Recall
10
Buffers• AbufferisasolutionthatresistsachangeinpH• Itistypicallyasolutioncontainingsignificantamountsofaweakacidorbaseanditssalt(whichcontainstheconjugateoftheacid/base)
• Anacidandabasearepresentinallbuffersolutionstoneutralizeanyaddedacidorbase
• Therefore,thepHshouldremainrelativelyconstant
CalculatingthepHofaBufferSoln’• TheCommonIonEffect:• Abuffersolutionalreadycontainstheconjugatebaseoftheweakacidinsolution:HC2H3O2(aq)+H2O(l)ßàH3O+
(aq)+C2H3O2-(aq)
NaC2H3O2(aq)àNa+(aq)+C2H3O2-(aq)
• AccordingtoLeChâtelier’sPrinciple,thepresenceoftheacetateionswillcausetheequilibriumofaceticacidtoshiftleft(theionizationofaceticacidissuppressed)
• Thiswillresultinalower[H3O+]andthusahigherpH
FindingpHofBuffers• CalculatethepHofabuffersolutionthatis0.100MinHC2H3O2and0.100MinNaC2H3O2(Ka=1.8x10-5)
1. Writetheequationfortheweakacid2. SetupanICEtable–includetheinitial
concentrationsofboththeacidandtheconjugatebase
3. SetupKaexpression4. Pluginandsolve.Approximatewhenever
possible(justbesuretoverify).5. Usextofind[H3O+]andpH
Shortcuts• Ifyoudon’tneedtoshowallofyourwork(suchasinMCproblems),formostcasesyoucansimplyuse:
[H3O+]=Ka[acid]/[base]
• Similarly,youcanusetheHenderson-HasselbachequationtoquicklycalculatethepHofabuffersolutionfromtheinitialconcentrationsofthebuffercomponentsaslongasthexissmallapproximationisvalid
pH=pKa+log[base]/[acid]
CalculatingpHChangesinBuffers• BufferscanresistpHchanges,butnotentirely–smallchangesstilloccur
• Canwecalculatethesechanges?• Yes!Withjustonesimpleextrastep① Usesimplestoichiometricrelationshipsto
determinetherelativechangesinthenumberofmoles.Remember–reactantsareconsumed,productsareproduced.
② Useyourvaluesfromstep1foryourinitialconcentrationsinyourICEtable.SolvetheICEtableaswehavealreadylearned.
CalculatingpHChangesinBuffers• Remember:buffersolutionscanalsobecreatedbyaddingabaseanditsconjugateacid(solongastheconjugateacidisanion).
• Ex/AddNH3andNH4Cl.• It’sbasicallythesameidea• TheonlydifferenceisthatifyouareusingashortcutyouneedtofirstcalculateKaorpKafromtheconjugateacidoftheweakbase(ifgivenKb).
BufferEffectiveness
17
BufferEffectiveness• Whatmakesabuffer“effective”?• Abuffershouldbeabletoneutralizesmalltomoderateamountsofaddedacidorbase
• Toomuchaddedacidorbasewilldestroyabuffer• Whatinfluencestheeffectivenessofabuffer?– Therelativeamountsofacid/baseanditsconjugate– Theabsoluteconcentrationsofacid/baseanditsconjugate
BufferEffectiveness• Let’sconsiderabuffersolutionthat’smadefromanacidanditsconjugatebase
• Thisbufferismosteffective(resistanttopHchanges)when:① [acid]and[conjugatebase]arealmostequal• Theacceptedrangeis0.10≤[base]/[acid]≤
10② Theconcentrationsarebotharehigh(the
bufferisconcentrated,notdilute)
BufferRange
20
BufferRange• Again,let’sconsiderabuffersolutionthat’smadefromanacidanditsconjugatebase
• Since0.10≤[base]/[acid]≤10,wecanusetheHenderson-HasselbachequationtodetermineanappropriaterangeofpHvaluesforaneffectivebuffer
• pH=pKa+log(0.10)andpH=pKa+log(10)• Soourrangeis:
pKa-1<pH<pKa+1
BufferRange• pKa-1<pH<pKa+1• Whatdoesthismean?• AweakacidwithapKaof5.0isusedtoprepareabufferintherangeof4.0–6.0.Youneedtocalculateandadjusttherelativeamountsof[conjugatebase]and[acid]togetanidealratio.
• TheclosertopHof5.0,themoreeffectivethebufferwillbe.
