Chapter-16-Acids-and-Bases-Chemistry-The-Molecular-Science第16章酸和碱的化学分子科学(77)课件

Chapter16: AcidsandBasesChemistry:TheMolecularScienceMoore,StanitskiandJursChapter16: AcidsandBasesCheArrheniusDefinitionArrhenius:anysubstancewhichionizesinwatertoproduce…BetterversionoftheArrheniusdefinition:Acid:hydroniumions(H3O+)inwaterareacidic.Base:hydroxideions(OH-)tons(H+)isanacid.

hydroxideions(OH-)isabase.

ArrheniusDefinitionArrhenius:Acid-BaseReactionsAstrongacidisanacidthationizescompletelyinwater(productfavored).Therearesixstrongacids:

HCl-hydrochloricacid HBr–hydrobromicacid HI–hydroiodicacidH2SO4-sulfuricacidHNO3–nitricacidHClO4–perchloricacidAcid-BaseReactionsAstrongacAcid-BaseReactionsStrongAcidsandBasesAstrongbase

isabasethatispresententirelyasions(productfavored).

ThehydroxidesofGroupIA,IIA(exceptBeandMghydroxides)arestrongbases.So,whyisNH3(aq)basic?Acid-BaseReactionsStrongAcidBrønsted-LowryConceptAnalternativedefinition:NH3(g)+H2O(l)NH4+(aq)+OH-(aq)

indirectlyproducedBase:H+acceptorAcid:H+donor

acid =proton(H+)donor

base =proton(H+)acceptorWorksfornon-aqueoussolutionsandexplainswhyNH3isbasic:Brønsted-LowryConceptAnalterWeakacidsandbasesdonotfullyionize(reactantfavored).Brønsted-LowryConceptStrongacidsandbasesalmostcompletelyionize(productfavored).HNO3(aq)+H2O(l) H3O+(aq)+NO3-(aq)Note:theproductsareanewacidandbasepair.LeChatelier–amoredilutesolution(morewater),willionizemore. HF(aq)+H2O(l) H3O+(aq)+F-(aq)Base:H+acceptorAcid:H+donorBase:H+acceptorAcid:H+donor100%3%WeakacidsandbasesdonotfuWater’sRoleasAcidorBaseWateractsasabasewhenanaciddissolvesinwater:HBr(aq)

+H2O(l) H3O+(aq)+Br-(aq)acid base acid baseWaterisamphiprotic

-itcandonateoracceptaproton(actasacidorbase).Butwateractsasanacidforsomebases: H2O(l)+NH3(aq)NH4+(aq)+OH-(aq)

acidbase acidbaseWater’sRoleasAcidorBaseWaPracticeCompletetheequations:(acidorbase?,mass/chargebalance?,singleordoublearrow?)HClO4+H2OCN-+H2OH2S+H2ONO2-+H2OCa(OH)2+H2OClO4-+H3O+HCN+HO-HS-+H3O+HNO2+HO-Ca+2+2OH-PracticeCompletetheequationsConjugateAcid-BasePairsMoleculesorionsrelatedbytheloss/gainofoneH+.ConjugateAcid ConjugateBaseH3O+ H2OCH3COOH CH3COO-NH4+ NH3H2SO4 HSO4-HSO4- SO42-HCl Cl-donateH+acceptH+NH4+andNH2-arenotconjugate(conversionrequires2H+)ConjugateAcid-BasePairsMolec0IdentifythebaseconjugatetoHC2H3O2(aq)andtheacidconjugatetoHCO3-(aq)160of200HCO3-+H2O OH- +HC2H3O2 +H2OH3O++ConjugateAcidBase:H+acceptorConjugateBaseAcid:H+donorConjugateAcid-BasePairsShowthatHCO3-isamphoteric.

C2H3O2-

H2CO30IdentifythebaseconjugatetPractice:ConjugateAcid-BasePairsWritetheequationfortheacidorbaseinwaterandidentifytheacid-basepairs.HBrHSO3-(asanacid)HSO3-(asabase)PH4+Practice:ConjugateAcid-BaseRelativeStrengthofAcids&BasesStrongacidsarebetterH+donorsthanweakacids.StrongbasesarebetterH+acceptorsthanweakbases.Strongacidshaveweakconjugatebases.Weakacidshavestrongconjugatebases.

