AP Chemistry Day 60

AP Chemistry Day 60

Friday, March 13th, 2020

Do-Now:1.  Takeoutapieceofpaperandtitleit“Ch.15

NotesPartC”2.  A50.0-mLsampleof0.200Msodium

hydroxideistitratedwith0.200Mnitricacid.CalculatethepH

a.  Afteradding30.00mLofHNO3b.  AttheequivalencepointAnswers:a.  pOHà1.30,sopH=12.70b.  pH=7.00

Announcements• QuizJ• LabcomingupJ• StudyforunittestJ• SolubilityequilibriawillbeincludedwithourElectrochemchapterinUnit10

CW/HWAssignments11.  Ch.15CNPartC12.  Ch.15BWPartB13.  FRQPacket

PLANNER •  Ch.15BookworkB:Pg.638#64,65,66,71,73,85,88•  STUDY!Irecommendpracticingproblemsassignedoronppts,thencomparewithnotestoseeifyougetthemcorrect.

Essentialknowledgestandards•  SAP–10.A.1:Theprotonationstateofanacidorbase(i.e.,therelativeconcentrationsofHAand

A–)canbepredictedbycomparingthepHofasolutiontothepKaoftheacidinthatsolution.WhensolutionpH<acidpKa,theacidformhasahigherconcentrationthanthebaseform.WhensolutionpH>acidpKa,thebaseformhasahigherconcentrationthantheacidform.

•  SAP–10.A.2:Acid–baseindicatorsaresubstancesthatexhibitdifferentproperties(suchascolor)intheirprotonatedversusdeprotonatedstate,makingthatpropertyrespondtothepHofasolution.

•  SAP–10.B.1:Abuffersolutioncontainsalargeconcentrationofbothmembersinaconjugateacid–basepair.Theconjugateacidreactswithaddedbaseandtheconjugatebasereactswithaddedacid.ThesereactionsareresponsiblefortheabilityofabuffertostabilizepH.

•  SAP–10.C.1:ThepHofthebufferisrelatedtothepKaoftheacidandtheconcentrationratiooftheconjugateacid–basepair.Thisrelationisaconsequenceoftheequilibriumexpressionassociatedwiththedissociationofaweakacid,andisdescribedbytheHenderson–Hasselbalchequation.Addingsmallamountsofacidorbasetoabufferedsolutiondoesnotsignificantlychangetheratioof[A–]/[HA]andthusdoesnotsignificantlychangethesolutionpH.ThechangeinpHonadditionofacidorbasetoabufferedsolutionisthereforemuchlessthanitwouldhavebeenintheabsenceofthebuffer.

•  SAP–10.D.1:Increasingtheconcentrationofthebuffercomponents(whilekeepingtheratiooftheseconcentrationsconstant)keepsthepHofthebufferthesamebutincreasesthecapacityofthebuffertoneutralizeaddedacidorbase.

•  SAP–10.D.2:Whenabufferhasmoreconjugateacidthanbase,ithasagreaterbuffercapacityforadditionofaddedbasethanacid.Whenabufferhasmoreconjugatebasethanacid,ithasagreaterbuffercapacityforadditionofaddedacidthanbase.

FLT•  Iwillbeableto:–  Explainresultsfromthetitrationofamono–orpolyproticacidorbasesolution,inrelationtothepropertiesofthesolutionanditscomponents.

–  ExplaintherelationshipbetweentheabilityofabuffertostabilizepHandthereactionsthatoccurwhenanacidorabaseisaddedtoabufferedsolution

–  IdentifythepHofabuffersolutionbasedontheidentityandconcentrationsoftheconjugateacid–basepairusedtocreatethebuffer

–  Explaintherelationshipbetweenthebuffercapacityofasolutionandtherelativeconcentrationsoftheconjugateacidandconjugatebasecomponentsofthesolution

bycompletingCh.15Notes

Ch.15:Acid-BaseEquilibria

Recall

8

Titrations•  Termstoknow:

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

–  Indicator=Substanceaddedthatwillundergoacolorchangeneartheequivalencepoint

– Endpoint=Whentheindicatorchangesthecolorofthesolution.Thismaynotmatchtheequivalencepoint,dependingontherangeofpHvalueswheretheindicatorchangescolor

– StandardSolution=solutionofknownconcentrationthatisslowlyaddedtosolutionofunknownconc’

Titrations•  AtitrationcurveisaplotofthepHofasolutionduringatitration

•  Itistypicallythevolumeadded(ourstandardsolution)vs.pH.

