Preview Chemistry IB Prepared by Sergey Bylikin, Brian Murphy, Alexandra Juniper (2018)

Preview Chemistry IB Prepared by Sergey Bylikin, Brian Murphy, Alexandra Juniper (2018) Preview Chemistry IB Prepared by Sergey Bylikin, Brian Murphy, Alexandra Juniper (2018) Preview Chemistry IB Prepared by Sergey Bylikin, Brian Murphy, Alexandra Juniper (2018) Preview Chemistry IB Prepared by Sergey Bylikin, Brian Murphy, Alexandra Juniper (2018) Preview Chemistry IB Prepared by Sergey Bylikin, Brian Murphy, Alexandra Juniper (2018)

O X F O R D I B P R E P A R E D CHEMIS T RY I B D I P L O M A Sergey Bylikin Brian Murphy Alexandra Juniper P R O G R A M M E O X F O R D I B P R E P A R E D CHEMIS T RY I B D I P L O M A P R O G R A M M E Sergey Bylikin Brian Murphy Alexandra Juniper problem Q1 and answer, Q12a; Great Clarendon Street, Oxford, OX2 6DP, United the by University University’s publishing Press in the Press is a objective of worldwide UK and in department excellence Oxford certain is a other of in the University research, registered of Oxford scholarship, trade mark of It and furthers Oxford University Press education Oxford University moral rights of the HP3 been in All 978 rights 19 842367 system, law, by in No organization should be address or part of this transmitted, writing licence or of under Enquiries sent publication may be reproduced,stored in must not this to p168: M16 p176: problem 1, p147: Sample M17 HP3 2, 2, TZ1, p139: M18 student HP3 TZ0, M16 p172: Q11b; 2, SP3 Q10; HP3 1, p168: problem M17 M17 3, problem Practice and by Sample Derived Practice Q14; Sample the Oxford terms in any agreed concerning Rights form University with or by Press, the means, as outside Oxford without expressly appropriate reproduction Department, any or the TZ1 Inspired SLP3; answer, Q14b; p163: Practice by M11 Sample N17 SP2 TZ1, student HP3 problem 2, TZ0, p168: TZ0, TZ1, p179: student M17 Practice TZ1, p183: Sample from M16 problem Derived Sample HP3 answer, Q11d; Practice p174: Q14d; HP3 Derived from Q13c(ii); M17 1, 1, HP3 p176: answer, Practice HP3 TZ1, TZ0, p174: N16 problem student student HP3 from Q12c; Derived 2, Practice N16 Sample from M16 problem p171: Sample Derived TZ0, p178: Q15; answer, Q9; HP3 M16 TZ0, p179: M17 problem N17 HP3 student TZ0, Q11; answer, Practice p186: problem From M17 2(a), HP3 p183: TZ1, Similar Q15d; to Practice M17 1, HP3 reprographics the University scope Press, of at circulate same this work condition on in any any other form and you Britain by Bell from M17 problem student 1, HP3 TZ2, p188: answer Q15; Sample Derived p191: from N17 HP3 student M17 TZ0, HP3 answer, TZ1, Q15(b); p188: problem Q16, and Practice N17 M17 HP3 1, HP3 problem TZ0, TZ2, p191: from M16 HP3 TZ0, Q10; Practice problem p191: Adapted and expanded prior permitted version of M17 HP3 Sample student TZ2, Q12; Example C.1.1, p192: Derived from N16 HP3 TZ0, Q17; by answer, p193: M17 HP3 TZ2, Q17(b); Practice problem p194: Derived rights the from M17 HP3 from M17 SP3 TZ1, Q20(c), and M16 HP3 TZ0, Q15(b)(ii); Example C.2.1, p195: Derived above TZ2, Q14(c); Sample student answer, p196: M16 HP3 TZ0, Q15; Practice the problem Great b; Sample Inspired must 1, HP3 answer, p197: TZ0, Derived Q17; p199: from Example M17 HP3 N16 C.3.2, TZ1, Q18; HP3 TZ2, p198: Q18; Inspired Practice Example by problem N17 C.3.1, HP3 p199: p198: TZ0, Q18; Derived Adapted Sample from M16 from student HP3 TZ0, acquirer Practice Inspired in Q4(b)(ii); problem Practice problem Q13f; Q14; Q17; Printed Practice problem a above impose TZ2, answer, p174: + Q13a, Q8c; M16 You Q1; 2018 reserved permission Q12a Practice Adapted retrieval SLP3 TZ0, Practice asserted p186: ISBN HP2 Q2; Practice Q9; answer, TZ1, TZ2, published M18 answer, TZ0, p183: First TZ2, TZ0, Q11c; have N13 TZ1, student Q10d; HP3 student countries 2018 authors p154: student SP3 The SP3 Sample SP3 M17 HP3 © p138: Kingdom N16 Oxford 1, M17 and Bain Ltd by problem N17 HP3 p199: TZ0, Q18; Adapted Sample from M17 student HP3 TZ2, answer, Q16; p201: Example M16 HP3 C.4.1, TZ0, p200: Q16; Glasgow Practice problem Derived from 1, N17 p201: HP3 Derived TZ0, Q19; from N17 Example SP3 C.5.1, TZ0, Q15; p202: Practice Derived problem from N17 2, SP3 p201: TZ0, Q13; Acknowledgements Sample The authors would like to thank the editing the IB team, Helen Payne and Ben their work on this book, and for their useful suggestions Sergey also Photo like to thank Dr Natalia Kalashnikova for her support and M17 Stock Photo; HP3 HP3 TZ0, p38 (T) Shutterstock Loraks/Shutterstock; p43 (T) tomaster/Shutterstock; Science Photo Library; (T) Shutterstock p102 (T) Charles DAM D Ropisme/Shutterstock; Jose Antonio Winters/Science p62 (T) Perez/Shutterstock; Photo Library; DAM Library Martyn p104 p116 (T) F Chillmaid/Science p192 (T) DAM Photo Nickolay Library; p162 (T) DAM Vinokurov/Shutterstock; Photo Charles D Library CNRI/Science Photo Photo Library; Library; p10 (T) p2 (T) DAM Science Alamy Stock Photo Fotomaton Energy by Rice / Alamy Stock University is p12 (B) under a 4.0 International License, except; p17 (T) Photo Photo p22 (T) DAM p55 (T) Ambelrip/Shutterstock; Shutterstock Oleksii p31 (T) problem Stock p85 Ropisme/Shutterstock; Photo; (T) p81 (T) Science Shutterstock Doug p79 Photo (T) Library Stock Shutterstock Photo CNP p91 (T) lbert p72 Photo Science Gschmeissner/Science Photo Photo Library; Library; p111 (T) p96 (T) Science Shutterstock Photo Photo Science from (T) (T) Shutterstock Alamy Artwork by Stock Smereka/Shutterstock; Aptara Photo p128 Corp 3QuarksMedia/Alamy (T) DAM and Stock publisher would like to by N16 to adapt guide The / references, questions Sample TZ1, student Q2; SPL1 the from International past adapted p27: Q1 M16 (c) SLP2 summarized answer, Practice TZ2, are p9: problem (i); TZ0, Q1 M16 4, Practice (a) problem derived 1, p42: from M17 Q1; Sample p16: M17 TZ1, Q5c; (i)-(iii); this book, Q1(c); SLP2 p21: Sample TZ2, answer, Example TZ0, Sample TZ0 answer Q1e; (1), p54: Example Q4 from p45: student Q8(a); answer, Sample HP2 p59: p67: student M16 HLP2 TZ0, Q4 N17 (ii); Library TZ1, Q8 answer, answer, (a) and p100: p84: (b); M17 Sample M16 problem HP2 2, TZ1, N17 SP1 Eye for their from M16 HLP2 HP2 N16 Q5e; Q2 TZ0 p78: answer, student Q8; p41: Q17; M16 p15: answer, Sample Practice problem TZ0, Q1a; from Sample (e) 3, p42: Sample M16 HP2 M17 student (i)-(iii); Q2; problem TZ2, student Q3a; M17 Sample Derived Q3d/e; student from N16 Sample answer, HP2 student TZ0, Q2 (b) answer, Sample HP2 p110: Q28; answer, Derived Sample from student M17 HP2 answer, Sample TZ0, from N16 student Q2; TZ0, answer, TZ2, Example S.2.2, M17 p115: TZ2, Q1a+Q1b; p136: from M16 TZ0, Q2; Q19; by M17 Sample from student HP3 student M17 Example p212: Q24a; N17 Example M16 Sample Q27; from HP3 HP3 D.1.1, HP3 p206: TZ0, answer, TZ2, Q16; answer, TZ2, Derived Q17; from Example p208: N17 Example p210: Q18; p212: TZ0, D.2.1, TZ0, student M16 Sample 1, p219: HP3 Derived TZ2, from Q22(b); 1, p220: Derived N16 Practice Derived Q23; p213: Q22; answer, HP3 student M16 Sample from Practice TZ0, N16 C.7.2, HP3 C.8.1, TZ0, p209: HP3 TZ0, problem from Practice Derived Practice p216: Q23; answer, HP3 TZ0, student N16 HP3 problem Practice D.7.1, HP3 TZ2, p223: Q25; derived Practice 2, M17 p224: HP3 Derived TZ2, Q2; problem from M17 problem from N17 problem N17 HP3 Practice p218: Q24; N17 Practice answer, TZ0, 1, 2, M17 problem from Sample problem 1, p226: Derived problem 2, p226: Inspired 2, Q21; p210: HP3 TZ1, p212: 2, HP3 p215: TZ0, Q18; Adapted TZ0, problem Q22(c); 2, Practice p217: HP3 TZ0, Q25(b); Adapted Practice p220: problem M17 HP3 2, Derived p219: TZ2, Q23; Q24(b); Sample student answer, Practice p222: p222: Derived p222: Derived from N16 from HP3 M17 TZ0, HP3 Q28(d), TZ1, and Q25(e); 1, HP3 TZ1, p224: Q28; Derived Sample from student N16 HP3 answer, TZ0, p223: Q26; Practice M17 HP3 student TZ1, Q28(b); answer, Example p226: N16 D.8.1, HP3 TZ0, p225: Q28; Derived Practice from HP3 SLP2 2, TZ2, p80: (iii); TZ0, problem Inspired student TZ0, M17 Q8 N16 M17 student (b); (2), 2, by M18 HP3 Example HP3 TZ2, D.9.2, TZ1, Q26(b), p228: Q29(c); and M16 Example HP3 Derived from D.9.1, TZ0, Q25(e); M16 p227: Practice HP3 TZ0, Derived Q23(c); from Sample answer, HP3 TZ2, p229: Q20(a) M17 (iii) HP3 and TZ1, (iv); Q25(d); Practice Practice problem problem 2, p229: 1, p229: Derived Derived from N16 from HP3 Q1, p235: Derived from N16 SP1 TZ0, Q1; Q2, p235: Derived from M16 SP1 TZ0, TZ0, Q3; Q3, Q5, p235: Q4 (d) Derived from student HLP2 answer, TZ1, answer, Q4 p103: (c); 18.3.2, N17 M17 M16 p71: N17 Q7, p235: Derived from N17 Q9, p236: Derived from M16 TZ1, HLP2 Q5d; Practice p110: TZ2, problem Derived Q2(b) (i) and from (ii); Sample HLP2 TZ2, student Q7 (a) answer, (i)-(iii)); Example TZ0, Adapted 22.1.2, Q1b; from p130: Example M16 TZ0, Inspired S.2.1, Q2; by p134: S.2.3, student answer, p138: N17 TZ0, Q2; p235: p235: Derived Derived from from N17 N17 SP1 SP1 TZ0, TZ0, Q7; Q10; M17 SP1 TZ0, SP1 Q12; TZ0, Q15; Q8, p236: Q10, Derived p236: from Derived M16 from SP1 M17 TZ0, SP1 