2022 H1 Theories of acids and bases notes (teacher)

St.Andrew’sJuniorCollegeH1Chemistry2022LectureNotesTheoriesofAcidsandBases AssessmentObjectives Candidatesshouldbeableto:(a) showunderstandingof,andapplytheArrheniustheoryofacidsandbases(b) showunderstandingof,andapplytheBronsted-Lowrytheoryofacidsandbases,includingtheconceptofconjugateacidsandbases;(c) explainqualitativelythedifferencesinbehaviourbetweenstrongandweakacidsandbasesintermsoftheextentofdissociation;(d) explainthetermspH;Ka ;Kb ;Kw [TherelationshipKw =Ka Kb isnotrequired];(e) calculate[H + (aq)]andpHvaluesforstrongacids,andstrongbases;(f) explainthechoiceofsuitableindicatorsforacid-basetitrations,givenappropriatedata,intermsofthestrengthsoftheacidsandbases;(g) (i)explainhowbuffersolutionscontrol pH;(ii)describeandexplaintheusesofbuffers,includingtheroleofH2 CO3 /HCO3 - incontrollingpHinblood. LectureOutline:1. Theoriesofacidsandbases2. Autoionisationofwater3. Dissociationofweakacidandbases4. Acid-BasePropertiesofSaltSolutions5. Indicatorsforacid-basetitrations6. MonoproticAcid-BaseTitrationCurves7. Buffersolutions References:1. ChemistrybySilberberg2. ChemistrybyRaymondChang3. ChemistryforAdvancedlevel byPeterCann4. ChemistryincontextbyHill andHolman5. A–level ChemistrybyRamsden 1

1. THEORIESOFACIDSANDBASES 1.1 ArrheniustheoryofacidsandbasesAnArrheniusacidisasubstancethatproducesH + inaqueoussolution.AnArrheniusbaseisasubstancethatproducesOH – inaqueoussolution. E.g.: HNO3 (l) + H2 O(l) NO3 – (aq) + H3 O + (aq)Note: H3 O + ishydroniumion, whichisthesameasH + (aq) NaOH(s) + aq Na + (aq) + OH – (aq) ● Anacid-basereactioninvolvestheformationofsaltandwater 1.2 Brønsted–LowrytheoryofacidsandbasesABrønsted–Lowryacidisaproton(H + )donor.ABrønsted–Lowrybaseisaproton(H + )acceptor. base conjugateacid E.g.: NH3 (aq) + H2 O(l) NH4 + (aq) + OH – (aq) Acid conjugatebase base conjugateacid E.g.: HCl(g) + NH3 (g) NH4 Cl(s) acid conjugatebase ● Anacid–basereactioninvolvesthetransferofaprotonfromtheacidtothebase.● When a Brønsted–Lowry acid (HA) loses a proton, the resulting product (A – ) iscalledtheconjugatebaseofHA. 2

● WhenaBrønsted–Lowrybase(B)acceptsaproton,theresultingproduct(BH + )iscalledtheconjugateacidofB.● Aconjugateacid-basepairdiffersbyaH + . Examplesofconjugatepairs:HCl/Cl – NH3 /NH4 + H2 O/OH – H2 O/H3 O + CH3 COOH/CH3 COO – CH3 NH2 /CH3 NH3 + Exercise1Identifytheacid,base,conjugateacidandconjugatebaseinthefollowingreactions: a) CH3 COOH + H2 O CH3 COO – + H3 O + acid base conjugatebase conjugateacidb) CH3 COOH + HCl CH3 COOH2 + + Cl – base acid conjugateacid conjugatebasec) HNO3 + H2 O NO3 – + H3 O + acid base conjugatebase conjugateacidd) HNO3 + H2 SO4 HSO4 – + H2 NO3 + base acid conjugatebase conjugateacid 👉Self–Check:Q1💪InSummary💪Arrhenius Brønsted-LowryAcids producesH + inwater donatesH + Bases producesOH – inwater accepts H + 3

1.3 StrengthofAcidsandBases Strongacidcompletelydissociatesinwater.E.g.: HCl(g) +aq H + (aq) + Cl – (aq)Mostmineral acidsarestrongacids.E.g.:HI,HBr,H2 SO4 ,HNO3 etc.representscompletedissociationAcid Weakacidpartiallydissociatesinwater.E.g.: CH3 COOH(l) + aq CH3 COO – (aq) + H + (aq)Mostorganicacidsareweakacids.E.g.:CH3 COOH,H2 CO3 ,HCNetc.⇌meansanequilibriumisestablishedbetweentheundissociatedmoleculesandthedissociatedions. Strongbasecompletelydissociatesinwater.E.g.: NaOH(s) +aq Na + (aq) + OH – (aq)ExamplesofstrongbasesareNaOH,KOH,Ba(OH)2 etc.Base Weakbasepartiallydissociatesinwater.E.g.: NH3 + H2 O NH4 + (aq) + OH – (aq)ExamplesofweakbasesareNH3 ,CH3 NH2 ,Na2 CO3 etc. 4