Strongacid +H2O H3O++weakconjugatebaseFullyionized,reversereactionessentiallydoesnotoccur.Theconjugatebaseisweak.Weaklyionized,reversereactionreadilyoccurs.Theconjugatebaseisstrong.

Weakacid+H2O H3O++strongconjugatebaseRelativeStrengthofAcids&BConjugateacid

ConjugatebaseH2SO4 HSO4-HBr Br-HCl Cl-HNO3 NO3-H3O+ H2OH2SO3 HSO3-HSO4- SO42-H3PO4 H2PO4-HF F-CH3COOH CH3COO-H2S HS-H2PO4- HPO42-NH4+ NH3HCO3- CO32-H2O OH-OH- O2-H2 H-CH4 CH3-AcidstrengthincreasingBasestrengthincreasingstongacidsstrongbaseextremelyweakacidsextremelyweakbasesRelativeStrengthofAcids&BasesConjugateacidConjugatebaHFisastrongeracidthanNH4+.

(NH3isastrongerbasethanF-)HFhasgreatertendencytoionizethanNH4+.(NH3morereadilyacceptsH+thanF-) ReactantFavoredNH4++F- NH3+HFProblemIsthefollowingaqueousreactionproductorreactantfavored?AcidstrengthincreasingBasestrengthincreasingConjacid.

Conj.baseH2SO4 HSO4-HBr Br-HCl Cl-HF F-NH4+ NH3OH- O2-H2 H-CH4 CH3-RelativeStrengthofAcids&BasesHFisastrongeracidthanNH4CarboxylicAcidsC3H7COOHCH3COOHCarboxylicAcidsC3H7COOHCH3COOAmines(bases)CH3NH2 (CH3)2NH

(CH3)3NprotonacceptorsAmines(bases)CH3NH2 (CH3)2NAutoionizationofWaterPurewaterconductsaverysmallelectricalcurrent.Autoionizationoccurs: H2O(l)+H2O(l)H3O+(aq)+OH-(aq)Kw=ionizationconstantforwaterHeavilyreactantfavored.Kw=[H3O+][OH-]

Kw=1.0x10-14(at25°C)BaseAcidAcidBaseAutoionizationofWaterPurewaIonizationConstantforWaterKwisaspecialequilibriumconstantfortheautoionizationofwater.ItisT-dependent.T=25°C(77°F)isusuallyusedasthestandardT. T(°C) Kw 10 0.29x10-14 15 0.45x10-14 20 0.68x10-14

25 1.01x10-14 30 1.47x10-14 50 5.48x10-14Kwincreaseswithtemperature.Isitendoorexothermic?IonizationConstantforWaterKAutoionizationofWaterHydroniumandhydroxideionsareproducedinequalnumbersinpurewater,Thus,theconcentrationsofH3O+andOH-inpurewaterareboth1.0x10-7M.Kw=[H3O+][OH-] 2H2O(l)H3O+(aq)+OH-(aq)AutoionizationofWaterHydroniAutoionizationofWaterH3O+andOH-arepresentinallaqueoussolutions.Neutralsolution. Purewater(@25°C):[H3O+]=10-7M=[OH-]AcidicsolutionIfacidisaddedtowater:[H3O+]isincreased,disturbingtheequilibrium: 2H2OH3O++OH-LeChatelier:equilibriumshiftstotheleft.[OH-]Equilibriumisreestablished:[H3O+]>10-7M>[OH-]Productofionconcentrationsisthesame[H3O+][OH-]=Kw=1.0x10-14

AutoionizationofWaterH3O+anAutoionizationofWaterBasicsolution.Ifbaseisaddedtowater.LeChatelier:equilibriumshiftstotheleft,[H3O+]Newequilibrium:[H3O+]<10-7M<[OH-][H3O+][OH-]=Kw=1.0x10-14at25°CExampleCalculatethehydroniumandhydroxideionconcentrationsat25°Cina6.0Maqueoussodiumhydroxidesolution. Kw=[H3O+][OH-]=1.0x10-14NaOH(aq)isstrong(100%ionized)so[OH-]=6.0M.

