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

This guide covers 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 with no new theory. It examines the techniques from Units 4 and 5, including Core Practicals 9 to 16, plus the tests for ions, gases and organic groups from Units 1 and 2.

See also the Unit 6 revision notes, the Unit 6 practice questions, the Unit 3 guide for AS skills, the course hub, the printable checklist and a free diagnostic.

What the specification says about Unit 6

Feature	Specification wording, summarised
Paper	A single written paper lasting 1 hour 20 minutes, with 50 marks; every question is compulsory
Share of grade	20% of IA2 and 10% of the full International A Level; all of it targets AO3 (experimental skills)
Question styles	Short-open, open-response and calculation questions
Mathematics	Level 2 (or higher) mathematics makes up at least 6 marks
Calculators	Allowed
Sessions	January, June and October

Settings may be new to you, so recall alone will not do. The specification names four areas: independent problem solving; scientific method (variables, data, evaluation, uncertainty, criticising a method); numeracy (graphs, processing, sensible significant figures, accuracy versus precision); and apparatus (choice, use, range, resolution, hazards).

Expect test observations from any unit, data from titrations, thermochemistry, equilibria and kinetics, and comments on preparations.

The eight core practicals

Core Practical	Outcome	Specification title, summarised
9a, 9b	11.12	Iodine-propanone rate by titration; an iodine clock reaction
10	11.13	Finding the activation energy of a reaction
11	14.23	Finding the Ka value for a weak acid
12	16.6	Investigating some electrochemical cells
13a, 13b	16.17	Redox titrations: iron(II) with manganate(VII); thiosulfate with iodine
14	17.33	Preparing a transition metal complex
15	19.12	Analysing some inorganic and organic unknowns
16	20.4	Preparing aspirin

The specification gives titles, not methods. The methods below are typical versions; yours may differ.

Core Practical 9a: iodination of propanone by titration

Purpose. To find the order with respect to iodine (outcome 11.6).

Method outline. Mix propanone and acid, add iodine and start a stopclock. At intervals, run a pipetted sample into sodium hydrogencarbonate solution, which removes the acid catalyst and stops the reaction. Then find how much iodine is left by titration against thiosulfate, with starch as the indicator late in the titration.

Worked example 1. 10.0 cm³ samples were titrated with 0.00500 mol dm⁻³ sodium thiosulfate. The titre was 15.80 cm³ at 2 minutes and 12.60 cm³ at 10 minutes, and a graph of all the titres against time was a straight line.

1. At 2 minutes: n(S₂O₃²⁻) = 0.00500 × 15.80/1000 = 7.90 × 10⁻⁵ mol, so n(I₂) = 3.95 × 10⁻⁵ mol (ratio 1 : 2).
2. [I₂] = 3.95 × 10⁻⁵ ÷ 0.0100 = 3.95 × 10⁻³ mol dm⁻³. The same steps at 10 minutes give 3.15 × 10⁻³ mol dm⁻³.
3. A straight line means a constant rate, so the reaction is zero order in iodine. Rate = (3.95 − 3.15) × 10⁻³ ÷ 8 = 1.0 × 10⁻⁴ mol dm⁻³ min⁻¹.

Errors. Record the time when each sample meets the hydrogencarbonate. Use a water bath to hold the temperature.

Core Practical 9b: the iodine clock

Purpose. To find the order with respect to a reactant from initial rates.

Method outline. Hydrogen peroxide oxidises iodide ions in acid. A small, fixed amount of thiosulfate removes iodine as it forms; once it is used up, starch turns blue-black. Each run reaches the same point, so rate is proportional to 1/t. Vary one volume per run and top up with water to the same total, which makes each concentration proportional to the volume used.

Worked example 2. With everything else fixed and a total volume of 50.0 cm³ each time:

Volume of H₂O₂ / cm³	10.0	15.0	20.0
Time to blue-black / s	74	49	37

Doubling the volume halves the time (74 ÷ 37 = 2.0); ×1.5 gives 74 ÷ 49 = 1.51 times the rate. Rate is proportional to [H₂O₂]: first order. Judging the colour change by eye is the main error; use a white tile.

