1
EXPERIMENTC2:BUFFERS&TITRATION
LearningOutcomes
Uponcompletionofthislab,thestudentwillbeableto:
1) PrepareabuffersolutionatagivenpHandconcentration.
2) Analyzethetitrationcurveforthetitrationofa:
a. Weakacidwithastrongbase
b. Weakbasewithastrongacid
Introduction
AbufferisasolutionthatresistschangestopHwhenastrongacidorbaseisadded
tothesolution.Therearetwocombinationsofsolutionsthatmayresultinabuffer.
1)asolutionpreparedbycombiningaweakacidandasaltofitsconjugatebaseor
2)asolutionpreparedbycombiningaweakbaseandasaltofitsconjugateacid.
Inabuffersolution,theweakacid(ortheweakbase)isinequilibriumwithits
conjugatebase(oracid).AssumethattheformulaoftheweakacidisHAandthe
conjugatebaseisA−;inabuffermadebycombiningHAandA−thefollowing
equilibriumwillbeestablished:
HA(aq) ⇔ H+(aq)+A−(aq)
ThepHoftheabovebuffercanbecalculatedbymanipulatingtheequilibrium
constantfortheequationgivenaboveinthefollowingmanner.
€
[H + ][A − ]
Ka =
[HA]
[HA]
[H + ] = K a × −
[A ]
⎛
[HA] ⎞
−log[H + ] = −log⎜ K a × − ⎟ [A ] ⎠
⎝
⎛ [HA] ⎞
−log[H + ] = −logK a − log⎜ − ⎟
⎝ [A ] ⎠
⎛ [A − ] ⎞
pH = pK a + log⎜
⎟
⎝ [HA] ⎠
TheaboveequationisreferredtoastheHendersonHasselbachequationandis
(only)usedtocalculatethepHofabuffer.ThepKaintheequationisthatofthe
€
2
weakacidfoundinthebuffersolution.[HA]and[A−]arerespectively,the
concentrationsoftheweakacidandtheconjugatebase.
TheexactsameequationcanalsobeusedtocalculatethepHofabuffermadefrom
acombinationofaweakbaseanditsconjugateacid.Inthisinstance,thepKawould
bethatoftheconjugateacidinthesolution.
Whenpreparingabuffer,itisimportanttochooseanacid/conjugatebasepairsuch
thatthedesiredpHofthebufferiswithinaboutoneunitofthepKaoftheacidinthe
buffer.Thisisdoneinordertomaximizethebuffercapacity-anindicatorofthe
extenttowhichthebufferisabletoresistchangestopHuponadditionofastrong
acidorbase.
Inordertoprepareabufferthefollowingpiecesofinformationareessential:
1. ThedesiredpH
2. Thetotalconcentrationoftheionsinthebuffer
3. Therequiredvolume
4. Theidentityoftheionsinthebuffer
BufferPreparation
Forinstance,assumethatthefollowingbufferistobeprepared:100.0mLof0.10M
phosphatebufferatapHof7.40
100.0mListhetotalvolumeofthebufferand0.10Misthetotalconcentrationof
thephosphateionsthatshouldbefoundinthebuffer.Inordertodeterminethe
identityofthephosphateionsthatwouldberequiredtopreparethisbuffer,one
mustconsiderthevariousformsofphosphate:H3PO4,H2PO4−,HPO42−,PO43−.
STEP1:Determinetheacid/conjugatebasepairsofthesedifferentformsof
phosphateandusethetableofequilibriumconstantvaluestoarriveatthepKaof
eachacid.ThisinformationisprovidedinTable1below.
ACID CONJUGATEBASE pKa
H3PO4
H2PO4−
2.16
−
2−
H2PO4 HPO4 7.21
HPO42−
PO43−
12.32
Table1
FromTable1,itisapparentthatthephosphateacidwithapKawithinoneunitofthe
pHofthedesiredbufferisH2PO4−.