Q13; TZ0, p236: Derived from M17 SP1 TZ2, Q17; Q12, TZ1, Q20; p236: Derived from M18 SP1 Q16; TZ1, Q13, p236: Derived from M17 SP1 Q14, p236: Derived from M16 SP1 SP2 Q20; Q15, p236: Derived from M17 SP1 TZ1, Q22; Q16, p236: Derived from M16 TZ0, Q22; SP1 TZ2, Q17, Q26; p236: Q19, Derived p236: from Derived M18 SP1 from TZ2, N17 SP1 Q25; TZ0, Q18, Q30; p236: Q20, Derived p237: from Derived from M16 SP1 TZ0, Q29; from M17 SP1 TZ1, Q3; from M16 SP1 TZ0, Q10; Q21, p237: Derived from M16 SP1 TZ0, Q4; Q22, p237: Derived SLP1; Q23, p237: Derived from M17 SP1 TZ2, Q6; Q24, p237: Derived HP2 from N17 SP1 Q25, TZ0, p237: Q14; Derived Q27, p237: from N16 Derived SP1 from TZ0, M18 Q12; SP1 Q26, TZ2, p237: Q21; Derived from M18 SP1 TZ2, Q27; Q29, p237: TZ1, Q29; Q31, Derived from N16 SP1 Q28, TZ0, Q25; HLP2 Q30, p237: Derived from M18 SP1 Q32, p238: Derived from M16 HP1 p237: Derived from N16 HP1 TZ0, Q6; Q33, p238: Derived from N17 HP1 TZ0, Q2; TZ0, Q2a; Q34, p238: Derived from N17 HP1 TZ0, Q11; Q35, p238: TZ0, Q15; Q37, Derived from N17 HP1 TZ0, HP2 N16 TZ0 Adapted Example Q4, student Q14; Q36, p238: Derived from M16 HP1 p238: Derived from HP1 TZ0, N17 HP1 Q18; Q38, p238: Derived from N17 HP1 TZ0, Q17; Q39, p238: Derived N17 from TZ0, TZ0, Q21; Q40, p238: Derived from M17 HP1 TZ2, Q20; Q41, p238: Derived 20.1.1, HP2 22.1.1, N17 Q5; Q6, p110: N17 p124: Q8; Sample HP2 Example Example TZ1, TZ1, HP2 M17 TZ0, N17 SP1 SP1 answer, (ii); p89: HP2 M17 p46: N10 Sample p109: from p42: M17 p61: and Derived Sample student (i) p235: HP2 answer, answer, p54: 3, answer, Sample HLP2 Example problem M17 M17 Sample Q21(c); M17 Q3b; Q1b(ii); problem student Sample p94: HP2 Q7b(ii); M16 TZ2, by past M17 HP1 TZ2, Q22; Q42, p238: Derived from M18 HP1 TZ1, Q26; Q43, p238: TZ0, from M18 HP1 TZ1, Q27; Q44, p239: Derived from N16 HP3 TZ0, Q32; Q45, p128: Derived from M18 HP1 TZ2, Q31; Q46, p239: Derived TZ2, Q36; Q48, p239: TZ1, Q37; Q50, from M17 HP1 TZ2, Q33; Q3; from Q47, p239: Q36; Q49, Derived from M17 HP1 p239: Derived from M17 HP1 p239: p137: Practice Derived from N16 HP1 TZ0, M16 TZ1, Adapted 2, HP3 kind p21: student SP2 answer Practice Sample Q4d(ii); Practice p126: p131: Sample problem M16 of p239: Adapted TZ0, Q25(e); Derived Q6; Q23; Q27(b); from p118: Q21(a); p215: Adapted TZ0, TZ0, Q26; from the corresponding student TZ1, answer, TZ0, Adapted SLP2, Practice TZ0, p103: TZ0, TZ1, Q10; Practice Q22; Inspired Adapted TZ2, TZ0, HP3 p237: student Practice Mopic/Shutterstock; content the Sample SLP2 Practice HP2 p48: answer, (e); HP3 p217: HP3 M17 Derived p80: Q22; Adapted Photo and and Sample (c) M18 student Q5a(ii); 8.1.2, M17 9.2.1, TZ0, answer answer, HP3 M17 student HP3 TZ0, HP3 M16 SP1 HP2 Sample from Photo/Shutterstock; Baccalaureate examinations for TZ0, M16 Adapted student TZ1, p206: p208: p210: M17 TZ0, TZ1, Q21; Derived here: SLP2 problem SP2 TZ0, Library; Q18; Partly HP3 p206: Library Q11, Practice 1, Alamy Q29; paper N16 OUP thank questions from problem Shut- M17 subject HP3 C.7.1, student N16 HP3 1, N16 student permission Derived Practice Russ/ (T) M17 The p205: Q17(c); 6.1 from p140 M17 N16 HP3 1, M17 student problem p125 C.6.1, TZ2, p12 M17 Science/Science M16 Sample Example Steve HP3 Commons Collection GIPhotoStock/Science Stevens/Shutterstock; Derived Humdan/Shut- Fedorenko/Shutterstock; Alamy p203: (T) problem terstock from from from Shutterstock; and problem Practice problem terstock; 1, Library from Creative Shutterstock Example M17 Example Derived from Attribution problem Science p211 Juhku/Shutterstock; Photo; licensed Q23; p205: Q21, Q21; problem Electromagnetic Practice (T) from (B) Q13; Science Sample Winters/Science TZ1, Adapted Inspired Science TZ0, Practice Q24(a); Photo/Shutterstock; TZ0, Pho- Adapted Photo SP3 DAM Q18; p47 HP3 answer, N16 p207: Alamy Yasni/Shutterstock; N16 suggestions credits: Cover: p202: Bylikin student would answer, Rout, from for student Q40; Q18(a), p252: Adapted from M17 HP3 TZ1, Q22 Derived from M18 HP1 C ontents Introduction iv Redox processes 18 Acids and bases (AHL) B.2 Proteins and enzymes 9.1 Oxidation and relationships reduction 9.2 1.1 Electrochemical cells 18.1 Lewis acids 55 58 and bases Introduction to the particulate The mole concept 10 Carbohydrates 172 105 B.5 Vitamins 174 B.6 Biochemistry and the environment Fundamentals of organic chemistry 10.2 62 19 Redox processes (AHL) B.8 Proteins and enzymes (AHL) 179 Nucleic acids (AHL) 184 Functional group B.9 chemistry 68 Biological pigments 19.1 Electrochemical (AHL) 111 B 10 2.1 The nuclear atom 10 2.2 Electron conguration 12 Stereochemistry in biomolecules 11 Measurement and (AHL) Periodicity 3.2 Periodic trends reactions C Energy 116 72 20.2 Synthetic routes 120 C.1 74 20.3 Stereoisomerism 122 Energy sources 192 C.2 Fossil fuels 194 C.3 Nuclear fusion 17 11.2 Graphical techniques 11.3 Spectroscopic 19 identication of organic compounds 189 Organic chemistry (AHL) 20.1 Types of organic Uncer tainties and errors in measurement and results Periodic table 20 data processing 11.1 3.1 186 Atomic structure cells 176 10 B.7 10.1 Reacting masses and volumes 169 B.4 104 Organic chemistry 18.3 pH curves 1.3 Lipids involving acids and bases chemical change B.3 18.2 Calculations nature of matter and 1.2 164 Stoichiometric and ssion 76 21 197 Measurement and C.4 Chemical bonding Solar energy 20 analysis (AHL) C.5 and structure 12 Atomic structure (AHL) Environmental impact—global 21.1 Spectroscopic warming 4.1 Ionic bonding and 201 identication of structure 22 4.2 Covalent bonding 23 4.3 Covalent structures 24 12.1 Electrons in atoms 79 organic compounds 125 C.6 Electrochemistry, rechargeable batteries and fuel cells (AHL) 13 4.4 Intermolecular forces The periodic table—the 22 Metallic bonding C.7 Nuclear fusion and nuclear ssion (AHL) questions (Section A) 20 29 C.8 13.1 203 28 transition metals (AHL) 4.5 Data-based and practical First-row d-block 22.1 Data-based elements 81 Coloured complexes 83 questions Photovoltaic and dye-sensitized solar 128 cells (AHL) 209 Energetics/ 13.2 22.2 Practical questions 133 thermochemistry D 5.1 Measuring energy changes 14 Chemical bonding and A Medicinal chemistry Materials D.1 31 Pharmaceutical structure (AHL) A.1 5.2 Hess’s Law 33 5.3 Bond enthalpies 34 introduction 14.1 and structure Chemical kinetics 6.1 Collision theory and rates of reaction 14.2 38 15 action 140 A.2 Hybridization D.2 Metals and inductively coupled plasma (ICP) 85 spectroscopy 142 A.3 Catalysts 145 A.4 Liquid crystals 147 D.3 Energetics/ A.5 Polymers 149 A.6 Nanotechnology 152 D.4 D.5 15.2 Equilibrium pH regulation of 217 Antiviral 219 Environmental Environmental 91 impact of some impact—plastics 7.1 215 medications A.7 Energy cycles 213 Opiates the stomach D.6 Equilibrium 15.1 Aspirin and penicillin 89 thermochemistry (AHL) 211 Fur ther aspects of covalent bonding products and drug Materials science 153 medications Entropy and 221 43 A.8 spontaneity Superconducting 93 D.7 Taxol—a chiral metals and x-ray auxiliary case crystallography (AHL) 155 study (AHL) 16 Chemical kinetics (AHL) A.9 Condensation D.8 polymers (AHL) 8.1 (AHL) 16.1 Rate expression A 10 47 Proper ties of acids and bases 48 8.3 The pH scale 50 8.4 Strong and weak 16.2 96 Activation energy 98 metals (AHL) 51 17 Equilibrium (AHL) Internal assessment 230 Practice exam biochemistry The equilibrium law 227 Introduction to 52 17.1 and analysis (AHL) 159 Biochemistry B.1 Acid deposition Drug detection impact—heavy mechanism B acids and bases 224 Environmental D.9 and reaction 8.5 Nuclear medicine 158 Theories of acids and bases 8.2 222 Acids and bases papers 235 Index 255 162 102 iii I N T R O D U CT I O N This book syllabus provides in full Chemistry coverage and of offers the IB support diploma Over view of the book structure to The students preparing for their examinations the book will help you revise the study book material, essential terms and concepts, internal skills and strengthen improve your IB examinations worked best All examples practices topics answers explain are to The and and warn marks is packed tips against illustrated questions why book exam by from may that past be errors The section is questions, dedicated which are to missed data-based the of the Numerous situations, data, and interpret suggest procedures of syllabus examples IB-style most of and 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and experimental and papers your in tackle to cover distinctive assessment how improvements Practice opportunities boost (first show SL data-based options section the statements are feature four understandings for practical core topics, examination largest syllabus which SL A separate (AHL) the practice follows student examinations, or that with common scored sections approach demonstrate annotated several your of to into assessment, level questions, problem-solving divided learn higher the is The a practice complete techniques that list may www.oxfordsecondary.com/ib-prepared-support The As to any study replace guide, your textbooks, th is cou rs e laboratory book is not ma teria ls , m an ua ls, in