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

12

TitrationofaStrongAcidwithaStrongBase

TitrationCurve:

TitrationofaStrongBasewithaStrongAcid

TitrationofaStrongBasewithaStrongAcid

•  ConsiderthepHcurveforthetitrationof100.0mLof0.50MNaOHwith1.0MHCl– OH–isinexcessbeforeequivalencepoint

– H3O+isinexcessaftertheequivalencepoint

TitrationofaWeakAcidwithaStrongBase

WeakAcidw/StrongBase•  Overall:

–  INITIALpH:Weakacidsdonotfullydissociate–weneedtodoanICEtabletodetermineinitialpH.Weexpectittobeweaklyacidic.

– BEFOREtheequivalencepoint:Thepresenceofboththeweakacidanditsconjugatebasecreateabuffersolution

– ATtheequivalencepoint:theacidandbasereactequivalently,buttheconjugatebaseisstillpresentandcontributestothepH.DetermineKbandpOHtodeterminepH.Weexpectaweaklybasicsolution.

– AFTERtheequivalencepoint:thepresenceofexcessOH-asyoucontinuetoaddbasemeansweweonlyneedtolookatthestrongbasecontributionto[OH-]

WeakAcidw/StrongBase•  Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).

Howmuchofouracidispresentinitially?•  25.0mL=0.0250L•  0.0250LHCHO2(0.100molHCHO2/1L)=•  0.00250molHCHO2(ourinitialamountpresent)


WhatvolumeofaddedNaOHcorrespondswiththeequivalencepoint?

NaOH(aq)+HCHO2(aq)àH2O(l)+NaCHO2(aq)•  0.00250molHCHO2(1molNaOH/1molHCHO2)•  0.00250molNaOH(1L/0.100molNaOH)=0.0250L

•  SoIneedtoadd0.0250L(or25.0mL)ofNaOHtoreachtheequivalence


Let’sfindtheinitialpH•  Onlytheacidispresentinitially…

HCHO2(aq)+H2O(l)ßàH3O+(aq)+CHO2

-(aq)

[HCHO2] [H3O+] [CHO2

-]ICE

[HCHO2] [H3O+] [CHO2-]

I 0.100M ≈0.00M 0.00MCE


I 0.100M ≈0.00M 0.00MC -x +x +xE


I 0.100M ≈0.000M 0.000MC -x +x +xE 0.100–x x x

WeakAcidw/StrongBaseLet’sfindtheinitialpH

HCHO2(aq)+H2O(l)ßàH3O+(aq)+CHO2

-(aq)

•  Sincex=4.24x10-3M•  Sincex=[H3O+],thepH=-log(4.24x10-3M)•  SopH=2.37


Let’sfindpHvaluesbeforeequivalence•  Beforeequivalence,someoftheacidisreactswiththeaddedbasetoproduceitsconjugatebase

•  Thepresenceofboththeacidanditsconjugatebasecreatesabuffersolution/region

•  ThismeanswecanuseourbuffershortcutequationstosolveforpH


Additionof5.0mLNaOH•  0.0050LNaOH(0.100molNaOH/1L)•  =0.00050molNaOH

OH-(strong

base)HCHO2(weak

acid)CHO2

-(conjugate)

Initial(moles)Addition

Afteraddition

OH-(strongbase)

HCHO2(weakacid)

CHO2-

(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00

AdditionAfteraddition

OH-(strongbase)

HCHO2(weakacid)

CHO2-


Addition 0.00050molAfteraddition

OH-(strongbase)

HCHO2(weakacid)

CHO2-


Addition 0.00050molAfteraddition ≈0.00mol 0.00200mol 0.00050mol


•  [CHO2-]=0.00050mol/(0.0250L+0.0050L)=

0.0167M•  [HCHO2]=0.0020mol/0.0300L=0.0667M•  pH=pKa+log[base]/[acid]•  pH=3.74+log(0.0167/0.0667)=3.14

OH-(strongbase)

HCHO2(weakacid)

CHO2-


Addition 0.00050molAfteraddition ≈0.00mol 0.00200mol 0.00050mol

OurDataTableSoFar…mLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL12.5mL15.0mL20.0mL25.0mL30.0mL35.0mL40.0mL

