Unit 5 (CH05): Practical & Synoptic Capstone Exam Practice | OxfordAQA A-Level

Hydrated ethanedioic acid, H₂C₂O₄ • xH₂O, is a weak diprotic solid acid used to standardise solutions of sodium hydroxide. A student is required to prepare 250.0 cm³ of a standard solution of hydrated ethanedioic acid and use it to determine the exact concentration of a commercial sodium hydroxide solution.

(a) Outline the precise experimental steps the student must follow to prepare exactly 250.0 cm³ of this standard solution, starting from a solid sample of hydrated ethanedioic acid in a weighing bottle. You must include details of the apparatus used and how the student ensures the concentration is accurate and uniform. [4 marks]

(b) The student weighs a sample of hydrated ethanedioic acid using an analytical balance. The recorded results are:
• Mass of empty weighing bottle + solid acid = 14.862 g
• Mass of bottle after transferring solid acid = 13.287 g
Each mass reading on this balance has an uncertainty of ±0.001 g. Calculate the percentage uncertainty in the mass of the hydrated acid transferred. Show your working. [2 marks]

(c) The student pours the 250.0 cm³ volumetric solution into a beaker. Using a pipette, 25.0 cm³ of this acid solution is transferred into a conical flask. The 250.0 cm³ volumetric flask used has a manufacturer uncertainty of ±0.15 cm³, and the 25.0 cm³ pipette has an uncertainty of ±0.03 cm³. Compare the percentage uncertainties of the volumetric flask and the pipette. [2 marks]

(d) The student titrates the 25.0 cm³ samples of the acid solution with the sodium hydroxide solution from a burette. The neutralisation reaction is:
\[ \text{H}_2\text{C}_2\text{O}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{C}_2\text{O}_4\text{(aq)} + 2\text{H}_2\text{O(l)} \] The student obtains a mean concordant titre of 21.40 cm³ of the sodium hydroxide solution. The mass of anhydrous ethanedioic acid, H₂C₂O₄ (M_r = 90.0), present in the 250.0 cm³ volumetric flask was exactly 1.125 g. Calculate the concentration, in mol dm^-3, of the sodium hydroxide solution. Give your answer to 3 significant figures. [3 marks]

(e) State a suitable indicator the student could use for this titration of a weak diprotic acid with a strong base, and state the color change at the endpoint. [1 mark]

Show Mark Scheme

(a) Standard solution preparation [4 marks]:

M1 (Weighing by difference): Weigh the weighing bottle with solid acid, transfer the solid into a beaker, and reweigh the empty weighing bottle (recording the difference in mass) [1].
M2 (Dissolution): Add deionised water (approx 100 cm³) to the beaker and stir thoroughly with a glass rod until all solid dissolves completely [1].
M3 (Quantitative transfer): Pour solution into a 250.0 cm³ volumetric flask using a funnel. Rinse the beaker, glass rod, and funnel with deionised water and add all washings to the flask [1].
M4 (Meniscus & Inversion): Add deionised water until the bottom of the meniscus touches the 250 cm³ calibration mark at eye level. Insert the stopper and invert the flask at least 10 times to ensure a uniform, homogeneous concentration [1].

Marking note: Reject filling past the line. Must mention inversion/homogenisation for M4.

(b) Percentage uncertainty in mass [2 marks]:

M1: Mass transferred = 14.862 - 13.287 = 1.575 g. Total absolute uncertainty for two balance readings = 2 × 0.001 = 0.002 g [1].
M2: Percentage uncertainty = (0.002 / 1.575) × 100% = 0.127% [1].

Marking note: Allow 0.13% or 0.1270%. If candidate calculates (0.001 / 1.575) × 100 = 0.0635%, award 1 mark max for ECF.

(c) Percentage uncertainty comparison [2 marks]:

M1: Volumetric flask % = (0.15 / 250.0) × 100 = 0.060%. Pipette % = (0.03 / 25.0) × 100 = 0.120% [1].
M2: The pipette has double the percentage uncertainty of the volumetric flask (0.120% vs 0.060%), so the pipette contributes the larger relative uncertainty [1].

