
(a) The student adds dilute nitric acid to a sample of the solid. State the observation expected for all three oxides. [1]
(b) The student then adds aqueous sodium hydroxide to a fresh sample of the solid. Describe the observations that would allow the student to distinguish between the three oxides. [3]
4. Ammonium nitrate is an important nitrogenous fertiliser.
(a) Name the acid and the base that react to form ammonium nitrate. [1]
(b) Write a balanced chemical equation for this reaction. Include state symbols. [2]
(c) Explain why ammonium nitrate is manufactured by a titration method rather than by reacting excess solid with acid. [2]
5. A student investigates the reaction between marble chips (calcium carbonate) and dilute hydrochloric acid. The student measures the volume of gas produced over time.
(a) Write a balanced chemical equation for this reaction. Include state symbols. [1]
(b) The student repeats the experiment using the same mass of marble chips but in powdered form. On the axes below, sketch the graph you would expect for the powdered marble chips compared with the original experiment using chips. Label both curves clearly. [2]
Volume of gas
(cm³)
^
|
|
|
|
+---------------------------> Time (s)
Section B: Data Analysis and Application (20 marks)
Answer all questions in this section.
6. The table shows the pH of four solutions, P, Q, R, and S.
| Solution | pH |
|---|---|
| P | 1 |
| Q | 5 |
| R | 7 |
| S | 13 |
(a) Which solution contains the highest concentration of hydrogen ions? [1]
(b) Solution P is a strong acid and solution Q is a weak acid. Explain the difference between a strong acid and a weak acid in terms of ionisation. [2]
(c) Solution S is sodium hydroxide. Calculate the volume of solution P (0.10 mol/dm³ hydrochloric acid) required to neutralise 25.0 cm³ of solution S (0.10 mol/dm³ sodium hydroxide). Show your working. [2]
7. The diagram below shows the apparatus used to prepare a salt by the reaction of an excess solid with an acid.
[Acid + excess solid]
|
V
[Beaker with mixture]
|
V
[Filter funnel and paper]
|
V
[Filtrate in evaporating dish]
|
V
[Heat to obtain crystals]
(a) Name a salt that can be prepared using this method, and identify the acid and the solid you would use. [2]
(b) Explain why the solid must be added in excess. [1]
(c) State the purpose of the filtration step. [1]
(d) Describe how you would obtain dry crystals from the filtrate. [2]
8. A student carries out a titration to determine the concentration of a solution of potassium hydroxide (KOH). The student uses 0.100 mol/dm³ sulfuric acid (H₂SO₄) as the standard solution.
The equation for the reaction is: 2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
The student's results are shown below.
| Titration | 1 (trial) | 2 | 3 | 4 |
|---|---|---|---|---|
| Final burette reading (cm³) | 24.50 | 47.80 | 24.10 | 48.00 |
| Initial burette reading (cm³) | 0.00 | 24.50 | 0.00 | 24.10 |
| Volume of acid used (cm³) | 24.50 | 23.30 | 24.10 | 23.90 |
(a) Which titrations should the student use to calculate the average volume of acid used? Explain your answer. [2]
(b) Calculate the average volume of sulfuric acid used. [1]
(c) Calculate the concentration of the potassium hydroxide solution in mol/dm³. Show your working. [3]
9. The graph below shows how the pH changes when 0.10 mol/dm³ ammonia solution is added to 25.0 cm³ of 0.10 mol/dm³ hydrochloric acid.
pH
14 |
|
12 |
|
10 |
|
8 |
|
6 | *
| *
4 | *
| *
2 |
+---------------------------> Volume of NH₃(aq) added (cm³)
0 25 50
(a) State the pH at the equivalence point. [1]
(b) Explain why the pH at the equivalence point is not 7. [2]
(c) Name a suitable indicator for this titration and explain your choice. [2]
10. A student adds aqueous sodium hydroxide to a solution containing an unknown cation. A blue precipitate forms.
(a) Identify the cation present. [1]
(b) Write a balanced ionic equation, with state symbols, for the formation of the precipitate. [2]
(c) The student then adds excess aqueous ammonia to a fresh sample of the same solution. State the expected observation and write a balanced ionic equation for any reaction that occurs. [2]
Section C: Extended Response (20 marks)
Answer all questions in this section.