Examples:PreparingaBuffer
23
PreparingaBuffer• Lookattheacidsbelow.WhichwouldyouchoosetocombinewithitssodiumsalttomakeasolutionbufferedatpH4.25?Forthebestchoice,calculatetheratiooftheconjugatebasetotheacidrequiredtoattainthedesiredpH.
a. Chlorousacid(HClO2)pKa=1.95b. Nitrousacid(HNO2)pKa=3.34c. Formicacid(HCHO2)pKa=3.74d. Hypochlorousacid(HClO)pKa=7.54
PreparingaBuffer• FormicacidisthebestchoicebecausethepKaistheclosesttothepHvalue
a. Chlorousacid(HClO2)pKa=1.95b. Nitrousacid(HNO2)pKa=3.34c. Formicacid(HCHO2)pKa=3.74d. Hypochlorousacid(HClO)pKa=7.54
PreparingaBuffer• Whatisthedesiredratio?Formicacid(HCHO2)pKa=3.74• UseyourequationpH=pKa+log[base]/[acid]
andplug-invaluesforpHandpKa.• So• 4.25=3.74+log[base]/[acid]• 0.51=log[base]/[acid]• 100.51=[base]/[acid]• 3.24=[base]/[acid]
TryThis:• Astudentisusinganaceticacid/sodiumacetatebuffersolution.ExplainhowtheyshoulddeterminetheratioofacidtoconjugatebaseneededtomaintainapHof5.00.
TryThis:• Lookattheacidsbelow.WhichwouldyouchoosetocombinewithitssodiumsalttomakeasolutionbufferedatpH7.35?Ifyouhave500.0mLofa0.10Msolutionoftheacid,whatmassofthecorrespondingsodiumsaltwouldyouneedtomakethebuffer?
a. Chlorousacid(HClO2)pKa=1.95b. Nitrousacid(HNO2)pKa=3.34c. Formicacid(HCHO2)pKa=3.74d. Hypochlorousacid(HClO)pKa=7.54
BufferCapacity
29
BufferCapacity• BufferCapacity=theamountofacidorbasethatwecanaddtoabufferwithoutcausingalargechangeinpH.
• Buffercapacityincreases…① Withincreasingabsoluteconcentrationsofthebuffer
components② Astherelativeconcentrationsofthebuffer
componentsbecomemoresimilartoeachother(theratioapproaches1)
BufferingCapacity
aconcentratedbuffercanneutralizemoreaddedacidorbasethanadilutebuffer
Think• A1.0-Lbuffersolutionis0.10MinHFand0.050MinNaF.Whichactiondestroysthebuffer?a. Adding0.050molHClb. Adding0.050molofNaOHc. Adding0.050molofNaFd. Noneoftheabove
SampleQuestion• Whichofthefollowingwillnotproduceabufferedsolution?a. 100mLof0.1MNa2CO3and50mLof0.1MHClb. 100mLof0.1MNaHCO3and25mLof0.2MHClc. 100mLof0.1MNa2CO3and75mLof0.2MHCld. 50mLof0.2MNa2CO3and5mLof1.0MHCle. 100mLof0.1MNa2CO3and50mLof0.1MNaOH
SampleQuestion• YouaregivenasolutionoftheweakbaseNovocain,Nvc(pHis11.00)– YouaddtothissolutionasmallamountofasaltcontainingtheconjugateacidofNovocain,NvcH+• Whichofthefollowingstatementsistrue?a. BoththepHandthepOHincreaseb. BoththepHandthepOHdecreasec. BoththepHandthepOHremainunchangedd. pHincreasesandthepOHdecreasese. pHdecreasesandthepOHincreases
SampleQuestion• ConsiderabufferedsolutionatpHof4.00madewithHF(Ka=7.2×10-4)andNaF– Whichofthefollowingstatementsistrue?
• Alloftheconcentrationsbelowrefertoequilibriumconcentrations
a. [HF]=[F–]b. [HF]=[H+]c. [HF]>[F–]d. [H+]=[F–]e. [HF]<[F–]
SampleQuestion• Asolutionwillhavethemosteffectivebufferingcapacitywhentheconcentrationsofthebufferingcomponentsare:a. Smallb. Largec. Concentrationdoesnotmatter
Titrations
Titrations• Recall:atypeofvolumetricanalysiswecanusetodeterminetheamountofacertainsubstanceisatitration.
• Often,welookatanacid-baseneutralizationrxn• Inatitration,astandardsolution(ofKNOWNconcentration)isaddedgraduallytoasolutionofunknownconcentrationuntilthechemicalreactioniscomplete.
• Sinceweknowtheconcentrationandthevolumeaddedofthestandardsolution,aswellasthevolumeoftheunknownsolution,wecancalculatetheconcentrationoftheunknownsolution.
