《无机化学简明教程》课件-Chapter 4 Acid-Base Equilibr

1Chapter4Acid-BaseEquilibriaThreeissues：溶液理论Acid-BaseTheoryAcid-BaseEquilibria21.Solutions

1.1Concentrationunits

(1)

Molarity(M)

(2)Molality(m)

3(3)Molefraction(X)XA+XB+…=1

4

1-2Colligativepropertiesofnonelectrolytesolutions

Thosesolutionpropertieswhichdependprimarilyontheconcentrationofsoluteparticlesratherthantheirnaturearecalledcolligativeproperties(依数性).Fourphysicalpropertiesofsolutionsarethesameforallnonvolatilesolutes:vaporpressureloweringboilingpointelevationfreezingpointdepressionosmoticpressure

56(1)Vaporpressurelowering

“在一定温度下，稀溶液的蒸气压等与纯溶液的蒸气压与溶剂摩尔数的乘积。”P=PA0XA=PA0(1-XB)7(2)Boilingpointelevation

ΔTb=T溶液-T溶剂

=Kb

mm:溶液的质量摩尔浓度单位：mol·kg-1,Kb:溶剂的沸点上升常数,单位K·kg·mol-1,是溶质的质量摩尔浓度为1mol·kg-1

时所引起的溶液沸点上升的数值。8(3)FreezingpointdepressionΔTf=Tf溶剂-Tf溶液=Kf

mKf：溶剂的摩尔凝固点下降常数，单位K·kg·mol-1,即m=1mol·kg-1时所引起的溶液凝固点的下降值。例如：Kf（H2O）=1.68K·kg·mol-19例：汽车的散热冷却水，冬季加入乙二醇(或甲醇⁄甘油），可防止结冻，冻裂水箱。求：在1kg水中，加入甘油989g，可使溶液的凝固点下降多少度？（甘油的分子量：MC3H8O3=92g/mol）解：ΔTf=Kf

mKf（H2O）=1.86K·kg·mol-1

m=(989

92)⁄1kg=10.75mol·kg-1

ΔTf=1.86

10.75=20(K)ΔTf=273-Tf溶液Tf溶液=253K10（4）Osmoticpressure

111213Л=nRT⁄V=CRTP=nRT⁄V141-3强电解质溶液互吸理论

问题：电解质溶液“依数性”反常解释：完全电离，且离子互吸

15强电解质溶液的互吸理论要点：（1）强电解质在水中完全电离，电离度а=100%.（2）（只考虑正负离子静电作用，不考虑离子的水合，离子溶剂作用）形成“离子氛”每个离子周围异号离子多于同号离子，净结果是在每个离子周围形成球形对称的“离子氛”，离子氛的电荷在数值上等于中心离子的电荷，符号相反。16

（3）活度：溶液中实际发挥作用的离子浓度，叫有效浓度，或称活度。

（4）“活度”的根源在于离子间作用力（m、Z越大，作用力越大）用离子强度来衡量

单位：mol/kg17（5）25℃,稀溶液（I

0.01m）1923年，德拜-休克尔的极限公式(Debye–Hückel’slimitinglaw)

18例1：0.1mol·kg-1盐酸和0.1molkg-1CaCl2溶液等体积混和，计算该溶液的I及aH+。解：mH+=0.05mol·kg-1mca2+=0.05mol·kg-1mcl-=0.15mol·kg-1∴I=1／2∑mizi2=1／2(0.05×12+0.05×22+0.15×12)=0.2mol·kg-1㏒f±(HCl)=-0.509∣1×1∣0.2.1／2=-0.2276f±=0.519=f+=f-∴aH+=f+mH+／m0=0.59×0.05=0.0295，误差较大。

㏒f±(HCl)=-0.509∣1×1∣0.2.1／2/（1+0.2.1／2）=-0.1573f±=0.70∴aH+=0.70×0.05=0.03519例2：I=1.0×10-4mol·kg-1,计算NaCl溶液，MgSO4溶液的f±。

解：一价离子NaCl㏒f±=-0.059∣1×1∣(1.0×10–4)1／2=-0.00509,f±=0.99二价离子MgSO4㏒f±=-0.059∣2×2∣(1.0×10–4)1／2=-0.02036,f±=0.95三价离子

㏒f±=-0.059∣3×3∣(1.0×10–4)1／2=-0.0458,f±=0.90四价离子

㏒f±=-0.059∣4×4∣(1.0×10–4)1／2=-0.0814,f±=0.83结论：I一定时，Z越小，f越大

1.20例3：分别计算浓度为

1.0×10-2mol·kg-1

，1.0×10-3mol·kg-1和1.0×10-4mol·kg-1NaCl溶液中Na+,Cl-的活度系数及活度。

解：I=Σ0.5miZi2=0.5×0.01×12+0.5×0.01×12=0.01mol·kg-1

logf±=-0.509∣1×1∣×0.01½=-0.0509∴f±=0.89=f+=f-

aNa+=aCl-=0.89×0.01=0.0089同样可得，m=1.0×10-3mol·kg-1

即I=1.0×10-3mol·kg-1时，f±=0.96m=1.0×10-4mol·kg-1时，I=1.0×10-4mol·kg-1，f±=0.99结果表明，m越稀，I越小，

