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Chemistry 9701/52 May/June 2021: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 5 Planning, Analysis and Evaluation, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_s21_qp_529701_s21_ms_529701/52/M/J/21

Paper
Paper 5 Planning, Analysis and Evaluation
Total marks
30
Questions
2
Code on the paper
9701/52/M/J/21
9701_s21_ms_52.pdf

9701/52 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 114 marks

Question

A student is asked to find the enthalpy change for the reaction between anhydrous magnesium sulfate and water. MgSO4(s) + 7H2O(l) → MgSO4•7H2O(s) This enthalpy change cannot be measured directly. (a) Predict whether the enthalpy change for this reaction is positive or negative. Explain the reason for your prediction. prediction …… explanation …… [1] (b) The student decided to do two separate experiments. Experiment 1 To find the enthalpy change of solution of anhydrous magnesium sulfate, MgSO4(s), 0.0250 moles of MgSO4(s) are dissolved in 50.0 cm3 distilled water. Experiment 2 To find the enthalpy change of solution of hydrated magnesium sulfate, MgSO4•7H2O(s), 0.0250 moles of MgSO4•7H2O(s) are dissolved in 50.0 cm3 distilled water. The results for Experiment 1 are shown in the graph of temperature against time on page 3. (i) Draw and extrapolate the cooling curve back to 180 seconds. Determine the temperature change during the reaction. temperature change = …… °C [1] (ii) The anhydrous magnesium sulfate was not added when the timing started. Explain why. . …… . …… [1] 29.0 28.0 27.0 26.0 25.0 temperature / °C 24.0 23.0 22.0 21.0 20.0 19.0 0 60 120 180 240 300 360 420 480 time / s (iii) 3.01 g (0.0250 mol) of anhydrous magnesium sulfate is weighed. Outline the next steps that should be taken in order to obtain the results in Experiment 1. Write your answer using a series of numbered steps. . …… . …… . …… . …… . …… . …… . …… [4] (c) (i) The student realised that when dissolving 0.0250 moles MgSO4•7H2O(s), the amount of water present in the compound alters the total volume of water used in Experiment 2. Calculate the volume of distilled water needed to make the total volume of water 50.00 cm3 in Experiment 2. Give your answer to the nearest 0.05 cm3. Assume that 1.00 cm3 of distilled water has a mass of 1.00 g. [Ar: O, 16.0; H, 1.0] volume of distilled water = …… cm3 [1] (ii) State which piece of apparatus should be used to measure the volume of distilled water in (c)(i). Explain your answer. apparatus . …… explanation …… [2] (d) The temperature change when 0.0250 moles of MgSO4•7H2O(s) is added to the water is very small. Suggest how the student should modify the experimental procedure to make the temperature change larger. …… [1] (e) (i) The energy released by 0.0250 moles of MgSO4(s), in Experiment 1, is 2125 J. The energy absorbed by 0.0250 moles of MgSO4•7H2O(s), in Experiment 2, is 477.5 J. Calculate the enthalpy change, ∆H, for the reaction. Include a sign in your answer. Give your answer to one decimal place. MgSO4(s) + 7H2O(l) → MgSO4•7H2O(s) enthalpy change, ∆H = …… kJ mol–1 [2] (ii) The student noticed that some MgSO4(s) in Experiment 1 was left undissolved. State and explain the effect this would have on the value of the enthalpy change for the reaction in Experiment 1. effect …… explanation …… [1] [Total: 14]