t e n d e d s uch p a st the as is pa pe r s the IB Chemis try s yl l abu s presented you own will many notes need of To to which s u cceed use are a in b roa d a va ilable the internal the that this critical boo k part preparation for of w ill ran ge of online na viga te y our th e studies, ex a m les s The explains and process y ou (SL) and how stres sf u l and three papers that level internal to report criteria and more The final you select in and section this a the AHL assessed four in options subtopics outlines will have suitable a data, the to nature carry topic, draw suitable achieve at the contains papers book opportunity higher the and section experimental your out collect conclusions format to satisfy and the the highest grade These to test same 1, IB-style and papers practice written will yourself time 3, give before provide exclusively you the an actual additional exam practice (HL) problems complete SL of your and level material t h roug h ma kin g for must the Each a u t h ors DP Chemistry assessment standard of assessment examination students re s ourc e s, efficient All series investigation and marking this a reviews paper exami n a t i on, present hope as of and of your section part a nd The markschemes, options second for every topic of core and options assessment material take assessment day and external in the DP iv Papers Paper a at the Chemistry part and day assessment table as or top grade, of are two marks their external usually later are page from of sat The to give (lowest) one internal combined V on as your to and shown overall (highest) The answers and and examination solutions papers to are all practice given online problems at www.oxfordsecondary.com/ib-prepared-support Blank answer available at sheets the for same examination address papers are also Assessment over view SL Assessment Description HL Topics marks Internal Experimental work with a written repor t Paper Multiple-choice questions — weight 24 Core: marks 20% 30 weight 24 20% 20% 40 20% 1–11 (SL) Paper Shor t- and extended-response questions 50 40% 95 36% 1–21 (AHL) Section A : data-based and practical — 15 15 questions 20% Paper Section B: shor t- and extended-response 24% Option of your 20 questions The final points IB from diploma theory of 30 choice score is calculated knowledge and by combining extended essay grades for six subjects with up to three additional components Command terms Command terms command For example, levels of are term the detail, (“discuss”), as pre-defined specifies the command from a shown and terms single in words type the state, word, next and phrases depth of outline, short the used explain sentence in all response or and IB Chemistry expected discuss numerical from require value questions you in a answers (“state”) to and problems particular with Each question increasingly comprehensive higher analysis table Question Possible answer State the eect of increasing temperature Rate increases on the reaction rate Outline how an increase in temperature For most reactions, the rate approximately doubles when temperature increases by aects the reaction rate 10 degrees As temperature increases, the average speed and thus kinetic energy of particles also Explain why an increase in temperature increase The particles collide with one another more frequently and with a greater force increases the reaction rate As a result, the frequency of successful collisions increases, so the rate increases Both factors increase the rate by increasing the frequency of successful collisions However, an increase in temperature increases the frequency and intensity of all Discuss the eects of increasing collisions (successful and unsuccessful) but has no eect on the activation energy temperature and the presence of a catalyst In contrast, a catalyst has no eect on the frequency or intensity of collisions but on the reaction rate lowers the activation energy by providing an alternative reaction pathway and thus allowing slow-moving par ticles to collide successfully Thus, the same macroscopic eect is achieved by dierent microscopic changes A list of commonly Understanding examination questions in the used exact Therefore, this command meaning you terms of should in Chemistry frequently explore this used examination command table and use it questions terms is is given essential regularly as a for in the your reference table success when below in the answering book Command term Denition Annotate Add brief notes to a diagram or graph Calculate Obtain a numerical answer showing your working Comment Give a judgment based on a given statement or result of a calculation Compare Give an account of the similarities between two or more items Compare and contrast Give an account of similarities and dierences between two or more items Construct Present information in a diagrammatic or logical form Deduce Reach a conclusion from the information given Describe Give a detailed account Determine Obtain the only possible answer Discuss Oer a considered and balanced review that includes a range of arguments, factors or hypotheses Distinguish Make clear the dierences between two or more items Represent by a labelled, accurate diagram or graph, drawn to scale, with plotted points (if Draw appropriate) joined in a straight line or smooth curve Continued on page VI v INTRODUCTION Command term Denition Estimate Obtain an approximate value Explain Give a detailed account including reasons or causes Formulate Express precisely and systematically a concept or argument Identify Provide an answer from a number of possibilities Justify Give valid reasons or evidence to suppor t an answer or conclusion Label Add labels to a diagram List Give a sequence of brief answers with no explanation Outline Give a brief account or summary Predict Give an expected result Represent by means of a diagram or graph (labelled as appropriate), giving a general idea of the Sketch required shape or relationship State Give a specic name, value or other brief answer without explanation Suggest Propose a solution, hypothesis or other possible answer A complete list of command terms is available in the subject guide Preparation and exam strategies In addition study and Get ready drink night’s exam the for Organize and physical it can are some adequate ventilation and Bookmark useful Plan studies your simple online and well, A by rules good before your arrange tasks Create make them into an offline agenda the that Make a list importance smaller, sure before by the Find a easily for you distractions organized Use this Get ready reference understanding on the Check more vi as Work systematically topics your during your Break studying complete tasks up your preparation large time your and way where s k il ls point the and textbook a nd on lin e for exams Practice under Chemistry and past use efficiently papers the the papers as at a Solve you the can end as than worked looking at reference answering time constraint data booklet many Take of every before future rather definitions solve questions Optimize terms as is topics in ex t r a imp rovemen t notes the to problem to how carefully , task the ga p s Sp en d Make Try and a this problems trial exam book your exam paying Keep possible units approach extra your attention answers Double-check Label Use axes exam in as all to short all from and and this questions command clear numerical graphs tips Read values annotate book of th rou g h iden tify an d information f ir st words practice for understanding points Learn and y o ur own on key and parts each Focus exam-style from deadline book follow Recite and answer quickly material manageable your can of your example diagrams should actively using the lighting, files Read using you memorizing score Avoid computer eat stress exercise environment with papers sleep, your important improve study place and there enough reduce particularly your your and is as temperature Have water thinking day , suggestions, itself study of sleep comfortable Keep above exam plenty positive to the you r t i me re qui re d res ou rc e s f or Do not panic concentrate realistic these goals goals improve a positive things and Be performance to Take on you work prepared and your learn future attitude can and improve systematically to reflect from your results on Set to achieve your errors in order Key features of the book Each chapter “You should outline the typically know” covers and one “You understandings and core should or option be able applications and topic, to” and starts checklists skills sections with These of the IB Theoretical concepts and key diploma Chemistry syllabus Some assessment statements have been definitions are discussed at a reworded or combined together to make them more accessible and level sufficient for answering simplify the navigation These changes not affect the coverage of typical examination questions key syllabus material, which is always explained within the chapter Many concepts are illustrated Chapters contain the features outlined on this page by diagrams, tables or worked examples Most definitions are Example