chapter 12: water as solvent; acids and baseskuwata/classes/2009-10/env chem/ch_1… · chapter 12:...
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Chapter 12: Water as Solvent; Acids and Bases
Preview We turn from water as a resource, to the properties of water as a solvent, and
consider • Hydrogen bonding, and the formation of clathrates and of biological
structures • Dissolved ions • Strong and weak and bases acids • Acid rain
12.1 Unique Properties of Water Water is the most commonplace of substances, and yet its properties are unique
among chemical compounds. Perhaps most remarkable are its high melting and boiling temperatures. Table 12.1 lists the melting and boiling points for the hydrides from carbon to fluorine, all of which are elements in the first row of the periodic table. Compared to the neighboring hydrides, water has by far the highest melting and boiling points. One characteristic that makes Earth uniquely suitable for the evolution of life is its surface temperature which, over most of the planet, lies within the liquid range of water.
12.S1HydrogenbondingThehightemperaturerequiredforboilingimpliesthatliquidwaterhasa
highcohesiveenergy;themoleculesassociatestronglywithoneanother.Onesourceofthiscohesioniswater’shighdipolemoment(1.85Debye*).TheO–Hbondsarehighlypolar;negativechargebuildsupontheoxygenendwhilepositivechargebuildsuponthehydrogenend.Thedipolestendtoalignwithoneanother,increasingthecohesiveenergy.Butifthedipole‐dipoleinteractionweretheonlyfactor,thenHFwouldhaveahigherboilingpointthanwater,becauseithasahigherdipolemoment(1.91Debye).Thegreatercohesionofwaterderives,inaddition,fromitsabilitytoformanetworkofhydrogenbonds(H‐bonds).‐‐‐
*Thedipolemoment,µ,isdefinedasthechargetimesthedistanceseparatingthecharge.1Debye(D) = 3.336x10–30coulomb(C)xmeters(m).Forexample,foroneelectron(1.60x10–19C)separatedfromaprotonby1Å(10–10m),µ =1.6x10–29Cx.m=4.80D.
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‐‐‐H‐bondsarethestrongelectrostaticinteractionbetweenanHatomonan
electronegativeatom,X[onewhichhasastrongattractionforelectrons]andalonepaironanotheratom,Y.WeuseadottedlinetorepresenttheH‐bond:
X‐H........:Y‐
TheelectronegativityofXisimportant,sinceitdeterminestheextenttowhichthebondingelectronsarepulledtowardXandawayfromH.ThushydrocarbonsdonotformH‐bondstoasignificantextent,becauseCandHhavesimilarelectronegativities,resultinginanessentiallyneutralHatom.Theelementstotherightofcarbonintheperiodictable–N,OandF‐areincreasinglyelectronegative,becausetheirincreasingnuclearchargedrawsthevalenceelectronsclosertothenucleus.TheyaretheonesthatcandonateH‐bonds,andtheyarealsotheelementsthatretainlonepairswhentheyformcompounds.
Thesameprinciplesapplytoelementsofthesecondandhigherrowsoftheperiodictable,butH‐bondingisweakerfortheseelements[e.g.P,SandCl]sincetheyarelesselectronegativethanthoseinthefirstrow[becausethenucleusisfartherfromthevalenceelectrons].ThemostimportantH‐bondsinnaturearethoseinwhichXandYareNand/orO.
H‐bondingisveryimportantinbiologicalmolecules,forwhichNandOatomsoccuratkeypointsinthestructure.TheH‐bondshelpmaintainthethree‐dimensionalshapeofthesemolecules,andareresponsiblefordetermininghowtheyinteractwithoneanother.PerhapsthemostimportantH‐bondsarethosewhichestablishcomplementarityofthebasesintheDNAdoublehelix[seep?],andtherebydeterminethegeneticcode.
Waterisabletodonatetwohydrogenbonds,onefromeachofitsHatoms,whilesimultaneouslyacceptingtwohydrogenbonds,onetoeachoftheelectronlonepairsontheoxygenatom.Thusliquidwaterformsathree‐dimensionalnetworkofH‐bonds,whereasHF,havingonlyonedonorsite,islimitedtoforminglinearH‐bondchains(Figure12.1).
Figure12.1LinearH‐bondinginHFversusnetworkH‐bondinginH2O.