Core Practical 10: activation energy

Purpose. To find Ea from rates at several temperatures (outcome 11.10). The Arrhenius equation is given when needed: ln k = −Ea/(RT) + constant.

Method outline. Time a fixed amount of reaction at four or five temperatures, warming the solutions separately before mixing and recording the mixture’s temperature. As 1/t is proportional to k, ln(1/t) against 1/T has gradient −Ea/R.

Worked example 3. R = 8.31 J mol⁻¹ K⁻¹.

T / K	295	305	315	325
t / s	148	72	37	19
(1/T) / 10⁻³ K⁻¹	3.390	3.279	3.175	3.077
ln(1/t)	−5.00	−4.28	−3.61	−2.94

1. The best-fit line passes through (3.390 × 10⁻³, −5.00) and (3.077 × 10⁻³, −2.95). Gradient = 2.05 ÷ (−3.13 × 10⁻⁴) = −6550 K.
2. Ea = −gradient × R = 6550 × 8.31 = 54 400 J mol⁻¹ = 54 kJ mol⁻¹.

Errors. The mixture cools after mixing, and the shortest times have the largest timing error.

Core Practical 11: Ka for a weak acid

Purpose. To find Ka, using the fact that pH = pKa when half the acid has been neutralised (outcome 14.21).

Method outline. Calibrate a pH meter with buffers. Titrate a portion of the acid with sodium hydroxide (phenolphthalein). Add exactly half that volume of alkali to a fresh, equal portion and measure the pH. Now [HA] = [A⁻], so Ka = 10^(−pH); no concentration is needed.

Worked example 4. The pH at half-neutralisation is 3.86.

1. Ka = 10^(−3.86) = 1.38 × 10⁻⁴ mol dm⁻³.
2. Second method: 0.0500 mol dm⁻³ of the acid has pH 2.58, so [H⁺] = 2.63 × 10⁻³ and Ka ≈ (2.63 × 10⁻³)² ÷ 0.0500 = 1.38 × 10⁻⁴ mol dm⁻³. They agree.
3. A pH reading of ±0.01 moves Ka between 1.35 × 10⁻⁴ and 1.41 × 10⁻⁴, about ±2%.

Errors. No calibration, reading before the value settles, or an unrinsed electrode.

Core Practical 12: electrochemical cells

Purpose. To compare measured cell emfs with values calculated from E⦵ data (16.5 to 16.8).

Method outline. A cleaned metal dips into a solution of its ions; for two ions of one element, use platinum in a solution of both. Connect the two half-cells through filter paper soaked in potassium nitrate solution, and read the emf on a voltmeter of very high resistance, which draws almost no current.

Worked example 5. Use E⦵: Cu²⁺/Cu = +0.34 V; Fe³⁺/Fe²⁺ = +0.77 V.

1. The iron half-cell is more positive, so it is the positive electrode. E⦵cell = 0.77 − 0.34 = +0.43 V.
2. Cell diagram: Cu(s) | Cu²⁺(aq) ‖ Fe³⁺(aq), Fe²⁺(aq) | Pt(s).
3. A reading of 0.40 V suggests non-standard concentrations or temperature, or an oxide film on the copper.

Core Practicals 13a and 13b: redox titrations

13a. 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Acidify with dilute sulfuric acid. Manganate(VII) is its own indicator: the end point is the first permanent pale pink. Read the top of the meniscus, because the dark solution hides the bottom.

Worked example 6. 1.40 g of steel wire reacts with excess dilute sulfuric acid, giving Fe²⁺(aq). The solution goes into a 250.0 cm³ volumetric flask and is diluted to the line. Each 25.0 cm³ aliquot needs a mean of 24.80 cm³ of 0.0200 mol dm⁻³ KMnO₄. Fe = 55.8.

