Download - Experiment 8 ∙ Acid–base titration - Boston Universityquantum.bu.edu/courses/ch131-summer-1-2018/lab/exp08-Acid-base titration.pdf · Experiment 8 ∙ Acid–base titration 8‐2

Experiment8∙Acid–basetitration 8‐1

Name___________________________________________________________LabDay__________LabTime_________Experiment8∙Acid–basetitrationPre‐labquestionsAnswerthesequestionsandhandthemtotheTFbeforebeginningwork.(1)Whatisthepurposeofthisexperiment?__________________________________________________________________________________________________________________________________________________________________________________________________________________(2)WhatisthemolarconcentrationofH3O+inasolutionwhosepHisequalto2?__________________________________________________________________________________________________________________________________________________________________________________________________________________(3)Youwilltitrateasolutionofanunknownacid(HX(aq))byaddingNaOH(aq).Supposethat25mLof0.5MNaOH(aq)isneededtoreachtheequivalencepoint.HowmanymolesofHXwerepresentinitially?Showthecalculation.(4)Youwilltitrateacidsolutionsbyadding0.2–0.3‐mLor2–3‐mLportionsofNaOH(aq).Whenshouldyouaddthesmallervolume?__________________________________________________________________________________________________________________________________________________________________________________________________________________(5)TheconcentrationofanaqueoussolutionofNaOHcannotbeaccuratelydeterminedbydissolvingaknownmassofNaOH(s)inaknownvolumeofwater:Why?__________________________________________________________________________________________________________________________________________________________________________________________________________________


Experiment8

Acid–basetitration

MathematicaldevelopmentAcids and bases are indispensable fixtures of daily life. Acidscanberecognizedby theirsharp taste.The tartnessof lemonjuice is causedbycitricacid, the sournessofwineby tartaricacidandmalonicacid.Bases,ontheotherhand,arecharacter‐izedbyabitter taste.Moderately concentratedaqueous solu‐tions of bases such as household ammonia (NH4OH(aq)) ex‐hibitaslimy,slippery feel.Manybasesarephysiologicallyac‐tiveandacutelytoxic.Cigarettesmokerscravethe“buzz”fromthe base nicotine. The sentence of execution exacted on theGreekphilosopher Socrates (his crimewas that of corruptingthe young) consisted of his being forced to drink hemlock,whichcontainsthedeadlybaseconiine.pKaandpHAcidsandbasesarepartners.Anacid,genericallyrepresentedbyHA,isaproton(H+)donor,whereasabase,genericallyrep‐resented by B, is a proton acceptor. When acids and basescombine,theyswapaproton:

HA+B A–+HB+Strong acids (e.g., sulfuric acid) have a very pronounced ten‐dencytotransferaprotontoabasewhereasweakacids(e.g.,water) give up a proton much less readily. Acid strength is


quantifiedbymeasuringtheequilibriumconstantof thereac‐tionofanacidwithwater,whichactsasabase:

HA(aq)+H2O(l) A–(aq)+H3O+(aq)

�

Keq = Ka =[A−(aq)][H3O

+(aq)][HA(aq)]

This equilibrium constant (called the “acid‐dissociation con‐stant”)isgiventhespecialsymbolKa.AstrongacidexhibitsalargevalueofKawhereastheKaofaweakacidisgenerallylessthanunity(1).Becauseacidstrengthvariesoveranenormousintervalofmorethan60ordersofmagnitude,rangingfromthe“superacid”hydroiodicacid(HI,Ka=1011)totheexceedinglyweakacidmethane(CH4,Ka=10–50),andbecausewe like toavoid scientific notation, it is usual to refer to an acid’s pKaratherthantoitsKa.ThepKaisrelatedtotheKabythedefini‐tion

pKa=–log10KaThus, HI (Ka = 1011) has a pKa of –11, whereas CH4 (Ka =10–50)hasapKaof50. Thepracticeofreferringtoanacid’spKaratherthantoitsKaisanalogoustothecustomofreferringtothemolarconcen‐trationofH3O+(aq)bypH:

pH=–log10[H3O+(aq)]Asolutionthatis1MinH3O+(aq)hasapHofzero,whereasasolutionthatis10–14MinH3O+(aq)hasapHof14. The relationshipbetweenpHandpKa is expressedby theHenderson–Hasselbalchequation(Eqn.8‐1):

