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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)
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
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
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
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
[HCHO2] [H3O+] [CHO2-]
I 0.100M ≈0.00M 0.00MC -x +x +xE
[HCHO2] [H3O+] [CHO2-]
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
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Let’sfindpHvaluesbeforeequivalence• Beforeequivalence,someoftheacidisreactswiththeaddedbasetoproduceitsconjugatebase
• Thepresenceofboththeacidanditsconjugatebasecreatesabuffersolution/region
• ThismeanswecanuseourbuffershortcutequationstosolveforpH
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
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-
(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00
Addition 0.00050molAfteraddition
OH-(strongbase)
HCHO2(weakacid)
CHO2-
(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00
Addition 0.00050molAfteraddition ≈0.00mol 0.00200mol 0.00050mol
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
• [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-
(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00
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
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Let’sfindthepHattheequivalencepoint• Atequivalence,equalamountsofacidandbasereact,soIshouldonlyhavetheconjugatebasepresent
• Since0.00250molofeachsubstancereacts,Iproduce0.00250molofmyconjugatebase(stoichiometricratios)
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
CHO2-(aq)+H2O(l)ßàHCHO2(aq)+OH-
(aq)• FindyourconcentrationoftheconjugatebaseusingtheequivalentnumberofmolesdividedbytheTOTALvolume
• So0.00250molCHO2-/(0.0250L+0.0250L)
• [CHO2-]=0.0500M
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
[CHO2-] [HCHO2] [OH-]
I
C
E
[CHO2-] [HCHO2] [OH-]
I 0.0500M 0.00M ≈0.00M
C
E
[CHO2-] [HCHO2] [OH-]
I 0.0500M 0.00M ≈0.00M
C -x +x +x
E
[CHO2-] [HCHO2] [OH-]
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
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Let’sfindthepHAFTERequivalencepoint• SinceI’maddingmorestrongbase,andalloftheacidalreadyreacted,IamonlylookingatOH-
• Thestrongbaseoverridestheweakbaseintermsofcontributions,soweonlylookatNaOH
WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Afteradding30.0mL• 0.0300LNaOH(0.100molNaOH/1L)=0.00300molNaOHadded
OH-(strongbase)
HCHO2(weakacid)
CHO2-
(conjugate)Initial(moles)
AdditionAfteraddition
OH-(strongbase)
HCHO2(weakacid)
CHO2-
(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00mol
AdditionAfteraddition
OH-(strongbase)
HCHO2(weakacid)
CHO2-
(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00mol
Addition 0.00300molAfteraddition
OH-(strongbase)
HCHO2(weakacid)
CHO2-
(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00mol
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-
(conjugate)Initial(moles) ≈0.00mol 0.00250mol 0.00mol
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
added35.0mLNaOH0.00100molNaOHxspH=12.22
AddingNaOHtoHCHO2
added12.5mLNaOH0.00125molHCHO2pH=3.74=pKahalf-neutralization
initialHCHO2solution0.00250molHCHO2pH=2.37
added5.0mLNaOH0.00200molHCHO2pH=3.14
added10.0mLNaOH0.00150molHCHO2pH=3.56
added15.0mLNaOH0.00100molHCHO2pH=3.92
added20.0mLNaOH0.00050molHCHO2pH=4.34
added40.0mLNaOH0.00150molNaOHxspH=12.36
added25.0mLNaOHequivalencepoint0.00250molCHO2
−[CHO2
−]init=0.0500M[OH−]eq=1.7x10-6pH=8.23
added50.0mLNaOH0.00250molNaOHxspH=12.52
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