CalculatepHfor10.0,12.5,15.0,&20.0mLmLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL12.5mL15.0mL20.0mL25.0mL30.0mL35.0mL40.0mL

CalculatepHfor10.0,12.5,15.0,&20.0mLmLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL 3.5612.5mL 3.7415.0mL 3.9220.0mL 4.3425.0mL30.0mL35.0mL40.0mL

NoticepH=pKaat½equivalencepointmLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL 3.5612.5mL 3.7415.0mL 3.9220.0mL 4.3425.0mL30.0mL35.0mL40.0mL


Let’sfindthepHattheequivalencepoint•  Atequivalence,equalamountsofacidandbasereact,soIshouldonlyhavetheconjugatebasepresent

•  Since0.00250molofeachsubstancereacts,Iproduce0.00250molofmyconjugatebase(stoichiometricratios)


CHO2-(aq)+H2O(l)ßàHCHO2(aq)+OH-

(aq)•  FindyourconcentrationoftheconjugatebaseusingtheequivalentnumberofmolesdividedbytheTOTALvolume

•  So0.00250molCHO2-/(0.0250L+0.0250L)

•  [CHO2-]=0.0500M


[CHO2-] [HCHO2] [OH-]

I

C

E


I 0.0500M 0.00M ≈0.00M

C

E


I 0.0500M 0.00M ≈0.00M

C -x +x +x

E


I 0.0500M 0.00M ≈0.00M

C -x +x +x

E 0.0500–x x x•  Kb=[HCHO2][OH-]/[CHO2

-]•  Kb=Kw/Ka=1.0x10-14/1.8x10-4=5.6x10-11•  Findx=[OH-]=1.7x10-6•  pOH=5.77àpH=8.23

OurdatatablesofarmLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL 3.5612.5mL 3.7415.0mL 3.9220.0mL 4.3425.0mL 8.2330.0mL35.0mL40.0mL


Let’sfindthepHAFTERequivalencepoint•  SinceI’maddingmorestrongbase,andalloftheacidalreadyreacted,IamonlylookingatOH-

•  Thestrongbaseoverridestheweakbaseintermsofcontributions,soweonlylookatNaOH


Afteradding30.0mL•  0.0300LNaOH(0.100molNaOH/1L)=0.00300molNaOHadded

OH-(strongbase)

HCHO2(weakacid)

CHO2-

(conjugate)Initial(moles)


OH-(strongbase)

HCHO2(weakacid)

CHO2-

(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00mol


OH-(strongbase)

HCHO2(weakacid)

CHO2-


Addition 0.00300molAfteraddition

OH-(strongbase)

HCHO2(weakacid)

CHO2-


Addition 0.00300molAfteraddition 0.00050mol ≈0.00mol 0.00250mol

WeakAcidw/StrongBaseAfteradding30.0mL

•  So[OH-]=0.00050molOH-/(0.0250L+0.0300L)•  So[OH-]=0.0091M•  pOH=-log(0.0091M)=2.04•  pH=14–2.04=11.96(verybasic)

OH-(strongbase)

HCHO2(weakacid)

CHO2-


Addition 0.00300molAfteraddition 0.00050mol ≈0.00mol 0.00250mol

OurdatatablesofarmLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL 3.5612.5mL 3.7415.0mL 3.9220.0mL 4.3425.0mL 8.2330.0mL 11.9635.0mL40.0mL50.0mL

FinishthedatatableJmLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL 3.5612.5mL 3.7415.0mL 3.9220.0mL 4.3425.0mL 8.2330.0mL 11.9635.0mL40.0mL50.0mL

FinishthedatatableJmLNaOHadded pH

0.00mL 2.375.0mL 3.1410.0mL 3.5612.5mL 3.7415.0mL 3.9220.0mL 4.3425.0mL 8.2330.0mL 11.9635.0mL 12.2240.0mL 12.3650.0mL 12.52

added30.0mLNaOH0.00050molNaOHxspH=11.96


AddingNaOHtoHCHO2

added12.5mLNaOH0.00125molHCHO2pH=3.74=pKahalf-neutralization

initialHCHO2solution0.00250molHCHO2pH=2.37

added5.0mLNaOH0.00200molHCHO2pH=3.14





added25.0mLNaOHequivalencepoint0.00250molCHO2

−[CHO2

−]init=0.0500M[OH−]eq=1.7x10-6pH=8.23


41

TryThis•  A40.0mLsampleof0.100MHNO2istitratedwith0.200MKOH.Calculate:a.  Thevolumerequiredtoreachtheequivalencepointb.  ThepHafteradding5.00mLofKOHc.  ThepHatone-halftheequivalencepoint

•  Answers:a.  20.0mLKOHb.  pH=2.86c.  pH=3.34

TitrationCurveofaWeakBasewithaStrongAcid

TheWeakvs.TheWeak?

