Cambridge A Level Chemistry 9701 — 2015 Oct/Nov Paper 3 · Variant 1
9701/31/O/N/15 · 40 marks · ≈45 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper12 pages












Mark scheme5 pages
Answers below. Sit the paper first if you are practising.





Paper as text
Question paper, page 1
READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Give details of the practical session and laboratory where appropriate, in the boxes provided. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions. Electronic calculators may be used. You may lose marks if you do not show your working or if you do not use appropriate units. Use of a Data Booklet is unnecessary. A copy of the Periodic Table is printed on page 12. Qualitative Analysis Notes are printed on pages 10 and 11. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. CHEMISTRY 9701/31 Paper 3 Advanced Practical Skills 1 October/November 2015 2 hours Candidates answer on the Question Paper. Additional Materials: As listed in the Confidential Instructions Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level This document consists of 12 printed pages and 1 insert. [Turn over IB15 11_9701_31/FP © UCLES 2015 *9899326369* Session Laboratory For Examiner’s Use 1 2 3 Total
Question paper, page 2
2 9701/31/O/N/15 © UCLES 2015 1 In this experiment you will determine the ionic equation for the reaction of acidified potassium manganate(VII) with potassium iodide. Excess potassium iodide is used and the reaction produces iodine. The amount of iodine produced is measured by titration with sodium thiosulfate. FA 1 is 0.0180 mol dm–3 potassium manganate(VII), KMnO4. FA 2 is 1.00 mol dm–3 sulfuric acid, H2SO4. FA 3 is 0.500 mol dm–3 potassium iodide, KI. FA 4 is 0.100 mol dm–3 sodium thiosulfate, Na2S2O3. starch indicator (a) Method ● Pipette 25.0 cm3 of FA 1 into a conical flask. ● Use the measuring cylinder to add 25 cm3 of FA 2 to the conical flask. ● Use the measuring cylinder to add 20 cm3 of FA 3 to the conical flask. ● Fill the burette with FA 4. ● Carry out a rough titration. When the colour of the mixture becomes yellow/orange, add a few drops of starch indicator. Then titrate until the mixture goes colourless. ● Record all your burette readings in the space below. The rough titre is … cm3. ● Carry out as many accurate titrations as you think necessary to obtain consistent results. ● Make sure any recorded results show the precision of your practical work. ● Record in a suitable form below all of your burette readings and the volume of FA 4 added in each accurate titration. Keep FA 1 and FA 2 for use in Question 3 and FA 4 for use in Question 2. [7] (b) From your accurate titration results, obtain a suitable value for the volume of FA 4 to be used in your calculations. Show clearly how you have obtained this value. Volume of FA 4 required is … cm3. [1] I II III IV V VI VII
Question paper, page 3
3 9701/31/O/N/15 © UCLES 2015 [Turn over (c) Calculations Show your working and appropriate significant figures in the final answer to each step of your calculations. (i) Calculate the number of moles of sodium thiosulfate in the volume of FA 4 calculated in (b). moles of Na2S2O3 = … mol (ii) Use the equation below to calculate the number of moles of iodine that reacted with the sodium thiosulfate in the titration. I2 + 2Na2S2O3 → Na2S4O6 + 2NaI moles of I2 = … mol (iii) Use information on page 2 to calculate the number of moles of potassium manganate(VII) in FA 1 used in the titration. moles of KMnO4 = … mol (iv) From your answers to (ii) and (iii), calculate the number of moles of iodine produced by the reaction of 2.00 moles of potassium manganate(VII) with excess potassium iodide. moles I2 = … mol (v) Using your answer to (iv), put a tick next to the ionic equation that represents the reaction between FA 1 and FA 3. 2MnO4 – + 2I– + 16H+ → I2 + 2Mn6+ + 8H2O … 2MnO4 – + 4I– + 16H+ → 2I2 + 2Mn5+ + 8H2O … 2MnO4 – + 6I– + 16H+ → 3I2 + 2Mn4+ + 8H2O … 2MnO4 – + 8I– + 16H+ → 4I2 + 2Mn3+ + 8H2O … 2MnO4 – + 10I– + 16H+ → 5I2 + 2Mn2+ + 8H2O … 2MnO4 – + 12I– + 16H+ → 6I2 + 2Mn+ + 8H2O …
Question paper, page 4
4 9701/31/O/N/15 © UCLES 2015 (vi) Prove that the iodide ion has been oxidised in the equation that you selected in (v). … … … [5] (d) (i) The error in calibration of the pipette you used is ±0.06 cm3. Calculate the percentage error when measuring FA 1, using the pipette. percentage error = … % (ii) A student suggested that the experiment would be more accurate if a pipette was used to measure solution FA 3. State and explain whether you agree with the student. … … … [2] [Total: 15]
Question paper, page 5