(d) Concentration of NaOH [3 marks]:

M1: Moles of H₂C₂O₄ in 250 cm³ = 1.125 / 90.0 = 0.0125 mol [1].
M2: Moles of acid in 25.0 cm³ aliquot = 0.0125 × (25.0 / 250.0) = 0.00125 mol. Moles of NaOH reacting = 2 × 0.00125 = 0.00250 mol [1].
M3: Concentration of NaOH = moles / volume = 0.00250 / (21.40 × 10^-3) = 0.117 mol dm^-3 [1].

Marking note: Allow ECF throughout. Final answer must be given to 3 significant figures (0.117). Omission of 1:2 ratio gives 0.0584 mol dm^-3 (scores M1 and M3, 2/3).

(e) Indicator and color change [1 mark]:

M1: Phenolphthalein. Color change: colorless to pale pink (or permanent faint pink) [1].

Marking note: Reject methyl orange (inappropriate pH transition range for weak acid - strong base titration). Reject clear for colorless.

Examiner tip: Always double the balance uncertainty when weighing by difference because two independent measurements (bottle before and bottle after) are taken. For weak acid - strong base titrations, the equivalence point occurs at pH 8 to 9, so phenolphthalein is mandatory.

Benzocaine (ethyl 4-aminobenzoate) is a local anaesthetic synthesised by the esterification of 4-aminobenzoic acid with ethanol:
\[ \text{H}_2\text{N-C}_6\text{H}_4\text{-COOH} + \text{CH}_3\text{CH}_2\text{OH} \overset{\text{H}_2\text{SO}_4}{\rightleftharpoons} \text{H}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + \text{H}_2\text{O} \] After synthesis, the crude benzocaine solid must be recrystallised to obtain a pure sample.

(a) Outline the experimental procedure the student should follow to recrystallise the crude benzocaine solid to obtain a pure, dry sample. Explain the scientific reason for each step in your procedure. [4 marks]

(b) The student wishes to verify the purity of the recrystallised benzocaine by measuring its melting point. Describe how the student would set up and carry out an accurate melting point determination, and state two observations that would confirm the sample is pure. [3 marks]

(c) In an experiment, a student starts with 5.48 g of 4-aminobenzoic acid (M_r = 137.0) and an excess of ethanol. After recrystallisation and drying, the student collects 4.29 g of pure benzocaine (M_r = 165.0). Calculate the percentage yield of benzocaine obtained. Show your working clearly. [3 marks]

(d) During recrystallisation, the filtered crystals are washed with a small volume of ice-cold solvent. Explain why the wash solvent must be:
(i) Ice-cold.
(ii) Used in a minimum volume. [2 marks]

(e) Thin-layer chromatography (TLC) is performed on both the crude product and the purified product. State how the TLC results would confirm that recrystallisation successfully removed impurities. [1 mark]

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(a) Recrystallisation protocol [4 marks]:

M1: Dissolve the crude solid in the minimum volume of hot solvent (to create a concentrated, saturated solution so maximum crystallisation occurs on cooling) [1].
M2: Filter the hot solution through fluted filter paper or pre-warmed funnel (to remove any insoluble impurities) [1].
M3: Allow filtrate to cool slowly to room temperature and then in an ice bath (to allow crystals of pure product to precipitate while soluble impurities remain dissolved in solution) [1].
M4: Filter crystals under reduced pressure using a Buchner flask and funnel, wash with a small volume of ice-cold solvent, and dry between filter papers or in a desiccator/warm oven [1].

Marking note: Each step must have its scientific purpose stated to achieve the mark.

(b) Melting point determination [3 marks]:

M1: Pack a small amount of dry powder into a capillary tube sealed at one end (to a depth of 2-3 mm) [1].
M2: Place in a melting point apparatus (or oil bath/Thiele tube) and heat slowly near the expected melting point, recording the temperature at which melting begins and ends [1].
M3 (Purity observations): The sample melts sharply over a narrow range (within 1 to 2 °C) AND the melting point matches the accepted literature value for benzocaine (88-90 °C) [1].

(c) Percentage yield calculation [3 marks]:

M1: Moles of 4-aminobenzoic acid = 5.48 / 137.0 = 0.0400 mol [1].
M2: Theoretical mass of benzocaine = 0.0400 × 165.0 = 6.60 g [1].
M3: Percentage yield = (4.29 / 6.60) × 100% = 65.0% [1].

Marking note: Allow ECF throughout. Award full 3 marks for correct final answer 65.0%.

