June 2022: Paper 3, Unified chemistry

Mark scheme

Examiner's comment

Candidates found the explanation difficult, and the responses showed some misconceptions. For example, many suggested that NH3 and NH4NO3 both have hydrogen bonds. Those identifying that NH4NO3 has ionic bonding usually compared the greater strength of ionic bonding over intermolecular forces in NH3. Unfortunately, many candidates described the ionic bonds as acting between molecules. This question proved to be one of the most difficult on the paper. Understanding could be improved by first considering the following. • What is the type of bonding? • What are particles in the structure? Candidates need to be very careful when describing the two types of structure containing covalent bonds: • Simple molecular with strong covalent bonds within the molecules and weaker intermolecular bonds between the molecules in the structure Giant covalent with strong covalent bonds between the atoms in the structure. than in NH3, or that NH4NO3 has more hydrogen bonds. This response was given 0 marks.

Notes from the mark scheme and report

• NH3 simple molecular with intermolecular forces AND NH4NO3 ionic
• ionic bonds stronger/need more energy to break than intermolecular forces
• A good understanding of structure and bonding continues to be difficult for candidates, demonstrated by many incorrect explanations for the different boiling points. This is a key misconception.

Mark scheme

Examiner's comment

Considering the large number of candidates describing NH4NO3 as molecular in Question 1 (a) (i), most candidates identified here that NH4NO3 contains NH4+ and NO3– ions. Most then went on to show that the nitrogen atoms in the ions have different oxidation numbers: –3 and +5 respectively. Candidates were only given marks if both signs had been included and this was usually the case. –4 and +6 were common incorrect responses, presumably by ignoring the charges on the ions. Comparatively few candidates grouped the nitrogen atoms together and suggested that they had the same oxidation number.

Notes from the mark scheme and report

• N in NH4+ is –3 AND N in NO3– is +5
• both signs essential

Mark scheme

Examiner's comment

Question 1 (b) presented candidates with a mass spectrum of two elements in an alloy. This novel question was answered well, and the introductory part (i) provided candidates with a hint of how to approach the harder part (ii). Almost all candidates analysed the percentages to show that the brass sample contains 66% Cu and 34% Zn, the 34 being required in Question 1 (b) (ii).

Notes from the mark scheme and report

• Cu 66% (45.64 + 20.36) and Zn 34%

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

Most candidates produced a stock expression for a relative atomic mass calculation with the numerator comprising the four zinc isotopes with their respective percentage abundances, equating to 2224.28. The second mark required candidates to realise that the numerator needed to be divided by the total zinc percentage abundance of 34% to produce a value of 65.42. Less successful responses divided by 100% to obtain a value of 22.24. Almost all candidates gave their answers to the required two decimal places. The second mark proved to be a very good discriminator, with some candidates usually obtaining both marks with less successful responses still able to obtain the first mark for the numerator.

Notes from the mark scheme and report

• common errors: 22.24 (÷100 instead of ÷34, 1 mark), 64.23 (all 6 isotopes, no marks)
• divide by Zn % (34)
• 2 d.p. required (65.42)
• ECF from (b)(i)

Skills: M0.2 Use ratios, fractions and percentages; M1.2 Find arithmetic means

Mark scheme

Examiner's comment

This question tested candidates’ ability to predict splitting patterns of an organic compound in a proton NMR spectrum. The 2 marks were given for the interacting protons in environments 1 and 2, and in environments 3 and 4. Most candidates predicted the correct triplet/quartet splitting pattern for the common CH3CH2 group. The splitting pattern for the less common CH2CHO was more difficult with many predicting at least one singlet. As with Question 1 (b) (ii), the second mark proved to be a very good discriminator.

Notes from the mark scheme and report

• 1 triplet, 2 quartet, 3 doublet, 4 triplet
• marked in pairs (1 and 2
• 3 and 4)
• ALLOW numbers 3 for triplet etc

Mark scheme

Examiner's comment

This novel question required candidates to apply their knowledge and understanding of proton NMR spectroscopy to explain different chemical shifts. Most candidates used the Data Sheet in their answers, and most were then able to relate the chemical shift at 𝛿 = 2.5 ppm to an adjacent C=O group. Fewer candidates were then able to interpret the unexpected chemical shift at 𝛿 = 3.6 ppm to the presence of two adjacent C=O groups. Less successful responses often resorted to the Data Sheet a second time, suggesting the presence of an adjacent HC–O group, despite none being present in the structure. It was encouraging to see how many candidates were able to explain the two chemical shifts correctly, suggesting that most candidates are comfortable with this concept.