�

[A−(aq)][HA(aq)]

=10pH−pKa (Eqn.8‐1)

Eqn. 8‐1 states that there is more HA(aq) than A–(aq) whenpH<pKaandmoreA–(aq)thanHA(aq)whenpH>pKa.When


pH=pKa,[HA(aq)]=[A–(aq)](seeFigure8‐1foraspecificex‐ampleofthisbehavior).Acid–basetitrationcurvesIn this experiment you titrate an aqueous solution of an un‐knownacid(HX(aq))byaddinganaqueoussolutionofthebaseOH–(aq);thebalancedreactionis

HX(aq)+OH–(aq)X–(aq)+H2O(l)When the pH of the solution containing the unknown acid isplottedasafunctionofthevolumeofOH–(aq)added,thedataformsanacid–basetitrationcurve.Suchacurvehasthreere‐gions:(1)a fairly flat“buffered”regionat lowpH;(2)asteep“equivalence point” region at intermediate pH; (3) a secondfairlyflat“unbuffered”regionathighpH.Figure8‐2showsthetitrationcurvethatresultswhena25.00‐mLsamplecontaining0.815 g of an unknown acid HX is titrated against 0.500M

Figure8‐1Therelativeamountsofaceticacid(abbreviatedAcOH,pKa=4.75)andacetateion(abbreviatedAcO–)asafunctionofpHascalculatedbyEqn.8‐2.AcOHdominateswhenpH<4.75whereasAcO–dominateswhenpH>4.75.WhenpH=4.75,thereareequalamountsofAcOHandAcO–.


NaOH(aq).Let’sdiscussthechemistrythattakesplaceineachofthesethreeregionsandlet’salsodemonstratehowthemo‐larmassMandthepKaoftheunknownacidcanbeextractedfromthedata.BufferedregionatlowpHEvenbeforeanyOH–(aq)isaddedtothesolution,someoftheunknownacidreactswithwater(aweakbase)accordingtothebalancedequation

HX(aq)+H2O(l)X–(aq)+H3O+(aq)to the extent that the solution in Figure 8‐2 exhibits a pH of2.73,thatis,[H3O+(aq)]=10–2.73M=1.87×10–3M.WhenOH–(aq)isaddedtothesolution,thereaction

HX(aq)+OH–(aq)X–(aq)+H2O(l)

Figure8‐2Titrationof25.00mLofasolutioncontaining0.815gofanunknownacidHXagainst0.500MNaOH(aq).Attheequivalencepoint(locatedatthemidpointofthesteeplyincreasingregionofthecurve),thecumulativevolumeVeofOH–(aq)addedcontainsasmanymolesofOH–asthereweremolesofHXinitiallypresent.ThepKaofHXequalsthepHatVe/2.


begins tooccur.ThepH increases,butonlymodestlybecausethe simultaneous presence of HX(aq) and X–(aq) produces abuffersolutionthatresistschangesinpH.EquivalencepointregionatintermediatepHEventually somuchOH–(aq) is added to the solution that thesupplyofHX(aq)becomesexhaustedandthebufferingcapac‐ityof thesolutioncollapses; thus, thepHrisessharply.At theequivalence point, which is located at the midpoint of thesteeplyincreasingregionofthetitrationcurve,thecumulativevolumeVeofOH–(aq)addedcontainsasmanymolesofOH–asthere were moles of HX initially present. In Figure 8‐2 theequivalencepointoccurswhen13.8mLof0.500 MOH–(aq)isaddedandthesolutionhasapHof8.2. Locating the equivalence point allows us to calculate themolarmassMandthepKaofanunknownacidHX.Thenum‐berofmolesofOH–needed toreach theequivalencepoint inFigure8‐2isgivenby

�

0.0138LOH–⎛

⎝ ⎜

⎞

⎠ ⎟ 0.500molOH–

LOH–⎛

⎝ ⎜

⎞

⎠ ⎟ =0.00690molOH–

implyingthatthe0.815gofHXinitiallypresentcorrespondsto0.00690molofHX.ThemolarmassofHXisthus