5 9701/31/O/N/15 © UCLES 2015 [Turn over 2 In this experiment you will investigate how the rate of reaction between sodium thiosulfate and hydrochloric acid is affected by the concentration of the acid. When aqueous thiosulfate ions react with hydrogen ions, H+, in any acid, a pale yellow precipitate of sulfur is formed. The ionic equation for this reaction is given below. S2O3 2–(aq) + 2H+(aq) → S(s) + SO2(aq) + H2O(l) The rate of the reaction can be determined by measuring the time taken to produce a fixed quantity of sulfur. FA 4 is 0.10 mol dm–3 sodium thiosulfate, Na2S2O3. FA 5 is 0.20 mol dm–3 hydrochloric acid, HCl. (a) Method Record all your measurements, in an appropriate form, in the space below. Experiment 1 ● Use the larger measuring cylinder to transfer 40 cm3 of FA 4 into the 100 cm3 beaker. ● Rinse the larger measuring cylinder thoroughly with water, then add 30 cm3 of FA 5 to the beaker and start timing immediately. ● Stir the mixture once and place the beaker on top of the printed insert page provided. ● Look down through the solution in the beaker at the print on the insert. ● Stop timing as soon as the precipitate of sulfur makes the print on the insert invisible. ● Record the reaction time to the nearest second. ● Empty and rinse the 100 cm3 beaker. ● Dry the outside of the beaker ready for Experiment 2. Experiment 2 ● Rinse the larger measuring cylinder, then use it to transfer 40 cm3 of FA 4 into the 100 cm3 beaker. ● Use the smaller measuring cylinder to add 10 cm3 of distilled water to the beaker. ● Use the same measuring cylinder to add 20 cm3 of FA 5 to the mixture in the beaker and start timing immediately. ● Stir the mixture once and place the beaker on top of the printed insert page provided. ● Stop timing as soon as the print on the insert becomes invisible. ● Record the reaction time to the nearest second. ● Empty and rinse the 100 cm3 beaker. ● Dry the outside of the beaker ready for Experiment 3. Experiment 3 ● Carry out the reaction using a mixture of 40 cm3 of FA 4, 20 cm3 of distilled water and 10 cm3 of FA 5. ● Measure and record the reaction time to the nearest second. [4] I II III IV
Question paper, page 6
6 9701/31/O/N/15 © UCLES 2015 (b) (i) The ‘rate of reaction’ can be represented by the formula below. ‘rate of reaction’ = reaction time 1000 Use this formula to calculate the ‘rate of reaction’ for Experiments 1 and 3. Give the unit. ‘rate of reaction’ for Experiment 1 … unit … ‘rate of reaction’ for Experiment 3 … unit … (ii) Calculate the initial concentrations of hydrochloric acid in the reaction mixtures in Experiments 1 and 3. initial concentration of HCl in Experiment 1 = … mol dm–3 initial concentration of HCl in Experiment 3 = … mol dm–3 (iii) How is the ‘rate of reaction’ affected by the concentration of hydrochloric acid in the mixture? … … (iv) Predict how the reaction time measured in Experiment 1 would have been affected if the experiment had been carried out using 0.20 mol dm–3 sulfuric acid instead of 0.20 mol dm–3 hydrochloric acid. Explain your answer. … … … (v) Predict how the reaction time measured in Experiment 3 would have been affected if the experiment had been carried out in a 250 cm3 beaker instead of a 100 cm3 beaker. Explain your answer. … … [5] [Total: 9]
Question paper, page 7
7 9701/31/O/N/15 © UCLES 2015 [Turn over 3 Qualitative Analysis At each stage of any test you are to record details of the following. ● colour changes seen ● the formation of any precipitate ● the solubility of such precipitates in an excess of the reagent added Where gases are released they should be identified by a test, described in the appropriate place in your observations. You should indicate clearly at what stage in a test a change occurs. No additional tests for ions present should be attempted. If any solution is warmed, a boiling tube MUST be used. Rinse and reuse test-tubes and boiling tubes where possible. Where reagents are selected for use in a test, the name or correct formula of the element or compound must be given. (a) FA 6 is a sodium compound containing one anion listed on page 11. Dissolve the FA 6 provided in about 15 cm3 of distilled water in a boiling tube. Carry out the following tests and record your observations in the table below. test observations (i) To a 1cm depth of the solution of FA 6 in a test-tube, add a few drops of aqueous barium chloride or aqueous barium nitrate, then add dilute hydrochloric acid. (ii) To a 1cm depth of the solution of FA 6 in a test-tube, add an equal volume of aqueous hydrogen peroxide, then add a few drops of aqueous barium chloride or aqueous barium nitrate, then add dilute hydrochloric acid.