TheH‐bondnetworkofwaterisdemonstratedstrikinglybythecrystalstructureofice,inwhicheachwatermoleculeisH‐bondedtofourothermoleculesinconnectedsix‐memberedrings(Figure12.2).Thesesix‐memberedringsproduceaveryopenstructure,whichaccountsforanotherunusualpropertyofwater,its
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expansionuponfreezing.Mostsubstancescontractuponsolidificationbecausethemoleculestakeupmoreroominthechaoticliquidstatethantheydointheclose‐packedsolid.Theliquidcontinuestoexpandasitisheatedduetotheincreasedmotionofthemolecules.Whenicemelts,however,theopenlatticestructurecollapsesonitselfandthedensityincreases.H‐bondednetworksstillexist,buttheyfluctuateveryrapidly,allowingtheindividualwatermoleculestobemobileandclosertogether.Asliquidwaterisheatedfrom0°C,theH‐bondednetworkisfurtherdisruptedandtheliquidwatercontinuestocontract.Whenthetemperaturereaches4°C,thedensityofliquidwaterreachesamaximum.Asthetemperaturerisesabove4°C,thewaterslowlyexpands,reflectingtheincreasingthermalmotion.Thelowerdensityoficerelativetowaterisofgreatecologicalsignificancetotheworld’stemperate‐zonelakesandrivers.Becausewintericefloatsontopofthewater,itprotectstheaquaticlifebelowthesurfacefromtheinhospitableclimateabove.Ificeweredenserthanwater,thelakesandriverswouldfreezefromthebottomup,creatingarcticconditions.
Figure12.2Thecrystalstructureofice.Thelargercirclesandsmallercirclesrepresentsoxygenatomsandhydrogenatoms,respectively.Oxygenatomswiththesamecolorshadingindicatetheylieinthesameplane.
12.S2.ClathratesandwatermiscibilityTheabilityofwatermoleculestohydrogenbondtooneanotheraccountsfor
theexistenceofcrystallineclathrates[alsocalledhydrates],compoundsinwhichwatermoleculescanpackthemselvesaroundnon‐polarmoleculesbyformingorderedarraysofH‐bonds.Thestructureofxenonhydrate,8Xe●46H2O,isshowninFigure12.3.Thecrystalsconsistofarraysofdodecahedraformedfrom20watermoleculesH‐ bondedtogether.WithinthedodecahedraaretheXeatoms,or
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nonpolarmoleculesofsimilarsize,suchasCH4orCO2.Byfillingthevoids,theguestmoleculesstabilizetheclathrateH‐bondnetwork.Theclathratesarestabletotemperaturesappreciablyhigherthanthemeltingpointofice.Forexample,methanehydratemeltsat18°Catatmosphericpressure.Cloggingofnaturalgaspipelinesbymethanehydratewasonceaproblemforgasdistribution.Itwassolvedbyremovingwatervaporbeforethegaswasfedintothelines.
Figure12.3Thestructureofaclathratecrystal,xenonhydrate;thexenonatomsoccupycavities(eightperunitcube)inahydrogen‐bondedthree‐dimensionalnetworkformedbywatermolecules(46perunitcube).
Asdiscussedearlier(p?)ithascometobeappreciatedthatmethanehydratesoccurverywidelyinnature,principallyinoceansedimentsorinthearcticpermafrost.Anaerobicmicrobesinsedimentsorinswampsdecomposeorganicmattertomethane[seep.?].Whenmethaneisreleasedintothesurroundingwater,itformsmethanehydrate,providedthetemperatureislowenough,and/orthepressureishighenough.
TheabilityofwatertoformextendedH‐bondnetworksaroundnonpolarmoleculesalsoexplainswhyoilandwaterdonotmix.Watermoleculesactuallyattracthydrocarbonmolecules;asshowninTable12.2,theenthalpyofdissolutioninwaterisnegativeforsimplehydrocarbons,meaningthatthemolecularcontactreleasesheat.Theheatreleasereflectsanattractiveforcecreatedbytheinteractionbetweenwaterandtheguestmolecule.Nevertheless,thefreeenergyvaluesarepositive;thereactionisdisfavoredbylargenegativeentropychanges(ΔG=Δ H – T Δ S ) . Thenegativeentropyimpliesthatmixingincreasesmolecularorder,consistentwiththewatermoleculespackingaroundthenonpolarsolute,astheydointheclathratestructures.Thus,oilandwaterdonotmixbecausethepackingtendencyofthewaterinhibitsit.
