AP Chem Week 26 Unit 9 Ch 15 B 2019flemingapchem.weebly.com/.../ap_chem_unit_9_ch_15_b_2019.pdf · 2019-03-04 · buffer in the range of 4.0 – 6.0. You need to calculate and adjust

Recall

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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

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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

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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

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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


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?


BufferCapacity

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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

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Titrations•  Termstoknow:– Titrant=standardsolution(knownM)– Analyte=substancebeinganalyzed(unknownM)– Equivalencepoint=WhenmolesofOH-=molesofH3O+(neutralizationhasoccurred)

–  Indicator=Substanceaddedthatwillundergoacolorchangeneartheequivalencepoint

– Endpoint=Whentheindicatorchangesthecolorofthesolution.Thismaynotmatchtheequivalencepoint,dependingontherangeofpHvalueswheretheindicatorchangescolor

– StandardSolution=solutionofknownconcentrationthatisslowlyaddedtosolutionofunknownconc’

s

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Titrations•  Acid-BaseTitrationMethodUsingIndicator:1.  Analytesolution(ofunknownM)isplacedina

flaskorbeaker2.  Asmallamountofindicatorisadded3.  Titrantisplacedinaburetteandslowlyaddedto

theanalyteandindicatormixture4.  Theprocessisstoppedwhentheindicatorcauses

achangeinthecolorofthesolution5.  Thechangeinvolumeisusedtodeterminethe

volumeoftheanalytesolution

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Titrations•  Note:SometitrationsrequireboilingtogetridofCO2produced,whichwillformcarbonicacid,bufferthesolution,andleadtoinaccuratedata

Titrations•  AtitrationcurveisaplotofthepHofasolutionduringatitration

•  Itistypicallythevolumeadded(ourstandardsolution)vs.pH.

•  Theshapeofthecurvedependsontheacid/basebeingused(inparticular,dependingontheacid/basestrength)aswellaswhetherornottheacidorbaseisourtitrant.

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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.


•  Let’sfindtheinitialpHvalue•  Forthetitrationofastrongacidandstrongbase,thepHinitial=pHstrongacid

•  [HCl]=[H3O+]=0.100M•  SopH=-log(0.100M)•  pH=1.000


•  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


•  Beforeequivalenceiswhenweaddlessthan25mLNaOH– calculatethenumberofmolesofaddedOH-fromtheinitialmolesofH3O+,andthendividebythetotalvolume

•  Let’slookatthepHafteradding5.00mLofNaOH


•  FindthepHafteradding5.00mLofNaOH•  0.00500LNaOH(0.100mol/L)=0.000500molNaOH

OH- H3O+Initial(moles) ≈0.00mol 0.00250mol

AdditionAfteraddition


Addition 0.000500molAfteraddition


Addition 0.000500molAfteraddition 0(allrxtsw/H3O+)


Addition 0.000500mol --Afteraddition 0(allrxtsw/H3O+) 0.00200mol


•  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


•  Afterequivalenceiswhenweaddmorethan25mLNaOH– calculate[OH-]bysubtractinginitialmolesofH3O+fromthenumberofmolesofaddedOH-,thendividebythetotalvolume

•  Let’slookatthepHafterIadd30.0mLofNaOH


•  Let’slookatthepHafteradding30.00mLofNaOH•  0.0300LNaOH(0.100mol/L)=0.00300molNaOH






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


•  Let’slookatthepHafterIadd30.0mLofNaOH•  So[OH-]=0.000500mol/(0.0250L+0.0300L)•  [OH-]=0.00909•  pOH=2.04•  pH=11.96


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


AddingNaOHtoHCl

25.0mL0.100MHCl0.00250molHClpH=1.00

added5.0mLNaOH0.00200molHClpH=1.18




added25.0mLNaOHequivalencepointpH=7.00



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

Documents

AP Chem Week 26 Unit 9 Ch 15 B 2019flemingapchem.weebly.com/.../ap_chem_unit_9_ch_15_b_2019.pdf · 2019-03-04 · buffer in the range of 4.0 – 6.0. You need to calculate and adjust