�

0.815gHX0.00690molHX

=118g/mol

Recall that Eqn. 8‐1 says that [HX(aq)] = [X–(aq)] whenpH=pKa.At theequivalencepoint (Ve), all of theHX initiallypresentisdepleted.WhenavolumeofNaOH(aq)equaltoVe/2isadded,one‐halfoftheunknownacidmoleculesinitiallypre‐sent have reacted to form X– but one‐half still exists as HX.Thus,thepKaofHXisequaltothepHofthesolutionatVe/2.ThepHatVe/2=6.9mLinFigure8‐2is4.8;thus,thepKaofHXis4.8.


UnbufferedregionBeyondtheequivalencepoint,OH–isbeingaddedtoasolutionthatnolongercontainsanyHX.Thereaction

HX(aq)+OH–(aq)X–(aq)+H2O(l)nolongeroccursandtheincreaseinpHisattributabletosim‐plyaddingOH–(astrongbase)tothesolution.ProcedurePreparationofH2C2O4(aq)Recordingmasstotwodecimalplaces,weighoutabout3gofoxalic acid dihydrate (H2C2O4·2H2O(s), M = 126.07g/mol)ontoataredweighingboat.TransfertheH2C2O4·2H2O(s)toa100‐mLvolumetricflask.Rinsetheweighingboatandtheneckoftheflaskwithdeionizedwaterfromawashbottle;youwantto get every last bit of H2C2O4·2H2O(s) into the volumetricflask.Fill the flaskhalf‐fullwithdeionizedwaterand swirl todissolve the solid; this may take some time and extensiveswirling.WhentheH2C2O4·2H2O(s)hascompletelydissolved,addwateruntilthesolutionreachestheneckoftheflask.Usingadroppercarefullyaddmorewateruntilthebottomoftheliq‐uidlevelreachestheringetchedontheneckoftheflask.Yes,ifyoupassthemark,youmustdiscardthesolutioninahazard‐ous‐waste container and begin again. Invert the flask severaltimestomix.CalculatethemolarityofMH2C2O4ofoxalicacidinthesolution.PreparationofNaOH(aq)Sodiumhydroxide(NaOH(s),M=40.00g/mol)pelletspickupunknown amounts ofwater and carbon dioxide from the air.Thus, it is impossible tomakeupanNaOHsolutionofknownconcentrationbysimplyweighingoutthesolidanddissolvingitinwater.TheexactconcentrationoftheNaOHsolutionisde‐termined by titrating it against theH2C2O4 (aq) solution youprepared.


Obtain a clean and dry plastic 500‐mL screw‐cap bottle.Weighthebottletotwodecimalplaces.Weighoutabout10gofsodiumhydroxidepellets;besuretorecordthemasstotwodecimalplaces.TransfertheNaOHpelletstotheplasticbottleandfillitwithdeionizedwater.Swirlandshakethebottleuntilthepelletsarecompletelydissolved.MeasuringpHusingtheBeckman340pHmeterTheprogressof theacid–base titrationsconducted in thisex‐periment ismonitoredusing theBeckman340pHmeter(seeFigure8‐3).FollowthesestepstousethepHmeter: •Turnthepoweron. •Pushthe“READ”buttonbeforetakingapHmeasurement.ThepHmeterisconnectedtoafragileandexpensiveelectrodethatwillprobablybe immered inayelloworredbuffersolu‐tion.Theelectrodecannoteverbeallowedtodryout!Keepit immersed in the redor yellowbuffer solutionwhennot inuse.Donotdroptheelectrode,knockitagainstthewallsorbottom of flasks, or use it as a stirrer! The yellow or redbuffer solution does not interfere with the experiment: anydropsoftheyelloworredbuffersolutionclingingtotheelec‐trodedonotneedtobeknockedofftheelectrodebeforeuse.TitrationofNaOH(aq)againstH2C2O4(aq)The exact concentration of theNaOH(aq) solution is nowde‐termined by titrating it against the H2C2O4(aq) solution youprepared.PrepareaburetteforNaOH(aq)solutionandfillthepreparedburettewiththatsolution.Recordtheinitialvolumereadingontheburettetotwodecimalplaces.Preparea25‐mLvolumetric pipette for H2C2O4(aq) solution. Using a pipettepump,transferexactly25.00 mLofH2C2O4(aq)solutionintoaclananddry125‐mLErlenmeyerflask.,immersetheelectrodeintheH2C2O4(aq)solutionandrecordthepH. Openthestopcockandadd2–3mLofNaOH(aq)solutiontotheH2C2O4(aq)intheErlenmeyerflask.Closethestopcockandswirl the solution. Record the pH and the volume reading ontheburettetotwodecimalplaces.Youshouldobserveamod‐estpHincreaseof0.1–0.2unitsbecauseyouareinthebuffered