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Thistendencyofwaterandnonpolarmoleculestoseparateisresponsiblefor
manyimportantstructuresinbiology.Forexample,proteinskeeptheirshapeinwaterbyfoldinginsuchawaythathydrophobic(“water‐hating”)alkylandarylgroupsoftheaminoacidsareintheinterior,awayfromthewater,whilehydrophilic(“water‐loving”)polarandchargedgroupsfacetheexterior,wheretheycontactthewater.Likewise,biologicalmembranesarebilayersoflipidmolecules(Figure12.4),withthehydrocarbonchainsofthelipidspackedtogetherawayfromthewaterintheinteriorofthebilayer.
Figure12.4a)Thehydrophilic‐hydrophobicbilayerstructureofabiologicalmembrane;b)molecularstructureofalipidmolecule.
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ThesolubilityoforganicmoleculesinwaterincreaseswhentheyhavefunctionalgroupscapableofformingH‐bondswithwater.Thusalcohols,ethers,acids,aminesandamidesareallsignificantlysolubleinwater,theextentofsolubilitydependingonthesizeofthehydrocarbonportionofthemolecule.Thesmallerthehydrocarbonportionthehigherthesolubility.Smallalcohols,likemethanolandethanol,arecompletelymisciblewithwater.The–OHgroupcanbothdonateandacceptH‐bondswithwatermolecules,providingastrongimpetusformixing.Ethersarenotcompletelymiscible,sincethe–O‐atomsubstituentcanacceptbutnotdonateanH‐bond.However,thesolubilityisconsiderablyenhancedbytheH‐bondacceptance,relativetothesolubilityofacomparablehydrocarbon.ThusthesolubilityofthegasolineadditiveMTBE[methyltertiarybutylether–seep?]is4700mg/l,comparedtoonly24mg/lfortheequivalenthydrocarbon,2,2’‐dimethylbutane.ThisiswhyMTBEisafarmoreseriousgroundwatercontaminantthanthegasolinetowhichithasbeenadded.
12.2 Acids, Bases, and Salts
12.S3.Ions,autoionization,andpHAnotheruniquepropertyofwateristheeasewhichwithitdissolvesionic
compounds.Theforcesbetweenionsofoppositechargeinacrystalareverystrong,andcostagooddealofenergytodisrupt.Thiscanbeseeninthehighmeltingpointofanioniccrystal,suchasordinarytablesalt.Yetsaltdissolvesreadilyinwater,becausethewatermoleculesstronglysolvateboththepositivesodiumionsandthenegativechlorideions.Thestronginterionicforcesarereplacedbyequallystrongsolvationforces.
Thestrongsolvationforcesstemfromwater’slargedipolemomentandH‐bondingability.AnionsinteractwiththepositiveendofthedipoleviaH‐bonds,whilecationsinteractwiththenegativeendviacoordinatebondsfromtheoxygenlone‐pairelectrons(Figure12.5).
Figure12.5Solvationofionsinwater.
Amongtheionsthatarestabilizedinwaterarethehydrogenandhydroxideions.StrongacidssuchasHCl,HNO3,andH2SO4releaseH + ionswhendissolvedinwater,whilestrongbasessuchasNaOHorKOHreleaseOH–.Theseionsarespecialbecausetheyreacttogethertoformwater
H+ + OH– = H2O [12.1] Reaction[12.1] isaneutralizationreaction;thestrongbaseneutralizesthe
strongacidandviceversa.Thereverseofthisreactionisautoionization;thewater
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canionizeitselfbecausetheionsarestabilizedbyotherwatermolecules.Thisreversereactiondoesnotproceedtoanygreatextent;thepositionofequilibriumforreaction[12‐1] liesfartotheright.Nevertheless,autoionizationisakeyfactorinacid‐basechemistrybecauseitsetstherangeofavailableacidandbasestrengthsinwater.
Incontrasttoreactionsinthegasphase,reactionsinaqueoussolutionareusuallyrapid.Thewatermoleculesareincontinuouscontactwithoneanother,andtheH‐bondsandcoordinatebondsarebrokenandreformedrapidly.Therearesomemetalionsforwhichthecoordinatebondstowaterarelong‐livedforspecialelectronicreasons,asinthecomplexionCr(H2O)63+.Butformostions,thebondsarebrokenandreformedveryrapidly,producinganaveragedsolvationenvironmentfortheion.Consequentlytheextentofreactionisgenerallysetbytheequilibriumconstant,ratherthanthekinetics,evenifthepositionofequilibriumliesfartotherightortotheleft.Wecanusetheexperimentallydeterminedequilibriumconstanttocalculatetheextentofreactionsfromvariousstartingconditions.Thisisaparticularlyusefulcapabilityinacid‐basereactions,whicharefundamentaltothechemistryoftheaqueousenvironment.