Question paper, page 8
8 9701/31/O/N/15 © UCLES 2015 test observations (iii) To a 1 cm depth of the solution of FA 6 in a boiling tube, add an equal volume of FA 2, sulfuric acid, then heat the mixture gently and cautiously. (iv) To a 1 cm depth of the solution of FA 6 in a test-tube, add an equal volume of aqueous sodium hydroxide, then add a few drops of FA 1, aqueous potassium manganate(VII), then add FA 2, sulfuric acid. (v) Identify the anion in FA 6, and state one piece of evidence for your identification. anion … evidence … … … (vi) Give the chemical equation for the reaction between FA 6 and hydrogen peroxide, H2O2, in test (ii). State symbols are not required. … [7]
Question paper, page 9
9 9701/31/O/N/15 © UCLES 2015 [Turn over (b) FA 7, FA 8, FA 9 and FA 10 each contain one cation from the list on page 10. You will attempt to identify the cations by testing with aqueous sodium hydroxide and aqueous ammonia. In each case, use a 1 cm depth of the solution in a test-tube. (i) Complete the table below. test observations FA 7 FA 8 FA 9 FA 10 add sodium hydroxide add aqueous ammonia (ii) Use your observations to identify, as far as possible, the cation present in each solution. If alternative identities are possible, state this clearly. FA 7 cation … FA 8 cation … FA 9 cation … FA 10 cation … (iii) Give the ionic equation for the reaction of one of your cations with a few drops of sodium hydroxide. State symbols are not required. … (iv) The precipitates obtained when alkalis are added to solutions of certain cations are sometimes difficult to see. Suggest how, using no additional apparatus, the experiment could be repeated in a way that would make these precipitates more visible. … … [9] [Total: 16]
Question paper, page 10
10 9701/31/O/N/15 © UCLES 2015 Qualitative Analysis Notes Key: [ppt. = precipitate] 1 Reactions of aqueous cations ion reaction with NaOH(aq) NH3(aq) aluminium, Al 3+(aq) white ppt. soluble in excess white ppt. insoluble in excess ammonium, NH4 +(aq) no ppt. ammonia produced on heating – barium, Ba2+(aq) no ppt. (if reagents are pure) no ppt. calcium, Ca2+(aq) white ppt. with high [Ca2+(aq)] no ppt. chromium(III), Cr3+(aq) grey-green ppt. soluble in excess giving dark green solution grey-green ppt. insoluble in excess copper(II), Cu2+(aq) pale blue ppt. insoluble in excess blue ppt. soluble in excess giving dark blue solution iron(II), Fe2+(aq) green ppt. turning brown on contact with air insoluble in excess green ppt. turning brown on contact with air insoluble in excess iron(III), Fe3+(aq) red-brown ppt. insoluble in excess red-brown ppt. insoluble in excess magnesium, Mg2+(aq) white ppt. insoluble in excess white ppt. insoluble in excess manganese(II), Mn2+(aq) off-white ppt. rapidly turning brown on contact with air insoluble in excess off-white ppt. rapidly turning brown on contact with air insoluble in excess zinc, Zn2+(aq) white ppt. soluble in excess white ppt. soluble in excess
Question paper, page 11