(d) Wash solvent rationale [2 marks]:

(i) Ice-cold: To minimise dissolution of the purified benzocaine crystals during washing, preventing yield loss [1].
(ii) Minimum volume: To wash away surface liquid and soluble impurities without redissolving significant mass of crystals [1].

(e) TLC confirmation [1 mark]:

M1: The purified sample will show only a single spot on the TLC plate, whereas the crude product will show two or more spots [1].

Examiner tip: Impure organic solids show a depressed and broad melting point range. A sharp melting range within 1 °C of the literature value is the primary experimental benchmark of purity. In recrystallisation questions, always justify "minimum volume of hot solvent".

The concentration of transition metal complexes in aqueous solution, such as the deep-blue tetramminecopper(II) complex, [Cu(NH₃)₄(H₂O)₂]²⁺(aq), can be determined with high precision using colorimetry.

(a) State the fundamental physical principle of colorimetry. Describe how a colorimeter is calibrated and used to determine the unknown concentration of copper(II) ions in an aqueous sample. [4 marks]

(b) The [Cu(NH₃)₄(H₂O)₂]²⁺ complex solution appears deep blue. State which color filter should be chosen for the colorimeter and explain why this filter gives the highest measurement sensitivity. [2 marks]

(c) A student prepares standard solutions using a 0.100 mol dm^-3 stock solution of copper(II) ions. The student pipettes a volume of the stock solution into a 50.0 cm³ volumetric flask and makes up to the mark with deionised water. Calculate the volume, in cm³, of the 0.100 mol dm^-3 stock solution required to prepare 50.0 cm³ of a 0.0150 mol dm^-3 standard solution. [2 marks]

(d) The student measures the absorbance of each standard solution and plots an absorbance against concentration calibration graph. The best-fit line passes through the origin and gives the equation:
\[ \text{Absorbance} = 6.25 \times \text{Concentration (mol dm}^{-3}\text{)} \] An unknown copper(II) solution is treated with excess ammonia, placed in the colorimeter cuvette, and gives an absorbance reading of 0.540. Calculate the concentration of copper(II) ions in the cuvette. Explain why a linear relationship is expected at low concentrations and why deviations occur at high concentrations. [3 marks]

Show Mark Scheme

(a) Colorimetry principles and procedure [4 marks]:

M1: Colorimetry measures the absorbance of light; light absorbance is directly proportional to the concentration of the colored absorbing species (Beer-Lambert law) [1].
M2: Zero the colorimeter with a cuvette containing a blank (deionised water or solvent) [1].
M3: Measure the absorbance of several standard solutions of known copper(II) concentration and plot a calibration curve of absorbance (y-axis) against concentration (x-axis) [1].
M4: Measure the absorbance of the unknown sample and read the corresponding concentration directly from the calibration curve (or calculate from the line equation) [1].

(b) Filter selection [2 marks]:

M1: Red filter (or orange/red filter, 600-650 nm) [1].
M2: Red is the complementary color to blue; the blue solution strongly absorbs red light and transmits blue light, producing the maximum change in absorbance per unit change in concentration [1].

(c) Volumetric dilution calculation [2 marks]:

M1: Use dilution formula: C₁V₁ = C₂V₂ ⇒ 0.100 × V₁ = 0.0150 × 50.0 [1].
M2: V₁ = (0.0150 × 50.0) / 0.100 = 7.50 cm³ [1].

(d) Concentration calculation and linearity [3 marks]:

M1: Concentration = Absorbance / 6.25 = 0.540 / 6.25 = 0.0864 mol dm^-3 [1].
M2: At low concentrations, the probability of photon absorption is directly proportional to the number of absorbing ions per unit path length (Beer-Lambert law holds) [1].
M3: At high concentrations, electrostatic interactions between ions, complex aggregation, or altered refractive index cause deviations from linearity [1].

Examiner tip: A common student trap is choosing a filter of the same color as the solution. Always select the complementary color: blue solutions absorb red/orange light; yellow solutions absorb violet/blue light. A blank cuvette must always be used to zero the instrument.

A student investigates the kinetics of the reaction between sodium thiosulfate and hydrochloric acid:
\[ \text{Na}_2\text{S}_2\text{O}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{S(s)} + \text{SO}_2\text{(g)} + \text{H}_2\text{O(l)} \] The reaction rate is studied using the "vanishing cross" method. Solid sulfur precipitates during the reaction and clouds the mixture. The time taken (\(t\), in seconds) for a black cross viewed through the conical flask to disappear is measured.