given in a grey side box like this one, and explained in the text Examples common offer solutions to problem-solving alternative answers and typical problems techniques explain how Many the and demonstrate examples marks are provide awarded This feature highlights the Nature of science relates a chemistry concept to the overarching essential terms and statements principles of the scientific approach that have appeared in past markschemes, warns against common errors and shows how Approaches to learning gives advice on the development of to optimize your approach to communication, social, self-management, research or thinking skills par ticular questions AT L Sample student answers show typical student responses to IB-style Links provide a reference to questions (most of which are taken from past examination papers) In relevant material, within another each response, the correct points are often highlighted in green while par t of this book or the IB Chemistry incorrect or incomplete answers are highlighted in red Positive or data booklet, that relates to the text negative feedback on student’s response is given in the green and red in question pull-out boxes An example is given below You will icon Number The may on marks have their the student earned marks of on see the question available from a an has past exam right been IB paper when the adapted paper based response Examination Negative question feedback SAMPLE STUDENT ANSWER ▼ “a monomer” clear enough, single This answer could have achieved 1/2 two condensation reaction is an as reacting it not implies species a while a marks: condensation A is [2] Describe what is meant by a condensation reaction anabolic reacting must involve species, which at least can be reaction identical or different molecules or ions that builds a polymer from a monomer ▲ “water” the most although The Questions student’s not taken response from past Positive IB examinations will not have feedback the exam the molecule” paper is accepted common answer would as by-product, be “a small more accurate icon Practice problems Practice problems are given at the end of each chapter problem-solving skills Some questions introduce These are IB-style questions that provide you with factual or theoretical material from the syllabus that can an oppor tunity to test themselves and improve your be studied independently vii STO I C H I O M ET R I C T O P I C R E L AT I O N S H I P S I N T R O D U C T I O N PA R T I C U L AT E A N D atoms ratios of to different form properties ✔ a mixture is substances a elements their combine which constituent combination that mixtures can be M AT T E R C H A N G E retain their of two in xed differ ✔ deduce in and or equations from given reactants elements; or ✔ apply state ✔ explain symbols in equations; more individual homogeneous chemical products; observed changes in physical properties properties; and ✔ O F Y s be abe t: compounds, from T H E N AT U R E C H E M I C A L Y s kw: ✔ T O temperature during a change of state heterogeneous When substances are mixed together physically , they can be combined • Ceca stcetry is the in any proportion Mixtures can be homogeneous (with uniform relationship between the amounts properties throughout, for example, air) or heterogeneous (in which of the reactants and products in a the composition varies and components may be in different phases, chemical reaction like a mixture of gravel and water) Mixtures can usually be separated • Stcetrc ceffcets by physical processes such as filtration or distillation However, when describe the ratios in which substances react to give a chemical compound, their proportions are amounts of species react with fixed in a stoichiometric ratio and they can only be separated again by a one another chemical reaction Stoichiometric stoichiometric aA + in bB → which a xX with In some questions, state in + of the of A reacts chemical of central to chemistry For a general form: each with These type b moles react with coefficients of of B, stoichiometric species stoichiometric number are yY coefficients which correct same equation moles stoichiometric ratios calculations atom on is a, one said each b, x and y coefficients another to be are the show An the equation balanced, with the side symbols are required and you To formulate and balance stoichiometric equations quickly , it is useful will be penalized if these are not to memorize the formulas and charges of common ions (table 1.1.1) included Remember that the state symbol for water in the liquid Name phase is (l), not (aq): H Formula and charge Name Formula and charge nitrite NO nitrate NO sulfite SO O(l) + ammonium NH 2– carbonate CO hydrogencarbonate HCO 3 2– 3 Symbols and names of chemical 2– ethanedioate (oxalate) C 2– O sulfate SO 4 elements can be found in section 3– phosphate 2– PO thiosulfate S O of the data booklet ▲ Tabe 1.1.1 The names, formulas and charges of common polyatomic ions Chemical gas (g) equations and aqueous often include solution (aq), state symbols: which means solid (s), dissolved liquid in (l), water 1.2 ThE molE ConCEpT Exae 1.1.1 Formulate reaction H PO , a of in balanced equation, potassium aqueous including hydroxide, KOH, state with symbols, for phosphoric the acid, solution Solution First, write KOH the + H formulas PO Then balance sides are 3KOH the H → the K equal + of PO equation Do this PO reactants so by → + H K that PO products O the adjusting and numbers the + of atoms coefficients 3H on on both each side O Remember, the chemical formula Finally , add the state symbols Aqueous solutions are involved, so of a substance should never be (aq) is used for all species except water changed when balancing chemical 3KOH(aq) + H PO T O P I C (aq) → K PO (aq) + 3H equations, only its coefficient O(l) T H E M O L E C O N C E P T Y s kw: Y s be abe t: 12 ✔ masses and of atoms expressed are as measured relative relative atomic mass to (A C ) ✔ calculate and the molecules molar and masses formula of atoms, ions, units; r relative formula/molecular mass (M ), which ✔ r have no solve numerical relationships ✔ the problems involving the units; mole is a measure of the amount n, and refers to a very n, m and M; of ✔ substance, between large, calculate empirical and molecular formulas xed and percentage composition by mass from 23 number of entities (6.02 × 10 ); given ✔ molar mass (mass of one mole of a data substance), –1 M, ✔ an has a of order of to SI formula each molecular atoms In derived empirical atoms ✔ the is element formula each unit element determine the in is g a ; simplest ratio of the compound; the in mol actual a number of molecule stoichiometric ratios from observations, chemists • Reate atc ass (A ) is the ratio r need a way to calculate the amount of substance—the number of atoms, of the average mass of an atom of a molecules or ions in a known mass of that substance chemical element in a given sample to one-twelfth of the mass of a carbon-12 The masses of atoms of most elements have been measured with atom Since the value is relative, it has a high degree of accuracy For example, an atom of carbon has a no units The terms reate ecar –26 mass of 1.993 × 10 kg However, it is more convenient to express ass and reate fra ass masses of atoms and molecules as ratios relative to the mass of the (both M ) are used for molecules and r 12 C atom, which is defined as 12.00 on the relative scale These ratios ionic species, respectively are known as relative atomic mass (A ) and relative molecular mass r (M ), r • The at f sbstace, n, is respectively , and have no units the number of atoms, molecules or The SI (Système measurement the SI unit for International It has seven amount of d’Unités) base is units, substance, the one symbol metric of n system which One is of the mole, mole contains 23 6.02 × 10 collection elementary of 12 objects entities, This just as number one is the dozen fixed represents a numerical ions, expressed in moles, in a given quantity of the substance • The e (abbreviated to mol) is the SI unit for amount of substance • The Aar cstat, N , A 23 value of the Avogadro constant, N 6.02 × 10 −1 mol , is the number of A par ticles in mol Without units, it The mole applies to elementary entities (atoms, molecules, ions, is called the Aar ber electrons, other particles, or specified groups of such particles) T O P I C ACIDS AND T H E O R I E S O F You should no: ✔ a Brønsted–Lowry Brønsted–Lowry ✔ ✔ species and pair of called Acids acid base Lowry a and a species have is is a a can proton proton act as donor and a ✔ deduce acceptor; A N D both differing by acid–base opposing a Brønsted– single Brønsted–Lowry chemical ✔ bases; conjugate bases A C I D S B A S E S You should be able to: amphiprotic acids BASES proton deduce the a species given acids and bases in reactions; conjugate acid or conjugate base for is pair properties and neutralize each The Lewis theory of acids and other when mixed together The concepts of acidity and basicity can bases is discussed in topic 18.1 be extended and to all classes of chemical compounds, including salts oxides + According and bases to the Arrhenius produce theory , hydroxide (OH acids ) produce ions in protons aqueous (H ions) solutions The + A free proton, H more general Brønsted–Lowry theory defines acids as proton , cannot exist donors in aqueous solutions, as it and bases as proton acceptors For example, aqueous hydrogen immediately forms a coordinate chloride is HCl(aq) → both an Arrhenius and a Brønsted–Lowry acid: bond with water and produces a + H (aq) + Cl (aq) + hydronium ion, H O However, the + Gaseous ammonia is a Brønsted–Lowry base but not an H Arrhenius (aq) symbol is often used as a + base, as it can accept protons (for example, from hydrogen chloride) shor thand equivalent of H O (aq) but does not contain oxygen and thus cannot produce hydroxide ions: In the IB examinations, chemical + equations involving H + or H O NH (g) + HCl(g) → NH Cl(s) When forms an its acid loses conjugate a ions are equally acceptable proton, acid by it forms accepting its a conjugate base, while a base proton Example 8.1.1 The ionization of hydrogen cyanide in water proceeds as follows: + HCN(aq) + H O(l) H O (aq) + CN (aq) Any equilibrium involves two a) Identify this the Brønsted–Lowry acids and Brønsted–Lowry bases equilibrium in reactions, forward and reverse In this forward reaction, hydrogen cyanide loses a proton to water, so b) Identify the conjugate acid–base pairs in this equilibrium HCN(aq) acts as an acid and H O(l) Solution acts as a base In the reverse + a) Brønsted–Lowry acids: HCN(aq) and H O (aq) reaction, the hydronium ion loses a proton to the cyanide ion, so Brønsted–Lowry bases: CN (aq) and H O(l) + H O (aq) is the acid and CN (aq) + b) HCN(aq)/CN (aq) and H O (aq)/H O(l) is the base 47 ACIDS AND BA SE S Amphiprotic species, such as water, can both donate and accept protons: + H O(l) H (aq) + OH (aq) Remember that the species in a + H O(l) + H + (aq) H O (aq) conjugate acid–base pair differ by one proton For example, H SO Amphoteric species species amphoteric, can react with both acids and bases All amphiprotic species are amphiprotic are not conjugates, as and SO are but not all amphoteric For they differ by two protons example, zinc oxide can react with both acids and bases, so it is amphoteric: + ZnO(s) + 2+ (aq) 2H → Zn (aq) + H O(l) 2 ZnO(s) + 2OH (aq) + H O(l) → [Zn(OH) At is the not same time, ZnO ] (aq) cannot donate a proton (as it has none), so it amphiprotic Typical amphiprotic species are amino acids (topic B.2) and acid salts, such as NaHCO Example 8.1.2 (topic D.4) The Oxides and hydroxides of zinc, hydrogenphosphate ion, HPO , is amphiprotic Outline what is aluminium and transition metals meant are amphoteric (topic 3.2) when by it amphiprotic, behaves in giving this the formulas of both species it forms manner Solution The ion HPO is amphiprotic because it can both donate and accept a proton: HPO + (aq) H (aq) + PO HPO (aq) + (aq) + H (aq) H PO (aq) + Alternative equations can involve H O , OH or any other acids and bases, for example: HPO (aq) + OH (aq) PO HPO + H most with acids active carbonates participate metals, and O neutralization produce salts H O(l) H PO P R O P E R T I E S O F (aq) + H O(l) A C I D S A N D B A S E S You should be able to: in metal characteristic oxides, reactions ✔ hydroxides, deduce and reactions of balance equations for typical acids; hydrogencarbonates; ✔ ✔ (aq) You should no: ✔ + + (aq) T O P I C (aq) reactions and are exothermic and identify the different acid and base required to make salts; water ✔ explain how to concentration Neutralization reactions can be determine by an acid–base represented by unknown titration molecular and Neutralization reactions are ionic equations: utilized in acid–base titrations (topic 18.3) and used in medicine HCl(aq) + NaOH(aq) → NaCl(aq) + H O(l) ΔH = –57.6 kJ ΔH = –57.6 kJ for relieving hear tburn by antacids + H (aq) + OH (aq) → H O(l) (topic D.4) Measurement of enthalpy changes is discussed in The topic 5.1 same neutralization net ionic of any equation strong and + of 48 H (aq) or OH (aq) ions acid releases with the any same strong base amount of has heat the per mole PROPER TIE S OF ACIDS AND BA SE S Example 8.2.1 Potassium Identify its sulfate the acid molecular can and and be produced base net by required ionic a for neutralization this reaction reaction and formulate equations Solution Reactants: sulfuric acid, H SO Molecular equation: H SO , and potassium hydroxide, KOH (aq) + 2KOH(aq) → K SO (aq) + 2H O(l) + Net ionic equation: H (aq) + OH (aq) → H O(l) Other neutralization carbonates NH (aq) and + reactions involve HCl(aq) → NH Cl(aq) + H + (aq) + H → NH (aq) + H 2HCl(aq) → MgCl + + 2H O(l) O(l) (aq) + H MgO(s) oxides, + metal + (aq) MgO(s) ammonia, hydrogencarbonates: NH can O(l) 2+ (aq) → Mg (aq) + H O(l) Na CO (aq) + 2HCl(aq) → 2NaCl(aq) + CO (g) + H O(l) + CO (aq) + 2H (aq) → CO (g) + H NaHCO (aq) + HCl(aq) → O(l) NaCl(aq) + CO (g) + H O(l) + HCO (aq) + H (aq) → CO Most acids (topic 9.1), Mg(s) (g) + H + react with producing 2HCl(aq) metals above hydrogen → MgCl O(l) hydrogen gas (aq) and + H + Mg(s) + 2H in metal the activity series salts: (g) 2+ (aq) → Mg (aq) + H (g) All the above solutions of heat in reactions (table 8.2.1) reactions distinguished by can with the be Bases in used to acids Acidic use of reveal solutions can and acid–base the be basic presence of discovered solutions indicators or a pH acids by can the also meter It is impor tant to distinguish in release level and their macroscopic be (topic between changes at the molecular 8.3) effects If you are asked to state an observable change in the reaction ▼ Table 8.2.1 Common tests for acids between an acid and a carbonate, Reaent type Example Obser vation soluble base (alkali) NaOH(aq) or NH the answer “carbon dioxide forms” will score no marks, as we cannot (aq) heat released visually distinguish carbon dioxide CuO(s), Mg(OH) (s) or solid dissolves* insoluble metal oxide, hydroxide or carbonate CaCO from any other colourless gas A (s) correct answer must refer to the soluble metal carbonate or metal Na hydrogencarbonate CO (aq) or NaHCO (aq) bubbles of gas* bubbles of gas produced or, in the bubbles of gas, active metal case of an insoluble carbonate, to Mg(s) or Zn(s) metal dissolves* the dissolution of the solid *Heat may also be released 49 ACIDS T O P I C AND BA SE S T H E p H S C A L E You should no: You should be able to: + ✔ pH ✔ a = –log[H change of + ] and one pH [H pH ] unit = 10 + ; represents a tenfold ✔ solve problems ✔ explain the involving use of a pH [H ], meter [OH ] and pH; and + change ✔ pH in [H values and universal ]; distinguish alkaline between acidic, indicator neutral solutions; + ✔ the ionic product of water, = K [H ][OH ], is a a weak w 14 constant equal to 1.00 ì 10 ã The ionic product of ater, K , at 298 K Water is electrolyte Its dissociation is characterized by the  is the equilibrium constant for the ionic product of water, K : w dissociation of water: + O(l) H H + (aq) + OH (aq) K = [H ][OH ] w + = [H K ][OH ] w 14 In dilute solutions at room temperature (25C), = K 1.00 ì 10 w ã The concentration of water is + Therefore, included in the value for K it is sufficient to know either [H ] or [OH ] in a solution, w as • The pH scale describes acidity as the concentration of the other ion can be found from the expression K w the potential of hydroen: The acidity and basicity of solutions (table 8.3.1) over a broad range + pH = –log[H ] + of H and known OH as the concentrations potential of can hydrogen, be or characterized by a single value, pH: + pH The potential of hydroxide is similar to pH: pOH = –log[OH = Note –log[H that ] pH values ] From the unitless + Solution expression for K are [H ] and [OH pH ] , it follows that w + acidic [H ] > [OH ] ] = [OH ] ] < [OH ] < at 25°C, pH + pOH = 14 You can + neutral [H use this expression in any pH + basic [H > calculation, although problems involving pOH appear only in ▲ Table 8.3.1 Acidic, neutral and basic aqueous solutions at 25°C higher level topic 18.2 Example 8.3.1 Calculate acid, H the SO , pH and values for sodium 0.0100 mol hydroxide, dm solutions NaOH Remember that each mole of sulfuric acid produces two moles of Solution + + H (aq) ions Similarly, one mole of a Sulfuric acid: H SO strong diprotic base, such as barium → 2H (aq) + SO (aq) + [H ] = × [H SO hydroxide, Ba(OH) (aq) ] = × 0.0100 = 0.0200 mol dm , produces two pH moles of OH = –log(0.0200) ≈ 1.699 (aq) ions You should always write the dissociation + Sodium hydroxide: NaOH(aq) → Na (aq) equation for an electrolyte before [OH ] = [NaOH] = 0.0100 mol dm calculating the pH of its solution + [H 14 ] × [OH ] = 1.00 × 10 Another common error is the 14 + use of [OH ] instead of [H ] in 1.00 × 10 + 12 [H ] = pH calculations If you obtain a = numerical answer that seems solution), check your calculations once more 50 × 10 mol 12 pH unlikely (e.g., a pH < for a basic 1.00 0.0100 = –log(1.00 × 10 ) = 12.000 dm + OH (aq) of sulfuric The pH of a solution estimated using across whole the can be universal pH measured indicator, with which a digital pH