Thereforethebestcombinationofweakacidandconjugatebaseforthebuffer
wouldbe:
3
Weakacid=A=H2PO4−
(dihydrogenphosphate)
Conjugatebase=B=HPO42−
(monohydrogenphosphate)
STEP2:Nowthattheidentityofthephosphateionsinthebufferhasbeen
established,thenexttaskinhandistodeterminethemolarconcentrationsofthese
ionsinthebuffer.
Keepinmindthatthetotalconcentrationofthephosphatesinthedesiredbufferis
giventobe0.10M.Thisimpliesthat:
[A]+[B]=0.10
Equation1
InordertodeterminetheexactconcentrationsofAandB,itisnownecessarytouse
theHendersonHasselbachequation.
⎛ [B] ⎞
Equation2
pH = pK a + log⎜ ⎟ ⎝ [A] ⎠
Inequation2,forthepresentsituation,pH=7.40andpKa=7.21.Substitutethese
valuesinequation2:
€
⎛ [B] ⎞
Equation3
7.40 = 7.21+ log⎜ ⎟ ⎝ [A] ⎠
Manipulationofequation3resultsinthefollowing:
€
[B]=1.55[A] Equation4
Substituteequation4inequation1andsolvefor[A]and[B].
[A]=0.039M
[B]=0.061M
STEP3:Oncethemolarconcentrationoftheacidandconjugatebasepairhasbeen
determined,determinethegramsormLoftheacidandconjugatebaseneededfor
thebuffer.
Inthelaboratory,theweakacidtobeusedinthisbuffer,H2PO4−,islikelyfoundasa
solidinitssodiumsaltform:NaH2PO4andlikewisetheconjugatebaseHPO42−,is
likelyd=foundasasolidinitssodiumsaltform:Na2HPO4.So,theexactmassofthe
acidandconjugatebaseneededtoprepare100.0mLofthebuffermustbe
calculatedasfollows:
4
GramsofNaH2PO4= 0.039
moles
grams
× 0.1000L × 120
= 0.47grams L
mole
moles
grams
× 0.1000L × 142
= 0.87grams GramsofNa2HPO4= 0.061
L
mole
€
STEP4:Usingtheinformationfromsteps1,2,and3,thedesiredbuffercanbe
preparedasfollows:
€
“Combine0.47gramsofNaH2PO4and0.87gramsofNa2HPO4ina100-mL
volumetricflask.Addasmallamountofdeionizedwatertocompletelydissolvethe
solidsandaddwatertothegraduationmarkofthevolumetricflask.Carefullymix
thecontentsoftheflaskandmeasurethepHtoconfirmthatitisindeed7.40.”
EffectofacidonthepHofabuffer
Whenastrongacidisaddedtoabuffer,theconjugatebasepresentinthebuffer
neutralizesit.Theequilibriumbetweentheweakacidandtheconjugatebaseofthe
bufferisshifted.Theamountofconjugatebaseinthebufferdecreasesand
consequentlytheamountofweakacidincreases.Thisleadstoadecreaseinthe
overallpHofthebuffer.ThedecreaseinthepHwouldbefarmorepronouncedifthe
solutionwerenotabuffer,i.e.,ifnoconjugatebasewasavailabletoneutralizethe
addedacid.
Thefollowingcalculationdemonstratestheeffectofaddingastrongacidsuchas
hydrochloricacidonthepHofabuffersolution.
Example:CalculatethepHofthebufferpreparedearlier(100.0mLof0.10M
phosphatebufferatpH7.40)aftertheadditionof1.00mLof1.0MHCl.
NOTE:TheHClwillreactwiththeconjugatebaseinthebuffer(HPO42−).Therefore
thenewconcentrationsoftheweakacidandconjugatebaseinthebuffermustbe
calculated.
moles
= 3.9 Millimolesofweakacid(H2PO4−)presentinitially= 100.0mL × 0.039
L
moles
= 6.1
Millimolesofconjugatebase(HPO42−)presentinitially= 100.0mL × 0.061
L
moles
= 1.0 MillimolesofHCl(H+)addedtothebuffer=
€1.00mL × 1.0
L
€
€
5
Reaction:
Initialmillimoles:
Reaction:
Amountleft: Molarity:
H+(aq)+HPO42−!H2PO4− (NOTE:TheH+hereisfromHCl)
1.06.13.9
-1.0-1.0+1.0
05.14.9
5.1
4.9
101.0
101.0
⎛ 5.1 101.0 ⎞
Therefore:pH=7.21+log ⎜
⎟ =7.23
101.0 ⎠
⎝ 4.9 €
€
ThepHofthebufferchangedfrom7.40to7.23uponadditionof1.0mLofastrong
acidsuchas1.0MHCl.