1. n(MnO₄⁻) = 0.0200 × 24.80/1000 = 4.96 × 10⁻⁴ mol, so n(Fe²⁺) = 5 × 4.96 × 10⁻⁴ = 2.48 × 10⁻³ mol per aliquot.
2. Whole flask: 2.48 × 10⁻² mol; mass of iron = 2.48 × 10⁻² × 55.8 = 1.384 g.
3. Iron in the wire = 1.384 ÷ 1.40 × 100 = 98.8%.
4. Uncertainty (16.16): titre 0.40%, pipette 0.16%, flask 0.12%, mass (±0.01 g) 0.71%; total 1.4%, so 98.8 ± 1.4%. The shortfall from 100% lies inside this, so the data cannot prove the wire holds anything but iron. A larger sample would cut the biggest term, the mass.

13b. 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻. An oxidising agent releases iodine from excess potassium iodide. Run in thiosulfate until the brown fades to pale yellow, add starch (blue-black), and stop when the blue-black goes. Iodine is volatile, so titrate promptly. A worked standardisation is in the revision notes.

Core Practical 14: a transition metal complex

One suitable preparation is tetraamminecopper(II) sulfate. In a fume cupboard, pour concentrated ammonia into copper(II) sulfate solution. The first pale blue solid redissolves as more ammonia goes in, leaving an intensely blue solution (17.24). Ethanol lowers the complex’s solubility, so deep blue crystals form. Filter under reduced pressure, wash with ethanol and dry.

Worked example 7. 2.50 g of CuSO₄·5H₂O (Mr 249.6) gave 2.10 g of [Cu(NH₃)₄]SO₄·H₂O (Mr 245.6). One copper ion gives one formula unit.

1. n = 2.50 ÷ 249.6 = 1.00 × 10⁻² mol; theoretical mass = 1.00 × 10⁻² × 245.6 = 2.46 g.
2. Yield = 2.10 ÷ 2.46 × 100 = 85.4%.

Some complex stays in the filtrate. Concentrated ammonia is corrosive with an irritating vapour; ethanol is flammable.

Core Practical 15: unknowns

The Unit 3 guide lists the AS tests. The IA2 additions:

• Outcome 17.22, metal aqua ions with hydroxide and ammonia: Cu²⁺ pale blue precipitate, deep blue solution in excess ammonia; Fe²⁺ green precipitate, browning in air; Fe³⁺ red-brown precipitate; Cr³⁺ grey-green precipitate, dissolving in excess of either; Zn²⁺ white precipitate, dissolving in excess of either. The transition metals guide has all eight ions.
• Carbonyls (15.8): 2,4-DNPH gives an orange precipitate with any aldehyde or ketone; Tollens’ reagent gives a silver mirror with an aldehyde only; iodine with alkali gives a pale yellow precipitate of iodoform with ethanal and methyl ketones.

Worked example 8. Liquid Q has molecular formula C₄H₈O. With 2,4-DNPH it forms an orange solid; Tollens’ reagent stays unchanged; iodine in alkali gives a pale yellow solid. It is a ketone containing CH₃CO–: butanone, CH₃COCH₂CH₃.

Core Practical 16: aspirin

Warm 2-hydroxybenzoic acid with excess ethanoic anhydride and a drop of acid catalyst. Add cold water to destroy excess anhydride, cool in ice and filter under reduced pressure. Recrystallise from the minimum of hot solvent, dry, weigh and take the melting temperature (20.5).

Worked example 9. 2.208 g of 2-hydroxybenzoic acid (Mr 138.0) gives 2.13 g of dry aspirin (Mr 180.0). Mole ratio 1 : 1.

1. n = 2.208 ÷ 138.0 = 0.0160 mol; theoretical mass = 0.0160 × 180.0 = 2.88 g.
2. Yield = 2.13 ÷ 2.88 × 100 = 74.0%.
3. Use a data value of 136 °C. A crude sample melting over 128 to 133 °C is impure; after recrystallisation, 134 to 136 °C shows it is much purer.

Ethanoic anhydride is corrosive and its vapour irritates, so use a fume cupboard.

Common errors

• Changing a clock-reaction volume without adding water to keep the total fixed.
• Reading pH at the equivalence point instead of at half-neutralisation.
• Using a potassium chloride salt bridge with a silver half-cell (silver chloride precipitates).
• Acidifying a manganate(VII) titration with hydrochloric acid, which it oxidises.
• Adding starch at the start of a thiosulfate titration.

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.