(I=1.0×10-4mol·kg-1)，f

1.212Acid-BaseTheory

2-1ArrheniusDefinitionofAcidsandBasesAcidsarecompoundsthatalltheircationsinaqueoussolutionsareprotons(H+).Eg.HCl,HAc.Basesarecompoundsthatalltheiranionsinaqueoussolutionsarehydroxides(OH-).Eg.NaOH,KOH22Theacid/basestrengthcanbescaledbyionizationextent(α).Ifαofanacidequalsto100%,itiscalledstrongacid,itsionizationiscompletedanditiscalledstrongacid.Ifαofanacidismuchlessthan100%,itsionizationispartialanditiscalledweakacid.

23Theessenceofacidsandbasesreactionistheneutralizationreaction:H++OH-=H2O△rHmº=

-55.84KJ/mol

24Theheatofmixingdilutesaltsolutionsiszero.

Itcaneasilyexplainthattheheatofneutralizationreactionsofallstrongacidsandstrongbasesisequalto-55.84KJ/molandtheheatofneutralizationreactionsofallweakacidsandweakbasesislessthan55.84KJ/mol25Insum,itmaketheaqueoussolutionchemistrysystematicalandtheoretical.However,itlimitedtheconceptofacidsandbasesinaqueoussolutionsanditcannotexplainthesamekindofproblemsinnon-aqueoussolutionornon-solventsystem.262-2Bronsted-LowryDefinitionofAcidsandBasesAcidsarespeciesthatdonateaproton(H+).Basesarespeciesthatacceptaproton.Accordingtothisdefinition,acidandbasehaveconjugatedbaseandacid.Theyareconjugatedacid-basepair.Acid=Base+H+HCl=Cl-+H+NH4+=NH3+H+27CommentsAcid/Basecanbeamolecule,acationandananion.Acidandbasecanbetransformed.Acompoundthatcanactaseitheranacidorabaseiscalledamphiprotic.

Thereisnoconceptofsalts.28TheBronsted-Lowrydefinitionofacidsandbasesdoesnotencompassallchemicalcompoundsthatexhibitacidicandbasicproperties.292-3LewisAcidsandBasesALewisacidisanelectron-pairacceptor.ALewisbaseisanelectron-pairdonor.

30Lewisacid+Lewisbase=Acid-basecomplexEg.H++OH-=H2OH++NH3=NH4+Ag++2NH3=Ag(NH3)2+Cu+4NH3=Cu(NH3)42+BF3+F-=BF4-Al3++6H2O=Al(H2O)63+Ag++Cl-=AgClHCl+NH3=NH4Cl313.Acid-BaseEquilibria——TheProtonTransferinAqueousSolutionKeyissue：Calculationof[H+]323-1Theionproductforwater----KwH2O

H++OH-

rHm=55.84KJ/molKw=[H+][OH-]T

,Kw

33

Kwatdifferenttemperature

Temperature,

CKw01.153

10-15206.87

10-15221.0

10-14251.012

10-14505.31

10-1410054.95

10-14

34acidicsolution:[H+]>[OH-]neutralsolution:[H+]=[OH-]basicsolution:[H+]<[OH-]

ThepHscale:pH=-log[H+]

353-2Weak-acid/baseEquilibria

(1)TheWeak-acid/baseHAc+H2O

H3O++Ac-

rHm=-0.46KJ/molHAc

H++Ac-t0c000tec0-xxxKa=x2/(c0-x)36Ifc0/Ka>400,c0–x

c0,[H+]=

Kac0

=

Ka/c0

Astheinitialacidconcentrationdecreases,thepercentdissociationoftheacidincreases.(DilutionLaw)HAc+H2O

H3O++Ac-Asforweakbase,[OH-]==

Kbc0

37SummaryKa/Kbstandsforthedissociationextentofacid/base,Ka

,thestrengthoftheacid.Sincethe

rHmissmall,Ka/Kbchangeverylittlewiththetemperature.Themainfactorthataffectacid-baseequilibriaisconcentration.CommonIonEffectSaltEffect38Solution:[H+]=1.3

10-3M,

=1.3%[H+]=7.16

10-6M,

=71.6%(needtobecalculatedexactly)[H+]=9.0

10-6M,

=0.009%[H+]=1.9

10-3M,

=1.9%Example1:Calculatethe[H+]and

ofthefollowingsolutions:(a)0.1MHAc(b)1.0

10-5MHAc(c)AddNaAc(s)to0.1MHAc,keeping[NaAc]=0.20M(d)AddNaCl(s)to0.1MHAc,keeping[NaCl]=0.20M39(2)TheDissociationofPolyproticAcidsAcidswithmorethanoneionizableprotonarepolyproticacids.Diproticacid:H2S,H2SO3,H2CO3,H2C2O4Triproticacid:H3PO440H2S