€
EffectofbaseonthepHofabuffer
Whenastrongbaseisaddedtoabuffer,theweakacidpresentinthebuffer
neutralizesit.Theequilibriumbetweentheweakacidandtheconjugatebaseofthe
bufferisshifted.Theamountofconjugatebaseinthebufferincreasesand
consequentlytheamountofweakaciddecreases.Thisleadstoanincreaseinthe
overallpHofthebuffer.TheincreaseinthepHwouldbefarmorepronouncedifthe
solutionwerenotabuffer,i.e.,ifnoweakacidwasavailabletoneutralizetheadded
base.
Thefollowingcalculationdemonstratestheeffectofaddingastrongbasesuchas
sodiumhydroxideonthepHofabuffersolution.
Example:CalculatethepHofthebufferpreparedearlier(100.0mLof0.10M
phosphatebufferatpH7.40)aftertheadditionof1.00mLof1.0MNaOH.
NOTE:TheNaOHwillreactwiththeweakacidinthebuffer(H2PO4−).Thereforethe
newconcentrationsoftheweakacidandconjugatebaseinthebuffermustbe
calculated.
moles
= 3.9 Millimolesofweakacid(H2PO4−)presentinitially= 100.0mL × 0.039
L
moles
= 6.1
Millimolesofconjugatebase(HPO42−)presentinitially= 100.0mL × 0.061
L
moles
1.00mL × 1.0
= 1.0 MillimolesofNaOH(OH−)addedtothebuffer=
€
L
€
€
Reaction:
Initialmillimoles:
Reaction:
Amountleft: 6
OH−(aq)+H2PO4−!HPO42−+H2O
1.03.96.1
-1.0-1.0+1.0
02.97.1
2.9
7.1
Concentrations: 101.0
101.0
⎛ 7.1 101.0 ⎞
Therefore:pH=7.21+log ⎜
⎟ =7.60
⎝ 2.9 101.0
⎠
€
€
ThepHofthebufferchangedfrom7.40to7.60uponadditionof1.0mLofastrong
basesuchas1.0MNaOH.
€
Titration
Thetitrationofaweakacidwithastrongbasewillbeexaminedtheoreticallyinthis
section.Thecalculationswillthenbeusedtoplotatitrationcurve.Thetitration
curveprovidesseveralusefulpiecesofinformationandcanoftenbeplotted
qualitatively.
Example:CalculatethepHduringthetitrationof10.0mLof0.10Maceticacid(Ka=
1.8×10-5)aftertheadditionofthefollowingvolumes(inmL)of0.10Msodium
hydroxide:a)0.0b)2.5c)5.0d)7.5e)10.0andf)11.0
Inthefollowingcalculations,aceticacid(CH3COOH),aweakmonoproticacidwillbe
writtenasHA.
a) 0.0mLNaOHadded
Sincethetitrationhasnotbegunyetatthispoint,thepHofthesolutionissimplythe
pHof0.10Maceticacid.
HA(aq) ⇔ H+(aq) +
A−(aq)
Initialconcentration
0.10 ~0
0
€
Change
-x
x
x
Equilibriumconcentration 0.10–x
x
x
7
x2
x2
≈
0.1 − x 0.1
x = 0.00134 = [H + ]
pH = 2.87
K a = 1.8 × 10 −5 =
b) 2.5mLNaOHadded
€
TheOH−fromtheaddedNaOHwillreactwiththeaceticacid.