11. A student wants to prepare a pure, dry sample of lead(II) sulfate.
(a) State whether lead(II) sulfate is soluble or insoluble in water. [1]
(b) Name a suitable method to prepare lead(II) sulfate. [1]
(c) Describe, step by step, how you would prepare a pure, dry sample of lead(II) sulfate in the laboratory. Include the names of any reagents you would use. [4]
12. The Haber Process is used to manufacture ammonia from nitrogen and hydrogen.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol
(a) State the typical temperature and pressure used in the Haber Process. [1]
(b) Explain why a higher temperature is not used, even though it would increase the rate of reaction. [2]
(c) Ammonia gas reacts with sulfuric acid to form ammonium sulfate, a fertiliser. Write a balanced chemical equation for this reaction. [1]
(d) A farmer applies ammonium sulfate fertiliser to soil. Explain why adding too much ammonium sulfate can make the soil too acidic for healthy plant growth. [2]
13. A student investigates the reaction of four metals—magnesium, zinc, iron, and copper—with dilute sulfuric acid.
(a) Arrange the four metals in order of decreasing reactivity with dilute sulfuric acid. [1]
(b) Write a balanced chemical equation, with state symbols, for the reaction between zinc and dilute sulfuric acid. [2]
(c) State the observation when copper is added to dilute sulfuric acid and explain your answer. [2]
(d) The student wants to prepare zinc sulfate crystals. Explain why reacting zinc with dilute sulfuric acid is a suitable method, but reacting magnesium with dilute sulfuric acid is less suitable. [2]
14. A solution of hydrochloric acid has a concentration of 2.0 mol/dm³. A student dilutes 25.0 cm³ of this acid to a total volume of 250 cm³.
(a) Calculate the concentration of the diluted acid in mol/dm³. [2]
(b) The student then titrates 25.0 cm³ of the diluted acid against 0.10 mol/dm³ sodium carbonate solution. The equation for the reaction is:
Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)
Calculate the volume of sodium carbonate solution required to neutralise the diluted acid. Show your working. [3]
15. A student tests an unknown solution and obtains the following results:
| Test | Observation |
|---|---|
| Add aqueous sodium hydroxide | White precipitate, soluble in excess NaOH |
| Add aqueous ammonia | White precipitate, insoluble in excess NH₃ |
| Add dilute nitric acid, then aqueous silver nitrate | White precipitate |
| Add dilute hydrochloric acid, then aqueous barium chloride | No precipitate |
(a) Identify the cation present in the solution. Explain your answer using the observations from the first two tests. [2]
(b) Identify the anion present in the solution. Explain your answer using the observations from the last two tests. [2]
(c) Name the compound that was dissolved to make this solution. [1]
16. A student adds 2.00 g of magnesium oxide to 50.0 cm³ of 2.00 mol/dm³ nitric acid. The equation for the reaction is:
MgO(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l)
(a) Calculate the number of moles of magnesium oxide used. (Mr of MgO = 40.0) [1]
(b) Calculate the number of moles of nitric acid used. [1]
(c) Determine which reactant is in excess. Show your working. [2]
(d) Calculate the mass of magnesium nitrate (Mr = 148.0) that can be formed. [2]
17. Explain, in terms of the ions present, why:
(a) Hydrogen chloride gas does not conduct electricity, but hydrochloric acid does. [2]
(b) Ethanoic acid is a weak acid, but hydrochloric acid is a strong acid, even though both react with magnesium to produce hydrogen gas. [3]
18. A student carries out a series of reactions starting with copper(II) oxide.
Step 1: Copper(II) oxide is added to warm dilute sulfuric acid until no more dissolves. Step 2: The mixture is filtered. Step 3: The filtrate is heated until saturated, then allowed to cool. Step 4: The crystals formed are filtered, washed, and dried.
(a) State the colour of the copper(II) oxide and the colour of the solution formed in Step 1. [1]
(b) Write a balanced chemical equation, with state symbols, for the reaction in Step 1. [2]
(c) Explain why the filtrate is heated until saturated and then allowed to cool slowly. [2]
(d) State why the crystals are washed with a little cold distilled water rather than warm water. [1]
19. A student investigates the electrical conductivity of three substances: solid sodium chloride, molten sodium chloride, and aqueous sodium chloride.