[H3O+](6.0)=1.0x10-14

[H3O+]=1.7x10-15M [OH-]=6.0M2H2OH3O++OH-AutoionizationofWaterBasicsAutoionizationofWaterInaneutralsolution,theconcentrationsofH3O+(aq)andOH-(aq)areequal.Inanacidicsolution,theconcentrationofH3O+(aq)isgreaterthanthatofOH-(aq).Inabasicsolution,theconcentrationofOH-(aq)isgreaterthanthatofH3O+(aq).TherelativeconcentrationsofH3O+andOH-indicateacidic,neutralorbasiccharacterofasolution:AutoionizationofWaterInaneAt25oC,weobservethefollowingconditions.AcidorBaseInanacidicsolution,[H3O+]>1.0x10-7M.Inaneutralsolution,[H3O+]=[OH-]=1.0x10-7M.Inabasicsolution,[H3O+]<1.0x10-7M.At25oC,weobservethefolloThepHscaleCandescribeacidityintermsof[H3O+],butalogarithmicscaleismoreconvenient:pH=−log10[H3O+]At25°Caneutralaqueoussolutionhas:

pH=−log10[1.0x10-7]=−(−7.00)=7.00acidicsolutions:[H3O+]>1.0x10-7,pH<7.00basicsolutions:[H3O+]<1.0x10-7,pH>7.00ThepHscaleCandescribeacidiThepHofaSolutionForasolutioninwhichthehydronium-ionconcentrationis1.0x10-3,thepHis:NotethatthenumberofdecimalplacesinthepHequalsthenumberofsignificantfiguresinthehydronium-ionconcentration.ThepHofaSolutionForasoluExample–pHgiven[H3O+]?Asampleoforangejuicehasahydronium-ionconcentrationof2.9x10-4M.WhatisthepH?Example–pHgiven[H3O+]?AsExample-[H3O+]givenpH?ThepHofhumanarterialbloodis7.40(acidicorbasic?).Whatisthehydronium-ionconcentration?(Guesss….>or<1.0x10-7M?)10logx=xExample-[H3O+]givenpH?TheThepOHscalepOH=−log10[OH-]Aneutralsolution(25°C)has: pOH=−log10[1.0x10-7]=−(−7.00)=7.00Since Kw=[H3O+][OH-]=1.0x10-14−log(KW)=−log[H3O+]+(−log[OH-])=−log(1.0x10-14)pKw=pH+pOH=14.00ThepOHscalepOH=−log10[OH-]ExampleAnammoniasolutionhasahydroxide-ionconcentrationof1.9x10-3M.WhatisthepHofthesolution? wefirstcalculatethepOH: ThenthepHis:ExampleAnammoniasolutionhaspHCalculationsSolutionA: pOH=−log[OH-]=3.37 pH+pOH=pKw=14.00 pH=10.63SolutionB: pH=−log[H3O+]=8.12

AhashigherpH,BismoreacidicGiventwoaqueoussolutions(25°C). SolutionA:[OH-]=4.3x10-4M, SolutionB:[H3O+]=7.5x10-9M.WhichhasthehigherpH?Whichismoreacidic?pHCalculationsSolutionA: pOPractice-WSs[H+]Kw=1x10-14=[H+][OH-]

[OH-]pH

14=pH

+pOH

pOHpH=-log[H+][H+]=10-pH

pOH=-log[OH-][OH-]=10-pOH

Practice-WSs[H+]pHofAqueousSolutionsStomachFluidsLemonJuiceVinegarWineTomatoesBlackCoffeeMilkPurewaterBlood,seawaterSodiumbicarbonateBoraxsolutionMilkofMagnesiaDetergentsAqueousammonia1MNaOH[H3O+][OH-]100 10-1410-1 10-1310-2 10-1210-3 10-1110-4 10-1010-5 10-910-6 10-810-7 10-710-8 10-610-9 10-510-10 10-410-11 10-310-12 10-210-13 10-110-14 1ExampleBatteryAcidpH01234567891011121314BleachpHofAqueousSolutionsStomachMeasuringpHH3O+concentrationscanbemeasuredwithan:ElectronicpHmeter:fastandaccuratepreferredmethod.Acid-baseIndicator:substancethatchangescolorwithinanarrowpHrangemayhavemultiplecolorchange(e.g.bromthymolblue)one“color”maybecolorless(e.g.phenolphthalein)cheapandconvenient.MeasuringpHH3O+concentrationHA(aq)+H2O(l)H3O+(aq)+A-(aq)Theacidionizationconstantisusedtoreportthedegreeofionization:[A-][H3O+][HA]