Notes from the mark scheme and report

• environment 2 adjacent to one C=O (HC–C=O)
• environment 3 between two C=O
• DO NOT ALLOW H–C=O or HC–O (simply read from data sheet)
• IGNORE 'next to 2 Os'

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

This question required candidates to apply their knowledge and understanding of bond angles and electron pair repulsion of NH3 and NH4+ to amino acid salts. The best candidates rose to this challenge and secured all 3 marks for correct bond angles and explanations in terms of the numbers of bonded and lone pairs around the N atoms. Overall, candidates found this question quite difficult. Many different bond angles were predicted, with 120º being the commonest incorrect H–N–H bond angle in H2NCH2COONa. The explanation for 120º was in terms of three bonding pairs and no lone pairs. 104.5º was also seen, presumably relating H2N to H2O. The 109.5º bond angle was correct more often, as was its explanation in terms of 4 bonding pairs. Many successful responses showed working on diagrams in which bonded and lone pairs had been included. This strategy will have helped candidates in their conclusions.

Notes from the mark scheme and report

• 107° for NH2 (3 bonded pairs AND 1 lone pair) and 109.5° for NH3+ (4 bonded pairs)
• 'bonded' essential
• IGNORE electron regions/density
• IGNORE shapes

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures

Mark scheme

Examiner's comment

Candidates were asked to predict an unfamiliar equation from provided information and to draw structures of square planar complexes containing an amino acid. Candidates found the structures easier than the equation, with many drawing 3D structures with 2 out-wedges and 2 in-wedges and attaching the NH2 and COO groups correctly. It was also common to see a ‘criss-cross’ orientation, looking down on the complex, which is easier to draw. Many candidates connected the NH2 and COO groups next to, and across from, each another in the isomers. A common error was for candidates to rotate their first structure, to produce a second drawing of the first structure. Less successful responses often tried to attach NH2 and COO groups but with no CH2 between the groups to produce a cyclic attachment. A minority of candidates ignored ‘square planar’ and drew tetrahedral structures instead. The equation proved to be very difficult, the commonest error being omission of the ‘2’ balancing numbers for H2NCH2COOH and CH3COOH. The formulae for ethanol or propanoic acid were also often seen for ethanoic acid. Candidates are advised to check all formulae and then to check balancing, the golden rules for successfully constructing all equations.

Notes from the mark scheme and report

• 2H2NCH2COOH + Cu(CH3COO)2 → Cu(H2NCH2COO)2 + 2CH3COOH
• cis and trans square planar isomers with ligands shown as bidentate rings bonded through N and O of COO
• 1 mark if both drawn as tetrahedral
• IGNORE charges

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

Most candidates calculated the amounts of Cl2 and Ca(ClO)2 correctly as 17.5 mol and 8.75 mol respectively. Only the least successful did not use the equation’s stoichiometry to halve 17.5 to 8.75. For the final step in the calculation, candidates needed to convert 4.00 m3 into 4000 dm3 and to then determine the concentration to an appropriate number of significant figures and standard form. As all the data had been provided to 3 significant figures, the final answer was also required to 3 significant figures, as 2.09  10–3 mol dm–3. Answers such as 2.2  10–3, 2.1875  10–3 and 2.19  10–6 and 0.00219 illustrate errors in these areas.

Notes from the mark scheme and report

• common errors: 4.38 × 10^-3 (no ÷2) 2 marks
• wrong power of ten 2 marks
• 2.2 × 10^-3 2 marks (not appropriate SF)
• m3 to dm3 conversion
• 3 SF AND standard form

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Candidates were required to write a balanced equation from provided reactants and products and to use oxidation numbers to illustrate disproportionation. In the equation, the reactants and products were sometimes unbalanced, or incorrectly balanced. A common error was to balance the equation after first adding O2 as an extra reactant. Illustrating disproportionation proved to be easier with the oxidation number changes from the initial +1 being required. Otherwise, more successful responses sometimes missed out on marks if they omitted detail. For example, the oxidation number changes were stated but the candidate omitted to state which change was oxidation and which was reduction. The best responses identified Ca(ClO3)2 as the oxidation product and CaCl2 as the reduction product. One unexpected error on many scripts was for calcium to be identified as the element undergoing disproportionation, with oxidation number changes from +6 to +14 and +2. It was difficult to see why this happened, with Ca forming +2 compounds, but the error was common.

