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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).