Ka=(Wateromitted,asusual)StrongacidshavelargeKavaluesWeakacidhavesmallKavaluesIonizationConstantsofAcidsandBasesWhenanacidionizesinwater:HA(aq)+H2O(l)H3BaseIonizationConstantsForabaseinwater:Thebaseionizationconstant,Kb,is:

Kb=

[BH+][OH-][B] A-(aq)+H2O(l) HA(aq)+OH-(aq) Kb= [HA][OH-][A-] B(aq)

+H2O(l) BH+(aq)+OH-(aq)Ifthebaseisananion:BaseIonizationConstantsForaLargerKa=strongeracidLargerKb=strongerbaseIonizationConstantsLargerKa=strongeracidLargeLargerKa=strongeracidLargerKb=strongerbaseIonizationConstantsLargerKa=strongeracidLargeAcidandBaseStrengthSmallvalueofKaindicatesthatthenumeratorissmallerthanthedenominator.Equilibriumfavorsreactantsoronlyasmallamountdissociates–aweakacid.SmallvalueofKbindicatesthatthenumeratorissmallerthanthedenominator.Equilibriumfavorsreactantsoronlyasmallamountdissociates–aweakbase.AcidandBaseStrengthSmallvWritetheionizationequationandionizationconstantexpressionforthefollowingacidsandbases: a.)HF b.)HBrO c.)NH3 d.)CH3NH2ExampleWritetheionizationequationKaValuesforPolyproticAcidsSomeacidscandonatemorethanoneH+Formula Name AcidicH’sH2S HydrosulfuricAcid 2H3PO4 PhosphoricAcid 3H2CO3 CarbonicAcid 2HOOC-COOH Oxalicacid 2C3H5(COOH)3 Citricacid 3EachH+ionizationhasadifferentKa.The1stprotoniseasiesttoremoveThe2ndisharder,etc.KaValuesforPolyproticAcidsKaValuesforPolyproticAcidsPhosphoricacid(H3PO4)isweak.Ithasthreeacidicprotons:H3PO4(aq)+H2O(l)H3O+(aq)+H2PO4-

(aq)Ka=7.5x10-3H2PO4-(aq)+H2O(l)H3O+(aq)+HPO42-

(aq)Ka=6.2x10-8HPO42-(aq)+H2O(l)H3O+(aq)+PO43-

(aq)Ka=3.6x10-13KadecreasingH+ishardertoremoveKaValuesforPolyproticAcidsPractice–RelativeStrength(KaandKb)ThefollowingreactionsallhaveKeq>1. NO2-+HF HNO2+F- CH3COO-+HF CH3COOH+F- HNO2+CH3COO HNO2-+CH3COOHArrangethesubstancesbasedontheirrelativebasestrength.

F-

CH3COOH

HF

HNO2-

CH3COO-

HNO21–strongestbase2–intermediatebase3–weakestbase4–notaB-Lbase344214Practice–RelativeStrength(MolecularStructureandAcidStrengthWhatmakesastrongacid?AweakH–Abond,soH+canbeeasilyremoved!smallerbondenergylargeracidstrengthHX BondEnergy(kJ) KaHF weakacid 566 7x10-4HCl 431 1x107HBr 366 1x108HI 299 1x1010strongacidsConsiderthebinaryacids

HF,HCl,HBr,HI.H-A,bondenergydecreasesdownagroup.Acidstrengthincreases.MolecularStructureandAcidSMolecularStructureandAcidStrengthWhatmakesastrongacid?AweakH–Abond,soH+canbeeasilyremoved!Considerthebinaryacids

SiH4,PH3,H2S,HCl.Thehigherelectronegativity,acrossaperiod,makesthebondmorepolarandhencemoreionic–astrongeracid.MolecularStructureandAcidSOxoacidshydrogen,oxygenandoneotherelementH

O-EhighertheelectronegativityonE,strongertheacidasthisweakensthebondbetweentheOandHMolecularStructureandAcidStrengthHOCl HOBr HOIElectronegativity:Cl=3.0,Br=2.8,I=2.5Ka:3.5×10-8 2.5×10-9 2.3×10-11OxoacidsMolecularStructureanTheacidstrengthincreasesasthenumberofoxygenatomsbondedtoEincreases.StrongoxoacidhasatleasttwooxygenatomsperacidicHatoms.MolecularStructureandAcidStrengthHClO HClO2HClO3