Notes from the mark scheme and report

• 3Ca(ClO)2 → 2CaCl2 + Ca(ClO3)2
• Cl reduced +1 to –1 and Cl oxidised +1 to +5
• 1 of 2 redox marks if changes correct but oxidation/reduction mis-assigned or not stated
• signs required
• IGNORE ionic charges such as Cl5+

Mark scheme

Examiner's comment

Most candidates were able to obtain some credit for this problem. The equations proved to be among the easier marks, and it was encouraging to see how many candidates were able to use the E data to help them combine the half equations correctly. Most then cancelled down the common species present on both sides of the equation: H+, H2O and e–. Explanations were generally sound, either by comparing the E values, stating the direction of equilibrium shift, or by using cell potentials. The hardest marks proved to be the identification of the chromium ions that were the oxidising and reduction agents. Many candidates just referred to ‘chromium’ despite the chromium half equation containing two chromium ions, Cr2O72– and Cr3+. Candidates who had written correct equations sometimes assigned the wrong chromium ion as the oxidising or reducing agent. Mores successful responses constructed correct equations and stated that Cr2O72– was the oxidising agent as it oxidised C2H5OH and that Cr3+ was the reducing agent as it reduced FeO42–.

Notes from the mark scheme and report

• Cr2O7 2– is oxidising agent with C2H5OH (its E more positive, equilibrium shifts right)
• Cr3+ is reducing agent with FeO4 2– (E less positive, shifts left)
• two overall equations with H+, H2O and e– cancelled (1 mark only if uncancelled)
• 6 marking points max 5

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

Candidates needed to interpret the information provided and to use this as the basis for their redox equation. The clue of a yellow product proved to be very difficult to interpret as being sulphur. The equation then required H+ to be added as a reactant (‘acidified’ in the information) and H2O as the other product. Balancing required use of oxidation numbers. Candidates found this equation very difficult and relatively few correct equations were seen. The mark scheme did allow 1 mark for any correct five species but the correct equation proved to be challenging in demand. Question 3 (a*) This question was assessed by level of response (LoR). Candidates were required to describe key features in a procedure to prepare a pure organic liquid, including a labelled diagram for reflux, a calculation of the percentage yield and the procedural steps for purification. Levels were determined using these three features. Marks within a level were determined by communication. This question discriminated extremely well. Level 3 candidates would draw a clear diagram with all key items labelled and the set up being capable of being used safely. The percentage yield calculation would be correct, producing a percentage yield close to 45.2%. The steps for the purification: use of a separating funnel, drying and redistillation would be described in the correct order and with some detail. Level 2 candidates would have obtained some of the features required for Level 3 but there would be some key omissions or errors. The diagram may have been drawn clearly but labelling may have been incomplete or a thermometer with bung may have been inserted into the top of the condenser, a very hazardous arrangement. The calculation would be attempted but with some errors, such as omitting to use the density, or using a mixture of moles and masses. The purification steps may have been described but in the wrong order. Purification steps would be incomplete, perhaps only including distillation. Level 1 candidates often drew a diagram resembling a tube above a flask, with water often flowing in the wrong direction. The percentage yield may have been a simple mass ratio with no moles being used. A significant number of candidates described the purification steps for an organic solid, including recrystallisation. The preparation of an organic liquid is a key practical procedure that will have been experienced by students during their A Level studies (PAG 5). The overall standard of drawing diagrams was poor, an area that needs improvement.

Notes from the mark scheme and report

• 5H2S + 2MnO4– + 6H+ → 2Mn2+ + 5S + 8H2O
• first mark for any five correct species (yellow precipitate is sulfur)
• IGNORE electrons and state symbols

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme, with the levels and indicative points

Examiner's comment

Showed limited knowledge and understanding of practical scenarios and drawing labelled diagrams of apparatus, e.g. • Question 3 (a): Diagram of reflux • Question 3 (a): Purification of an organic liquid There was no evidence that any time constraints had led to a candidate underperforming or of scripts where there were no responses to many questions.