MillimolesofHApresent=10.0mL×0.10M=1.0
MillimolesofOH−added=2.5mL×0.10M=0.25
HA(aq) +
OH−(aq)!
A−(aq) +
H2O(l)
Initialmilimoles
1.0 0.25
Amountreacted
-0.25 -0.25 0.25
Finalmillimoles
0.75 0
0.25
0.75
0.25
Concentrations
12.5
12.5
Atthispointinthetitration,sinceaceticacid(HA)ispresentincombinationwithits
conjugatebase(A−),thesolutionisabuffer.ThepHofthesolutionmaythereforebe
€
€
calculatedusingtheHendersonHasselbachequation.
⎛ 0.25 12.5 ⎞
pH=4.74+log ⎜
⎟ =4.26
⎝ 0.75 12.5 ⎠
c) 5.0mLNaOHadded
€
TheOH−fromtheaddedNaOHwillreactwiththeaceticacid.
MillimolesofHApresent=10.0mL×0.10M=1.0
MillimolesofOH−added=5.0mL×0.10M=0.50
HA(aq) +
OH−(aq)!
A−(aq) +
H2O(l)
Initialmilimoles
1.0 0.5
Amountreacted
-0.5 -0.5 0.5
8
Finalmillimoles
0.5
Concentrations
0.5
15.0
0
0.5
0.5
15.0
Atthispointinthetitration,sinceaceticacid(HA)ispresentincombinationwithits
conjugatebase(A−),thesolutionisabuffer.ThepHofthesolutionmaythereforebe
€
€
calculatedusingtheHendersonHasselbachequation.
⎛ 0.5 15.0 ⎞
pH=4.74+log ⎜
⎟ =4.74
⎝ 0.5 15.0 ⎠
NOTE:Here,theamountofNaOHaddedisexactlyonehalfoftheamountofacetic
acidpresent.ThispointinthetitrationisreferredtoastheHALFEQUIVALENCE
€
POINT.ThepH,asshownabove,atthehalfequivalentpointinthesetypesof
titrationsisalwaysequaltothepKaoftheweakacid.
d) 7.5mLNaOHadded
TheOH−fromtheaddedNaOHwillreactwiththeaceticacid.
MillimolesofHApresent=10.0mL×0.10M=1.0
MillimolesofOH−added=7.5mL×0.10M=0.75
HA(aq) +
OH−(aq)!
A−(aq) +
H2O(l)
Initialmilimoles
1.0 0.75
Amountreacted
-0.75 -0.75 0.75
Finalmillimoles
0.25 0
0.75
0.25
0.75
Concentrations
17.5
17.5
Atthispointinthetitration,sinceaceticacid(HA)ispresentincombinationwithits
conjugatebase(A−),thesolutionisabuffer.ThepHofthesolutionmaythereforebe
€
€
calculatedusingtheHendersonHasselbachequation.
⎛ 0.75 17.5 ⎞
pH=4.74+log ⎜
⎟ =5.22
⎝ 0.25 17.5 ⎠
€
9
e) 10.0mLNaOHadded
TheOH−fromtheaddedNaOHwillreactwiththeaceticacid.
MillimolesofHApresent=10.0mL×0.10M=1.0
MillimolesofOH−added=10.0mL×0.10M=1.0
NOTE:Thisistheequivalencepointinthistitration.
HA(aq) +
OH−(aq)!
A−(aq) +
H2O(l)
Initialmilimoles
1.0 1.0
Amountreacted
-1.0 -1.0 1.0
Finalmillimoles
0
0
1.0
1.0
= 0.05M Concentrations
20.0
Attheequivalencepointofthetitration,alltheaceticacidandtheaddedNaOHare
completelyconsumed.Theonlyremainingspeciesinthesolutionistheconjugate
€
base,CH3COO−(A−).Thissolutionisnolongerabuffer.ThepHofthesolutionmust
becalculatedbyconsideringtheequilibriumestablishedbytheconjugatebase.