12.S4ThepHScaleForsolutionsofastrongacid,HX,theconcentrationofH+ionsisthesameas
theanalyticalconcentration(CHX),i.e.thenumberofmolesperliter[symbolizedM]ofdissolvedHX.Wewrite
[H+] = CHX M
Likewiseforsolutionsofastrongbase,MOH
[OH-] = CMOH M
Because[H+]isgenerallylessthan1M,andfrequentlymuchless,ithasbecomestandardtoexpresstheconcentrationasthenegativelogarithm,symbolizedby“p”:
pH = -log[H+] Forexample,whentheHXconcentrationis10‐3M[i.e.1millimolar],thepHis
3,andwhenitistwicethismuch,thepHis2.7[because–log[2x10‐3]=0.3‐3=2.7].Inthesameway
pOH = -log[OH-]
Ifwenowconsidertheautoionizationreaction[reverseofequation[12.1]]H2O=H++OH‐ [12.2]
wecanwritetheequilibriumexpressionas
K=[H+][OH–]/[H2O] [12.3]
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Weseethat[H+]and[OH‐]arereciprocallyrelated
[H+]=K[H2O]/[OH‐] or pH=‐log(K[H2O])–pOH [12.4]
Inanaqueoussolutiontheconcentrationofthewatermoleculesisessentiallyconstant;aliterofwaterweighs1000g,andcontains1000g/18gmol‐1=55.5moles.[Theaddedweightfromthedissolvedionsisasmallfractionof1000g,unlessthesolutionisveryconcentrated].Thereforeitiscustomarytoinclude[H2O]inthe‘effective’equilibriumconstant.Fortheautoionizationequilibrium,theeffectiveconstant,K[H2O],iscalledKw;itsexperimentalvalueis10‐14M2.Substitutingthisvalueineq.[12‐4],weseethat
pH=14–pOH [12.5]
Inotherwords,theautoionizationequilibriumrequiresthatthepHandpOH
addupto14.InpurewaterthereisnosourceofH+orOH‐otherthantheautoionization
reactionitself,whichrequiresthatoneH+beproducedforeveryOH‐.Inthatcase[H+]=[OH‐],andpluggingthisconditionintotheequilibriumexpression[12‐3]gives
[H+]2=10‐14M2,or[H+]=10‐7MThustheautoionizationreactionrequiresthatpurewaterhaveapHof7.
ThisisalsothepHofasolutioninwhichastrongacidhasbeenexactlyneutralizedbyastrongbase.ApHof7definesneutrality;pHvaluesbelow7areacidic,whilepHvaluesabove7arebasic,oralkaline.
12.S5Weakacidsandbases;buffersBecausewaterdissolvessomanydifferentsubstances,theonlyplaceinthe
environmentwherepurewaterisfoundisatthestartofthehydrologicalcycle,inwatervapor.Evenraindropsincludeothersubstances;asdetailedinPartII,theformationofraindropsinsaturatedairisnucleatedbyatmosphericparticles,especiallynitratesandsulfates.Inaddition,thewaterincloudvaporandraindropsisinequilibriumwithotherconstituentsoftheatmosphere,especiallycarbondioxide.ThesesubstancestendtolowerthepHofrainwatersignificantlybelowneutrality.
Strongacidstransferaprotoncompletelytowater,butmanyacidicsubstancesholdtheprotonmoreorlesstenaciously,andthetransfertowatermaybeincomplete.Forageneralizedacid,HA,theextentofthetransferdependsontheequilibriumconstant,Ka,fortheaciddissociationreaction
HA=H++A– [12.6] Ka=[H+][A–]/[HA] [12.7]Kaisoftenreferredtoastheacidityconstant.IfKaissmall,thenonlyasmall
fractionofHAwillbedissociated,creatingaconsiderabledifferencebetweentheconcentrationofacidmoleculesandtheconcentrationofhydrogenions.