11 9701/31/O/N/15 © UCLES 2015 2 Reactions of anions ion reaction carbonate, CO3 2– CO2 liberated by dilute acids chloride, Cl –(aq) gives white ppt. with Ag+(aq) (soluble in NH3(aq)) bromide, Br –(aq) gives cream ppt. with Ag+(aq) (partially soluble in NH3(aq)) iodide, I –(aq) gives yellow ppt. with Ag+(aq) (insoluble in NH3(aq)) nitrate, NO3 –(aq) NH3 liberated on heating with OH–(aq) and Al foil nitrite, NO2 –(aq) NH3 liberated on heating with OH–(aq) and Al foil; NO liberated by dilute acids (colourless NO → (pale) brown NO2 in air) sulfate, SO4 2–(aq) gives white ppt. with Ba2+(aq) (insoluble in excess dilute strong acids) sulfite, SO3 2–(aq) SO2 liberated with dilute acids; gives white ppt. with Ba2+(aq) (soluble in excess dilute strong acids) 3 Tests for gases gas test and test result ammonia, NH3 turns damp red litmus paper blue carbon dioxide, CO2 gives a white ppt. with limewater (ppt. dissolves with excess CO2) chlorine, Cl 2 bleaches damp litmus paper hydrogen, H2 “pops” with a lighted splint oxygen, O2 relights a glowing splint sulfur dioxide, SO2 turns acidified aqueous potassium manganate(VII) from purple to colourless
Question paper, page 12
12 9701/31/O/N/15 © UCLES 2015 To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Group 140 Ce Cerium 58 141 Pr Praseodymium 59 144 Nd Neodymium 60 Pm Promethium 61 150 Sm Samarium 62 152 Eu Europium 63 157 Gd Gadolinium 64 159 Tb Terbium 65 163 Dy Dysprosium 66 165 Ho Holmium 67 167 Er Erbium 68 169 Tm Thulium 69 173 Yb Ytterbium 70 175 Lu Lutetium 71 Th Thorium 90 Pa Protactinium 91 U Uranium 92 Np Neptunium 93 Pu Plutonium 94 Am Americium 95 Cm Curium 96 Bk Berkelium 97 Cf Californium 98 Es Einsteinium 99 Fm Fermium 100 Md Mendelevium 101 No Nobelium 102 Lr Lawrencium 103 1.0 H Hydrogen 1 6.9 Li Lithium 3 23.0 Na Sodium 11 24.3 Mg Magnesium 12 40.1 Ca Calcium 20 45.0 Sc Scandium 21 47.9 Ti Titanium 22 50.9 V Vanadium 23 52.0 Cr Chromium 24 54.9 Mn Manganese 25 55.8 Fe Iron 26 58.9 Co Cobalt 27 58.7 Ni Nickel 28 63.5 Cu Copper 29 65.4 Zn Zinc 30 69.7 Ga Gallium 31 27.0 Al Aluminium 13 10.8 B Boron 5 12.0 C Carbon 6 14.0 N Nitrogen 7 16.0 O Oxygen 8 19.0 F Fluorine 9 28.1 Si Silicon 14 31.0 P Phosphorus 15 32.1 S Sulfur 16 35.5 Cl Chlorine 17 39.9 Ar Argon 18 20.2 Ne Neon 10 4.0 He Helium 2 72.6 Ge Germanium 32 74.9 As Arsenic 33 79.0 Se Selenium 34 79.9 Br Bromine 35 83.8 Kr Krypton 36 39.1 K Potassium 19 87.6 Sr Strontium 38 88.9 Y Yttrium 39 91.2 Zr Zirconium 40 92.9 Nb Niobium 41 95.9 Mo Molybdenum 42 Tc Technetium 43 101 Ru Ruthenium 44 103 Rh Rhodium 45 106 Pd Palladium 46 108 Ag Silver 47 112 Cd Cadmium 48 115 In Indium 49 119 Sn Tin 50 122 Sb Antimony 51 128 Te Tellurium 52 127 I Iodine 53 131 Xe Xenon 54 137 Ba Barium 56 139 La Lanthanum 57 * 178 Hf Hafnium 72 181 Ta Tantalum 73 184 W Tungsten 74 186 Re Rhenium 75 190 Os Osmium 76 192 Ir Iridium 77 195 Pt Platinum 78 197 Au Gold 79 201 Hg Mercury 80 204 Tl Thallium 81 207 Pb Lead 82 209 Bi Bismuth 83 Po Polonium 84 At Astatine 85 Rn Radon 86 Rf Rutherfordium 104 Db Dubnium 105 Sg Seaborgium 106 Bh Bohrium 107 Hs Hassium 108 Mt Meitnerium 109 Uun Ununnilium 110 Uuu Unununium 111 Uub Ununbium 112 Uuq Ununquadium 114 Uuh Ununhexium 116 Uuo Ununoctium 118 Fr Francium 87 Ac Actinium 89 9.0 Be Beryllium 4 I II III IV V VI VII 0 85.5 Rb Rubidium 37 133 Cs Caesium 55 Ra Radium 88 a X b a = relative atomic mass X = atomic symbol b = proton (atomic) number Key *58-71 Lanthanides 90-103 Actinides The Periodic Table of the Elements *
Mark scheme, page 1
® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2015 series 9701 CHEMISTRY 9701/31 Paper 3 (Advanced Practical Skills 1), maximum raw mark 40 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.