A−(aq) +
H2O(l) ⇔ HA(aq) +
OH−(aq)
Initialconcentration
0.05 0
0
€
Change
-x
x
x
Equilibriumconcentration 0.05–x
x
x
Sincethisistheequilibriumestablishedbythebase,theKbofthebasemustfirstbe
calculatedinordertosolvefor“x”.
K a × K b = KW = 1.0 × 10 −14
1.0 × 10 −14
x2
x2
−10
Kb =
= 5.6 × 10 =
≈
1.8 × 10 −5
0.05 − x 0.05
x = 5.27 × 10 −6 = [OH − ]
pOH = 5.28
pH = 8.72
€
10
NOTE:Attheequivalencepointinthetitrationofaweakacidwithastrongbase,the
speciespresentinthesolutionistheconjugatebaseoftheweakacid.Thereforethe
pHwillalwaysbegreaterthe7(basic).
f) 11.0mLNaOHadded
TheOH−fromtheaddedNaOHwillreactwiththeaceticacid.
MillimolesofHApresent=10.0mL×0.10M=1.0
MillimolesofOH−added=11.0mL×0.10M=1.1
NOTE:Sincethispointisbeyondtheequivalencepoint,theaddedNaOHisthe
excessreagentandtheaceticacidisthelimitingreagent.
HA(aq) +
OH−(aq)!
A−(aq) +
H2O(l)
Initialmilimoles
1.0 1.1
Amountreacted
-1.0 -1.0 1.0
Finalmillimoles
0
0.1 1.0
0.1
1.0
Concentrations
21.0
21.0
Atthispointinthetitration,twobasesarepresentinthesolution:1)thestrong
baseNaOHand2)theconjugatebaseofaceticacid,A−.Theconjugatebaseofacetic
€
€
acidisaweakerbasethanthestrongbaseNaOH.Insolutionsthatcontainastrong
baseandaweakbase,thepHofthesolutioncansimplybedeterminedfromthe
concentrationofthestrongbase.ThecontributiontothepHfromtheweakbaseis
negligibleincomparisontothestrongbase.
⎛ 0.1 ⎞
pOH=-log ⎜
⎟ =2.32
⎝ 21.0 ⎠
Therefore,pH=11.68
€
11
Thedatafromtheabovecalculationissummarizedinthefollowingtable:
Volumeofbaseadded(mL) pH
0.0
2.87
2.5
4.26
5.0
4.74
7.5
5.22
10.0
8.72
11.0
11.68
AcompletelabeledtitrationcurveisshowninFigure1below.
TitrationCurve:Aceticacidvs.NaOH
14
12
Buffer:CH3COOH+CH3COO−
10
pH
8
6
pKa
4
2
0
0
2
4
6
8
10
12
VolumeofNaOH(mL)
HalfEquivalencepoint
FIGURE1
Equivalencepoint
12
ExperimentalDesign
Inthefirstpartoftheexperiment,eachstudentwillberequiredtoprepareabuffer.
ThepHofthepreparedbufferwillbemeasuredusingapHmeterandadjustedto
thedesiredvalueusingeitheranacidorabase.Astrongacidandastrongbasewill
beaddedtothisbufferandthemeasuredpHineachinstancewillbecomparedwith
thecalculatedvalueofthepH.
Inthesecondpartofthisexperiment,eachstudentwilltitrateaweakbasewitha
strongacidandrecordthepHduringthecourseofthetitration.Adetailedtitration
curvewillbeplottedandlabeledusingthedatafromthetitration.
ReagentsandSupplies
0.10Maceticacid,solidsodiumacetate,solidsodiumphosphate,solidsodium
monohydrogenphosphate,0.10Mammonia,solidammoniumchloride,0.10M
hydrochloricacid,0.10Msodiumhydroxide,1.0Mhydrochloricacid,1.0Msodium
hydroxide
pHmeterkit,50-mLvolumetricflask,25-mLburette
(SeepostedMaterialSafetyDataSheets)
Procedure
13
PART1:PREPARATIONOFABUFFER
1. Obtainone50-mLvolumetricflaskandapHmeterkitfromthestockroom.
2. Prepare50mLof0.05M(totalconcentrationofions)ofoneofthefollowing
threebuffers(asassignedbytheinstructor):
i. pH=5.0
ii. pH=10.0
iii. pH=12.0
Beforepreparingthebuffercompletethefollowingsteps:
a. Choosetheappropriateacid/conjugatebasecombinationfromtheprovided
reagents.
b. Calculatetheamountsofacidandconjugatebase(ingramsormL)neededto
preparethebuffer.
c. Verifystepsaandbwiththeinstructor.