gradually probe changes STRONg AND wE Ak ACIDS AND BA SE S or colour range The progress of an acid–base + For a solution with known pH, the concentration mol of H (aq) + dm can be found by using the formula [H ions titration can be monitored with in pH ] = 10 a pH probe connected to a data logger or computer Alternatively, the equivalence point of the Example 8.3.2 titration can be determined using A solution of potassium hydroxide has a pH of 11.70 Calculate the an acid–base indicator (topic concentration, in mol dm , of potassium hydroxide in this solution 18.3) You should be familiar with these techniques and be Solution + KOH(aq) → K (aq) + OH × 10 able to describe the purpose (aq) and practical details of typical + [H 11.70 ] = 12 10 ≈ 2.0 mol dm titration experiments 14 × 1.00 10 [OH ] = = 5.0 × 10 mol dm 12 × 2.0 10 c(KOH) = [OH T O P I C ] = 5.0 × 10 mol dm S T R O N G A N D A N D strong extent and of weak their You should be able to: acids and ionization bases in differ aqueous in the ✔ distinguish solutions; and bases reactions ✔ a strong acid is a good proton donor and has ✔ conjugate a strong a weak Strong base such solutions terms and the strong of the and rates electrical weak of their acids typical conductivities of solutions base; a good conjugate acids, aqueous is between in a their weak A C I D S B A S E S You should no: ✔ W E A K as proton acceptor and has acid hydrogen while weak chloride, acids, HCl, such as dissociate ethanoic completely acid, CH in COOH, dissociate reversibly: Strong acids and bases dissociate + HCl(aq) → H (aq) + Cl (aq) irreversibly, which is shown by the straight arrow (→) in equations + CH COOH(aq) H (aq) + CH COO (aq) The dissociation schemes of weak In addition to HCl, strong acids include sulfuric acid, H SO HNO , perchloric acid, HClO HBr, and Strong some , chloric acid, HClO hydrogen bases iodide, include hydroxides of all HI nitric acid, acids and bases must include the hydrogen bromide, equilibrium sign ( Almost hydroxides group , , metals of all other Group (Ca to Ba) acids are metals All (Li other ) weak to Cs) bases and are weak + NaOH(aq) → Na (aq) + OH (aq) + NH (aq) + H With the NH than of stepwise the first (aq) + OH (aq) exception dissociate less O(l) sulfuric The proton acid, second all and because the polyprotic following acid acids protons anion exerts are weak and dissociate a stronger + electrostatic attraction for the next leaving H (aq) cation 51 ACIDS AND BA SE S Example 8.4.1 Deduce the scheme of the successive dissociation of phosphoric The terms “strong” and “weak” acid, H PO refer to the nature of acids and bases but not to their Solution concentrations A solution of a H + PO (aq) H (aq) + H PO (aq) strong acid can be dilute, while + H PO (aq) H 2− (aq) + HPO (aq) a solution of a weak acid can 2− HPO be concentrated Colloquial + (aq) H 3− (aq) + PO (aq) expressions, such as “strong Note that equilibrium signs are used in all equations, as phosphoric solution” or “weak solution”, acid is a weak acid must never be used in examination answers Relative by strengths comparing A solution weak of acid; the a of acids and properties strong a solution a weak acid of a bases of their will have strong can be determined solutions a base lower will of pH have equal than a a experimentally concentration solution higher pH of than a a When the strengths of acids and solution of base Both strong acids and strong bases have higher bases are compared, the solutions standard enthalpies of neutralization and produce solutions with used for pH, conductivity or higher electrical conductivities than weak acids and bases Strong acids reaction rate measurements must react with active metals, oxides, carbonates and hydrogencarbonates have equal concentrations Many faster than weak acids students lose marks by forgetting to mention this fact in examinations The electrical conductivities of solutions can be compared using a beaker Enthalpies of neutralization and reaction rates are discussed with a pair of electrodes that form a series circuit with a battery and a light in topics 5.1 and 6.1, respectively bulb The bulb will glow brighter when the beaker contains a solution with The reactions of acids with active more ions For precise measurements, the bulb can be replaced with a metals, oxides, carbonates and digital ammeter You should be able to describe the setting of a conductivity hydrogencarbonates are discussed experiment and interpret its results in topics 8.2, 8.5 and D.4 T O P I C A C I D D E P O S I T I O N You should no: ✔ rain is naturally approximately You should be able to: acidic 5.6 and because has of a pH of ✔ dissolved CO , deduce and combustion balance of the sulfur equations and nitrogen for the and the while ✔ acid and acid deposition deposition sulfur that is has caused dissolve a pH by in below oxides water of to 5.0; nitrogen form ✔ HNO subsequent formation distinguish between and , of the post-combustion acids; pre-combustion methods of reducing HNO , ✔ acid H SO and and SO deposition causes, H has as emissions well of sulfur soil, as and anthropogenic forests ✔ deduce the deposition and equations with the is naturally following acidic and has processes: Remember that carbon dioxide CO (g) CO (aq) is not responsible for acid CO deposition, as it cannot reduce the (aq) + H O(l) H CO (aq) + H CO 52 for reactive the reactions metals and of acid carbonates buildings Rainwater pH of rainwater below 5.0 oxides; natural damages watercourses ; (aq) H (aq) + HCO (aq) a pH of approximately 5.6 from Acid has deposition a significantly Acid with rain forms (g) + H pH than oxides O(l) HNO (g) + H SO lower when occurs of in the form normal nitrogen of acid rainwater and rain, which (typically < sulfur (table DEPOSITION 8.5.1) 5.0) react rainwater: 2NO SO commonly ACID O(l) H (g) + H (aq) + HNO SO O(l) → H (aq) (aq) SO Large quantities of sulfur dioxide (aq) are formed as a by-product of the + The acids produced in these reactions dissociate and form H (aq) ions, combustion of fossil fuels (topic C.2) which decrease the pH of rainwater Source O xide Natural Anthropoenic reaction of nitrogen with oxygen at high lightning strikes, biological temperature in car or jet engines: NO processes N (g) + O NO oxidation of NO in the atmosphere: 2NO(g) + O (g) 2NO(g) (g) 2NO (g) combustion of fossil fuels containing sulfur volcanic eruptions, forest res, SO biological processes (g) → SO impurities: S(s) + O SO oxidation of SO in the atmosphere: 2SO (g) + O (g) 2SO (g) (g) ▲ Table 8.5.1 Sources of oxides of nitrogen and sulfur Acid and rain can works these of damage art materials some made is of stone, marble calcium concrete or and limestone carbonate, CaCO , mortar The in main which buildings, component reacts with acids of in the rainwater, CaCO (s) + H Other for SO reduced (aq) effects and combustion methods of include coal) from and natural scrubbing of chemical acid lakes, sulfur products from of as productivity physical CaSO (s) + soil deposition rivers and oxides and and oil) of include of power the (physical fuels damage aquatic metals can be reduced (pre-combustion) of (chemical and to deforestation, Common removal sulfur O(l) plants Post-combustion of H corrosion from treatment removal + wetlands), (post-combustion) flotation (chemical (g) hydrodesulfurization gas CO (such emissions → negative environments The example: or by their pre-combustion elementary removal methods nitrogen of sulfur sulfur include oxides alkaline by their Catalytic conver ters use reactions with bases and water) and the use of catalytic converters palladium, platinum and other in cars Another technique, lime-injection fluidized bed combustion, precious metals as heterogeneous permits capture of oxides of sulfur immediately after they are released catalysts (topic A.3) by combustion of solid fuels Example 8.5.1 Sulfur dioxide equations, is how a major lime, cause CaO, and of acid rain limestone, Explain, CaCO , using can be chemical used to reduce sulfur dioxide emissions from coal power plants Solution Coal contains dioxide S(s) + traces upon O (g) → SO Both lime CaO(s) + of (s) sulfur, which forms sulfur (g) and SO limestone (g) → + react CaSO CaCO elementary combustion: with sulfur dioxide: (s) SO (g) → CaSO (s) + CO (g) 53 ACIDS AND BA SE S Sulfur can be removed from Calcium combustion products with the sulfite is non-volatile, so no sulfur is released into atmosphere calcium oxide (lime) Lime is Note that the last two reactions can occur either during the obtained by heating calcium combustion (when lime and limestone are added to coal) or after carbonate to drive off carbon the combustion (when the combustion products are bubbled dioxide, which is released into through a mixture of lime, limestone and water) In the second case, the atmosphere Thus, a solution lime is converted first into calcium hydroxide, which reacts with to one problem—acid deposition— sulfur dioxide: contributes to another— CaO(s) global warming + H O(l) → Ca(OH) Ca(OH) (aq) (aq) + SO (g) → CaSO (s) + H O(l) S AMPLE STUDENT ANS