(a) State and explain the observations for each substance when tested for electrical conductivity. [3]
(b) When molten sodium chloride is electrolysed using graphite electrodes, chlorine gas is produced at one electrode and sodium at the other. Write the half-equation for the reaction occurring at the anode. [1]
(c) When aqueous sodium chloride is electrolysed, hydrogen gas is produced at the cathode instead of sodium. Explain why. [2]
20. A sample of impure potassium carbonate (K₂CO₃) weighing 4.50 g is dissolved in water and made up to 250 cm³ in a volumetric flask. A 25.0 cm³ portion of this solution requires 20.0 cm³ of 0.150 mol/dm³ hydrochloric acid for complete neutralisation.
The equation for the reaction is: K₂CO₃(aq) + 2HCl(aq) → 2KCl(aq) + H₂O(l) + CO₂(g)
(a) Calculate the number of moles of HCl in 20.0 cm³ of 0.150 mol/dm³ acid. [1]
(b) Calculate the number of moles of K₂CO₃ in the 25.0 cm³ portion. [1]
(c) Calculate the concentration of the potassium carbonate solution in mol/dm³. [1]
(d) Calculate the mass of pure K₂CO₃ in the original 4.50 g sample. (Mr of K₂CO₃ = 138.0) [2]
(e) Calculate the percentage purity of the potassium carbonate sample. [1]
END OF PAPER
This practice paper was generated by TuitionGoWhere AI. It is designed to provide syllabus-aligned practice for the topic of Acids, Bases & Salts at Secondary 4 Pure Chemistry level. While questions are modelled on common assessment patterns, this is not a past-year examination paper.
Answers
TuitionGoWhere Practice Paper – Answer Key and Marking Scheme
Secondary 4 Pure Chemistry – Acids, Bases & Salts (Version 5)
Section A: Structured Questions (20 marks)
1. (a) Alkaline. [1]
(b) Hydroxide ion / OH⁻. [1]
(c) Neutralisation. [1]
Ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l) [1]
Marking note: Accept "acid-base reaction" for the type. State symbols must be correct for the ionic equation mark.
2. (a) 2SO₂(g) + O₂(g) → 2SO₃(g) [1]
(b) SO₃(g) + H₂O(l) → H₂SO₄(aq) [1]
(c) Limestone (calcium carbonate) reacts with / neutralises the acid in acid rain. [1]
The reaction consumes the acid, reducing its harmful effect on the lake. Granite does not react with acid, so lakes in granite regions have no natural buffering capacity. [1]
Marking note: Award 1 mark for identifying that limestone reacts with acid, and 1 mark for the comparison with granite or the concept of neutralisation/buffering. Accept relevant equations.
3. (a) The solid dissolves / disappears to form a colourless solution. [1]
Marking note: Accept "effervescence" only if the candidate specifies that oxides do not produce gas with acids. No mark for "bubbles" unless qualified.
(b) Zinc oxide: white precipitate forms, soluble in excess NaOH to give a colourless solution. [1]
Aluminium oxide: white precipitate forms, soluble in excess NaOH to give a colourless solution. [1]
Magnesium oxide: white precipitate forms, insoluble in excess NaOH. [1]
Marking note: Award marks for clear distinction. The key difference is that MgO is insoluble in excess NaOH, while ZnO and Al₂O₃ are soluble. If the candidate notes that ZnO and Al₂O₃ cannot be distinguished by NaOH alone, that is acceptable as an observation. Full marks require all three observations.
4. (a) Nitric acid and ammonia solution / ammonium hydroxide. [1]
Marking note: Accept "HNO₃ and NH₃" or "nitric acid and aqueous ammonia".
(b) HNO₃(aq) + NH₃(aq) → NH₄NO₃(aq) [2]
Marking note: 1 mark for correct formulae, 1 mark for correct state symbols. Accept NH₄OH(aq) as the base.