Mark scheme, page 2
Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9701 31 © Cambridge International Examinations 2015 Question Indicative material Mark Total 1 (a) I Initial and final readings and titre value given for rough titre and initial and final readings for two (or more) accurate titrations (minimum of 2 × 2 box) 1 II Titre values recorded for accurate titrations and Appropriate headings for the accurate titration table and cm3 units. • initial / start burette reading / volume / value • final / end burette reading / volume / value (not amount) • titre or volume / FA 4 and used / added • unit: / cm3 or (cm3) or in cm3 (for each heading) 1 III All accurate burette readings are to the nearest 0.05 cm3. Do not award this mark if: • 50(.00) is used as an initial burette reading • more than one final burette reading is 50.(00) • any burette reading is greater than 50.(00) • there is only one accurate titration. 1 IV There are two uncorrected accurate titres within 0.10 cm3 • Do not award this mark if, having performed two titres within 0.10 cm3, a further titration is performed which is more than 0.10 cm3 from the closer of the initial two titres, unless a further titration, within 0.10 cm3 of any other, has also been carried out. • Do not award the mark if any “accurate” burette readings (apart from initial 0 cm3) are given to zero dp 1 Examiner rounds any burette readings to the nearest 0.05 cm3, checks subtractions and then selects the “best” titres using the hierarchy: • two (or more) accurate identical titres, then • two (or more) accurate titres within 0.05 cm3, then • two (or more) accurate titres within 0.10 cm3, etc These best titres are used to calculate the mean titre, expressed to nearest 0.01 cm3. Examiner calculates the difference (δ) between the mean titres obtained by the candidate and the Supervisor. Accuracy marks are awarded as shown. Award V, VI and VII if δ ⩽ 0.20 (cm3) Award V and VI if 0.20 < δ ⩽ 0.30 Award V, only, if 0.30 < δ ⩽ 0.50 Spread penalty: if the two “best” (corrected) titres used by the Examiner were ⩾ 0.50 cm3 apart, cancel one accuracy mark. 3 [7]
Mark scheme, page 3
Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9701 31 © Cambridge International Examinations 2015 Question Indicative material Mark Total (b) Candidate must take the average of two (or more) titres that are within a total spread of not more than 0.20 cm3. Working / explanation must be shown or ticks must be put next to the two (or more) accurate readings selected. The mean should be quoted to 2 dp, and be rounded to nearest 0.01 cm3. Two special cases, where the mean need not be to 2 dp: • Allow mean expressed to 3 dp only for 0.025 or 0.075 (e.g. 26.325 cm3) • Allow mean if expressed to 1 dp, if all accurate burette readings were given to 1 dp and the mean is exactly correct. (e.g. 26.0 and 26.2 = 26.1 is allowed) (e.g. 26.0 and 26.1 = 26.1 is wrong – should be 26.05) Note: the candidate’s mean will sometimes be marked correct even if it was different from the mean calculated by the Examiner for the purpose of assessing accuracy. 1 [1] (c)(i)(ii) Correctly calculates • n(thio) = 0.10 × (b)/1000 • n(I2) = 0.5 × (i) Both answers must be given to 3 or 4 significant figures 1 (iii) Correctly calculates n(KMnO4) = 0.025 × 0.018 = 0.00045 or 0.000450 or 0.0004500 1 (iv) Correct expression, with answer given to 2, 3 or 4 sig fig n(I2) = (ii)/(iii) × 2 Theoretical answer = 5.0 (for 2.0 mol KMnO4) 1 (v) Correct equation ticked, corresponding to (iv) 1 (vi) Allow any one of the following answers: • An iodide ion loses one electron • 2I– – 2e– → I2 (ionic equation must be correctly balanced) • Oxidation number of iodine increases from –1/ 1– (in iodide ion) to 0 (in iodine) 1 [5] (d) (i) % error = 0.06/25 × 100 = 0.24 % 1 (ii) The student is wrong, since KI / FA 3 is in excess. 1 [2] Qn 1 [Total: 15] 2 (a) I Table for readings, headings and correct units: Headings: • volume of FA 5 / acid • (volume of water – if included must be correct) • Time Units allow Vol or / cm3 etc. 1
