AP Chemistry Day 60
Transcript of AP Chemistry Day 60
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AP Chemistry Day 60
Friday, March 13th, 2020
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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
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Announcements• QuizJ• LabcomingupJ• StudyforunittestJ• SolubilityequilibriawillbeincludedwithourElectrochemchapterinUnit10
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CW/HWAssignments11. Ch.15CNPartC12. Ch.15BWPartB13. FRQPacket
PLANNER • Ch.15BookworkB:Pg.638#64,65,66,71,73,85,88• STUDY!Irecommendpracticingproblemsassignedoronppts,thencomparewithnotestoseeifyougetthemcorrect.
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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.
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FLT• Iwillbeableto:– Explainresultsfromthetitrationofamono–orpolyproticacidorbasesolution,inrelationtothepropertiesofthesolutionanditscomponents.
– ExplaintherelationshipbetweentheabilityofabuffertostabilizepHandthereactionsthatoccurwhenanacidorabaseisaddedtoabufferedsolution
– IdentifythepHofabuffersolutionbasedontheidentityandconcentrationsoftheconjugateacid–basepairusedtocreatethebuffer
– Explaintherelationshipbetweenthebuffercapacityofasolutionandtherelativeconcentrationsoftheconjugateacidandconjugatebasecomponentsofthesolution
bycompletingCh.15Notes
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Ch.15:Acid-BaseEquilibria
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Recall
8
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Titrations• Termstoknow:
– Titrant=standardsolution(knownM)– Analyte=substancebeinganalyzed(unknownM)– Equivalencepoint=WhenmolesofOH-=molesofH3O+(neutralizationhasoccurred)
– Indicator=Substanceaddedthatwillundergoacolorchangeneartheequivalencepoint
– Endpoint=Whentheindicatorchangesthecolorofthesolution.Thismaynotmatchtheequivalencepoint,dependingontherangeofpHvalueswheretheindicatorchangescolor
– StandardSolution=solutionofknownconcentrationthatisslowlyaddedtosolutionofunknownconc’
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Titrations• AtitrationcurveisaplotofthepHofasolutionduringatitration
• Itistypicallythevolumeadded(ourstandardsolution)vs.pH.
• Theshapeofthecurvedependsontheacid/basebeingused(inparticular,dependingontheacid/basestrength)aswellaswhetherornottheacidorbaseisourtitrant.
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TitrationofaStrongAcidwithaStrongBase
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TitrationCurve:
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TitrationofaStrongBasewithaStrongAcid
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TitrationofaStrongBasewithaStrongAcid
• ConsiderthepHcurveforthetitrationof100.0mLof0.50MNaOHwith1.0MHCl– OH–isinexcessbeforeequivalencepoint
– H3O+isinexcessaftertheequivalencepoint
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TitrationofaWeakAcidwithaStrongBase
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WeakAcidw/StrongBase• Overall:
– INITIALpH:Weakacidsdonotfullydissociate–weneedtodoanICEtabletodetermineinitialpH.Weexpectittobeweaklyacidic.
– BEFOREtheequivalencepoint:Thepresenceofboththeweakacidanditsconjugatebasecreateabuffersolution
– ATtheequivalencepoint:theacidandbasereactequivalently,buttheconjugatebaseisstillpresentandcontributestothepH.DetermineKbandpOHtodeterminepH.Weexpectaweaklybasicsolution.
– AFTERtheequivalencepoint:thepresenceofexcessOH-asyoucontinuetoaddbasemeansweweonlyneedtolookatthestrongbasecontributionto[OH-]
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WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Howmuchofouracidispresentinitially?• 25.0mL=0.0250L• 0.0250LHCHO2(0.100molHCHO2/1L)=• 0.00250molHCHO2(ourinitialamountpresent)
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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
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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
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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
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WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Let’sfindpHvaluesbeforeequivalence• Beforeequivalence,someoftheacidisreactswiththeaddedbasetoproduceitsconjugatebase
• Thepresenceofboththeacidanditsconjugatebasecreatesabuffersolution/region
• ThismeanswecanuseourbuffershortcutequationstosolveforpH
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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
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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
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OurDataTableSoFar…mLNaOHadded pH
0.00mL 2.375.0mL 3.1410.0mL12.5mL15.0mL20.0mL25.0mL30.0mL35.0mL40.0mL
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CalculatepHfor10.0,12.5,15.0,&20.0mLmLNaOHadded pH
0.00mL 2.375.0mL 3.1410.0mL12.5mL15.0mL20.0mL25.0mL30.0mL35.0mL40.0mL
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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
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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
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WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Let’sfindthepHattheequivalencepoint• Atequivalence,equalamountsofacidandbasereact,soIshouldonlyhavetheconjugatebasepresent
• Since0.00250molofeachsubstancereacts,Iproduce0.00250molofmyconjugatebase(stoichiometricratios)
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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
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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
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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
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WeakAcidw/StrongBase• Let’sconsiderthetitrationof25.0mLof0.100MHCHO2with0.100MNaOH(theKaofformicacidis1.8x10-4).