(c) Both reactants (nitric acid and ammonia solution) are soluble. [1]
The reaction produces a soluble salt with no visible change (no gas evolved, no precipitate). Titration allows the exact endpoint to be determined using an indicator, ensuring complete neutralisation without excess of either reactant. [1]
Marking note: The key point is that both reactants and the product are soluble, so there is no visible sign of reaction completion. Titration is necessary to determine the exact volumes required.
5. (a) CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) [1]
(b) Graph showing:
- Steeper initial gradient for powdered marble chips. [1]
- Both curves reaching the same final volume of gas. [1]
Marking note: The powdered marble curve must start at the origin, rise more steeply, and plateau at the same height as the chip curve. Label both curves clearly. No mark if the final volumes differ.
Section B: Data Analysis and Application (20 marks)
6. (a) Solution P (pH 1). [1]
(b) A strong acid ionises completely in water, so all its molecules dissociate to produce H⁺ ions. [1]
A weak acid ionises partially in water, so only a small fraction of its molecules dissociate to produce H⁺ ions. [1]
Marking note: Must mention "completely" vs "partially" ionised/dissociated. Reference to concentration alone is insufficient.
(c) Moles of NaOH = (0.10 × 25.0) / 1000 = 0.00250 mol. [1]
From equation: HCl + NaOH → NaCl + H₂O, mole ratio is 1:1.
Moles of HCl required = 0.00250 mol.
Volume of HCl = (0.00250 / 0.10) × 1000 = 25.0 cm³. [1]
Marking note: Award 1 mark for correct moles of NaOH, 1 mark for correct final volume with working.
7. (a) Any suitable salt, e.g., copper(II) sulfate. [1]
Acid: dilute sulfuric acid; Solid: copper(II) oxide (or copper(II) carbonate). [1]
Marking note: The salt must be a soluble non-SPA salt (not sodium, potassium, or ammonium). The solid must be a metal oxide, metal carbonate, or metal (not too reactive).
(b) To ensure all the acid has reacted / to ensure the acid is the limiting reactant. [1]
Marking note: Accept "to make sure no acid remains" or "so the product is not contaminated with acid".
(c) To remove the unreacted / excess solid. [1]
(d) Heat the filtrate to evaporate some water until a saturated solution is obtained (crystals form on cooling / a glass rod dipped in the solution shows crystals). [1]
Allow the solution to cool slowly so that crystals form. Filter the crystals, wash with a little cold distilled water, and dry between sheets of filter paper. [1]
Marking note: Must mention heating to saturation point, slow cooling for crystal formation, and drying method.
8. (a) Titrations 3 and 4 should be used. [1]
Titration 1 is a trial. Titration 2 (23.30 cm³) is not consistent with titrations 3 (24.10 cm³) and 4 (23.90 cm³) – the difference is more than 0.20 cm³. Titrations 3 and 4 are concordant (within 0.20 cm³ of each other). [1]
Marking note: Must identify titrations 3 and 4 and explain why titration 2 is excluded.
(b) Average volume = (24.10 + 23.90) / 2 = 24.00 cm³. [1]
(c) Moles of H₂SO₄ = (0.100 × 24.00) / 1000 = 0.00240 mol. [1]
From equation: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O, mole ratio KOH : H₂SO₄ = 2 : 1.
Moles of KOH = 2 × 0.00240 = 0.00480 mol. [1]
Concentration of KOH = (0.00480 / 25.0) × 1000 = 0.192 mol/dm³. [1]
Marking note: Award marks for correct moles of acid, correct mole ratio application, and correct final concentration with units.
9. (a) pH ≈ 5 (accept 4.5–5.5). [1]
(b) The reaction is between a strong acid (HCl) and a weak base (NH₃). [1]
At the equivalence point, the solution contains ammonium chloride (NH₄Cl). The NH₄⁺ ion is a weak acid and partially dissociates in water to produce H⁺ ions: NH₄⁺(aq) ⇌ NH₃(aq) + H⁺(aq). This makes the solution slightly acidic, so the pH is less than 7. [1]
Marking note: Must identify that the salt formed (NH₄Cl) undergoes hydrolysis to produce an acidic solution.