*HClO4Ka:3.5×10-8

1.1×10-2

≈103≈108TheacidstrengthincreasesasProblemSolvingUsingKaandKbExample:KafrompHLacticacidismonoprotic.ThepHofa0.100Msolutionwas2.43at25°C.DetermineKaforthisacid.HA(aq)+H2O(l)H3O+(aq)+A-(aq)UsingthepHinformation:-log[H3O+]=2.43 [H3O+]=10-2.43=0.0037MThesehydroniumionsareproducedby:H2O(l)+H2O(l)H3O+(aq)+OH-(aq)andfromtheautoionizationofwater:ProblemSolvingUsingKaandK

Ka

= =1.4x10-4(0.0037)(0.0037)(0.100–0.0037)HA(aq)+H2O(l)H3O+(aq)+A-(aq)[]initial 0.100 1.0x10-70 0 (waterautoionization)

change -0.0037 +0.0037 +0.0037[]equil 0.100–0.0037 0.0037 0.0037Ka=[H3O+][A-][HA]ProblemSolvingUsingKaandKb Ka= =1.4x10-ProblemSolvingUsingKaandKbExample:pHfromKaDeterminethepHofa0.100Mpropanoicacidsolutionat25°C.Ka=1.4x10-5.What%ofacidisionized?C2H5COOH+H2O(l)H3O+(aq)+C2H5COO-(aq)

Ka= =1.4x10-5[H3O+][C2H5COO-][C2H5COOH]ProblemSolvingUsingKaandKHA(aq)+H2O(l)H3O+(aq)+A-(aq)[]initial 0.100 0* 0

change -x +x +x[]equil 0.100–x x* x*Ignorethewatercontribution

1.4x10-5= x2(0.100–x)[H3O+][A-][HA]Ka=1.4x10-5=ProblemSolvingUsingKaandKbDeterminethepHof0.001Mpropanoicacidsoln.at25°C.Ka=1.4x10-5.What%ofacidisionized?HA(aq)+H2O(l)H3ProblemSolvingUsingKaandKbBecauseKaissmall,assumex<[HA]Ka=1.4x10-5= ≈x2(0.100–x)x20.100x=(1.4x10-5)(0.100)=0.00118M(0.00118<<0.100M;theapproximationisgood)pH=-log(0.00118)=2.93%-ionized= x100=1.18%0.001180.100<5%rulecanassume[HA]eq~[HA]0ProblemSolvingUsingKaandKWhatisthepHofa0.20Msolutionofpyridine,C5H5N,inaqueoussolution?TheKbforpyridineis1.4x10-9.Example3–calculatingpHfromknownKbInitial0.20

00Change-x

+x+xEquilibrium0.20-x

xxWhatisthepHofa0.20MsolExample3–calculatingpHfromknownKbWhatisthepHofa0.20Msolutionofpyridine,C5H5N,inaqueoussolution?TheKbforpyridineis1.4x10-9.Example3–calculatingpHfroExample3–calculatingpHfromknownKbWhatisthepHofa0.20Msolutionofpyridine,C5H5N,inaqueoussolution?TheKbforpyridineis1.4x10-9.SolvingforxwegetExample3–calculatingpHfroExample3–calculatingpHfromknownKbWhatisthepHofa0.20Msolutionofpyridine,C5H5N,inaqueoussolution?TheKbforpyridineis1.4x10-9.SolvingforpOHExample3–calculatingpHfroPracticeKafrompHPara-aminobenzoicacid(PABA),HC7H6NO2,isusedinsomesunscreenagents.Asolutionismadebydissolving0.263molofPABAinenoughwatertomake750.0mLofsolution.Thesolution’spH=2.59.WhatistheKaforPABA?(1.9x10-5)pHfromKa

Barbituricacid(Ka=1.1x10-4)isusedinthemanufactureofsomesedatives.WhatisthepHfora0.673Msolutionofbarbituricacid?(2.07)PracticeKafrompHRelationshipbetweenKaandKbvaluesForanacid-baseconjugatepair:HAandA-[HA][OH-][A-]KaxKb

=[H3O+][A-][HA]=[H3O+][OH-]=KwKaKbWhentworeactionsareadded,theirequilibriumconstantsaremultiplied(chapter14).RelationshipbetweenKaandKbRelationshipbetweenKaandKbvaluesPhenol,C6H5OH,isaweakacid,Ka=1.3x10-10at25°C.CalculateKbforthephenolateionC6H5O-

