2022 H1 Theories of acids and bases Self check and Tutorial (teacher)

St.Andrew’sJuniorCollegeH1Chemistry2022PhysicalChemistryTutorialTheoriesofAcidsandBasesSELF-CHECKQUESTIONS 1. Identifytheconjugateacid-basepairinthefollowingreaction.HCl +NaHCO3 → H2 CO3 + NaCl 2. CalculatethepHofsolutionswiththefollowingH3 O + concentrationsinmol dm –3 (a) 6.80x10 –3 (b) 3.20x10 –5 3. CalculatethepHofsolutionswiththefollowingOH – concentrationsinmol dm –3 (a) 3.70x10 –10 (b) 6.40x10 –5 4. CalculatetheH3 O + andOH – concentrationsinsolutionswiththefollowingpHvalues:(a) 9.21 (b) 13.7 5. CalculatethepHandpOHofthefollowingsolutions:(a) 25cm 3 of0.200mol dm –3 ofhydrochloricacidmadeupto500cm 3 ofsolution.(b) 1.00 g of calciumhydroxide dissolved in water and made up to 250 cm 3 ofsolution 6. Explainthefollowingterms:(i)pH (ii) Ka (iii) Kb (iv) Kw 7. Calculatethedegreeof dissociationoftheweakmonobasicacidslistedbelow:(a) asolutionof0.0100mol dm –3 CH3 COOHhasapHof3.38.(b) asolutionof0.200mol dm –3 HCNhasapHof5.05 8. Calculatethepercentagedissociationoftheweakmonoacidicbaseslistedbelow:(a) asolutionof0.0100mol dm –3 CH3 NH2 has[OH – ]of4.78x10 –7 mol dm –3 .(b) asolutionof0.0500mol dm –3 C6 H5 NH2 has[OH – ]of9.65x10 –5 mol dm –3 1

9. (a)Stateifeachofthefollowingsaltsisacidic,basicorneutral.A–acidicsalt;B–basicsalt;N–neutral saltSalt A,BorN Salt A,BorN(i) KBr (iv) (C6 H5 NH3 )2 SO4 (ii) C6 H5 ONa (v) (CH3 )2 NH2 Br(iii) NaCN (vi) CH3 COOK (b) (i)ForC6 H5 O – Na + ,writeanequationtoillustrateitsacidity/basicity.(ii)ForCH3 NH3 Cl,writeanequationtoillustrateitsacidity/basicity. 10. Indicateifthesolutioncontainingthespecifiedspeciesisabuffersolution.Ifitis,stateifitisacidicorbasicbuffer.X:notbuffer A:acidicbuffer B:basicbuffer Solution XorAorB(a) CH3 COOH/CH3 COONa(b) CH3 COOH/NaCl(c) HCl/NaCl(d) HNO3 /NaNO3 (e) HCN/KCN(f) NH3 /NH4 NO3 (g) NaOH/Na2 SO4 ANSWERS2 (a)2.17(b)4.49 3 (a)4.57(b)9.81 4 (a)[H3 O + ]=6.17x10 –10 mol dm −3 ;[OH - ]=1.62x10 –5 mol dm −3 (b)[H3 O + ]=2.00x10 –14 mol dm −3 ;[OH - ]=0.500mol dm −3 5 (a)pH=2;pOH=12.(b)pOH=0.967;pH=13.0 7 (a) α=0.04(b) α=4.46x10 -5 8 α=4.78x10 −3 %α=0.193% 9 (a)(i)N(ii)B(iii)B(iv)A(v)A(vi)B 10 (a)A(b)X(c)X(d)X(e)A(f)B(g)X 2