Notes from the mark scheme and report

• reflux diagram: condenser above flask, water in at bottom and out at top
• mass = 6.10 × 1.268 = 7.73 g, yield 45.2% (35.6% if density omitted)
• separating funnel - organic layer is the BOTTOM layer (denser), dry with named anhydrous salt, redistil at 102 °C
• bare 'separate, dry, redistil' not detailed

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; PAG5 Synthesis of an organic liquid

Mark scheme

Examiner's comment

This question is one of two on this paper assessing understanding of unfamiliar organic reaction mechanisms. The stem to the question includes important information and clues that should have then guided candidates towards this unfamiliar mechanism (which is related to the familiar nucleophilic substitution of haloalkanes). The two prompts for Step 1 and Step 2 are critical but many candidates did not use these, instead inventing their own mechanisms. However, there were many successful responses seen that gained the full 4 marks. and dipoles. Step 1 is an attempt to show the alcohol OH group accepting a proton, but a curly arrow shows the movement of an electron pair. It cannot travel from a + charge to a lone pair. The intermediate shown does contain the correct atoms but the + charge has been omitted from the O atom. Marks were given for the curly arrow from the Br– lone pair and from the C–O bond. However, the candidate has drawn the bonds with very short lines making it all too easy for a curly arrow to be shown imprecisely. This response was given 2/4 marks.

Notes from the mark scheme and report

• step 1 curly arrow from lone pair on O to H+ giving intermediate with + charge on O (–OH2+)
• step 2 curly arrow from Br– lone pair/charge to C and from C–O bond to O (or via carbocation)
• max 2 for mechanism without positively charged intermediate
• IGNORE dipoles
• In organic chemistry mechanisms, a curly arrow shows the movement of an electron pair and demonstrates the direction of electron flow in organic reactions.
• A curly arrow must start from: • A lone pair or negative charge and go to an atom to show where a bond forms • A bond to show where a bond breaks.

Mark scheme

Examiner's comment

This question asked candidates to explain the different rates from a novel experiment carried out at 50ºC and 90ºC, and to predict the ∆H sign for the forward reaction. Candidate explanations for the rates were often superficial, solely in terms of greater energy at 90ºC. Many responses referred neither to the different frequency of collisions nor the greater number of particles exceeding the activation energy at 90ºC. Most candidates predicted that ∆H would have a negative sign. Candidates were expected to link the evidence from the absorbance data in the graph to less [Co(H2O)6]2+ being present at 90ºC. When experimental information has been presented, candidates are advised to look for the evidence responsible in their explanations.

Notes from the mark scheme and report

• at 90 °C faster rate AND more frequent collisions
• more particles have energy ≥ Ea
• equilibrium [Co(H2O)6]2+ is lower at 90 °C so forward reaction exothermic/∆H negative
• IGNORE more successful collisions

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

Almost all candidates realised that Cl– ions would react with the added AgNO3 at time = t1.

Notes from the mark scheme and report

• Cl– reacts with Ag+ to form AgCl (precipitate)
• IGNORE chlorine for chloride

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme

Examiner's comment

In contrast with Question 4 (a), most candidates did interpret the graphical information provided and related this to the reduced concentration of CoCl42– ions and the increased concentration of [Co(H2O)6]2+ ions. Most candidates also referred to Equilibrium 4.1 to conclude that the equilibrium shifts to the right. Only the very best candidates recognised that the increase in Cl– concentration following the initial addition of AgNO3 was 4 times greater than the increase in the concentration of Co(H2O)6]2+, arising from the 4 : 1 ratio in the stoichiometry in the equation.

Notes from the mark scheme and report

• Cl– increase is 4 × the change in the complex ions
• equilibrium shifts to the right (to replace Cl–)

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Mark scheme, with the levels and indicative points

Examiner's comment

This question was assessed by level of response (LoR). Candidates were required to process raw experimental results to determine one enthalpy change, and then to determine a second enthalpy change by using an energy cycle. Levels were determined by the accuracy of the candidates’ processing of the results, calculations and use of the energy cycle. Marks within a level were determined by communication. This question discriminated extremely well. Level 3 candidates used the mass of solution as 106.6 g with mc∆T to obtain an energy change of 8.24 kJ. They then divided this value by the moles of CaCl2 that reacted (0.0835 mol) to obtain the enthalpy change of –98.7 kJ mol–1. Finally, they constructed an energy cycle which they then used to obtain the second enthalpy change of –1566 kJ mol–1. Level 2 candidates determined the first enthalpy change but may have used the approximate mass of 100 g for the mass of solution from the experimental method. Their energy cycle was often incorrect or absent, with the second enthalpy change calculated incorrectly. Level 1 candidates often calculated the initial energy change using mc∆T but made little further correct progress. Less successful responses used the solid mass of calcium chloride (9.28 g) instead of the mass of the solution in their mc∆T calculation. Overall, mathematical skills were displayed well but some basic errors were made, particularly with subtractions. This may have been the result of mis-keying values into a calculator and believing the answer displayed. calcium chloride and the first enthalpy change using a correct method. This response has not been given marks for the communication strand of Level 3 because there is nothing to indicate what the numbers refer to. The response could be summarised as a mass of numbers scrawled across the page. Unfortunately, this is the pattern of many responses. So, this response was given 3/6 marks.