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WorkedProblem12.1:pHofaWeakAcidQ:WhatisthepHofa0.1Msolutionofaceticacid,whichhasaKaof10–4.75
M2?A:DissociationofHAproducesanequalnumberofH+andA‐ions,and
therefore[H+]=[A‐].Substitutingthisequalityineq.[12‐7],andrearranging,gives[H+]2=Ka[HA] [12.8]
IfonlyasmallfractionofHAisdissociated,then[HA]isessentiallythesame
astheanalyticalconcentration,CHA[thiswillbethecaseprovidedKa<<CHA],andtherefore
[H+]2~KaCHA [12.9] InthisproblemCHA=0.1M,andKa=10–4.75M2(whichismuchsmallerthan
CHA*)so [H+]2=10–4.75x10–1.0=10–5.75
[H+]=10–2.88MorpH=2.88 Therefore,a0.1MsolutionofaceticacidhasapHcloseto3.0.* [footnote:*Sincethevalueof[H+]isonly1%ofthevalueofCHA,the
approximationthat(HA)=CHAisagoodoneinthiscase.Whenthisapproximationispoor,thenallowancemustbemadefortheloweringofHAduetotheionizationreaction.SinceeverymoleculeofHAthationizesproducesoneH+ion,CHA=[HA]+[H+],or[HA]=CHA–[H+].Substitutingthisintoequation[12‐7]andrearranging,wehave
[H+]2+Ka[H+]–KaCHA=0whichmaybesolvedwiththequadraticformula.]
========================================= IfHAholdsitsprotontenaciously,thentheanion,A‐,willhavesome
tendencytoremoveaprotonfromwater.ThusifasaltofA‐isaddedtowater,thereaction
A‐+H2O=HA+OH‐ [12.10]whichiscalledahydrolysis,orabasedissociationreaction,willproceedtosomeextent.SinceOH‐isproduced,A‐isactingasabase.Assumingthattheequilibriumdoesnotliecompletelytotheright,A‐isaweakbase.Theequilibriumconstant,calledthebasicityconstant,is
Kb=[HA][OH‐]/[A‐] [12.11]
[Asinthecaseoftheautoionizationreactionthewaterconcentration,[H2O],beingconstant,islumpedintotheequilibriumconstant.]
Theacidandbasedissociationreactionsarelinkedbytheautoionizationreaction.Thiscanbeseenbyaddingreaction[12‐6]toreaction[12‐10]toyieldreaction[12‐2].Whenreactionsareadded,theequilibriumconstantsmultiply,i.e.
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KaKb=Kw[Thisresultcanbecheckedbysubstitutingtheequilibriumexpressions,eq.[12‐7],[12‐11]and[12‐3].]ThusdeterminingKaalsospecifiesKb.
WorkedProblem12.2:pHofaWeakBaseQ:WhatisthepHofa0.1Msolutionofsodiumacetate?A:SincebasedissociationproducesoneOH‐foreveryHA[eq.[12‐10],then
[OH‐]=[HA]forasaltofA‐dissolvedinwater.Then[OH‐]canbecalculatedbysubstitutionineq.[12‐11]andrearranging,
[OH‐]2=Kb[A‐]~KbCA‐ [12.12]
providedthatKb<<CA‐,sothatthedifferencebetween[A‐]andCA‐isnegligible.InthisproblemCA‐=0.1M,andKb=Kw/Ka=10‐14/10‐4.75=10‐9.25M(which
ismuchlessthanCA‐),sothat[OH‐]2=10‐9.25x10‐1.0=10‐10.25
or [OH‐]=10‐5.12M.Since[H+]=Kw/[OH‐],thenpH=14–5.12=8.88.Thusa0.1MsodiumacetatesolutionhasapHcloseto9,afairlyalkalinevalue.=========================================
12.S6ConjugateAcidsandBases;BuffersAweakacid,HA,anditsanion,A‐,constituteaconjugateacidbasepair.The
twoarethesamemolecularentity,differingonlybyoneproton.Theactualchargeisunimportantinthiscontext.Thebasecouldequallywellbeneutral,andthenitsconjugateacidwouldhaveapositivecharge.Forexample,theammoniumionandammonia,NH4+andNH3,constituteaconjugateacidbasepair.