KaxKb=1.0x10-14Kb= =7.7x10-51.0x10-141.3x10-10RelationshipbetweenKaandKbAcidBaseNeutralizationReactionsSalt=ioniccompoundformedinacid+basereactionStrongacid+strongbase

formneutralsalts.HCl(aq)+NaOH(aq) NaCl(aq)+H2O(l)H+(aq)+Cl-(aq)+Na+(aq)+OH-(aq) Na+(aq)+Cl-(aq)+H2O(l)HX(aq)+MOH(aq) MX(aq)+H2O(l)acid base saltNetionicequation:H+(aq)+OH-(aq) H2O(l) H3O+(aq)+OH-(aq) 2H2O(l)AcidBaseNeutralizationReactH3O+(aq)+OH-(aq) H2O(l)+H2O(l)

Keq=1/Kw=1x1014strongbasestrongacidweakbaseweakacidNa+andCl-arespectatorions.Na+isthev.weakconjugateacidofav.strongbase.Cl-isthev.weakconjugatebaseofav.strongacid.FinalsolutionhaspH=7.SaltsofStrongBasesandStrongAcidsH3O+(aq)+Cl-(aq)+Na+(aq)+OH-(aq) Na+(aq)+Cl-(aq)+2H2O(l)100%H3O+(aq)+OH-(aq) H2O(SaltsofStrongBasesandWeakAcidsWeakacid+strongbase

formbasicsalts.CH3COOH(aq)+NaOH(aq) CH3COONa(aq)+H2O(l)Netionic:CH3COOH(aq)+OH-(aq) CH3COO-+H2O(l)CH3COOH(aq)+Na+(aq)+OH-(aq) Na+(aq)+CH3COO-(aq)+H2O100%finalsolutionSaltsofStrongBasesandWeakacetateisaweakbase(muchstrongerthanwater)solutionisbasic(pH>7.0).CH3COO-(aq)+H2O(l) CH3COOH(aq)+OH-(aq) CalculatepHofsolutionfromKbofCH3COO-.Ahydrolysisreaction–waterisbrokenapart.CH3COOH(aq)+OH-(aq) CH3COO-(aq)+H2O(l)baseacidbaseacidSaltsofStrongBasesandWeakAcids100%acetateisaweakbase(muchspHofaSaltSolutionForaweakacid+strongbase,pHdependsonKb LargerKb=strongerbaseExampleWhatisthepHofa1.50MaqueoussolutionofNa2CO3?Kb(CO32-)=2.1x10-4.(orequalvolumesof1.5MH2CO3and3.0MNaOHaremixed,whatisthepHofthemixture?)pHofaSaltSolutionForaweaNa+istheconjugateacidofNaOH(strongbase).Itremains100%ionizedNa+hasnoeffectonpH.CO32-istheconjugatebaseofHCO3-(weakacid)MostCO32-ionsundergohydrolysisGeneratesHCO3-andOH-ChangesthepHCO32-(aq)+H2O(l)HCO3-(aq)+OH-(aq)WhatisthepHofa1.50MaqueoussolutionofNa2CO3?Kb=2.1x10-4

pHofaSaltSolutionNa+istheconjugateacidofNCO32-(aq)+H2O(l)HCO3-(aq)+OH-(aq)[]initial 1.50 0 0

change -x +x +x[]equil 1.50-x x xKb=2.1x10-4= = ≈[HCO3-][OH-][CO32-]x2(1.50–x)x21.50x=1.77x10-2 pOH=−log(1.77x10-2)=1.75 pH=14.00-1.75=12.25WhatisthepHofa1.50MaqueoussolutionofNa2CO3?Kb=2.1x10-4

pHofaSaltSolutionCO32-(aq)+H2O(l)HSaltsofWeakBasesandStrongAcidsNH3(aq)+HCl(aq) NH4+(aq)+Cl-(aq)Theproductsoftheneutralizationreactionformanacidicsolution: NH4+(aq)+H2O(l) NH3(aq)+H3O+(aq)NH3(aq)+H3O+(aq)+Cl-(aq) NH4+(aq)+Cl-(aq)+H2O(l)Netionic:NH3(aq)+H3O+(aq) NH4+(aq)+H2O(l)ThefinalpHdependsontheKa,largerKa=moreacidicSaltsofWeakBasesandStrongpHofaSaltSolutionFindthepHofa0.132MNH4Br.Isthissolutionacidic,basicorneutral?Ka(NH4+)=5.6x10-10.ThesolutioncontainsNH4+ionsandBr-ions.NH4+istheconjugateacidofNH3(aweakbase).ItwillpartiallyionizeinsolutiongivingNH3andH3O+Formsanacidicsolution.Br-istheconjugatebaseofHBr