Figure8‐3TheBeckman340pHmeteranditsas‐sociatedelectrode.

SomeofthepHmeasure‐

ments,especiallythefirstfew

inthebufferedregion,are

subjecttodrift.Althoughan‐

noying,driftdoesnotcom‐

promisethequalityofthedata

aslongasyouobservea

steadyincreaseinpHas

NaOH(aq)solutionisadded.


region of the titration curve. Continue adding 2–3 mL ofNaOH(aq), closing the stopcock, swirling the solution aftereachaddition,recordingpHandrecordingthevolumereadingontheburette. Eventually, theadditionof2–3mLofNaOH(aq)willcausethepH to increaseby0.5unitsormore:youareentering theequivalencepointregionof thetitrationcurve.Drasticallyde‐creasethevolumeoftheNaOH(aq)additionstothesolutioninthe Erlenmeyer flask: try adding no more than 0.2–0.3mL,swirling, recording pH, and recording the volume reading ontheburetteasbefore. If adding0.2–0.3mLofNaOH(aq)doesnot result in apHincrease of 0.5 units ormore, youmay have recorded an ex‐perimentalartifactthatwewillcalla“falsejump”:stopadding0.2–0.3 mL of NaOH(aq) and go back to adding 2–3mL ofNaOH(aq), swirling, recording pH, and recording the volumereadingontheburetteasbefore.Torepeat:Theonlytimetheadditionof0.2–0.3mLofNaOH(aq)isappropriateisafteryouseeapHjumpof0.5unitormore.IfthepHeverjumpsby0.5unitorless,yournextadditionofNaOH(aq)shouldbe2–3mL. Eventually,theadditionof0.2–0.3mLNaOH(aq)willbringaboutaverylargepHincrease(≥ 5units):youhavepassedtheequivalencepointofthetitrationcurveandarenowintheun‐buffered region. Once the equivalence point is exceeded, it’ssafe to go back to adding 2–3mL of NaOH(aq), swirling, re‐cordingpH,andrecordingthevolumereadingontheburette.Getting plenty of data beyond the equivalence point iscritical to the success of the experiment. Do not abandonthetitrationafteryouhavepassedtheequivalencepoint. Repeat the titration on a fresh 25.00‐mL sample ofH2C2O4(aq). Youwant to calculate themolarity of NaOH(aq)fromanaverageofatleasttworuns.CalculatingthemolarityofNaOH(aq)After completing the first titration of H2C2O4(aq) againstNaOH(aq),plotthedatayoujustcollected:youwanttocalcu‐late the molarity of NaOH(aq).Do not wait to prepare theplotathomeafterlab!IfthemolarityoftheNaOH(aq)issig‐


nificantlydifferent from0.5M, you’vedonesomethingwrongandremedialactionmustbetakenimmediately. PlotthecumulativevolumeofNaOH(aq)addedalongthexaxis and pH along the y axis. Connect the points; the plotsshould resemble Figure 8‐4. Draw a line through the datapoints that lie on the gently sloping segmentsof the titrationcurveatlowpHandathighpH.Thetwolinesshouldideallybeparalleltoeachother.Nowdrawathirdlineparalleltoandlo‐catedmidway between the first two. The point at which thecenterlineintersectsthetitrationcurvecorrespondstoVe,thevolumeofNaOH(aq)attheequivalencepoint.InFigure8‐4,Veisequalto24.5mL. BecausetwomolesofNaOHareneededtoreactcompletelywithonemoleofH2C2O4accordingtothebalancedequation