Animportantcharacteristicofaconjugateacid‐basepairisthatamixtureofthetwopartnershasapHthatisclosetothenegativelogarithmoftheacidityconstant,calledthepKa.Thiscanbeseenbyrearrangingtheequilibriumexpression,eq.[12‐7]
[H+]=Ka[HA]/[A‐] or pH=pKa–log[HA]/[A‐] [12.13]
If[HA]=[A‐],thenthepHisexactlypKa.Moreover,thepHwillbeclosetothe
pKa,aslongas[HA]/[A‐]isnotfarfromunity.Evenifthisratioreachesavalueof10or0.1,thepHwilldeviatefrompKabyonlyoneunit.ThusthepHisbufferedagainstlargechanges.AmixtureofHAandA‐constitutesabuffersolution,becauseitresistslargepHchangeswhensmallamountsofotheracidsorbasesareaddedtothesolution.
Thebufferingeffectisbestillustratedbyatitrationcurveoftheacid.Figure12.6showsatitrationcurvefora0.1Maceticacidsolution,towhichsuccessiveamountsofastrongbase[e.g.NaOH]areadded.AtthestartofthetitrationthepHiscloseto3,whileattheendpoint,whenalloftheHAchasbeenconvertedtoNaAc,thepHiscloseto9.Atthehalfwaypoint[HAc]=[Ac‐],andpH=pKa=4.74.Fromthe10%tothe90%pointsofthetitration,thepHvariesonlybyminusandplusone
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unitfromthisvalue.Butastheendpointisapproached,thebuffercapacityoftheHAc/Ac‐mixtureisusedup,andthepHshootsupto9andbeyond.Withonlya10%excessofthebasetitrant[OH‐]=10‐2M],thepHreaches12.
Figure12.6ThechangeinpHduringthetitrationofaweakacidwithastrongbase;50.0mLof0.100Maceticacidistitratedwith0.100MNaOH.
12S.7WaterintheAtmosphere:AcidRainAlthoughpurewaterisneutral,andhasapHof7.0,rainwaterisnaturally
acidicbecauseitisinequilibriumwithcarbondioxide.Whendissolvedinwater,carbondioxideformscarbonicacid,aweakacid:
CO2+H2O=H2CO3 [12.14]Wecangaugetheamountofcarbonicacidfromtheequilibriumconstantfor
reaction[12‐7]Ks=[H2CO3]/PCO2=10–1.5M/atm [12.15] wherePCO2isthepartialpressureofCO2,inatmospheres.*[footnote:*Actually,dissolvedCO2ismostlyunhydrated;onlyasmall
fractionexistsasH2CO3molecules:CO2 (aq) + H2O (l) = H2CO3 (aq) Kh = [H2CO3]/[CO2] = 10–2.81
The total concentration of dissolved CO2 is:
[CO2]T = [CO2] + [H2CO3] = [H2CO3][1 + Kh–1] = 102.81[H2CO3]
MostequilibriummeasurementsaremadewithrespecttototaldissolvedCO2 , and[H2CO3]istobeunderstoodas[CO2]T.]‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
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SincetheatmosphericconcentrationofCO2iscurrently379ppm,PCO2=379x10–6or10–3.4atm(atsealevel),andtherefore[H2CO3]=KsPCO2=10–1.5∞ 10–3.4=10–4.9M.
Rainwateristhusadilutesolutionofcarbonicacid.Theaciddissociatespartiallytohydrogenandbicarbonateions:
H2CO3=H++HCO3–Ka=10–6.4M [12.16]
FromKawecancalculate[H+],asinWorkedProblem12.1
[H+]2=Ka[H2CO3]=10–6.4x10–4.9=10–11.3
[H+]=10–5.7andpH=5.7
Thus,atmosphericCO2depressesthepHofrainwaterby1.3unitsfromneutrality.Evenintheabsenceofanthropogenicemissions,rainwaterisnaturallyacidic(althoughitcanoccasionallybeneutraloralkalineasaresultofcontactwithalkalinemineralsinwindblowndust,orwithgaseousammoniafromsoilsorfromindustrialemissions).
Totheacidifyingeffectofcarbondioxidemustbeaddedthecontributionsfromotheracidicconstituentsoftheatmosphere,particularlyHNO3andH2SO4.Theseacidscanbothbeformednaturally:HNO3derivesfromNOproducedinlightningandforestfires,whileH2SO4derivesfromvolcanoesandfrombiogenicsulfurcompounds.Atthenaturalbackgroundconcentrations,theseacidsrarelyinfluencerainwaterpHsignificantly.