Self–CheckAnswers: 1. Identify the Bronsted acid and its conjugate base and the Bronsted base and itsconjugateacidinthefollowing:HCl + NH3 → NH4 + + Cl − acid base conjugateacid conjugatebase 2 ApplyingpH=-log[H + ]orpH=-log[H3 O + ](a)2.17(b)4.49 3 ApplyingpOH=-log[OH - ]FollowedbypH=14–pH(a)4.57(b)9.81 4 (a)[H3 O + ]=10 -9.21 =6.17x10 –10 mol dm −3 [OH - ]=10 -14 /[H3 O + ]= 1.62x10 –5 mol dm −3 (b)[H3 O + ]=10 -13.7 =2.00x10 –14 mol dm −3 [OH - ]=10 -14 /[H3 O + ]= 0.500mol dm −3 5 (a)[HCl]=[H + ]=[(25/1000)x0.2]/(500/1000)=0.01mol dm -3 pH=-log[H + ]=2pOH=14–2=12. (b)moleofcalciumhydroxide=1.00/[40.1+2(17)]=0.0135molMol ofOH - ions=0.0270[OH - ]=0.0270/250x1000=0.108mol dm -3 pOH=-log[OH - ]=0.967pH=14–0.967=13.0 6 (i)pH=−lg[H + ] (ii)ForaweakacidHA,HA(aq)+H2 O(l) A − (aq)+H3 O + (aq) Ka = moldm −3 . (iii)ForaweakbaseB,B(aq)+H2 O(l) BH + (aq)+OH − (aq) Kb = moldm −3 . (iv) H2 O(l) H + (aq)+OH − (aq)Kw =[H + ][OH − ]mol 2 dm −6 . 7 degreeofdissociation,α= 3

(c) α=10 -3.38 /0.01=0.04(d) α=10 -5.05 /0.2=4.46x10 -5 8 α= x100%(a) α=4.78x10 −7 /0.01x100%=4.78x10 −3 %α=9.65x10 −5 /0.05x100%=0.193% 9 (a)(i)N(ii)B(iii)B(iv)A(v)A(vi)B(b) (i) C6 H5 O – +　H2 O C6 H5 OH +　OH – (ii) CH3 N + H3 +　H2 O CH3 NH2 +　H3 O + 10 (a)A(b)X(c)X(d)X(e)A(f)B(g)X 4

DISCUSSIONQUESTIONS Theoriesofacidsandbases1. [2013/P3/Q2](a) Explainwhat ismeant bytheterms Brønsted-Lowry baseand conjugateacid-basepair.Brønsted-Lowrybaseisaprotonacceptor.Inanacid-basereaction,aH + istransferredfromanacidtoabase. Forexample:HA H + +A – HAandA – isaconjugateacid-basepair,whichdiffersbyoneH + .Examiner’scomments:Most candidates gave a correct definition of a Bronsted-Lowry base as a protonacceptor. Some defined a Lewisbase(electron-pair donor) andaminoritydefinedaLewis acid. Far fewer were able to offer an adequate description of a conjugateacid-base pair. Several wroteabout strongacidsbeingassociatedwithweakbases,or vice versa, without focussing on the fact that the pair differ by the presence orabsenceof aproton. (b) Theammoniamoleculecanreactinvariousways:asanacid,base,oxidisingagentorreducingagent.Studythefollowingreactionsanddecideinwhichwayammoniaisreactingineachcase. Explainyouranswersfully. (i) NH3 + H2 O NH4 OH (ii) NH3 + HNO2 N2 + 2H2 O (iii) NH3 + NaH NaNH2 + H2 (iv) 2NH3 + NaOCl N2 H2 + NaCl + H2 O (i) Ammoniaisreactingasabase. NH3 acceptsaH + fromH2 OtoformNH4 + .(ii) Ammonia is reacting as a reducing agent. NH3 decreases the oxidationnumberofNfrom+3inHNO2 to0inN2 .(iii) Ammoniaisreactingasanacid.NH3 losesaH + toH - toformNH2 − .(iv) Ammoniaisreactingasareducingagent.NH3 decreasestheoxidationnumberofClfrom+1inOCl – to–1inCl – . Examiner’scomments: 5