Titrations• EquipmentRequired:• Beaker/flask• Measuringpipetteorburet• pHmeterand/orindicator
40
Titrations• Termstoknow:– Titrant=standardsolution(knownM)– Analyte=substancebeinganalyzed(unknownM)– Equivalencepoint=WhenmolesofOH-=molesofH3O+(neutralizationhasoccurred)
– Indicator=Substanceaddedthatwillundergoacolorchangeneartheequivalencepoint
– Endpoint=Whentheindicatorchangesthecolorofthesolution.Thismaynotmatchtheequivalencepoint,dependingontherangeofpHvalueswheretheindicatorchangescolor
– StandardSolution=solutionofknownconcentrationthatisslowlyaddedtosolutionofunknownconc’
s
43
Titrations• Acid-BaseTitrationMethodUsingIndicator:1. Analytesolution(ofunknownM)isplacedina
flaskorbeaker2. Asmallamountofindicatorisadded3. Titrantisplacedinaburetteandslowlyaddedto
theanalyteandindicatormixture4. Theprocessisstoppedwhentheindicatorcauses
achangeinthecolorofthesolution5. Thechangeinvolumeisusedtodeterminethe
volumeoftheanalytesolution
45
46
Titrations• Note:SometitrationsrequireboilingtogetridofCO2produced,whichwillformcarbonicacid,bufferthesolution,andleadtoinaccuratedata
Titrations• AtitrationcurveisaplotofthepHofasolutionduringatitration
• Itistypicallythevolumeadded(ourstandardsolution)vs.pH.
• Theshapeofthecurvedependsontheacid/basebeingused(inparticular,dependingontheacid/basestrength)aswellaswhetherornottheacidorbaseisourtitrant.
49
TitrationofaStrongAcidwithaStrongBase
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• Wewillneedtodeterminethevolumerequiredtoreachtheequivalencepoint
• WewillneedtodeterminethepHatvariousvolumesbefore,during,andaftertheequivalencepoint
• Keepinmind:IamaddingNaOHtoHCl– BEFOREequivalence–thereisexcessH3O+– AFTERequivalence–thereisexcessOH-
StrongAcidw/StrongBase• Forastrongacidandstrongbase:– BEFOREequivalence-sinceH3O+isinexcess,youcalculatethenumberofmolesofaddedOH-fromtheinitialmolesofH3O+,andthendividebythetotalvolume
– ATequivalence–H3O+andOH-arecompletelyneutralized.Onlywatercontributesto[H3O+],sothepH=7.
– AFTERequivalence–sinceOH-isinexcess,calculate[OH-]bysubtractinginitialmolesofH3O+fromthenumberofmolesofaddedOH-,thendividebythetotalvolume.DeterminepHfrompOHordetermine[H3O+]from[OH-]todeterminepH.
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• Let’sfindtheinitialpHvalue• Forthetitrationofastrongacidandstrongbase,thepHinitial=pHstrongacid
• [HCl]=[H3O+]=0.100M• SopH=-log(0.100M)• pH=1.000
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• Wewantmolesacidàmolesofbase• 25.0mL=0.0250L• 0.0250L(0.100mol/LHCl)=0.00250molHCl• 0.00250molHCl=0.00250molOH-atequivalence• 0.00250molOH-(1L/0.100molNaOH)=0.0250LNaOH
• Theequivalencepointwillbereachedwhenweadd25mLofNaOH
• ThatmeansthepHwillbe7
OurDataTableSoFar…mLNaOHadded pH
0.00mL 1.005.00mL10.00mL15.00mL20.0mL25.0mL 7.0030.0mL35.0mL40.0mL50.0mL
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• Beforeequivalenceiswhenweaddlessthan25mLNaOH– calculatethenumberofmolesofaddedOH-fromtheinitialmolesofH3O+,andthendividebythetotalvolume
• Let’slookatthepHafteradding5.00mLofNaOH
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• FindthepHafteradding5.00mLofNaOH• 0.00500LNaOH(0.100mol/L)=0.000500molNaOH
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
AdditionAfteraddition
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
Addition 0.000500molAfteraddition
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