H2C2O4(aq)+2NaOH(aq)→Na2C2O4(aq)+2H2O(l)

Figure8‐4Hypotheticaldataofthetitrationof25.00mLof0.247MH2C2O4(aq)against0.504MNaOH(aq).Drawlinesthroughthedatapointslyingontheslopingsegmentsoftheti‐trationcurveatlowpHandathighpH.Drawathirdlineparalleltoandmidwaybetweenthefirsttwo.ThepointatwhichthecenterlineintersectsthetitrationcurvecorrespondstoVe,thevolumeofNaOH(aq)attheequivalencepoint.Inthefigure,Veisequalto24.5mL.


the molarityMNaOH of NaOH(aq) is computed using the for‐mula

�

MNaOH =2MH2C2O4VH2C2O4

Ve (Eqn.8‐2)

where MH2C2O4 is the molarity of the H2C2O4(aq) solution,VH2C2O4 is the volume ofH2C2O4(aq) taken for titration (i.e.,0.02500L if you follow instructions) andVe is the volume inlitersofNaOH(aq)requiredtoreachtheequivalencepoint.TitrationofanunknownacidWeighaplasticweighingboattotwodecimalplaces.Recordingmasstotwodecimalplaces,transfer2–5gofanunknownacidtotheweighingboat.Besuretorecordthecodenumberoftheunknown.Transfertheunknownacidtoa100‐mLvolumetricflask. Rinse theweighing boat and the neck of the flaskwithdeionizedwaterfromawashbottle;youwanttogeteverylastbitofunknownacidintothevolumetricflask.Filltheflaskhalf‐fullwith deionizedwater and swirl to dissolve the solid; thismaytakesometimeandextensiveswirling.Whenthethesolidhascompletelydissolved,addwateruntilthesolutionreachestheneckoftheflask.Usingadroppercarefullyaddmorewateruntil thebottomoftheliquidlevelreachestheringetchedontheneckof the flask.Yes, ifyoupass themark,youmustdis‐card the solution in a hazardous‐waste container and beginagain.Inverttheflaskseveraltimestomix. Using the same technique that you employed in the titra‐tion of NaOH(aq) against H2C2O4(aq), titrate 25.00‐mL sam‐ples of the unknown acid solution against NaOH(aq) at leasttwice.YouwanttocalculatethemolarmassMandthepKaofthe unknown acid from an average of at least two runs. Ofcourse,perform the second titrationona fresh sampleof theunknownacidsolution.Don’tforgettodecreasetheamountofNaOH(aq) added to the unknown acid solution as you ap‐proach the eqivalence point and be sure to record plenty ofdataafter theequivalencepoint isexceeded.Also,bemindfuloftheannoying“falsejump”phenomenon.


DataanalysisEmployingthethree‐linetechniquedescribedearliertodeter‐minethemolarityofNaOH(aq),forbothofyourrunsinwhichyou titrated theunknownacid solution find thevolumeVe inlitersofNaOH(aq)thatcorrespondstotheequivalencepointofthetitration. TheunknownacidHXandNaOHreactaccordingtothebal‐ancedequation

HX(aq)+NaOH(aq)→NaX(aq)+H2O(l)The number of moles nNaOH of NaOH required to reach theequivalencepointisthusgivenby

nNaOH=MNaOHVewhereMNaOHisthemolarityofNaOHinunitsofmolesperli‐ter. Because of the one‐to‐one stoichiometry of the reaction,nNaOHisalsoequaltothenumberofmolesofunknownacidinthe 25.00‐mL sample of unknown acid solution that you ti‐trated.But recall, however, that the25.00‐mL samples of un‐knownacidsolutionrepresentonly1/4ofthe100‐mLvolumeofunknownacidsolutionyouprepared;thus,thetotalnumberofmolesnHXofHXyouweighedoutisgivenby

nHX=4nNaOH=4MNaOHVeDividing themassmHXofunknownacidyoudissolvedby thetotal number of moles nHX of unknown acid gives themolarmassMHXoftheunknownacid.Insummary,then,

�

MHX =mHX

4MNaOHVe (Eqn.8‐3)

TofindthepKaoftheunknownacid,dividethevolumeVebytwo.ThepHthattheunknownacidsolutionexhibitsatVe/2correspondstothepKaoftheunknownacid.