wER Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride ▼ Not is not or accepted, an observable “bubbles” rst as would change—“gas” score This answer could have achieved 1/2 marks: the mark Hydrogen is released: 2+ + 2NH ▲ Correct [2] “hydrogen” + Mg → Mg + 2NH equation + H S AMPLE STUDENT ANS wER Suggest why the enthalpy change of neutralization of CH COOH is less negative than that of HCl ▲ Correct statement about This acid [2] the answer could have achieved 1/2 marks: strength Because CH COOH is a weaker acid than ▼ This so a for is a further two-mark question, explanation example, “some of is the required; HCl, so it dissociates dissociates released to fully by neutralization ionize CH only partially, while HCl energy is COOH” completely used would score the second mark Practice problems for Topic Problem Problem What occurs when solid calcium carbonate reacts with The hydrogencarbonate ion, HCO , par ticipates in the aqueous nitric acid? following equilibria: A Bubbles of nitrogen dioxide form HCO + (aq) + H B Bubbles of carbon dioxide form HCO (aq) + H C Bubbles of hydrogen form O(l) H O (aq) + CO O(l) H (aq) CO (aq) + OH (aq) Identify one conjugate acid–base pair and t o amphiprotic species in these reactions D Bubbles of both carbon dioxide and nitrogen dioxide form Problem Barium hydroxide, Ba(OH) , is a strong base State the Problem + What is the ratio of [H equation for the dissociation of barium hydroxide and ] in aqueous ammonia of pH 10 calculate the pH of its 2.00 × 10 + to [H mol dm solution ] in water at 298 K? Problem A : Acid deposition is responsible for accelerated corrosion B 10 : of active metals State the molecular and net ionic equations for the reaction of sulfuric acid from rainwater C : 10 with zinc metal 10 D : 10 54 T O P I C R E D OX PROCESSES O X I D AT I O N A N D Yo ho kow: ✔ oxidation and of oxygen or change Yo ho be abe to: reduction can gain/hydrogen in oxidation be dened loss, in electron terms ✔ deduce transfer or transition metals metals can have and the activity most variable ease ✔ the with series which Winkler demand method (BOD) pollution in ranks they a as a main-group oxidation metals undergo uses non- according to identify the and oxidation; of the compound of an atom in an ion name from of its a transition metal formula; the oxidized oxidizing species, and reduced reducing species agents in redox reactions; oxygen degree ✔ deduce of acidic redox or reactions neutral using half-equations in solutions; sample ✔ deduce the ✔ Redox state states; biochemical measure water oxidation compound; deduce ✔ ✔ the state; ✔ ✔ R E D U C T I O N (reduction–oxidation) reactions play a activity solve fundamental the role redox feasibility series or of a redox reaction titration reaction from data; problems in • O iatio tate, also known many chemical and biochemical processes as oiatio ber, is the Oxidation and reduction can be defined in several ways (table 9.1.1) hypothetical charge that an atom would have if all its polar covalent ▼ Tabe 9.1.1 Definitions of oxidation and reduction bonds were ionic I ter of… O iatio oxygen/hydrogen gain/loss gain of oxygen or loss of hydrogen • An oiizig aget increases the Retio gain of hydrogen or loss of oxygen oxidation state of another species while being reduced itself oxidation state increase in oxidation state decrease in oxidation state electron transfer loss of electrons gain of electrons • A reig aget decreases the oxidation state of another species while being oxidized itself In a redox which agent, Cl is reaction, reduced and (g) the + one The species reducing 2I species agent (aq) is oxidized that is the is reduced species → by 2Cl reacting is that with therefore is another, an oxidizing oxidized (aq) + I (aq) One example of a key redox oxidizing reducing reduced oxidized agent agent (gained (lost process in the core syllabus is the electrons) electrons) oxidation of alcohols in topic 10.2 O iatio tate The • • rules The state metals oxidation of any always states atom have a in +1 are an as follows: elementary oxidation state substance in is ions compounds Group and assigning oxidation Group and • for elements always have a +2 oxidation state in ions compounds 55 REDOX P R O C E SS E S • Aluminium, most • of member The oxidation non-metal a The in state oxidation state The of of (e.g., state (species oxygen oxidation other of group 3, has an oxidation state of +3 in hydrogen in HCl), is and +1 when when it hydrogen is is to bonded bonded a metal NaH) peroxides • a compounds to (e.g., • its group of is an is usually –O–O– The linkage), in main which exception the are oxidation state 17 oxygen with of fluorine elements is in (halogens), all the its compounds oxidation state For is the usually When writing oxidation states, in binary compounds (HI, NaCl, KBr) but can be positive in the charge precedes the number oxides, oxoanions and oxoacids (e.g., chlorine in HClO has a For example, the oxidation state +7 of magnesium in MgCO oxidation state) is +2, • In a neutral molecule the sum of the oxidation states of all the not 2+ This is a common error atoms is zero seen in answers to exam questions • In a polyatomic equals the ion overall the sum charge on of the the oxidation states of all the atoms ion The common oxidation states of transition metals are listed in Note section 14 of the data booklet variable For that transition oxidation example, FeCl Some oxoanions and their non- metals states and SO oxide, and main These are group states are called non-metals specified iron(III) by can have Roman chloride and numerals sulfur(IV) respectively There are also non-systematic names for oxoanions systematic names are listed in of main group non-metals, such as hypochlorite (ClO ), chlorite (ClO topic 4.1 ), chlorate (ClO ) and perchlorate (ClO ) Eape 9.1.1 Section of the data booklet Deduce the names of the transition metal compounds CuO and gives the names of the elements in Co(NO ) the periodic table Solution To to deduce work the out names the oxidation state: CuO: ( x + 2) of transition oxidation = 0, so x state = of metal the complexes transition you metal first Let x need be the +2 The answer “copper oxide” would Co(NO to both Cu O (copper(I) oxide) and ) not get any credit, as it could apply The : x + × ( 1) = 0, so x = +3 names are therefore copper(II) oxide and cobalt(III) nitrate CuO (copper(II) oxide) The ativity erie The activity undergo will The activity series is given in series ranks oxidation, occur from their Fe(s) + More and reactive compounds, metals enables in elements for order us to of the predict will ease with whether displace less which they redox reactions reactive elements example: section 25 of the data booklet CuSO (aq) → FeSO (aq) + Cu(s) Haf-eqatio A half-equation either For the example, above) is an oxidation can the be equation or that reduction reaction of represented iron by the includes process metal with following 2+ Fe(s) → Fe (aq) + 2e oxidation 2+ Cu 56 (aq) + 2e → Cu(s) for reduction electrons the and overall copper(II) two describes redox sulfate reaction (see half-equations: O x I d AT I O n And REducTIOn Eape 9.1.2 a) Deduce reaction the of balanced copper redox metal, equation Cu(s), with in an nitrate acidic ions, solution NO (aq), for to the give 2+ copper(II) ions, Cu (aq), and nitrogen dioxide, NO (g) b) Identify the oxidizing and reducing agents Solution a) Step 1: Deduce the oxidation states for all atoms present in each species Cu(s): (since (aq): NO copper for is oxygen; an x elementary + × ( 2) = substance) –1, so x = +5 for nitrogen 2+ Cu (aq): NO (g): +2 for oxygen; x + × ( 2) = 0, so x = +4 for nitrogen Step is 2: Deduce which species is oxidized and which species reduced N is Cu reduced is oxidized There Step 3: from is no from change Formulate reduction +5 to to in +4 (decrease +2 the (increase oxidation half-equations for in in oxidation state the oxidation for state) state) oxygen oxidation and processes: 2+ Oxidation: Cu(s) → Cu (aq) + 2e + Reduction: NO (aq) + 2H (aq) + e → NO (g) + H O(l) + Note that H equations Step lost 4: Balance equals (aq) in the ions acidic the and water are used to balance half- solutions half-equations number of so electrons that the number of electrons + 2H gained: 2+ Oxidation: Cu(s) Reduction: 2NO → Cu (aq) + 2e + (aq) + 4H (aq) + 2e → 2NO Step 5: Add the (g) half-equations to give the overall O(l) equation: Note that the oxidizing agent is + Cu(s) + 2NO (aq) + 4H 2+ (aq) → Cu (aq) + 2NO (g) + 2H O(l) the nitrate ion, NO (aq) No marks b) Oxidizing agent (undergoes reduction): nitrate ions, NO (aq) would be awarded for the answers “nitrogen” or “N” Reducing agent (undergoes oxidation): copper metal, Cu(s) Reo titratio Titrations involving redox reactions require the same types of Acid–base titrations are calculations as acid–base titrations, starting from the balanced discussed in topics 1.3 and 18.3 equation the If method The are you based half-equations given not on know the for some you reagents typically ions (oxidizing + (aq) + 8H out using used in redox titrations 2+ (aq) + 5e → Mn ions (oxidizing (aq) + 4H