Addition 0.000500molAfteraddition 0(allrxtsw/H3O+)
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
Addition 0.000500mol --Afteraddition 0(allrxtsw/H3O+) 0.00200mol
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• NowwedividethemolesofH3O+bytheTOTALvolume
• So[H3O+]=0.00200molH3O+/(0.0250Linitial+0.00500Ladded)
• [H3O+]=0.0667MandpH=1.18OH- H3O+
Initial(moles) ≈0.00mol 0.00250molAddition 0.000500mol --
Afteraddition 0(allrxtsw/H3O+) 0.00200mol
OurDataTableSoFar…mLNaOHadded pH
0.00mL 1.005.00mL 1.1810.00mL15.00mL20.0mL25.0mL 7.0030.0mL35.0mL40.0mL50.0mL
UsethesametechniquetofindthepHfor10.00,15.00,and20.00mLofaddedNaOH
mLNaOHadded pH0.00mL 1.005.00mL 1.1810.00mL15.00mL20.0mL25.0mL 7.0030.0mL35.0mL40.0mL50.0mL
UsethesametechniquetofindthepHfor10.00,15.00,and20.00mLofaddedNaOH
mLNaOHadded pH0.00mL 1.005.00mL 1.1810.00mL 1.3715.00mL 1.6020.0mL 1.9525.0mL 7.0030.0mL35.0mL40.0mL50.0mL
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• Afterequivalenceiswhenweaddmorethan25mLNaOH– calculate[OH-]bysubtractinginitialmolesofH3O+fromthenumberofmolesofaddedOH-,thendividebythetotalvolume
• Let’slookatthepHafterIadd30.0mLofNaOH
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• Let’slookatthepHafteradding30.00mLofNaOH• 0.0300LNaOH(0.100mol/L)=0.00300molNaOH
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
AdditionAfteraddition
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
AdditionAfteraddition
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
Addition 0.00300mol --Afteraddition
OH- H3O+
Initial(moles) ≈0.00mol 0.00250mol
Addition 0.00300mol --
Afteraddition 0.000500(0.00250molreactswiththeH3O+
present,andthen0.000500isleft)
≈0.00mol
StrongAcidw/StrongBase• Let’slookathowwecanplotthetitrationcurveforthetitrationof25.0mLof0.100MHClwith0.100MNaOH.
• Let’slookatthepHafterIadd30.0mLofNaOH• So[OH-]=0.000500mol/(0.0250L+0.0300L)• [OH-]=0.00909• pOH=2.04• pH=11.96
OH- H3O+Initial(moles) ≈0.00mol 0.00250mol
Addition 0.00300mol --Afteraddition 0.000500 ≈0.00mol
OurDataTableSoFar…
mLNaOHadded pH0.00mL 1.005.00mL 1.1810.00mL 1.3715.00mL 1.6020.0mL 1.9525.0mL 7.0030.0mL 11.9635.0mL40.0mL50.0mL
FinishtheTableJ
mLNaOHadded pH0.00mL 1.005.00mL 1.1810.00mL 1.3715.00mL 1.6020.0mL 1.9525.0mL 7.0030.0mL 11.9635.0mL40.0mL50.0mL
FinishtheTableJ
mLNaOHadded pH0.00mL 1.005.00mL 1.1810.00mL 1.3715.00mL 1.6020.0mL 1.9525.0mL 7.0030.0mL 11.9635.0mL 12.2240.0mL 12.3650.0mL 12.52
added30.0mLNaOH0.00050molNaOHpH=11.96
added35.0mLNaOH0.00100molNaOHpH=12.22
AddingNaOHtoHCl
25.0mL0.100MHCl0.00250molHClpH=1.00
added5.0mLNaOH0.00200molHClpH=1.18
added10.0mLNaOH0.00150molHClpH=1.37
added15.0mLNaOH0.00100molHClpH=1.60
added20.0mLNaOH0.00050molHClpH=1.95
added25.0mLNaOHequivalencepointpH=7.00
added40.0mLNaOH0.00150molNaOHpH=12.36
added50.0mLNaOH0.00250molNaOHpH=12.52
TitrationCurve:
StrongAcidw/StrongBase• Notes:– ThepHchangesVERYquicklyneartheequivalencepoint– ThismeansthatsmallamountsofaddedbasecauselargechangesinpH
TitrationofaStrongBasewithaStrongAcid
TitrationofaStrongBasewithaStrongAcid
• ConsiderthepHcurveforthetitrationof100.0mLof0.50MNaOHwith1.0MHCl– OH–isinexcessbeforeequivalencepoint
– H3O+isinexcessaftertheequivalencepoint
TryThis:• A50.0-mLsampleof0.200Msodiumhydroxideistitratedwith0.200Mnitricacid.CalculatethepH
a. Afteradding30.00mLofHNO3b. AttheequivalencepointAnswers:a. pOHà1.30,sopH=12.70b. pH=7.00