Name___________________________________________________________LabDay__________LabTime_________Experiment8∙Acid‐basetitrationLabreportform Page1(I.A)ReportthemassofH2C2O4·2H2O(s)(M=126.07g/mol)taken=______________________g(I.B)ShowthecalculationofMH2C2O4,theconcentrationofH2C2O4(aq)inunitsofmoleperliteroftheH2C2O4(aq)solutionyouprepared.(I.C)ReportthemassofNaOH(s)taken=_______________________________________________________g(II.A) On separate sheets, present plots of the data collected during the titrations ofH2C2O4(aq)againstNaOH(aq).PlotthecumulativevolumeofNaOH(aq)addedalongthexaxisandpHalongtheyaxis.Prepareaseparateplotforeachrun.Giveeachplotatrulyin‐formativetitle(i.e.,don’tjustcallit“Run1”),labeltheaxes,andincludeappropriateunitsanddivisionsofthoseaxes.Donotsubmitsmallplots:useawholesheetofpaper.Scalethehorizontal and vertical axes so that the data points occupymost of the area of the plot.DrawthethreelinesneededtodeterminetheequivalencepointVe(seeFigure8‐4)andin‐dicatethelocationofVeontheplot.(II.B) Report themolarityMNaOH of NaOH(aq) calculated according to Eqn. 8‐2 and themeanmolarityofNaOH(aq).

�

MNaOH =2MH2C2O4VH2C2O4

Ve (Eqn.8‐2)

Run MH2C2O4

[mol/L]VH2C2O4[L]

Ve[L]

MNaOH[mol/L]

1

2

mean


Name___________________________________________________________LabDay__________LabTime_________Experiment8∙Acid‐basetitrationLabreportform Page2(III.A)Reportthecodenumberoftheunknownacid=__________________________________________(III.B)ReportthemassmHXofunknownacidtaken=__________________________________________g(IV.A)Onseparatesheets,presentplotsofthedatacollectedduringthetitrationoftheun‐knownacidagainstNaOH(aq).PlotthecumulativevolumeofNaOH(aq)addedalongthexaxisandpHalongtheyaxis.Prepareaseparateplotforeachrun.Giveeachplotatrulyin‐formativetitle(i.e.,don’tjustcallit“Run1”),labeltheaxes,andincludeappropriateunitsanddivisionsofthoseaxes.Donotsubmitsmallplots:useawholesheetofpaper.Scalethehorizontal and vertical axes so that the data points occupymost of the area of the plot.DrawthethreelinesneededtodeterminetheequivalencepointVe(seeFigure8‐4)andin‐dicatethelocationofVeontheplot.IndicatethepKaoftheunknownontheplot.(IV.B)Report themolarmassMHXof theunknownacidcalculatedaccordingtoEqn.8‐3,thepKaoftheunknownacid,andthemeanvaluesofMHXandpKa.

�

MHX =mHX

4MNaOHVe (Eqn.8‐3)

Run mHX

[g]MNaOH[mol/L]

Ve[L]

MHX[g/mol]

pKa

1

2

mean

Postlabquestions(1)Circlethecompoundthatmostcloselymatchesyourunknownacid.Acid M pKa Acid M pKaBoricacid 61.8 9.14 trans‐Crotonicacid 86.1 4.69Malicacid 134.1 3.40 Glutamicacid 147.1 4.25L‐Ascorbicacid 176.1 4.17 Potassiumhydrogenphthalate 204.2 5.513‐Iodobenzoicacid 248.0 3.80

Download - Experiment 8 ∙ Acid–base titration - Boston Universityquantum.bu.edu/courses/ch131-summer-1-2018/lab/exp08-Acid-base titration.pdf · Experiment 8 ∙ Acid–base titration 8‐2

Top Related