Mark scheme, page 4
Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9701 31 © Cambridge International Examinations 2015 Question Indicative material Mark Total II Three reaction times all recorded to nearest second 1 III and IV Examiner to calculate the time differences between Expt 1 and Expt 2 (t2 – t1). Then calculate 10% of the time for Expt 1 to 1dp (x). If (t2 – t1).> x award III. Examiner to calculate the time differences between Expt 2 and Expt 3 (t3 – t2). Then calculate 20% of the time for Expt 2 to 1dp (y). If (t3 – t2).> y award IV. 2 [4] (b) (i) • Rates correctly calculated • All answers expressed to same sig fig (but not 1 sf) • Unit given… / s–1 or (s–1) 1 (ii) Correctly calculates • Expt 1, conc = 0.08571 (or 0.0857 or 0.086) mol dm–3 and • Expt 3, conc = 0.02857 (or 0.0286 or 0.029) mol dm–3 Both answers must be given to 2, 3 or 4 sig figs 1 (iii) Rate increases with (increase of) concentration 1 (iv) • Time is shorter for sulfuric acid • Sulfuric acid has a greater / doubled concentration of H+ ions. 1 (v) • time (for reaction) will be greater • less depth (of solution) in the 250 cm3 1 [5] Qn 2 [Total: 9] FA 6 is Na2SO3; FA 7 is CaCl2; FA 8 is MgSO4; FA 9 is Al2(SO4)3; FA 10 is MnSO4 3 (a) (i) Both observations required • white precipitate with Ba2+ ion • Precipitate dissolves / partially dissolves in (excess) HCl 1 (ii) Both observations required • white precipitate with Ba2+ ion • precipitate insoluble / no change with HCl 1 (iii) When heated, gas produced decolourises KMnO4 paper. 1 (iv) No change (when NaOH added) / no ppt / no reaction and green (solution) formed when KMnO4 added 1 Colourless solution(with acid) 1
Mark scheme, page 5
Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9701 31 © Cambridge International Examinations 2015 Question Indicative material Mark Total (v) Anion is sulfite and one piece of evidence • FA 6 with acid – SO2 / gas which decolourises KMnO4 is formed or • FA 6 with Ba2+ – white precipitate / BaSO3 formed which dissolves in acid / partially soluble in acid 1 (vi) Na2SO3 + H2O2 → Na2SO4 + H2O 1 [7] (b) (i) FA 7 FA 8 FA 9 FA 10 NaOH white ppt white ppt white ppt off-white / buff / beige / light brown ppt excess NaOH no change or insoluble in excess no change or insoluble in excess (ppt) dissolves or soluble in excess insoluble in excess or ppt darkens (owtte) NH3 no ppt or no reaction white ppt white ppt off-white / buff / beige / light brown ppt excess NH3 (ignore) no change or insoluble in excess no change or insoluble in excess insoluble in excess or ppt darkens (owtte) 5 (ii) Conclusions • FA 7 – calcium / Ca2+ or barium/Ba2+ • FA 8 – magnesium / Mg2+ • FA 9 – aluminium / Al 3+ • FA 10 – manganese(II)/Mn2+ Four correct = 2 marks Two or three correct = 1 mark 2 (iii) M2+ + 2OH– → M(OH)2 (for any divalent cation) or M3+ + 3OH– → M(OH)3 (for any trivalent cation) 1 (iv) Use higher concentration 1 [9] Qn 3 [Total: 16]
What you needed in this session
Cambridge’s own grade thresholds for 2015 Oct/Nov, Paper 3 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.