Some candidates correctly assigned the various roles of ammonia. Somecandidates did not explain their reasons sufficiently. A number of candidatesseemedtohaveconfusedoxidationwithreduction, or acidwithbase. Kw 2. N2013/I/11The dissociation constant, Kw , for the ionisation of water, H2 O H + + OH − , atdifferenttemperaturesisgivenbelow.Temperature/°C Kw /mol 2 dm −6 0 1.15x10 −15 25 1.00x10 −14 50 5.50x10 −14 Whatcanbededucedfromthisinformation?A Onlyat25°Care[H + ]and[OH − ]equal.B Theequilibriumliesfurthesttotherightat0°C.C Theforwardreactionisexothermic.D ThepHofwaterdecreasesastemperatureincreases. Ans:DA: Incorrect.H2 O(l) H + (aq) +OH − (aq)[H + ]=[OH − ]forpurewateratanytemperature B: Incorrect. K= . ComparingKw at0 O Cand50 O C, Kw at0 O Cissmaller. Thisimpliesthatatlower temperature, thebackwardreactionis favoured. Hence,theequilibriumpositionliestotheleftat0 O C C: Incorrect. K= . Comparing Kw at0 O Cand50 O C, Kw at50 O Cislarger. Thisimpliesthatathigher temperature, the forward reaction is favoured. Hence, the forward reaction isendothermic.D: Correct. Kw = [H + ][OH − ] As temperature increases, Kw increases, implying[H + ] increases. Thus,pHdecreases. 6

Calculationsof[H + ]andpHforstrongacidsandstrongbases3. [2012P1Q10]StomachjuiceshaveapHof1.0.Aspirinisamonobasic(monoprotic)acidrepresentedbyHA(Ka =1×10 –4 mol dm –3 )whichdissociatesintoionsH + andA – .WhataretherelativeconcentrationsofH + , A – andHAwhenaspirinfromatabletentersthestomach?A [HA] > [H + ] = [A – ]B [H + ] = [A – ] > [HA]C [H + ] > [A – ] > [HA]D [H + ] > [HA] > [A – ] Ans:DStomachjuice,pH=1[H + ]=0.1mol dm –3 SinceHAisaweakacid,itdissociatespartially.HA H + +A – high[H + ]will suppressthedissociationofaspirinthus[HA]>[A – ] 7

4 2018/P2Q7Hydrogenchloridedissolvesinwatertogivehydrochloricacid,whichisastrongacid.Hydrogenfluoridedissolvesinwatertogivehydrofluoricacid,whichisaweakacid.(a) (i) Explain the meaning of the terms acid, strong acid and weak acid, intermsoftheBronsted-Lowrytheoyofacids.Astrongaciddissociatesfully/completelyinwaterandisaprotondonor.Aweakaciddissociatespartiallyinwaterandisaprotondonor. (ii) Suggestareasonwhyhydrochloricacidisastrongeracidthanhydrofluoricacid.Chlorine has a larger atomic size than fluorine. There is less effective orbitaloverlapbetweenhydrogenandchlorine.ThebondstrengthinHCl isweakerandhencetheHCl bondismoreeasilybrokentoreleaseH + . (b) (i) DefinethetermpH.pH=-lg[H + ] (ii) CalculatethepHofa1.5x10 -4 mol dm -3 solutionofhydrochloricacid.[ANS:3.82]pH=-lg(1.5x10 -4 )=3.82 (iii) Bariumhydroxideisastrongbase.CalculatethepHofa0.075mol dm -3 solutionofbariumhydroxide.[ANS:13.2][OH - ]=2x0.075=0.15pOH=-lg0.15=0.824pH=14–0.824=13.2 8

5. J90/P2/2ApplejuicehasapHof3.5.(a) (i) DefinepH.pH=-log[H + ] (ii) Calculatethemolarconcentrationofhydrogenionsinapplejuice.[ANS:3.16x10 −4 moldm -3 ][H + ]=10 -3.5 =3.16x10 -4 mol dm -3 Applejuicecanbetitratedwithstandardalkali.A25.0cm 3 sampleof applejuicewasexactlyneutralisedby27.5cm 3 of0.10mol dm −3 sodiumhydroxideusingphenolphthaleinasindicator.(b) Assuming that apple juice contains a single acidwhichismonobasic, calculatethemolarconcentrationoftheacidinthejuice.[ANS:0.110moldm −3 ]Amtofacid=amtof