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TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityChapter4BufferSolutionTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversity—solutionsthatresistchangeinhydroniumion,H+,andthehydroxideion,OH-,concentration(andconsequentlypH)uponadditionofsmallamountsofacidorbase,orupondilution.4.1.1TheEssentialFeatureofaBufferBufferSolution:4.1CompositionandActionofBufferSolutionsTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversity•Buffersaremixturescontainingacommonion:eitherweakacidsandtheirconjugatebasesorweakbasesandtheirconjugateacid.•Twocommonbuffers:ammonium-ammonia,carbonate-bicarbonate•NH4+(aq)H+(aq)+NH3(aq)CO3-2(aq)+H2O(l)HCO3-1(aq)+OH-1(aq)TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityBuffersystemConjugateacidConjugatebasepKa(at25℃)HAc－NaAcHAcAc-4.76H2CO3－NaHCO3H2CO3HCO3-6.35H3PO4－NaH2PO4H3PO4H2PO4-2.16Tris·HCl－TrisTris·H+Tris7.85H2C8H4O4－KHC8H4O4H2C8H4O4HC8H4O4-2.89NH4Cl－NH3NH4+NH39.25CH3NH3+Cl-－CH3NH2CH3NH3+CH3NH210.63NaH2PO4－Na2HPO4H2PO4-HPO42-7.21BuffersystemsthatareusefulatvariouspHvaluesTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityTris:Tris(Hydroxymethy)methanamin三羟甲基氨基甲烷CCH2OHCH2OHHOH2CNH2TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityABufferedSolutionresistschangeinitspHwheneitherH+orOHareadded.1.0Lof0.50MCH3COOH+0.50MCH3COONapH=4.74Adding0.010molsolidNaOHraisesthepHofthesolutionto4.76,averyminorchange.TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityTheeffectofadditionofacidorbaseto…anunbufferedsolutionabufferedsolutionacidaddedbaseaddedacidaddedbaseaddedTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityBufferwithequalconcentrationsofconjugatebaseandacidOH-H3O+BufferafteradditionofH3O+H2O+CH3COOHH3O++CH3COO-BufferafteradditionofOH-CH3COOH+OH-H2O+CH3COO-4.1.2HowaBufferWorksTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityHAc+H2OH3O++Ac–H++Shiftleft+OH–H2OShiftrightAnti-acidAnti-baseanti-acidmechanismanti-basemechanismTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityTheamountsofweakacidandweakbaseinthebuffermustbesignificantlylargerthantheamountsofH3O+orOH-thatwillbeadded,otherwisethepHcannotremainapproximatelyconstant.Thusadditionoflimitedamountsofastrongacidorbaseiscounteractedbythespeciespresentinthebuffersolution,andthepHchangesverylittle.NosolutioncankeepthepHapproximatelyconstantifyouaddlargeramountsofeitheracidorbasethatarepresentintheoriginalbuffer.TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityForaHB-NaBbuffersystem,HB+H2OH3O++B–NaBNa++B–Ka=[H+][B–][HB][H+]=Ka[HB][B–]Apply–logonbothsidesofaboveequation,pH=pKa+lg[B–][HB]TheHenderson-HasselbalchEquation4.2TheHenderson-HasselbalchEquation—ThepHofbuffersolutionTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversitypH=pKa+lg[B–][HB]=pKa+lg[conjugatebase][conjugateacid]pKa：the–logofKaoftheconjugateacid[B–]、[HB]：equilibriumconcentration[B–]/[HB]：bufferratio[B–]+[HB]：totalconcentration[HB]=cHBcHB(dissociated)[B–]=cNaB+cHB(dissociated)cHBcNaBTheHenderson-HasselbalchEquationTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversitypH=pKa+lgcB–cHB=pKa+lgnB–/VnHB/VpH=pKa+lgnB–nHBIftheconcentrationsofconjugateacidandbaseusedareequal,i.e.cB－=cHB.pH=pKa+lgcB–•VB–cHB•VHB=pKa+lgVB–VHBthreedifferenttypesofofHenderson-HasselbalchEquationFightdilutionHBBVVlgpKapH-TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityCalculatingthepHofaBufferSolution-1PROBLEMSampleProblem4-1Abufferispreparedbymixingequalvolumesof0.1mol·L-1NaAcand0.2mol·L-1HAc.WhatisthepHofthefinalsolution?ThepKaofHAcis4.74.WhatwillthepHbeaftertheadditionof0.005molofNaOH(s)to500mLofthebuffersolutiondescribedabove?TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversitySOLUTIONBeforetheadditionof0.005molofNaOH(s)to500mLofthebuffersolution44.42.01.0lg74.4][][lgHAApKpHaCalculatingthepHofaBufferSolution-1SampleProblem4-1TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversitySOLUTIONAftertheadditionof0.005molofNaOH(s)to500mLofthebuffersolution50.455002.055001.0lg74.4)()()()(lg11mmolmLLmolmmolmLLmolOHnHAnOHnAnpKpHaCalculatingthepHofaBufferSolution-1SampleProblem4-1TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversityCalculatingthepHofaBufferSolution-2PROBLEMSampleProblem4-2CalculatethepHofabufferpreparedbymixingHAwithNaA.ThebuffercontainsHAataconcentrationof0.25mol·L-1.After0.20gNaOHisaddedto100mLofthisbuffer,thepHoftheresultingsolutionis5.60.Theacid-ionizationconstantofHAis5.0×10-6(pKa=5.3).TheCollegeofChemistryandChemicalEngineeringCentralSouthUniversitySOLUTIONHA(aq)+OH-(aq)A-(aq)+H2O(l)Initial(mol)0.0250.005(added)0.1Acchange(mol)-0.005-0.005+0.005Equil-(mol)0.0200.1+0.005AcStep1:calculatetheconcentration()ofconjugatebase,MA,inthebufferbeforetheadditionofNaOH,AcCalculatingthepHofaBufferSolution-2SampleProblem4-26.502.01.0005.0lg3.5][][lgAacHAApKpHTheCollegeofChemistryandChemicalEngineeringCentralSouthUniversitySOLUTION6.502.