Notes from the mark scheme and report

• q = 106.6 × 4.18 × 18.5 = 8.24 kJ (mass of final solution)
• n(CaCl2) = 9.28/111.1 = 0.0835
• ∆solH = –98.7
• ∆hydH(Ca2+) = –2223 + (–98.7) – 2(–378) = –1566
• 2 × hydration of Cl–
• one small error allowed e.g. using 97.32 g or 115.88 g for m

Skills: M0.0 Recognise and make use of appropriate units in calculations; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Most candidates attempted to write an equation for the combustion of the 1:1 molar mixture of octane and ethanol. The formulae of C8H18 and C2H5OH were usually seen although some candidates combined these as a ‘mixture formula’ of C10H24O (which was accepted). The balancing of the equation using 15½O2 was the hardest part of the equation and many different balancing numbers for O2 were seen (10CO2 and 12H2O where usually correct). Less successful responses often attempted a combustion equation using octane OR ethanol, but not both. This is not an easy equation to construct, and the context was novel. Overall candidates made a good attempt at this question.

Notes from the mark scheme and report

• C8H18 + C2H5OH + 15½O2 → 10CO2 + 12H2O
• ALLOW multiples

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This question took the novel context introduced in 5b a stage further by considering the energy released during the combustion of this fuel. Most candidates were able to obtain some credit, and many obtained the correct energy of 341,850 kJ. The commonest error was for candidates to assume that the 8 kg mixture would contain 4 kg of octane and 4 kg of ethanol, rather than an equal moles of each. Such an approach could still be partly given marks by ECF, provided that the method was sound and clear.

Notes from the mark scheme and report

• 160 g per mole pair so 50 mol of each
• 341 850 kJ
• common error 310 800 (equal masses 4 kg each rather than equal moles) scores 2
• IGNORE sign

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Most candidates added two asterisks to the diagram of vitamin C for the possible chiral centres. More successful responses usually realised that two chiral centres would give rise to 22 = 4 optical isomers, with 2 optical isomers being the commonest error.

Notes from the mark scheme and report

• 4 (two chiral centres)

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

Most candidates realise that hydrogen bonds would be formed from the OH groups in vitamin C to water. Candidates are advised to read the question carefully as the word ‘extremely’ was a hint that ‘many’ OH groups would be needed in the explanation. The most successful responses quoted that hydrogen bonds would form between the 4 OH groups in vitamin C and water. Some candidates stated that O atoms in vitamin C would be involved. This was not given marks as not all O atoms in vitamin C are a part of OH groups and capable of hydrogen bonding.

Notes from the mark scheme and report

• hydrogen bonding (with water) AND many/four OH groups
• DO NOT ALLOW OH–

Mark scheme

Examiner's comment

More successful responses determined that x = 15 and y = 31. The key to success here was to subtract the formula of vitamin C from the formula of the ester and to add the formula of water: C22H38O7 – C6H8O6 + H2O → C16H32O2 → C15H31COOH. A significant number of candidates did obtain one of these values, with 15/16 and 29/30/32 being common incorrect answers. Omitting part(s) of sequence above would result in these incorrect numbers.

Notes from the mark scheme and report

• x = 15, y = 31 (C22H38O7 + H2O – C6H8O6 = C16H32O2)

Skills: M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

This novel mechanism assessed a candidate’s understanding of curly arrows, and four curly arrows were needed. One mark was available for one correct curly arrow, usually from the H2O: or from the C=O. 2 marks were given for three or four correct curly arrows. The two curly arrows within the ring structure proved to be the most difficult. The question discriminated extremely well: many candidates were able to secure one mark with the most able being given both marks. A candidate showing all four curly arrows correctly demonstrated an excellent understanding of curly arrows.