Mark scheme

1(a) negative 1 AND bond formation is happening (between MgSO4 and H2O) 1(b)(i) 8.05(°C) 1 1(b)(ii) to allow the temperature of the water to equilibrate with room temperature 1 1(b)(iii) M1: (measure) 50 cm3 (distilled) water and transfer into a polystyrene / polyfoam / plastic cup 4 M2: measure the temperature of the water at (30 second) intervals (until constant / before adding MgSO4) M3: MgSO4 added at 180 s / 3 minutes M4: stir and measure the temperature of the water at (30 second) intervals (until at least 8 mins have elapsed since time 0) 1(c)(i) 0.025 × 7 × 18 = 3.15 1 AND 50.00 – 3.15 = 46.85 (cm3) 1(c)(ii) M1: burette 2 M2: (apparatus) measures to 0.05 (cm3) 1(d) use a larger number of moles (of hydrated magnesium sulfate/solid) 1 OR use a smaller volume of distilled water 1(e)(i) M1: 2 ΔHexp1 = –2125 / 1000 × 0.025 = –85.0 and ΔHexp2 = 477.5 / 1000 × 0.025 = 19.1 M2: ΔH = ΔHexp1 – ΔHexp2 –85.0 – (+)19.1 = –104.1 kJ mol–1 1(e)(ii) less exothermic 1 AND fewer moles reacted

Question 216 marks

Question

A student is asked to determine the acid dissociation constant, Ka, for butanoic acid, CH3CH2CH2COOH. [CH3CH2CH2COO–][H+] Ka = [CH3CH2CH2COOH] The student is told to measure the pH of eight buffer solutions containing CH3CH2CH2COOH and the salt sodium butanoate, CH3CH2CH2COO–Na+. The salt provides butanoate ions, CH3CH2CH2COO–, as the base. Each buffer solution contains a different ratio of [CH3CH2CH2COOH] to [CH3CH2CH2COO–]. Each buffer solution is prepared by mixing different volumes of distilled water, 2.00 mol dm–3 CH3CH2CH2COOH and 2.00 mol dm–3 CH3CH2CH2COO–Na+. (a) Pure butanoic acid must be kept away from naked flames. Explain why. …… [1] (b) (i) What is the maximum volume of a 2.00 mol dm–3 solution that can be prepared using 55.0 g of CH3CH2CH2COO–Na+(s)? [Ar: Na, 23.0; O, 16.0; C, 12.0; H, 1.0] volume = …… cm3 [1] (ii) A student is given 55.0 g of CH3CH2CH2COO–Na+(s) in a beaker. Describe the next steps the student should take to make a 2.00 mol dm–3 solution of the volume calculated in (b)(i). Give the name and capacity of any key apparatus which should be used. Write your answer as a series of numbered steps. . …… . …… . …… . …… . …… . …… . …… . …… . …… [3] (iii) Complete the table to show the volumes of CH3CH2CH2COO–Na+(aq) and distilled water that would be needed to provide the stated number of moles of CH3CH2CH2COOH and CH3CH2CH2COO– for a 50.0 cm3 buffer solution. volume of volume of volume of moles of 2.00 mol dm–3 moles of 2.00 mol dm–3 distilled water CH3CH2CH2COOH CH3CH2CH2COOH CH3CH2CH2COO– CH3CH2CH2COO–Na+ / cm3 / cm3 / cm3 0.050 25.0 0.005 0.050 25.0 0.008 0.050 25.0 0.010 0.050 25.0 0.025 0.030 15.0 0.050 0.010 5.0 0.050 0.006 3.0 0.050 0.004 2.0 0.050 [2] (c) The pH of each buffer solution is measured. The value of the pH is recorded in the table along with the number of moles of the CH3CH2CH2COOH and CH3CH2CH2COO– in each 50 cm3 solution of buffer. [acid] (i) Complete the table. Give your values of –log to two decimal places. [base] moles of acid, moles of base, ratio of [acid] –log pH CH3CH2CH2COOH CH3CH2CH2COO– [acid]/[base] [base] 0.050 0.005 10.00 3.82 0.050 0.008 6.25 4.01 0.050 0.010 5.00 4.12 0.050 0.025 2.00 4.70 0.030 0.050 0.60 5.04 0.010 0.050 0.20 5.52 0.006 0.050 0.12 5.74 0.004 0.050 0.08 5.91 [1] [acid] (ii) Plot a graph of pH (y-axis) against –log (x-axis) on the grid on page 9. [base] Use a cross (×) to plot each data point. Draw a line of best fit. [2] (iii) Circle the point on the graph you consider to be most anomalous. Suggest one reason why this anomaly may have occurred during this experimental procedure. Assume no error was made in recording the pH. . …… . …… [2] (iv) When the concentration of acid is equal to the concentration of base, pH = pKa. Use this information and your graph to find the value for the pKa. pKa = …… [1] 6.0 5.8 5.6 5.4 5.2 5.0 pH 4.8 4.6 4.4 4.2 4.0 3.8 3.6 –1.4 –1.2 –1.0 –0.8 –0.6 –0.4 –0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 [acid] –log [base] (v) Use your answer to (c)(iv) to calculate the value of Ka. State the units of Ka. Give your answer to three significant figures. pKa = –log Ka Ka = …… units = …… [2] (d) The value of pKa is lower when the experiment is repeated at a higher temperature. What does this tell you about the enthalpy of dissociation of CH3CH2CH2COOH? Explain your answer. …… [1] [Total: 16]