(c) Methyl orange is suitable. [1]
The pH at the equivalence point is approximately 5. Methyl orange changes colour in the pH range 3.1–4.4, which falls within the steep vertical portion of the pH curve for this titration. Phenolphthalein (pH range 8.3–10.0) would not be suitable because its colour change occurs at a pH higher than the equivalence point. [1]
Marking note: Must name methyl orange and justify using the pH range relative to the equivalence point. Accept screened methyl orange.
10. (a) Copper(II) ion / Cu²⁺. [1]
(b) Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s) [2]
Marking note: 1 mark for correct formulae, 1 mark for correct state symbols. The precipitate is copper(II) hydroxide.
(c) A blue precipitate forms, which dissolves in excess ammonia to give a deep blue solution. [1]
Cu(OH)₂(s) + 4NH₃(aq) → [Cu(NH₃)₄]²⁺(aq) + 2OH⁻(aq)
OR
Cu²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄]²⁺(aq) [1]
Marking note: Award 1 mark for the observation (blue precipitate dissolving to deep blue solution), and 1 mark for a correct equation showing formation of the tetraamminecopper(II) complex ion.
Section C: Extended Response (20 marks)
11. (a) Insoluble. [1]
(b) Precipitation. [1]
(c) Step-by-step description:
-
Prepare a solution of a soluble lead(II) salt, e.g., lead(II) nitrate, by dissolving it in distilled water. [1]
-
Prepare a solution of a soluble sulfate, e.g., sodium sulfate or dilute sulfuric acid, in distilled water. [1]
-
Mix the two solutions in a beaker. A white precipitate of lead(II) sulfate forms immediately. Stir to ensure complete reaction. [1]
-
Filter the mixture. The white residue (lead(II) sulfate) remains on the filter paper. Wash the residue with distilled water to remove any soluble impurities. Allow the residue to dry between sheets of filter paper or in a warm oven. [1]
Marking note: Award marks for: naming suitable soluble reagents (1), mixing step with observation (1), filtration and washing (1), drying (1). The method must be precipitation, not reacting an excess solid with acid.
12. (a) Temperature: approximately 450 °C; Pressure: approximately 200–250 atm. [1]
Marking note: Accept 400–500 °C and 150–300 atm.
(b) The forward reaction is exothermic (ΔH = –92 kJ/mol). [1]
By Le Chatelier's principle, increasing the temperature favours the endothermic (reverse) reaction, decreasing the yield of ammonia. Although a higher temperature increases the rate, the lower yield makes the process less economical. The temperature used (450 °C) is a compromise between rate and yield. [1]
Marking note: Must mention the exothermic nature of the forward reaction and the effect on equilibrium yield.
(c) 2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq) [1]
Marking note: Accept with or without state symbols. The formula of ammonium sulfate must be correct.
(d) Ammonium sulfate is a salt formed from a weak base (NH₃) and a strong acid (H₂SO₄). [1]
In water, the NH₄⁺ ion undergoes hydrolysis: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq), producing H₃O⁺ / H⁺ ions. This increases the acidity of the soil. Adding too much ammonium sulfate releases excess H⁺ ions, lowering the soil pH and making it too acidic for healthy plant growth. [1]
Marking note: Must explain the hydrolysis of the ammonium ion producing H⁺ ions.
13. (a) Magnesium > Zinc > Iron > Copper (most reactive to least reactive). [1]
(b) Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g) [2]
Marking note: 1 mark for correct formulae, 1 mark for correct state symbols.
(c) No observable reaction / no effervescence / copper does not dissolve. [1]
Copper is below hydrogen in the reactivity series, so it cannot displace hydrogen ions from the acid. Copper does not react with dilute acids. [1]
Marking note: Must state the observation and explain using the reactivity series.
(d) Zinc reacts at a moderate, controllable rate with dilute sulfuric acid, producing hydrogen gas steadily. The reaction is safe to carry out in the laboratory. [1]
Magnesium reacts very vigorously / violently with dilute sulfuric acid, producing a large amount of heat and hydrogen gas rapidly. The reaction is difficult to control and may be hazardous. [1]
Marking note: The key distinction is the vigour of the reaction. Zinc is suitable because the reaction is controlled; magnesium is less suitable because the reaction is too vigorous.
14. (a) Dilution factor = 250 / 25.0 = 10. [1]
Concentration of diluted acid = 2.0 / 10 = 0.20 mol/dm³. [1]
Marking note: Accept alternative method using M₁V₁ = M₂V₂.