O(l) agent): + O it dichromate(VI) work agent): Cr can above below manganate(VII) MnO equation, half-equations (aq) + 14H (aq) Manganate(VII) ions, MnO 3+ + 6e → 2Cr (aq) + 7H O(l) , are also known as permanganate ions thiosulfate ions (reducing agent): 2S O (aq) → S O (aq) + 2e 57 REDOX P R O C E SS E S The Winkler method uses thiosulfate ions in a redox reaction to measure • Bioheia oyge ea (BOd) the amount of dissolved oxygen in water (see practice problem 2) is the amount of oxygen required to Since oxygen in water is used by microorganisms to break down oxidize organic matter in a sample of (oxidize) waterborne organic pollutants, the biochemical oxygen demand water at a specified temperature over (BOD) a period of five days acts A high as a measure concentration of of the degree dissolved of pollution oxygen implies in a a water low sample BOD, which • The Wiker etho measures means low pollution levels The more pollutants are present, the more the BOD of water by redox titration bacteria T O P I C thrive and in electrochemical occurs the at the in by cells, cathode reduction and always oxidation ✔ construct at and electrolytic voltaic cells, chemical spontaneous energy reactions is explain released converted higher the demand becomes C E L LS annotate voltaic and cells; how electricity to in conducted a redox a voltaic in an reaction cell is and electrolytic used how to produce current is cell; energy; ✔ in the anode; electrical ✔ so Yo ho be abe to: ✔ ✔ oxygen, E L E C T R O C H E M I C A L Yo ho kow: ✔ use electrolytic converted to cells, electrical chemical energy energy by distinguish is in electron electrochemical ow and ion ow cell; driving ✔ non-spontaneous an between write a one-line cell diagram notation for a reactions voltaic ✔ cell; deduce molten There are convert and two types chemical electrolytic energy (in a of products to electrical which of the electrolysis of a salt electrochemical energy cells, the convert non-spontaneous cells—voltaic energy (in electrical a cells, which spontaneous energy to process), chemical process) Votai e In voltaic (galvanic) oxidation each are prevents instead which energy all electrode by ions a the occurs, so reduced cathode in the bulb the is positive, – an and to to in In the from if the is by a a salt This in wire – the half-cells, The electrodes bridge separation one half-cell There supplies may be and a electrical cells at the cathode, spontaneous reduction anode solution cell occurs cells, reaction separate circuit the the voltaic from two occurring along reduction where redox in solutions circuit are a electrolyte the travel voltaic the of occur complete the electrons and parts reaction cells, half-cell withdraws and batteries anode two wire, move) electrochemical at a electrons light Commercial oxidation 58 or the processes spontaneous forces voltmeter an together the cells, reduction containing linked (through In and occurs, electrode; negative the in redox reaction species other (figure and that words, 9.2.1) is the 9.2 e E l E c T R O c H E mIc A l c E ll s e V + Zn The command term “annotate” anode Cu salt cathode bridge ) ( is often used in examination (+) questions It means that you should add appropriate labels to Figre 9.2.1 The 2+ Cu a diagram or graph In drawing (aq) Daniell cell, in which electrochemical cells, make sure the redox reaction + + that you label all relevant par ts on between zinc metal and the diagram, including the anode, 2+ copper ions is used to Zn (aq) anode cathode (oxidation) (reduction) the cathode and the direction of produce electricity electron flow Cell in a and diagram single each cathode line The phase is diagram notation salt the written Daniell 2+ Zn(s) In | Zn fuel released (aq) in in by on cell a is a way to represented single the in quick figure by vertical right-hand 9.2.1 represent two line, | side is voltaic vertical By For written a lines, ||, convention, example, as cell the the cell follows: 2+ cells, oxidized bridge boundary always for provides a a || Cu fuel (aq) that voltaic the form hydrogen–oxygen Cu(s) would cell, of | so normally that a electricity fuel cell, in be large rather an combusted proportion than acidic heat of in air the One is instead energy example is is the electrolyte: + Oxidation: 2H (g) → 4H (aq) + 4e + Reduction: O (g) + 4H (aq) + 4e → 2H O(l) Fuel cells are discussed in more Overall: 2H (g) + O (g) → 2H O(l) detail in option C.6 Eetroyti e Electrolytic cells use non-spontaneous is negative and Electrolysis the can an external redox be anode used source reactions is to In of electrical electrolytic energy cells the to drive cathode positive extract metals from their salts (example 9.2.1) Eape 9.2.1 Lead can lead(II) be produced bromide, by the electrolysis of molten PbBr negative cell) electrode The Br ions (the are cathode, negative in so an will electrolytic move a) Explain electric b) how molten lead bromide conducts current Identify reduction the towards b) electrodes occur during at the which oxidation and State at each a half-equation for positive Reduction place at the in the reaction any electrode (the anode) electrochemical cathode electrolysis and oxidation cell at takes the anode 2+ c) c) the Reduction: + Pb 2e → Br (g) Pb(l) occurring Oxidation: → 2Br + 2e electrode d) d) e) Deduce Outline the the overall what you cell Overall reaction: PbBr would (l) → Pb(l) + Br observe e) A silvery-grey at the molten metal, (g) Pb(l), will form during cathode, and a brown gas will form at the electrolysis anode, f) cell reaction Suggest why this experiment might not be for a school (g) The characteristic odour of bromine gas suitable Br will be noticeable laboratory f) The reaction should be done in a fume cupboard Solution as a) Molten lead bromide, PbBr , contains ions bromine is both corrosive and toxic Lead(II) that bromide are free to move only in the molten state is also toxic The 2+ Pb ions are positive so will move towards the 59 REDOX P R O C E SS E S s AmPlE sTudEnT Ans WER The activity series lists the metals in order of reactivity m ot reative ni Ag eat reative a) Identify the strongest reducing agent in the given list [1] 2+ b) A voltaic cell is made up of a Mn |Mn half-cell and a 2+ Ni | Ni half-cell Deduce the equation for the cell reaction [1] ) The voltaic cell stated in par t (b) is par tially shown below Draw and label the connections needed to show the direction of electron movement and ion ow between the two half-cells Mn(s) Ni(s) 2+ Ni This ▼ Mn here, is oxidized and reducing ▼ This most therefore is answer 2+ (aq) Mn could have (aq) achieved 2/4 marks: easily the strongest (a) Ag (b) Ni(s) agent equation is the wrong 2+ way not Ni round be in and the arrow reversible—as the activity Mn series, + Mn 2+ (aq) Mn(s) + Ni (aq) should is above Mn will 2+ undergo oxidation and Ni will be reduced: electron (c) 2+ Mn(s) + Ni 2+ (aq) → Ni(s) + Mn (aq) V salt movement bridge wire ion ▲ A perfect answer, with all ow: the anions key parts of the voltaic cell labelled cations correctly Ni(s) Mn(s) 2+ Ni 60 2+ (aq) Mn (aq) [2] 9.2 E l E c T R O c H E mIc A l c E ll s Pratie probe for Topi Probe Probe The following reactions occur spontaneously Deduce the names of the following compounds: MnO , Fe(OH) , Cr (SO ) Fe(s) + NiCl (aq) → FeCl (aq) + Ni(s) Probe Zn(s) + FeCl (aq) → ZnCl (aq) + Fe(s) The Winkler method uses redox reactions to determine the concentration of oxygen present in water Ni(s) + PbCl (aq) → NiCl (aq) + Pb(s) A 0.100 dm water sample was taken from a river and What is the ireaig order of the reactivity of the metals? analysed using this method The reactions taking place A Fe < Ni < Zn < Pb B Pb < Ni < Fe < Zn c Ni < Zn < Pb < Fe d Zn < Fe < Ni < Pb are summarized as follows: step 1: Probe 2+ 2Mn (aq) + 4OH (aq) + O (aq) → 2MnO (s) + 2H O(l) Zinc can be produced by the electrolysis of molten step 2: zinc chloride, ZnCl + MnO (s) + 2I (aq) + 4H 2+ (aq) → Mn (aq) + I (aq) + a) Explain how molten zinc chloride conducts 2H O(l) electric current step 3: I (aq) + 2S O 2 (aq) → 2I (aq) + S O (aq) b) Identify the electrodes at which oxidation and a) Identify the oxidizing and reducing agents in reduction occur during the electrolysis each step ) State a half-equation for the reaction occurring at b) 0.00150 mol of I (aq) formed in step Calculate each electrode the amount, in mol, of oxygen, O (aq), dissolved in the ) Deduce the overall cell reaction water sample e) Outline what you would observe during the electrolysis 61 ... D I P L O M A P R O G R A M M E Sergey Bylikin Brian Murphy Alexandra Juniper problem Q1 and answer, Q12a; Great Clarendon Street, Oxford, OX2 6DP, United the by University University’s publishing... chemical by the equilibrium equilibrium is homogeneous constant (K reaction; ), c ✔ derive new K expressions for equations c which is dened by the equilibrium law; with ✔ Le Châtelier ’s equilibrium... Nickolay Library; p162 (T) DAM Vinokurov/Shutterstock; Photo Charles D Library CNRI/Science Photo Photo Library; Library; p10 (T) p2 (T) DAM Science Alamy Stock Photo Fotomaton Energy by Rice /