(a) State the key apparatus components required to carry out this reaction and record the reaction time. Describe how the apparatus is assembled. [3 marks]

(b) Explain why the reciprocal of time, \(1/t\), can be taken as a direct measure of the initial rate of reaction. State the fundamental assumption required for this approximation. [2 marks]

(c) The student repeats the experiment at three different temperatures, keeping the volumes and concentrations of all reactants constant. The results are recorded in the table below:

Temperature (\(T\) / K)	Time (\(t\) / s)	\(1/T\) (\(\text{K}^{-1}\))	\(\ln(1/t)\)
293	58.0	\(3.41 \times 10^{-3}\)	\(-4.06\)
313	15.0	\(3.19 \times 10^{-3}\)	\(-2.71\)
333	4.5	\(3.00 \times 10^{-3}\)	\(-1.50\)

The Arrhenius relationship can be expressed as: \[ \ln\left(\frac{1}{t}\right) = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \text{constant} \] Using the data for 293 K and 333 K, calculate the activation energy, \(E_a\), of this reaction in kJ mol^-1.
The gas constant, \(R = 8.31\text{ J K}^{-1}\text{mol}^{-1}\). [4 marks]

(d) The student uses a digital stopwatch that measures time to the nearest 0.01 s. Explain why the actual experimental uncertainty in the measurement of \(t\) is much larger than the precision of the stopwatch, and describe one experimental modification that would improve the reliability and accuracy of this measurement. [3 marks]

Show Mark Scheme

(a) Apparatus setup [3 marks]:

M1: Conical flask containing the reaction mixture placed directly on top of white paper / tile marked with a clear black cross [1].
M2: Stopwatch / digital timer started immediately upon adding the hydrochloric acid and swirling the flask [1].
M3: Eye positioned vertically above the mouth of the conical flask to observe the cross from a consistent viewpoint until obscured [1].

(b) 1/t rate approximation [2 marks]:

M1: Rate is defined as change in concentration divided by time: \(\text{Rate} \propto \frac{\Delta[\text{S}]}{t}\) [1].
M2: The amount of sulfur precipitate needed to completely obscure the cross is fixed and constant for every run (\(\Delta[\text{S}] = \text{constant}\)), so rate is proportional to \(1/t\) [1].

(c) Activation energy calculation [4 marks]:

M1 (Gradient calculation): \[ \text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1.50 - (-4.06)}{(3.00 - 3.41) \times 10^{-3}} = \frac{2.56}{-0.41 \times 10^{-3}} = -6244\text{ K} \quad \mathbf{[1]} \]
M2 (Relation to Arrhenius equation): \[ \text{Gradient} = -\frac{E_a}{R} \implies E_a = -\text{Gradient} \times R \quad \mathbf{[1]} \]
M3 (Energy in Joules): \[ E_a = -(-6244) \times 8.31 = +51888\text{ J mol}^{-1} \quad \mathbf{[1]} \]
M4 (Conversion to kJ mol^-1): \[ E_a = +51.9\text{ kJ mol}^{-1} \quad (\text{accept } 51.8 \text{ to } 52.0\text{ kJ mol}^{-1}) \quad \mathbf{[1]} \]

Marking note: Allow ECF if gradient is calculated with an arithmetic error. Negative final answer loses M4.

(d) Experimental uncertainty and improvement [3 marks]:

M1: Human reaction time when starting and stopping the stopwatch introduces an uncertainty of approx ±0.2 s, which exceeds the 0.01 s stopwatch resolution [1].
M2: The endpoint judgment is subjective; deciding the exact instant the cross is completely obscured varies between trials and observers [1].
M3 (Improvement): Use a light sensor connected to a computer / datalogger placed on one side of the beaker with a light source on the opposite side to record light transmission objectively [1].

Examiner tip: Never confuse instrument resolution with experimental uncertainty. A stopwatch measuring to 0.01 s does not eliminate human reaction time or endpoint subjectivity. When calculating activation energy from Arrhenius gradients, always remember to convert J mol^-1 to kJ mol^-1.

A commercial liquid plant food contains iron(II) ions, Fe²⁺(aq). A student determines the concentration of iron(II) ions in the plant food by titrating acidified 25.0 cm³ samples with a standard 0.0200 mol dm^-3 solution of potassium manganate(VII), KMnO₄.

