Acids and Bases

AcidsandBasesCopyright©TheMcGraw-HillCompanies,Inc.Permissionrequiredforreproductionordisplay.ArrheniusacidisasubstancethatproducesH+(H3O+)inwaterArrheniusbaseisasubstancethatproducesOH-inwater4.3Ifitcanbebothacidandbasethen……itisamphiprotic.HCO3HSO4H2O©2012PearsonEducation,Inc.ABrønstedacidisaprotondonorABrønstedbaseisaprotonacceptoracidbaseacidbase15.1acidconjugatebasebaseconjugateacidConjugateAcidsandBases•ThetermconjugatecomesfromtheLatinword“conjugare,”meaning“tojointogether.”•Reactionsbetweenacidsandbasesalwaysyieldtheirconjugatebasesandacids.©2012PearsonEducation,Inc.AcidandBaseStrength•Strongacidsarecompletelydissociatedinwater.–Theirconjugatebasesarequiteweak.•Weakacidsonlydissociatepartiallyinwater.–Theirconjugatebasesareweakbases.©2012PearsonEducation,Inc.OHH+OHHOHHHOH-+[]+Acid-BasePropertiesofWaterH2O(l)H+(aq)+OH-(aq)H2O+H2OH3O++OH-acidconjugatebasebaseconjugateacid15.2autoionizationofwaterH2O(l)H+(aq)+OH-(aq)TheIonProductofWaterKc=[H+][OH-][H2O][H2O]=constantKc[H2O]=Kw=[H+][OH-]Theion-productconstant(Kw)istheproductofthemolarconcentrationsofH+andOH-ionsataparticulartemperature.At250CKw=[H+][OH-]=1.0x10-14[H+]=[OH-][H+][OH-][H+][OH-]SolutionIsneutralacidicbasic15.2WhatistheconcentrationofOH-ionsinaHClsolutionwhosehydrogenionconcentrationis1.3M?Kw=[H+][OH-]=1.0x10-14[H+]=1.3M[OH-]=Kw[H+]1x10-141.3==7.7x10-15M15.2pH–AMeasureofAciditypH=-log[H+][H+]=[OH-][H+][OH-][H+][OH-]SolutionIsneutralacidicbasic[H+]=1x10-7[H+]1x10-7[H+]1x10-7pH=7pH7pH7At250CpH[H+]15.315.3pOH=-log[OH-][H+][OH-]=Kw=1.0x10-14-log[H+]–log[OH-]=14.00pH+pOH=14.00ThepHofrainwatercollectedinacertainregionofthenortheasternUnitedStatesonaparticulardaywas4.82.WhatistheH+ionconcentrationoftherainwater?pH=-log[H+][H+]=10-pH=10-4.82=1.5x10-5MTheOH-ionconcentrationofabloodsampleis2.5x10-7M.WhatisthepHoftheblood?pH+pOH=14.00pOH=-log[OH-]=-log(2.5x10-7)=6.60pH=14.00–pOH=14.00–6.60=7.4015.3StrongElectrolyte–100%dissociationNaCl(s)Na+(aq)+Cl-(aq)H2OWeakElectrolyte–notcompletelydissociatedCH3COOHCH3COO-(aq)+H+(aq)StrongAcidsarestrongelectrolytesHCl(aq)+H2O(l)H3O+(aq)+Cl-(aq)HNO3(aq)+H2O(l)H3O+(aq)+NO3-(aq)HClO4(aq)+H2O(l)H3O+(aq)+ClO4-(aq)H2SO4(aq)+H2O(l)H3O+(aq)+HSO4-(aq)15.4HF(aq)+H2O(l)H3O+(aq)+F-(aq)WeakAcidsareweakelectrolytesHNO2(aq)+H2O(l)H3O+(aq)+NO2-(aq)HSO4-(aq)+H2O(l)H3O+(aq)+SO42-(aq)H2O(l)+H2O(l)H3O+(aq)+OH-(aq)StrongBasesarestrongelectrolytesNaOH(s)Na+(aq)+OH-(aq)H2OKOH(s)K+(aq)+OH-(aq)H2OBa(OH)2(s)Ba2+(aq)+2OH-(aq)H2O15.4F-(aq)+H2O(l)OH-(aq)+HF(aq)WeakBasesareweakelectrolytesNO2-(aq)+H2O(l)OH-(aq)+HNO2(aq)15.4StrongAcidWeakAcid15.4WhatisthepHofa2x10-3MHNO3solution?HNO3isastrongacid–100%dissociation.HNO3(aq)+H2O(l)H3O+(aq)+NO3-(aq)pH=-log[H+]=-log[H3O+]=-log(0.002)=2.7StartEnd0.002M0.002M0.002M0.0M0.0M0.0MWhatisthepHofa1.8x10-2MBa(OH)2solution?Ba(OH)2isastrongbase–100%dissociation.Ba(OH)2(s)Ba2+(aq)+2OH-(aq)StartEnd0.018M0.018M0.036M0.0M0.0M0.0MpH=14.00–pOH=14.00+log(0.036)=12.615.4HA(aq)+H2O(l)H3O+(aq)+A-(aq)WeakAcids(HA)andAcidIonizationConstantsHA(aq)H+(aq)+A-(aq)Ka=[H+][A-][HA]KaistheacidionizationconstantKaweakacidstrength15.515.5WhatisthepHofa0.5MHFsolution(at250C)?HF(aq)H+(aq)+F-(aq)Ka=[H+][F-][HF]=7.1x10-4HF(aq)H+(aq)+F-(aq)Initial(M)Change(M)Equilibrium(M)0.500.00-x+x0.50-x0.00+xxxKa=x20.50-x=7.1x10-4Kax20.50=7.1x10-40.50–x0.50Ka1x2=3.55x10-4x=0.019M[H+]=[F-]=0.019MpH=-log[H+]=1.72[HF]=0.50–x=0.48M15.5WhencanIusetheapproximation?0.50–x0.50Ka1Whenxislessthan5%ofthevaluefromwhichitissubtracted.x=0.0190.019M0.50Mx100%=3.8%Lessthan5%Approximationok.WhatisthepHofa0.05MHFsolution(at250C)?Kax20.05=7.1x10-4x=0.006M0.006M0.05Mx100%=12%Morethan5%Approximationnotok.Mustsolveforxexactlyusingquadraticequationormethodofsuccessiveapproximation.15.5Solvingweakacidionizationproblems:1.IdentifythemajorspeciesthatcanaffectthepH.•Inmostcases,youcanignoretheautoionizationofwater.•Ignore[OH-]becauseitisdeterminedby[H+].2.UseICEtoexpresstheequilibriumconcentrationsintermsofsingleunknownx.3.WriteKaintermsofequilibriumconcentrations.Solveforxbytheapproximationmethod.Ifapproximationisnotvalid,solveforxexactly.4.Calculateconcentrationsofallspeciesand/orpHofthesolution.15.5WhatisthepHofa0.122MmonoproticacidwhoseKais5.7x10-4?HA(aq)H+(aq)+A-(aq)Initial(M)Change(M)Equilibrium(M)0.1220.00-x+x0.122-x0.00+xxxKa=x20.122-x=5.7x10-4Kax20.122=5.7x10-40.122–x0.122Ka1x2=6.95x10-5x=0.0083M0.0083M0.122Mx100%=6.8%Morethan5%Approximationnotok.15.5Ka=x20.122-x=5.7x10-4x2+0.00057x–6.95x10-5=0ax2+bx+c=0-b±b2–4ac2ax=x=0.0081x=-0.0081HA(aq)H+(aq)+A-(aq)Initial(M)Change(M)Equilibrium(M)0.1220.00-x+x0.122-x0.00+xxx[H+]=x=0.0081MpH=-log[H+]=2.0915.5percentionization=IonizedacidconcentrationatequilibriumInitialconcentrationofacidx100%ForamonoproticacidHAPercentionization=[H+][HA]0x100%[HA]0=initialconcentration15.5NH3(aq)+H2O(l)NH4+(aq)+OH-(aq)WeakBasesandBaseIonizationConstantsKb=[NH4+][OH-][NH3]KbisthebaseionizationconstantKbweakbasestrength15.6Solveweakbaseproblemslikeweakacidsexceptsolvefor[OH-]insteadof[H+].15.615.7IonizationConstantsofConjugateAci