Mark scheme

2(a) butanoic acid is flammable 1 2(b)(i) 250.(0) cm3 1 2(b)(ii) M1: dissolve the solid / sodium butanoate (55.0 g) (in the beaker) (using a small volume of) (distilled) water M2: transfer / add to a (250 cm3 volumetric) flask AND rinse (with distilled water) M3: top the 250 cm3 volumetric flask up to mark (with distilled water) 2(b)(iii) 2 moles of volume of moles of volume of volume of acid acid base base water 0.050 25.0 0.005 2.5 22.5 0.050 25.0 0.008 4.0 21.0 0.050 25.0 0.010 5.0 20.0 0.050 25.0 0.025 12.5 12.5 0.030 15.0 0.050 25.0 10.0 0.010 5.0 0.050 25.0 20.0 0.006 3.0 0.050 25.0 22.0 0.004 2.0 0.050 25.0 23.0 M1: column 4 M2: column 5 2(c)(i) 1 ratio of –log [acid] moles of acid moles of base pH [acid] / [base] [base] 0.050 0.005 10.00 –1.00 3.82 0.050 0.008 6.25 –0.80 4.01 0.050 0.010 5.00 –0.70 4.12 0.050 0.025 2.00 –0.30 4.70 0.030 0.050 0.60 (+)0.22 5.04 0.010 0.050 0.20 (+)0.70 5.52 0.006 0.050 0.12 (+)0.92 5.74 0.004 0.050 0.08 (+)1.10 5.91 2(c)(ii) M1: 8 points plotted correctly 2 M2: straight line of best fit drawn which must extend to cover all data points 2(c)(iii) M1: Plot at –0.3, 4.70 2 M2: (actual) moles of acid (in the buffer) is lower (than recorded in the table) OR (actual) moles of base (in the buffer) is higher (than recorded in the table) OR concentration of acid (in buffer) lower (than they think for this reading) OR concentration of base (in buffer) higher (than they think for this reading) OR volume of acid used (to make the buffer) is lower (than recorded in table / 25 cm3) OR volume of base used (to make the buffer) is higher (than recorded in table / 12.5 cm3) 2(c)(iv) pKa = 4.82 1 2(c)(v) M1: Ka = 1.51 × 10–5 (must be 3 SF) 2 M2: mol dm–3 2(d) endothermic 1 AND Ka increases (with increasing temperature)

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On this page, free: the mark scheme for AS & A Level Chemistry 9701/52, May/June 2021 (9701_s21_ms_52), and the question paper (9701_s21_qp_52), to read or download, with each question's marking points written out underneath.

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Chemistry 9701/52 for May/June 2021 is out of 30 marks, across 2 questions.

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