HS-+H+Ka1=1.3

10-7HS-

S2-+H+Ka2=7.1

10-15Ka1>Ka2,Successiveaciddissociationconstantstypicallydifferbyseveralordersofmagnitude.ThisfactsimplifiespHcalculationsinvolvingpolyproticacids，

becausewecanusuallyneglecttheH+comingfromthesubsequentdissociations.41Solution:H2S

HS-+H+t00.100te0.1-x

0.1xxx=[H+]=[HS-]=

Kac0=1.14

10-4MHS-

S2-+H+t01.14

10-401.14

10-4te1.14

10-4-yy1.14

10-4+y

1.14

10-4

1.14

10-4

Ka2=y=7.1

10-15MH2O

H++OH-x+y+[OH-]

x[OH-][OH-]=Kw/x=8.8

10-11MExample2:CalculatethesaturatedH2Saqueoussolution(0.1M).42Conclusion:[H+]=

Ka1c0

433-3.HydrolysisofSaltsWhensaltsdissolveinwater,thepHofthesolutionisaffected.NaCl,NaAc,NH4Cl,NH4Ac,NaHCO344(1)NaAcAc-+H2O

HAc+OH-t0c000tec0-xxxKh=Kb=[HAc][OH-][H+]/[Ac-][H+]=Kw/Ka=5.6

10-10

x=[OH-]=

Khc0=

Kwc0/Kah=[OH-]/c0=

Kw/Kac045Solution:(a)Ac-+H2O

HAc+OH-t00.0100te0.01-x

0.01xxx=[OH-]=

Khc0=

Kwc0/Ka

=2.37

10-6M[H+]=4.23

10-9M,pH=8.37h=[OH-]/c0=0.0237%Example3:CalculatethepHandhofthefollowingsaltsolutions:0.010MNaAc(Ka=1.8

10-5)0.010MNaCN(Ka=4.93

10-10)pH=10.65,h=4.5%结论：Ka越小（酸越弱），[OH-]和h越大。46NaAc[OH-]=

Khc0=

Kwc0/Ka

Na2CO3[OH-]=

Kh1c0=

Kwc0/Ka2

NH4Cl[H+]=

Khc0=

Kwc0/Kb

NH4Ac[H+]=

KwKa/Kb

47(2)NaHCO3NaHCO3(c0)

Na+(c0)+HCO3-(

c0)Twotrends:HCO3-

H++CO32-Ka2=5.6

10-11HCO3-+H2O

OH-+H2CO3Kh2=Kw/Ka1=2.4

10-8[OH-]=[H2CO3]–[CO32-]Kw/[H+]=[H+][HCO3-]/Ka1-Ka2[HCO3-]/[H+][H+]2[HCO3-]/Ka1=Kw+Ka2[HCO3-][H+]2=Ka1(Kw+Ka2[HCO3-])/[HCO3-]

[H+]=

Ka1(Kw+Ka2c0)/c0IfKa2c0>>Kw,[H+]=

Ka1Ka2

48

Solution:(a)[H+]=

Ka1Ka2(b)[H+]=

Ka1Ka2(c)[H+]=

Ka2Ka3=1.66

10-10pH=9.78Example4:CalculatethepHof0.1Msolutions:(a)

NaHCO3(b)

NaH2PO4(c)

Na2HPO449Practice:RankthefollowingsaltsinorderofincreasingpHoftheiraqueoussolutions:(a)

KNO3,K2SO3,K2S,Fe(NO3)2(b)NH4NO3,NaHSO4,NaHCO3,Na2CO3

50(3)Thefactorsthataffectthehydrolysisofsalts:C:h=[OH-]/c0=

Kh/c0c0

,h

T:

rHm>0

T

,Kh

Applications:可通过控制溶液pH来控制水解。

Eg.SnCl2/concentratedHClKCN/KOH51

3-4

BufferSolution(1)Definition,ComponentsandPrinciple

TheCommon–IonEffectAbufferisasolutionofaconjugateacid-basepairinwhichthesecomponentsoccurinsimilarconcentrations.52HAc–Ac-

HAc

H++Ac-t0ca0csteca–x

caxcs+x

csKa=xcs/ca[H+]=x=Kaca/cspH=pKa–lgca/cs

TheHenderson–HasselbalchEquation

53CT=ca+csca/cs[H+]Add0.01molacidTheincreasedpercentof[H+]GivenV=1L,

nT=na+nsca/cs=na/ns[H+]=Kana/ns[H+]’[H+]’=Ka(na+0.01)/(ns-0.01)([H+]’-[H+])/[H+]2mol1:1Ka1.02Ka2%2mol1:100