Let’sfindthepHAFTERequivalencepoint• SinceI’maddingmorestrongbase,andalloftheacidalreadyreacted,IamonlylookingatOH-
• Thestrongbaseoverridestheweakbaseintermsofcontributions,soweonlylookatNaOH
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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
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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
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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
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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
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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
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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
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TryThis• A40.0mLsampleof0.100MHNO2istitratedwith0.200MKOH.Calculate:a. Thevolumerequiredtoreachtheequivalencepointb. ThepHafteradding5.00mLofKOHc. ThepHatone-halftheequivalencepoint
• Answers:a. 20.0mLKOHb. pH=2.86c. pH=3.34
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TitrationCurveofaWeakBasewithaStrongAcid
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TheWeakvs.TheWeak?
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Weakw/Weak• Whathappensifwetitrateaweakbasewithaweakacid(orviceversa)?
• Theequivalencepointcanvary,dependingontherelativestrengthsoftheacidandbase(forexampleitmaybe<7,=7,or>7).
• Wemightnotseethe“steepness”weusuallyseewithequivalenceaswell(depending)
• Thesetypicallyaren’tactuallyperformedinexperiments,justknowtheyvary
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PolyproticAcids
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TitrationofaPolyproticAcid• ifKa1>>Ka2,therewillbetwoequivalencepointsinthetitration– theclosertheKa’saretoeachother,thelessdistinguishabletheequivalencepointsare
titrationof25.0mLof0.100MH2SO3with0.100MNaOH
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MonitoringpHandIndicators
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MonitoringpHChanges• WecanmonitorthepHchangesofanacid-basechangeusing:① ApHmeter.② Anacid-baseindicator• Theacid-baseindicatormarkstheendpointofatitrationby
changingcolor• Theendpointisnotnecessarilythesameastheequivalence
point,butyouwanttochooseanindicatorthatwillbecloseenoughsothatwecanhavenegligibleerror
• Whatyoureallyneedtoknow:• Foraspecifictitrationofanacidbyabase,anindicatoris
selectedthathasapKaoneunitabovethepHvalueoftheequivalencepoint.
• ThisisbecausethetransitionrangeformostindicatorsispKa+/-1
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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
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51
Phenolphthalein
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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-
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MonitoringaTitrationwithanIndicator
• formosttitrations,thetitrationcurveshowsaverylargechangeinpHforverysmalladditionsofbaseneartheequivalencepoint
• anindicatorcanthereforebeusedtodeterminetheendpointofthetitrationifitchangescolorwithinthesamerangeastherapidchangeinpH– pKaofHInd≈pHatequivalencepoint
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Acid-BaseIndicators
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SampleQuestion• Whichofthefollowingstatementsistrue?
a. Atthestoichiometricpointofthetitrationofaweakacidwithastrongbase,thepHoftheresultingsolutionisgreaterthan7.00
b. ThereisnoOH–inastronglyacidicsolutionc. AsolutionwithapHof2.00istwiceasacidicasa
solutionwithapHof1.00d. Adilutesolutionofanacidisthesameasaweakacid
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SampleQuestion• Anunknownweakmonoproticacid,HA,istitratedtotheendpointwith25.0mLof0.100MNaOHandthen13.0mLof0.100MHClisadded– pHoftheresultingsolutionis4.7
• Whichofthefollowingstatementsistrue?a. AtpH4.7,halfoftheconjugatebaseA–hasbeenconvertedtoHAb. pKaoftheacidis4.7c. pKaoftheacidislessthan4.7d. pKaoftheacidisgreaterthan4.7
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SampleQuestion• InthetitrationofaweakacidHAwith0.100MNaOH,thestoichiometricpointisknowntooccuratapHvalueofapproximately10– Whichofthefollowingindicatoracidswouldbebesttousetomarktheendpointofthistitration?a. IndicatorA,Ka=10–14b. IndicatorB,Ka=10–11c. IndicatorC,Ka=10–8d. IndicatorD,Ka=10–6