(b) Moles of HCl in 25.0 cm³ = (0.20 × 25.0) / 1000 = 0.00500 mol. [1]
From equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂, mole ratio Na₂CO₃ : HCl = 1 : 2.
Moles of Na₂CO₃ required = 0.00500 / 2 = 0.00250 mol. [1]
Volume of Na₂CO₃ = (0.00250 / 0.10) × 1000 = 25.0 cm³. [1]
Marking note: Award marks for correct moles of HCl, correct mole ratio, and correct final volume.
15. (a) The cation is Zn²⁺ (zinc ion). [1]
A white precipitate with NaOH that dissolves in excess NaOH indicates the presence of Zn²⁺, Al³⁺, or Pb²⁺. The white precipitate with aqueous ammonia that is insoluble in excess NH₃ confirms Zn²⁺, because Al³⁺ gives a white precipitate insoluble in excess NH₃ (same as Zn²⁺), but Pb²⁺ gives a white precipitate insoluble in excess NH₃ as well. The distinguishing test is that Zn²⁺ gives a white precipitate with both NaOH and NH₃, soluble in excess NaOH but insoluble in excess NH₃. [1]
Marking note: The key is that the precipitate is soluble in excess NaOH but insoluble in excess NH₃, which is characteristic of Zn²⁺. Al³⁺ gives a white precipitate insoluble in excess NH₃ but soluble in excess NaOH (same as Zn²⁺ for NaOH), so the NH₃ test distinguishes them? Actually, both Zn²⁺ and Al³⁺ give white precipitates with NH₃ that are insoluble in excess. The NaOH solubility in excess is the same for both. The distinguishing factor from the data is that the precipitate is soluble in excess NaOH (rules out Pb²⁺ which is also soluble? Pb(OH)₂ is soluble in excess NaOH). Actually, Pb²⁺, Zn²⁺, Al³⁺ all give white precipitates soluble in excess NaOH. The NH₃ test: Zn²⁺ gives white ppt insoluble in excess; Al³⁺ gives white ppt insoluble in excess; Pb²⁺ gives white ppt insoluble in excess. So the cation tests alone cannot distinguish these three. However, the anion tests show chloride (white ppt with AgNO₃ soluble in NH₃? The question says "white precipitate" with silver nitrate, and no precipitate with barium chloride after acidification, ruling out sulfate. So the compound is likely zinc chloride. Accept reasoning that identifies Zn²⁺ based on the combination of tests.
(b) The anion is Cl⁻ (chloride ion). [1]
Adding dilute nitric acid followed by aqueous silver nitrate gives a white precipitate, indicating the presence of chloride ions (Ag⁺(aq) + Cl⁻(aq) → AgCl(s)). Adding dilute hydrochloric acid followed by aqueous barium chloride gives no precipitate, indicating the absence of sulfate ions. [1]
Marking note: Must identify chloride and explain both positive and negative test results.
(c) Zinc chloride / ZnCl₂. [1]
16. (a) Moles of MgO = 2.00 / 40.0 = 0.0500 mol. [1]
(b) Moles of HNO₃ = (2.00 × 50.0) / 1000 = 0.100 mol. [1]
(c) From equation: MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O, mole ratio MgO : HNO₃ = 1 : 2.
0.0500 mol MgO requires 0.100 mol HNO₃. Available HNO₃ = 0.100 mol. [1]
The amounts are exactly stoichiometric. Neither reactant is in excess; both are completely used up. [1]
Marking note: Award 1 mark for correct mole ratio comparison, 1 mark for correct conclusion.
(d) Moles of Mg(NO₃)₂ formed = moles of MgO reacted = 0.0500 mol. [1]
Mass of Mg(NO₃)₂ = 0.0500 × 148.0 = 7.40 g. [1]
Marking note: Award 1 mark for correct moles, 1 mark for correct mass with units.