Notes from the mark scheme and report

• arrow from lone pair on H2O to H of enol O–H, O–H bond pair to form C=O, C=C pair moving along, ring C=O pair to O giving O–
• 3 or 4 correct arrows = 2 marks, 1 correct = 1 mark
• IGNORE incorrect/'double' curly arrows

Mark scheme

Examiner's comment

Most candidates had learnt a standard method for calculating the pH of a weak acid. with the correct answer of 2.16 being seen on very many scripts. Success required conversion of 0.150 moles of vitamin C in 250 cm3 to its concentration as 0.600 mol–3, calculation of [H+] using [H+] =  (Ka  [HA]) and determination of pH using –log[H+]. Common errors usually resulted from one mistake and could still be rewarded with 2 of the available 3 marks. Examples are shown below. • pH = 4.32 No square root of (7.94  10–5  0.600) • pH = 2.46 No conversion of 0.150 mol to 0.600 mol dm–3  (7.94  10–5  0.150) • pH = 2.76 ÷ 4 instead of  4 for concentration  (7.94  10–5  0.0375) pH calculations are common in A Level Chemistry There are four different types, and it is essential that the standard methods for determination of [H+] in the calculations are learnt: • pH of strong acids • pH of weak acids, using Ka and [HA] • pH of strong bases, using Kw and [OH–] • pH of buffers, using Ka and [HA]/[A–] It is extremely likely that at least one of these types of pH calculation will feature in at least one of the A Level units. This question was a standard titration calculation, set in a practical context. As with Question 6 (c) (i), this assessed one of the important principles encountered in A Level Chemistry. Success required three main stages: • Calculation of the amount, in mol, of I2 used in the titration of 25.0 cm3 of the diluted orange juice. • Determination of the amount in mol of vitamin C in the 250 cm3 solution (effectively scaling by 10) • Use of this value with the molar mass of vitamin C to determination of the mass of vitamin C, in mg, in the 150 cm3 serving of the orange juice. Most candidates calculated the amount of I2 (and vitamin C) in the titre as 2.16  10–5 mol. The next scaling stage by 10 to 2.16  10–4 mol introduced a problem. Many candidates were distracted by the ‘150 cm3’ serving of orange juice and they scaled further by a factor of 150/250 to give 1.296  10–4 mol. Candidates scaling corrected by 10 usually used the vitamin C molar mass of 176 g mol–1 to determine the correct mass of 38 mg in the orange juice serving. The common incorrect mass of 22.8 mg resulted from the extra scaling by 150/250 described above. This could still be given 3 of the available 4 marks as only one error had been made. Some candidates worked out an incorrect molar mass for vitamin C, with 174 often seen. Provided that this value was used with a correct method, ECF could still be applied for the final mass. If any of your students’ results are not as expected, you may wish to consider one of our post-results services. For full information about the options available visit the OCR website. We send a weekly roundup to tell you about important updates. You can also sign up for your subject specific updates. If you haven’t already, sign up here. Attend one of our popular CPD courses to hear directly from a senior assessor or drop in to a Q&A session. Most of our courses are delivered live via an online platform, so you can attend from any location. Please find details for all our courses on the relevant subject page on our website or visit OCR professional development. ExamBuilder is the question builder platform for a range of our GCSE, A Level, Cambridge Nationals and Cambridge Technicals qualifications. Find out more. ExamBuilder is free for all OCR centres with an Interchange account and gives you unlimited users per centre. We need an Interchange username to validate the identity of your centre's first user account for ExamBuilder. If you do not have an Interchange account please contact your centre administrator (usually the Exams Officer) to request a username, or nominate an existing Interchange user in your department. Review students' exam performance with our free online results analysis tool. It is available for all GCSEs, AS and A Levels and Cambridge Nationals. It allows you to: • review and run analysis reports on exam performance • analyse results at question and/or topic level • compare your centre with OCR national averages • identify trends across the centre • facilitate effective planning and delivery of courses • identify areas of the curriculum where students excel or struggle • help pinpoint strengths and weaknesses of students and teaching departments. Find out more. Post-results services Keep up-to-date OCR Professional Development Signed up for ExamBuilder?

Notes from the mark scheme and report

• common errors: 4.32 no square root
• 2.46 no ×4 (0.150 used)
• 2.76 ÷4
• concentration 0.600 mol dm–3
• 2 d.p. required (2.16)
• pH ECF only if [H+] derived from Ka and concentration

Skills: M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Notes from the mark scheme and report

• common errors: 22.81 mg scaling by 150/250 (3 marks)
• 42.24 mg using 25.0 instead of 22.50 (3 marks)
• 63.36 mg scaling by 250/150
• 1:1 ratio, ×10 for 250 cm3 flask which contains the whole 150 cm3 serving
• answer 38 mg

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities