Edexcel International A Level Chemistry: Practical Skills in Chemistry II (YCH11) – 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.

These questions target Unit 6: Practical Skills in Chemistry II from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. Unit 6 is an International A2 (IA2) unit covering Units 4 and 5 practical work, including Core Practicals 9 to 16, and earlier tests. Learn the methods in the study guide and recap with the revision notes. The course hub, checklist and free diagnostics help you plan.

Data for this set: Fe = 55.8. R = 8.31 J mol⁻¹ K⁻¹. Use any other values given in a question.

Questions

1. In Core Practical 9a, why is each sample run into sodium hydrogencarbonate solution before titration? [2]

2. A 5.00 cm³ quenched sample from the iodination of propanone needs 9.20 cm³ of 0.0100 mol dm⁻³ sodium thiosulfate. Calculate [I₂] in the mixture. [3]

3. In a hydrogen peroxide-iodide clock, only the volume of dilute sulfuric acid is varied. Water is added so the total volume is always 40.0 cm³.

Volume of acid / cm³	8.0	12.0	16.0
Time / s	61	60	61

(a) Deduce, with reasoning, the order of reaction in H⁺ ions. [3] (b) Explain why water is added. [1]

4. A reaction takes 96 s to reach a fixed point at 300 K and 31 s at 320 K. Use ln(t₁/t₂) = (Ea/R)(1/T₁ − 1/T₂).

(a) Calculate the activation energy in kJ mol⁻¹. [3] (b) State why the temperature is measured in the reaction mixture, not the water bath. [1] (c) Explain why results at five temperatures give a more reliable Ea than two. [2]

5. 25.0 cm³ of a weak acid, HA, needs 23.80 cm³ of sodium hydroxide to reach the end point. 11.90 cm³ of the same alkali is added to a second 25.0 cm³ portion, and the pH is 4.21.

(a) Explain why pH = pKa for the second mixture. [2] (b) Calculate Ka. [1] (c) The pH meter reads to ±0.02. Calculate the range of possible Ka values. [2]

6. Use E⦵: Ni²⁺/Ni = −0.25 V; Ag⁺/Ag = +0.80 V.

(a) Calculate E⦵cell for the nickel-silver cell. [1] (b) Write the cell diagram. [2] (c) Explain why the salt bridge contains potassium nitrate, not potassium chloride. [1] (d) Why is a high-resistance voltmeter used? [1]

7. Three iron(II) tablets are dissolved in dilute sulfuric acid; the solution is transferred to a 100.0 cm³ volumetric flask and diluted to the mark. 20.0 cm³ portions are titrated with 0.00400 mol dm⁻³ KMnO₄. Titres: rough 14.10, then 13.60, 13.70 and 13.65 cm³.

(a) Find the mean of the concordant titres. [1] (b) Calculate the mass of iron in one tablet, in mg. [4] (c) Why is hydrochloric acid not used to acidify? [1] (d) State the colour change at the end point. [1]

8. 5.00 cm³ of a hydrogen peroxide disinfectant is diluted to 100.0 cm³. A 10.0 cm³ aliquot reacts with excess acidified potassium iodide: H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O. The iodine released needs 16.80 cm³ of 0.0500 mol dm⁻³ sodium thiosulfate.

(a) Describe how starch is used to find the end point. [2] (b) Calculate the concentration of H₂O₂ in the undiluted disinfectant. [3]

9. Tetraamminecopper(II) sulfate, [Cu(NH₃)₄]SO₄·H₂O (Mr 245.6), is prepared from 3.00 g of CuSO₄·5H₂O (Mr 249.6). 2.38 g of dry product is obtained.

(a) Describe what is seen as concentrated ammonia is added dropwise to copper(II) sulfate solution, then in excess. [2] (b) Calculate the percentage yield. [2] (c) Give one hazard of concentrated ammonia and a precaution. [1] (d) Why are the crystals washed with ethanol, not water? [1]

10. (a) A solution of salt A gives a white precipitate with NaOH(aq) that dissolves in excess; the precipitate with NH₃(aq) also dissolves in excess. With dilute nitric acid then silver nitrate, a cream precipitate forms. Identify A. [2] (b) Liquid B (C₃H₆O) forms an orange solid with 2,4-DNPH and produces a silver mirror when warmed with Tollens’ reagent. Identify B, explaining each result. [3] (c) State what you would see if B were warmed with iodine in alkali. [1]

11. 3.588 g of 2-hydroxybenzoic acid (Mr 138.0) is warmed with 5.5 cm³ of ethanoic anhydride (Mr 102.0; 1.08 g per cm³). 2.95 g of dry aspirin (Mr 180.0) is obtained. The mole ratio of acid to anhydride is 1 : 1.

(a) Show that the anhydride is in excess. [2] (b) Calculate the percentage yield. [2] (c) Describe how the crude aspirin is recrystallised. [3] (d) The product melts at 135 to 136 °C; the data value is 136 °C. What does this show? [1] (e) Ethanoic anhydride is corrosive and its vapour irritates. State a precaution. [1]

Answers

1. Sodium hydrogencarbonate neutralises the acid catalyst [1], so the reaction stops and the iodine left matches the time of quenching [1]. [2] Examiner insight: Name the catalyst being removed; “to slow the reaction” misses the point.

2. n(S₂O₃²⁻) = 0.0100 × 9.20/1000 = 9.20 × 10⁻⁵ mol [1]; n(I₂) = 4.60 × 10⁻⁵ mol [1]; [I₂] = 4.60 × 10⁻⁵ ÷ 0.00500 = 9.20 × 10⁻³ mol dm⁻³ [1]. [3] Examiner insight: Put the 1 : 2 ratio on a separate line; a reader can then follow your method even if a later number is wrong.

3. (a) Doubling the acid volume doubles [H⁺], yet 1/t stays between 0.0164 and 0.0167 s⁻¹ [1]. The rate does not depend on [H⁺] [1]: zero order [1]. (b) It makes the total volume the same in every run, so [H⁺] varies in direct proportion to the acid volume [1]. Examiner insight: Times that differ by a second are within timing error; say so rather than claiming a small trend.

4. (a) ln(96/31) = 1.130 [1]; 1/300 − 1/320 = 2.083 × 10⁻⁴ K⁻¹ [1]; Ea = 8.31 × 1.130 ÷ (2.083 × 10⁻⁴) = 45 100 J mol⁻¹ = 45 kJ mol⁻¹ [1]. (b) The mixture can be at a different temperature from the bath, and it is the mixture’s temperature that controls the rate [1]. (c) A graph uses every point, so one poor reading matters less [1], and anomalies show up off the line [1]. Examiner insight: Convert temperatures to kelvin before taking 1/T; using °C gives an Ea that is badly wrong.

5. (a) Half of the acid has been converted to its anion, giving [HA] = [A⁻] [1]. In Ka = [H⁺][A⁻]/[HA] the ratio cancels, so Ka = [H⁺] and pKa = pH [1]. (b) Ka = 10^(−4.21) = 6.17 × 10⁻⁵ mol dm⁻³ [1]. (c) pH 4.19 gives 6.46 × 10⁻⁵ [1]; pH 4.23 gives 5.89 × 10⁻⁵ mol dm⁻³ [1]. Examiner insight: An “explain” needs the reason: show that [HA] = [A⁻] makes the ratio cancel, not just the statement pH = pKa.

6. (a) 0.80 − (−0.25) = +1.05 V [1]. (b) Ni(s) | Ni²⁺(aq) ‖ Ag⁺(aq) | Ag(s): nickel on the left as the negative electrode [1], with correct phase lines and the salt bridge as a double line [1]. (c) Chloride would precipitate silver chloride, lowering [Ag⁺] [1]. (d) It draws almost no current, so it measures the full emf [1]. Examiner insight: Put the more positive half-cell on the right; a reversed diagram gives a negative emf that contradicts your part (a).

7. (a) All three titres are within 0.10 cm³: mean = 13.65 cm³ [1]. (b) n(MnO₄⁻) = 0.00400 × 13.65/1000 = 5.46 × 10⁻⁵ mol [1]; n(Fe²⁺) = 5 × 5.46 × 10⁻⁵ = 2.73 × 10⁻⁴ mol [1]; in 100.0 cm³ = 1.365 × 10⁻³ mol [1]; mass per tablet = 1.365 × 10⁻³ × 55.8 ÷ 3 = 0.0254 g = 25.4 mg [1]. (c) Manganate(VII) would oxidise chloride ions, so the titre would be too large [1]. (d) Colourless (pale yellow) to the first permanent pale pink [1]. Examiner insight: Leave out the rough titre and say why; check which titres are concordant before averaging.

8. (a) Run in thiosulfate until the brown iodine has faded to pale yellow, then add starch, which turns blue-black [1]; the end point is the first drop that removes the blue-black colour completely [1]. (b) n(S₂O₃²⁻) = 0.0500 × 16.80/1000 = 8.40 × 10⁻⁴ mol; n(I₂) = n(H₂O₂) = 4.20 × 10⁻⁴ mol per aliquot [1]; in 100.0 cm³, which came from 5.00 cm³ of disinfectant: 4.20 × 10⁻³ mol [1]; c = 4.20 × 10⁻³ ÷ 0.00500 = 0.840 mol dm⁻³ [1]. Examiner insight: Label both scale steps: aliquot to flask, then flask to the 5.00 cm³ sample.

9. (a) A pale blue precipitate forms [1], which redissolves in more ammonia, leaving a deep blue solution [1]. (b) Theoretical = (3.00 ÷ 249.6) × 245.6 = 2.952 g [1]; yield = 2.38 ÷ 2.952 × 100 = 80.6% [1]. (c) Corrosive, with an irritating vapour: use a fume cupboard and wear gloves and goggles [1]. (d) The complex is soluble in water but much less soluble in ethanol, so less is lost [1]. Examiner insight: Name the specific hazard and a precaution that matches it; “wear goggles” alone does not deal with a vapour.

10. (a) Zn²⁺: a white hydroxide that dissolves in excess of both reagents [1]. Br⁻ from the cream precipitate: zinc bromide [1]. (b) Orange precipitate: a carbonyl group [1]. Silver mirror: an aldehyde [1]. B is propanal, CH₃CH₂CHO [1]. (c) No pale yellow precipitate, because propanal has no CH₃CO– group [1]. Examiner insight: Tie each observation to a deduction; a correct name with no reasoning does not show how the evidence was used.

11. (a) n(acid) = 3.588 ÷ 138.0 = 0.0260 mol [1]; n(anhydride) = (5.5 × 1.08) ÷ 102.0 = 0.0582 mol, more than 0.0260 mol [1]. (b) Theoretical = 0.0260 × 180.0 = 4.68 g [1]; yield = 2.95 ÷ 4.68 × 100 = 63.0% [1]. (c) Dissolve in the minimum volume of hot solvent [1]; filter hot to remove insoluble impurities, then cool so crystals form [1]; collect the crystals by suction filtration, rinse them with a small amount of ice-cold solvent and let them dry [1]. (d) The narrow range at the data value shows the aspirin is pure [1]. (e) Measure and use it in a fume cupboard, wearing gloves [1]. Examiner insight: “Minimum” and “hot” both matter in recrystallisation; too much solvent leaves product dissolved and lowers the yield.

Where marks are usually lost

• Forgetting the 1 : 2 (iodine-thiosulfate) or 1 : 5 (manganate-iron) ratio.
• Averaging a rough titre with concordant ones.
• Using °C in 1/T, or giving Ea in a unit the question did not ask for.
• Taking pKa from the end-point pH rather than at half the end-point volume.
• Naming a hazard without a precaution that fits it.
• Reversing the electrodes in a cell diagram.

Next steps

• Recap with the Unit 6 revision notes.
• Reread any weak area in the Unit 6 study guide.
• Browse every unit on the course hub and tick off the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Unit 6: Practical Skills in Chemistry II, with Core Practicals 9 to 16 from Units 4 and 5.