17. (a) Hydrogen chloride gas consists of covalent HCl molecules. There are no mobile ions or free electrons to carry charge, so it does not conduct electricity. [1]
Hydrochloric acid is HCl gas dissolved in water. In water, HCl molecules dissociate completely to form H⁺(aq) and Cl⁻(aq) ions. These mobile ions can carry charge, so the solution conducts electricity. [1]
(b) Ethanoic acid (CH₃COOH) is a weak acid because it ionises partially in water. Only a small fraction of CH₃COOH molecules dissociate to form H⁺ and CH₃COO⁻ ions. The concentration of H⁺ ions is relatively low. [1]
Hydrochloric acid (HCl) is a strong acid because it ionises completely in water. All HCl molecules dissociate to form H⁺ and Cl⁻ ions. The concentration of H⁺ ions is high for the same acid concentration. [1]
Both acids react with magnesium because both contain H⁺ ions that can be reduced to hydrogen gas: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). However, the reaction with ethanoic acid is slower because the concentration of H⁺ ions is lower. [1]
Marking note: Must explain the difference in ionisation extent and link to H⁺ ion concentration. The final mark is for explaining why both react despite the difference in strength.
18. (a) Copper(II) oxide is black. The solution formed is blue. [1]
(b) CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) [2]
Marking note: 1 mark for correct formulae, 1 mark for correct state symbols.
(c) Heating the filtrate evaporates water, increasing the concentration of copper(II) sulfate until the solution becomes saturated. [1]
Allowing the saturated solution to cool slowly enables large, well-formed crystals of copper(II) sulfate to form. Rapid cooling would produce small, irregular crystals. Slow cooling allows the ions to arrange themselves in an orderly crystal lattice. [1]
Marking note: Must explain both the purpose of heating (to achieve saturation) and slow cooling (for crystal formation).
(d) Washing with cold distilled water removes any soluble impurities from the surface of the crystals without dissolving a significant amount of the crystals themselves. Warm water would dissolve more of the crystals, reducing the yield. [1]
Marking note: The key point is that copper(II) sulfate is more soluble in warm water than in cold water.
19. (a) Solid sodium chloride: does not conduct electricity. The ions (Na⁺ and Cl⁻) are held in fixed positions in the giant ionic lattice and cannot move. [1]
Molten sodium chloride: conducts electricity. The ions are free to move and carry charge. [1]
Aqueous sodium chloride: conducts electricity. The ions are free to move in the solution and carry charge. [1]
Marking note: Must state observation and explanation for each state. The explanation must refer to mobile ions.
(b) 2Cl⁻(l) → Cl₂(g) + 2e⁻ [1]
Marking note: Accept Cl⁻ → ½Cl₂ + e⁻. State symbols not strictly required for molten electrolysis but good practice.
(c) In aqueous sodium chloride, both Na⁺ and H⁺ ions are attracted to the cathode. [1]
H⁺ ions are lower in the electrochemical series (less reactive) than Na⁺ ions, so H⁺ ions are preferentially discharged: 2H⁺(aq) + 2e⁻ → H₂(g). Sodium ions remain in solution. [1]
Marking note: Must refer to the selective discharge series / electrochemical series and explain that H⁺ is preferentially discharged over Na⁺.
20. (a) Moles of HCl = (0.150 × 20.0) / 1000 = 0.00300 mol. [1]
(b) From equation: K₂CO₃ + 2HCl → 2KCl + H₂O + CO₂, mole ratio K₂CO₃ : HCl = 1 : 2.
Moles of K₂CO₃ in 25.0 cm³ = 0.00300 / 2 = 0.00150 mol. [1]
(c) Concentration of K₂CO₃ = (0.00150 / 25.0) × 1000 = 0.0600 mol/dm³. [1]
(d) Moles of K₂CO₃ in 250 cm³ = 0.0600 × (250 / 1000) = 0.0150 mol. [1]
Mass of pure K₂CO₃ = 0.0150 × 138.0 = 2.07 g. [1]
Marking note: Award 1 mark for correct moles in the original solution, 1 mark for correct mass.
(e) Percentage purity = (2.07 / 4.50) × 100% = 46.0%. [1]
Marking note: Accept 46%. The answer should be given to three significant figures or as appropriate from the data.
END OF ANSWER KEY
This answer key provides model answers and marking guidance. In an actual examination, markers would use a detailed mark scheme with additional acceptable responses. Students should use this to self-assess their understanding and identify areas for improvement.