Weakw/Weak•  Whathappensifwetitrateaweakbasewithaweakacid(orviceversa)?

•  Theequivalencepointcanvary,dependingontherelativestrengthsoftheacidandbase(forexampleitmaybe<7,=7,or>7).

•  Wemightnotseethe“steepness”weusuallyseewithequivalenceaswell(depending)

•  Thesetypicallyaren’tactuallyperformedinexperiments,justknowtheyvary

PolyproticAcids

TitrationofaPolyproticAcid•  ifKa1>>Ka2,therewillbetwoequivalencepointsinthetitration–  theclosertheKa’saretoeachother,thelessdistinguishabletheequivalencepointsare

titrationof25.0mLof0.100MH2SO3with0.100MNaOH

MonitoringpHandIndicators

MonitoringpHChanges•  WecanmonitorthepHchangesofanacid-basechangeusing:① ApHmeter.② Anacid-baseindicator•  Theacid-baseindicatormarkstheendpointofatitrationby

changingcolor•  Theendpointisnotnecessarilythesameastheequivalence

point,butyouwanttochooseanindicatorthatwillbecloseenoughsothatwecanhavenegligibleerror

•  Whatyoureallyneedtoknow:•  Foraspecifictitrationofanacidbyabase,anindicatoris

selectedthathasapKaoneunitabovethepHvalueoftheequivalencepoint.

•  ThisisbecausethetransitionrangeformostindicatorsispKa+/-1

Indicators•  ManydyeschangecolordependingonthepHofthesolution•  Thesedyesareweakacids,establishinganequilibriumwith

theH2OandH3O+inthesolutionHInd(aq)+H2O(l)⇔Ind-(aq)+H3O+

(aq)

•  ThecolorofthesolutiondependsontherelativeconcentrationsofInd-:HInd–  whenInd-:HInd≈1,thecolorwillbemixofthecolorsofInd-andHInd

–  whenInd-:HInd>10,thecolorwillbemixofthecolorsofInd-

–  whenInd-:HInd<0.1,thecolorwillbemixofthecolorsofHInd

51

Phenolphthalein

MethylRed

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N NH

NaOOC

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

NaOOC

H3O+ OH-

MonitoringaTitrationwithanIndicator

•  formosttitrations,thetitrationcurveshowsaverylargechangeinpHforverysmalladditionsofbaseneartheequivalencepoint

•  anindicatorcanthereforebeusedtodeterminetheendpointofthetitrationifitchangescolorwithinthesamerangeastherapidchangeinpH– pKaofHInd≈pHatequivalencepoint

54

Acid-BaseIndicators

SampleQuestion•  Whichofthefollowingstatementsistrue?

a.  Atthestoichiometricpointofthetitrationofaweakacidwithastrongbase,thepHoftheresultingsolutionisgreaterthan7.00

b.  ThereisnoOH–inastronglyacidicsolutionc.  AsolutionwithapHof2.00istwiceasacidicasa

solutionwithapHof1.00d.  Adilutesolutionofanacidisthesameasaweakacid

SampleQuestion•  Anunknownweakmonoproticacid,HA,istitratedtotheendpointwith25.0mLof0.100MNaOHandthen13.0mLof0.100MHClisadded– pHoftheresultingsolutionis4.7

• Whichofthefollowingstatementsistrue?a.  AtpH4.7,halfoftheconjugatebaseA–hasbeenconvertedtoHAb.  pKaoftheacidis4.7c.  pKaoftheacidislessthan4.7d.  pKaoftheacidisgreaterthan4.7

SampleQuestion•  InthetitrationofaweakacidHAwith0.100MNaOH,thestoichiometricpointisknowntooccuratapHvalueofapproximately10– Whichofthefollowingindicatoracidswouldbebesttousetomarktheendpointofthistitration?a.  IndicatorA,Ka=10–14b.  IndicatorB,Ka=10–11c.  IndicatorC,Ka=10–8d.  IndicatorD,Ka=10–6

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AP Chemistry Day 60