(a) Write the ionic half-equations for:
(i) The reduction of acidified manganate(VII) ions, MnO₄^-(aq), to manganese(II) ions.
(ii) The oxidation of iron(II) ions, Fe²⁺(aq), to iron(III) ions.
(iii) Deduce the balanced overall ionic redox equation for the titration reaction. [3 marks]

(b) The student acidifies the iron(II) solution in the conical flask using an excess of dilute sulfuric acid. Explain why dilute sulfuric acid is suitable for this titration, and explain why hydrochloric acid cannot be used to acidify the solution. [3 marks]

(c) In the titration, a 25.0 cm³ sample of the acidified plant food solution requires exactly 18.65 cm³ of the 0.0200 mol dm^-3 KMnO₄ solution to reach the equivalence point. Calculate the concentration of iron(II) ions in the original commercial plant food solution, in g dm^-3. Give your answer to 3 significant figures.
(A_r of Fe = 55.8). [4 marks]

(d) Describe how the student would identify that the equivalence point has been reached in this titration, and explain why no external indicator is needed. [2 marks]

Show Mark Scheme

(a) Redox half-equations and overall equation [3 marks]:

(i) Reduction half-equation: \[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \mathbf{[1]} \]
(ii) Oxidation half-equation: \[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad \mathbf{[1]} \]
(iii) Overall balanced redox equation: \[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \quad \mathbf{[1]} \]

Marking note: State symbols not required. Electrons must cancel in the overall equation.

(b) Acidification rationale [3 marks]:

M1: Dilute sulfuric acid provides the H⁺ ions required for the reduction of MnO₄^- without participating in any redox reaction itself (sulfate ions are not oxidised or reduced under these conditions) [1].
M2: Hydrochloric acid cannot be used because chloride ions (Cl^-) are oxidised by the strong oxidising agent MnO₄^- to chlorine gas (Cl₂) [1].
M3: This reaction consumes additional KMnO₄ titrant, resulting in an erroneously high titre and false overestimation of iron(II) concentration (plus releases toxic chlorine gas) [1].

(c) Iron(II) concentration calculation [4 marks]:

M1 (Moles of KMnO₄): \[ \text{Moles of MnO}_4^- = 0.0200 \times \left(\frac{18.65}{1000}\right) = 3.73 \times 10^{-4}\text{ mol} \quad \mathbf{[1]} \]
M2 (Moles of Fe²⁺ in 25.0 cm³): \[ \text{Moles of Fe}^{2+} = 5 \times 3.73 \times 10^{-4} = 1.865 \times 10^{-3}\text{ mol} \quad \mathbf{[1]} \]
M3 (Concentration in mol dm^-3): \[ \text{Concentration of Fe}^{2+} = \frac{1.865 \times 10^{-3}}{0.0250} = 0.0746\text{ mol dm}^{-3} \quad \mathbf{[1]} \]
M4 (Concentration in g dm^-3): \[ \text{Concentration in g dm}^{-3} = 0.0746 \times 55.8 = 4.16\text{ g dm}^{-3} \quad (\text{allow } 4.16 \text{ to } 4.17) \quad \mathbf{[1]} \]

Marking note: Allow ECF at each stage. Final answer must be given to 3 significant figures.

(d) Endpoint detection [2 marks]:

M1: The endpoint is signaled by the first appearance of a permanent pale-pink color in the conical flask [1].
M2: No indicator is required because potassium manganate(VII) is self-indicating: deep purple MnO₄^- ions are reduced to virtually colorless Mn²⁺ ions until all Fe²⁺ is consumed, at which point the first excess drop of MnO₄^- tints the solution permanent pale pink [1].

Examiner tip: In manganate(VII) redox titrations, the stoichiometric ratio between MnO4- and Fe2+ is 1:5. Forgetting this 1:5 factor is the most common mathematical mistake. Never overshoot the endpoint to dark purple; the true endpoint is the first faint, permanent pale pink.

Unit 5: Practical & Synoptic Capstone Exam Practice

Section B: Structured Practical Questions

Question 1: Volumetric Analysis & Titration

Question 2: Organic Preparation, Recrystallisation & Purity

Question 3: Transition Metals & Colorimetry

Question 4: Kinetics & Experimental Design

Question 5: Synoptic Inorganic Redox Titration