(astrongacid)Br-ionsstayfullyionizedDoesnotchangethepH.pHofaSaltSolutionFindtheNH4+(aq)+H2O(l)NH3(aq)+H3O+(aq)[]initial 0.132 0 0

change -x +x +x[]equil 0.132-x x x

x=8.57x10-6 pH=−log(8.57x10-6)=5.07

Ka=5.6x10-10= = ≈[NH3][H3O+][NH4+]x2(0.132–x)x20.132pHofaSaltSolutionpHofa0.132Maq.solutionofNH4Br?Isthisacidic,basicorneutral?

Ka(NH4+)=5.6x10-10.Acidic!NH4+(aq)+H2O(l)NSaltsofWeakBasesandWeakAcidsThemostdifficult.Considerqualitativeresults.e.g.F-(aq)+H2O(l) HF(aq)+OH-(aq)NH3(aq)+HF(aq) NH4+(aq)+F-(aq) NH4+(aq)+H2O(l) NH3(aq)+H3O+(aq)Ka(NH4+)=5.6x10-10;Kb(F-)=1.4x10-11Ka(weakacid)>Kb(weakbase) Theammoniumreactionismorefavorable. Theacidwins!Thesolutionwillbeacidic!SaltsofWeakBasesandWeakAAcids,BasesandSaltsStrongacid

+

strongbase

→saltsolution,

pH=7Strongacid

+

weakbase

→saltsolution,

pH<7Weakacid

+

strongbase

→saltsolution,

pH>7Weakacid

+

weakbase

→saltsolution,

pH=?

NeedK’s “strongestwins”Acids,BasesandSaltsStrongaNeutralIonsinWater(Table16.5)Anions Cl- NO3- Br- ClO4- I-

Cations Li+ Ca+2 Na+ Sr+2 K+ Ba+2NeutralIonsinWater(Table1WhatisthepHofa0.10MNaCNsolutionat25oC?TheKbforCN-is2.5x10-5.ExampleSodiumcyanidegivesNa+ionsandCN-ionsinsolution.OnlytheCN-ionhydrolyzes.pH=11.2WhatisthepHofa0.10MNaCWhatisthepHofa0.15MNH4NO3solutionat25oC?TheKaforNH4+is5.6x10-10.ExampleNH4NO3givesNH4+ionsandNO3-ionsinsolution.OnlytheNH4+ionhydrolyzes.WhatisthepHofa0.15MNH4Thecommon-ioneffectistheshiftinanionicequilibriumcausedbytheadditionofasolutethatprovidesanioncommontotheequilibrium.TheCommonIonEffectConsiderasolutionofaceticacid(HC2H3O2),inwhichyouhavethefollowingequilibrium.Thecommon-ioneffectisthesTheCommonIonEffectTheequilibriumcompositionwouldshifttotheleft(LeChatelierPrinciple)andthedegreeofionizationoftheaceticacidwoulddecrease.IfweweretoaddNaC2H3O2tothissolution,itwouldprovideC2H3O2-ionswhicharepresentontherightsideoftheequilibrium.Thisrepressionoftheionizationofaceticacidbysodiumacetateisanexampleofthecommon-ioneffect.TheCommonIonEffectTheequilWhatisthepHofasolutionthatis0.025Minformicacid?TheKaforformicacidis1.7x10-4.

WhatisthepHofthesamesolutionafter0.018Minsodiumformate,NaCH2Oisadded?ExampleInitialChangeEquilibrium0.025 0 -x +x +x0.025-x x 0.018+xFromNaCH2O0.018WhatisthepHofasolutiontInitial0.025

00.018Change-x

+x+xEquilibrium0.025-x

x0.018+xExamplex[H3O+]=2.4×10-4

pH=-log(H3O+)=-log(2.4x10-4)=3.62Initial0.02500.018Change-x+xChapter16: AcidsandBasesChemistry:TheMolecularScienceMoore,StanitskiandJursChapter16: AcidsandBasesCheArrheniusDefinitionArrhenius:anysubstancewhichionizesinwatertoproduce…BetterversionoftheArrheniusdefinition:Acid:hydroniumions(H3O+