Inpollutedareas,however,theconcentrationsoftheseacidscanbemuchhigherandcanreducethepHofrainwatersubstantiallyoverextendedregions,producingwhatisknownasacidrain.ItisnotuncommoninpollutedareastofindthepHofrainwaterinthepH5–3.5range.Somefogs,whichcanbeincontactwithpollutantsovermanyhours,havehadpHreadingsaslowas2.0,adegreeofacidityequivalenttoa0.01Msolutionofastrongacid.Moreover,theacidraincanfallquitefarfromthesourcesofpollution,duetolong‐rangeatmospherictransport.Inparticular,acidrainisapressingproblemforareasdownwindofcoal‐firedpowerplants,whosetallsmokestacksamelioratelocalpollutionbyloftingSO2andNOemissionshighintotheair.Thus,powerplantsinwesternandcentralEuropeaffecttherainfallingonScandinavia,andpowerplantsintheU.S.MidwesternindustrialbeltsimilarlyaffectrainfallingonnortheasternregionsoftheUnitedStatesandonsoutheasternCanada(Figure12.7),althoughemissionsofSO2havebeenreducedinrecentdecades(seeFigure2.x,p.?).Whileacidrainisaphenomenoninthehydrosphere,itdependsuponatmosphericconditions—theextentofacidicemissionsandtheprevailingweatherpatterns.
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Figure12.7Contourlinesoflong‐term,volume‐weightedaveragesofpHinprecipitationina)Europe,1978‐1982,andb)easternNorthAmerica,1980‐1984.Sources:EMEP/CCC(1984).SummaryReport,Report2/84(Lillestrøm,Norway:NorwegianInstituteforAirResearch);NationalAcidPrecipitationAssessmentProgram(NAPAP)(1987).NAPAPInterimAssessment(Washington,DC:U.S.GovernmentPrintingOffice).
12.S8PolyproticAcidsCarbonicacidactuallyhastwoionizableprotons,buttheydissociate
successively,andwithverydifferentKavalues: H2CO3=H++HCO3–Ka1=10–6.40 [12.17]and HCO3–=H++CO32‐Ka2=10‐10.33 [12.18]
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Becausethebicarbonateionisnegativelycharged,itholdsitsprotonmoretenaciouslythandoescarbonicacid,andtheseconddissociationreactionmakesnosignificantcontributiontotheH+concentrationofacarbonicacidsolutionortothatofaH2CO3/HCO3‐buffer.However,wecanneutralizebothprotonsonH2CO3withastrongbase,orequivalentlystartwithasolutionofacarbonatesalt,andthenthebasicproperties ofCO32‐comeintoplay:CO32‐ +H2O=H C O 3–+OH‐ Kb=Kw/Ka=10‐14/10‐10.33=10‐3.67 [12.19]
Forasolutionwhichis0.1Minacarbonatesalt(justasinWorkedProblem:12.2foracetate)
[OH‐]2=Kb[CO32‐]~10‐3.67x10‐1=10‐4.67 and [OH‐]=10‐2.34,givingpH=11.66,ahighlyalkalinevalue.
Also,abufferhavingequimolarHCO3–andCO32‐has
pH=pKa–log[HCO3‐]/[CO32‐]=10.33However,ifonlyoneofthetwoH2CO3protonsisneutralized,or,equivalently,ifwestartwithasolutionofabicarbonatesalt,thenthingsareslightlymorecomplicated,sinceHCO3‐isbothaconjugatebase[ofH2CO3]andaconjugateacid[ofCO32‐].TotheextentthatHCO3‐dissociatesaproton[viareaction12‐18],theprotonistakenupbyanotherHCO3‐ion[viathereverseofreaction12‐17].Consequentlythemainacid‐basereactioninabicarbonatesolutionis
2HCO3‐=H2CO3+CO32‐ [12.20]
Thisreactionisobtainedbysubtractingreaction[12.17]fromreaction[12.18],andconsequentlytheequilibriumconstantisgivenby
Ka2/Ka1=[H2CO3][CO32‐]/[HCO3‐]2Fromequation[12‐20],[H2CO3]=[CO32‐]forasolutioncontainingonly
bicarbonateinitially.Then
Ka2/Ka1 = [H2CO3]2 /[HCO3-]2 or [HCO3
-]/[H2CO3] = [Ka1/Ka2]1/2
TocalculatethepH,wecaninsertthisratiointotheequilibriumexpression
forreaction[12‐17]:
[H+] = Ka1[HCO3-]/[H2CO3] = Ka1[Ka2/Ka1]1/2 = [Ka1Ka2]1/2
or pH=[pKa2+pKa1]/2=[10.33+6.40]/2=8.36 [12.21]Equation[12‐21]canbegeneralizedtoanydiproticacid:thepHofasolution
ofthemonoproticformisjustthegeometricmeanofthetwopKavalues.Therecanalsobemorethantwoionizableprotons.Forexample,phosphoric
acidhasthree:
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H3PO4=H++H2PO4‐ pKa1=2.17 [12.22] H2PO4‐=H++HPO42‐ pKa2=7.31 [12.23] HPO42‐=H++PO43‐ pKa3=12.36 [12.24]
Asinthecaseofcarbonicacid,thepHofasolutionofH3PO4orofPO43‐canbecalculatedjustasitwouldbeforamonoproticacidorbase,whileforsolutionsofthediproticandmonoproticforms,thepHisthegeometricmeanofthebracketingpKavalues,i.e.forH2PO4‐
pH=[pKa1+pKa2]/2=4.74
while for HPO42-
pH=[pKa2+pKa3]/2=9.83
Thephosphoricacidsystemhasthreebufferzones,correspondingtothe
threepKavalues,i.e.solutionshavingcomparableconcentrationsofH3PO4andH2PO4‐,orofH2PO4‐andHPO42‐,orofHPO42‐andPO43‐.