3. CalibratethepHmeteraspreviouslyindicated(SeeExperiment…….).
4. MeasurethepHofthepreparedbufferandadjustthepHtotherequiredvalue
using1.0MHClor1.0MNaOH.
5. OncethebufferhasbeenbroughttotherequiredpHvalue,showtheinstructor
thevalueofthepHonthepHmeter.
6. Transferexactlytwo25mLpotionsofthebufferintotwo100-mLbeakers.
7. Performthefollowingcalculations:
a. CalculatethepHofthepreparedbufferif0.25mLof1.0MHClisaddedto25
mLofthebuffer.
b. CalculatethepHofthepreparedbufferif0.25mLof1.0MNaOHisaddedto
25mLofthebuffer.
8. Toone25-mLpotionofthebuffer,add0.25mLof1.0MHCl.Thoroughlymixthe
contentswithastirringrod.MeasurethepHandcomparethemeasuredpHwith
thevaluecalculatedin7a.
9. Tothesecond25-mLpotionofthebuffer,add0.25mLof1.0MNaOH.
Thoroughlymixthecontentswithastirringrod.MeasurethepHandcompare
themeasuredpHwiththevaluecalculatedin7b.
10. Discardthebufferandanyotherwastegeneratedasdirectedbytheinstructor.
14
PART2:TITRATIONOFAWEAKBASEWITHASTRONGACID
1. Obtaina25-mLburetteandapHmeterfromthestockroom.
2. Obtain20mLof0.10Mammoniasolutionand30mLof0.10MHClsolution
fromthereagentstation.
3. Addthe20mLof0.10Mammoniasolutionintoa125-mLErlenmeyerflask.
4. Rinseandcleantheburettewithdeionizedwaterandconditionandfillthe
burettewith0.10MHCl.
5. MeasurethepHoftheammoniasolutionintheflaskpriortobeginningthe
titration.
6. Addexactly1.0mLof0.10MHClsolutionfromtheburetteintotheErlenmeyer
flaskcontainingtheammoniasolution.
7. ThoroughlymixthecontentsoftheErlenmeyerflaskandmeasureandrecord
thepH.
8. Repeatsteps6and7untilatotalof25mLofHClhasbeenadded.
9. Discardallsolutionsandanyotherwastethathasbeenaccumulatedintoawaste
containerprovidedbytheinstructor.
15
DataTable&Analysis
PART1:PREPARATIONOFABUFFER
Requiredbuffer(volume,concentration,pH)
ChosenAcid/Conjugatebasepair
CALCULATIONS(FORAMOUNTSOFACIDANDBASE)
Amountofacid
Amountofconjugatebase
MeasuredpHofthebuffer
16
CalculatethepHofthepreparedbufferif0.50mLof1.0MHClisaddedto50
mLofthebuffer.
CalculatedpHvalue
MeasuredpHvalue
CalculatethepHofthepreparedbufferif0.50mLof1.0MNaOHisaddedto50
mLofthebuffer.
CalculatedpHvalue
MeasuredpHvalue
17
PART2:TITRATIONOFAWEAKBASEWITHASTRONGACID
VolumeofHCladded(mL)
pH
0.00
Plotalabeledtitrationcurveusingthedataobtainedandindicatethe
following:
a. Chemicalspeciesbeforethetitrationwasbegun.
b. Bufferregion
c. Chemicalcompositionofthebuffer
d. Halfequivalencepointandequivalencepoint
e. Chemicalcompositionbeyondtheequivalencepoint.
18