1.4 ThepHscaleConcentrationofH + canbeusedasameasureofacidityandalkalinityofasolution.However,[H + ]canbeverysmall values.Hence,wecanexpressthesevaluesaspH.pH= –log10 [H + ] or [H + ] = 10 –pH [H + ] inmoldm –3 AsolutionwithlowpHhasahigh[H + ],andismoreacidicthanasolutionwithhigherpH. Similarly,pOH= –log10 [OH – ]Note: The logarithmused are to the base 10 (not to the base e), so make sure whendoingcalculations, youpressthelogor lgbuttononyour calculator (not thelnbutton) 1.5 Basicityofacid● HClandCH3 COOHaremonoprotic(monobasic)acidsaseachacidcanonlyloseoneH + .● H2 SO4 isadiprotic(dibasic)acidasitcanlosetwoH + .H2 SO4 + H2 O H3 O + + HSO4 - conjugatebaseHSO4 - + H2 O H3 O + + SO4 2- conjugatebase ● Converselyforbases,NaOHisamonoacidicbaseandCa(OH)2 isadiacidicbase. 2. AUTOIONISATIONOFWATER Waterdissociatesveryslightlytogiveions.H2 O(l) H + (aq) +OH – (aq) When[H + ]=[OH – ],thesolutionisneutral.When[H + ]>[OH – ],thesolutionisacidic.When[H + ]<[OH – ],thesolutionisalkaline. We name the equilibriumconstant for theautoionisationof water asionicproduct of water,Kw . 5

Note: [H2 O] isnot reflectedinKw pKw =–log10 Kw =–log10 ([H + ][OH – ])=–log10 [H + ] –log10 [OH – ]pKw = pH+ pOHpH= pKw –pOHNote: Derivationof pKw isnotrequired At25 o C inDataBooklet pKw =14pKw = pH+ pOH14 = pH+pOH Note: Derivationof pKw isnot required Inpurewaterat25 o C, H2 O(l) H + (aq) +OH – (aq)[H + ]=[OH – ]=1.00x10 –7 mol dm –3 pH=pOH=– lg(1.00x10 –7 ) = 7.0 Theautoionisationofwatermoleculesisanendothermicprocess.H2 O(l) H + (aq) +OH – (aq) ΔH >0 Kw = . When temperature increases, both kf and kb increases. Since the forwardreaction is endothermic and is favoured, kf increases more than kb . Hence, Kw increaseswhentemperatureincreases. Table1:Variationofionicproductofwater, Kw ,withtemperatureTemperature/ o C Kw / mol 2 dm –6 20 0.68x10 –14 30 1.47x10 –14 40 2.92x10 –14 💪Insummary💪 6

Exercise2[2018P1Q14]Theionicproductofwater, Kw ,isaffectedbytemperature.Temperature/°C Kw x10 −14 /mol 2 dm −6 10 0.29340 2.92 Whichstatementdescribeswhathappensasthetemperatureofwaterisincreasedfrom10°Cto40°C?A pHofwaterdecreasesand[H + ]=[OH − ]B pHofwaterdecreasesand[H + ]isgreaterthan[OH − ]C pHofwaterincreasesand[H + ]=[OH − ]D pHofwaterincreasesand[H + ]islessthan[OH − ] Ans:A H2 O(l) H + (aq) +OH – (aq)At40 o CKw = [H + ][OH – ] =2.92x10 –14 [H + ]=[OH – ][H + ] 2 =2.92x10 –14 [H + ]=1.709x10 –7 moldm –3 pH=6.77pHof purewater at 40 o CislessthanpH7, but it DOESNOTmeanthat thewaterisnotneutral!Aneutral solutionhaspH7onlyat25 o C. As Kw changeswithtemperature,pHofaneutralsolutioncanchangewithtemperature.Hence,aslongas[H + ]=[OH − ],asolutionisneutral. 7

StepsinCalculatingpHofStrongAcid/Base1. Writeequationtoillustratecompletedissociation(ifnecessary).2. Determine[H + ]foracids(or[OH − ]forbases)3. DeterminepHorpOHusingAcidspH=−log10 [H + ] BasespOH=−log10 [OH – ]pH=14–pOH at25 o C Exercise3CalculatethepHof a)0.0100mol dm –3 sulfuricacidb)0.00250mol dm –3 aqueoussodiumhydroxidea) H2 SO4 2H + + SO4 2– 0.0100moldm –3 0.0200moldm –3 pH=–lg(0.0200)=1.70 b) NaOH Na + + OH – 0.00250moldm –3 0.00250moldm –3 pOH=–lg(0.00250)=2.602pH =pKw –pOH=14–2.602=11.4 8

Exercise4The concentration of OH – ions in a certain household ammonia cleaning solution is0.0025mol dm –3 .CalculatethepHofthesolution.[OH – ] = 0.0025mol dm 3 Since[H + ][OH – ]=10 –14 [H + ] = == 4x10 –12 mol dm −3 pH =–lg(4x10 –12 )=11.4 👉Self–Check:Q2–5 3 DISSOCIATIONOFWEAKACIDSANDBASES 3.1 Dissociationconstantsofacids/bases, Ka / Kb I.Aciddissociationconstant, Ka foraweakacidForaweakmonobasic(monoprotic)acid,thedissociationcanberepresentedasfollows:HA(aq) H + (aq) + A – (aq) We name the equilibrium constant for the dissociation of an acid itsaciddissociationconstant, Ka . Ka = Units: mol dm –3 pKa = –log10 Ka Note: [H2 O] is ignored in Ka as it is present in large amount and its concentration remainsalmost constant 9

▪ Ka istemperaturedependent.Table2: Dissociationconstantsof acidsinwater at 25 o CAcids Formula Ka /moldm –3 pKa Ethanoicacid CH3 COOH 1.74x10 –5 4.75 strongeracidPhenol C6 H5 OH 1.28x10 –10 9.89 weakeracid ▪ Ka and pKa gives a good measurement of the strength of an acid, as it indicates theextent to which the acid is dissociated. The higher the Ka or the lower the pKa , thestrongertheacid. Polybasic(polyprotic)acids,suchasH3 PO4 ,havemorethanonedissociationconstant. Table3: Dissociationconstantsof phosphoricacid, H3 PO4 (triprotic)Dissociation Equilibrium Ka /moldm –3 pKa Ka1 H3 PO4 (aq)+