WorkedProblem12.3:PhosphateProtonationQ:GiventhepKavaluesabove,whatisthedominantformofthephosphate
ionina)alakewithpH=5.0,andb)seawaterwithpH=8.0,andwhatistheratioofthisformtothenextmostabundantform?
A:ThelakepHishigherthanpKa1butlowerthanpKa2,somostofthephosphatewillbepresentasH2PO4‐.FortheseawaterHPO42‐dominates,becausethepHisbetweenpKa2andpKa3.SincethelakepHisclosertopKa2thanpKa1,thenextmostabundantformisHPO42‐,whilefortheseawaterH2PO4‐isthenextmostabundantform,sincethepHisclosertopKa2thantopKa3.Theratioscanbeobtainedfromtheequilibriumexpressionforthesecondionization,[12.23]
[HPO42‐]/[H2PO4‐]=Ka2/[H+]
AtpH=5.0,thisratiois10‐7.31/10‐5.0=10‐2.31=0.0049,whileatpH=8.0,theratiois10‐7.31/10‐8.0=100.69=4.9[or[H2PO4‐]/[HPO42‐]=0.20].
Problems:
1. H2Sboilsat– 61οC,H2Seat–42οC,andH2Teat–2οC.Basedonthistrend,atwhattemperaturewouldoneexpectwatertoboil?Whydoeswaterboilatamuchhighertemperature?
2. ThepHofaqueous0.20Mchlorousacid(HClO2),aweakacid,was
measuredtobe1.1.WhatarethevaluesofKaandpKaofchlorousacid?3. ThepHofaqueous0.20Mpropylamine(C3H7NH2),aweakbase,was
measuredtobe11.96.WhatarethevaluesofKbandpKbofpropylamine?
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4. Calculatetheinitialconcentrationofanaqueoussolutionoftheweak
acidhydrocyanicacid(HCN),withapHof6.4.ThepKaofhydrocyanicacidis9.31.5. Calculatetheinitialconcentrationofanaqueoussolutionoftheweak
basehydrazine,NH2NH2,withapHof10.20.ThepKbofhydrazineis5.77. 6. AsolutionmadeofequalconcentrationsoflacticacidandsodiumlactatewasmeasuredtohaveapH=3.08.a)calculatethepKaandKaoflacticacid.b)CalculatethepHofthesolutioniftherewastwiceasmuchacidassaltinthesolution. 7. CalculatetheratioofmolaritiesofPO42‐andHPO4‐ionsinabuffersolutionwithapHof11.0.
8. IftheconcentrationofatmosphericCO2weretodoublefromitscurrentvalueof370ppm,whatwouldbethecalculatedpHofrainwater(assumingthatCO2weretheonlyacidicinput)?WithrespecttorisingCO2levels,dowehavetobeconcernedaboutenhancedacidityofraininadditiontopotentialclimatewarming?
9.Writethestepwiseprotontransferequilibriaforthedeprotonationofthefollowingchemicalspecies: a)hydrosulfuricacid;H2S
b)arsenicacid;H3AsO4 b)succinicacid;(CH2)2(COOH)2 10.Selenousacid,H2SeO3,isadiproticacidwithpKavaluesof2.46and7.31.EstimatethepHof0.200MNaHSeO3(aq).