分析化学中的化学平衡

第四章分析化学中的化学平衡

ChemicalEquilibriuminAnalyticalChemistry

滴定分析中化学平衡四大平衡体系：酸碱平衡配位平衡氧化还原平衡沉淀平衡四种滴定分析法：酸碱滴定法配位滴定法氧化还原滴定法沉淀滴定法4A

Chemical

equilibriuminsolution滴定分析中化学平衡4B

Distributionfractionδ

平衡浓度及分布分数4C

Calculationofprotonconcentration酸碱溶液的H+浓度计算4D

Buffersolution缓冲溶液4A

ChemicalequilibriuminsolutionModerntheoryofacidsandbasesBrösted-Lowrytheory——质子酸碱理论Conjugateacid

Conjugatebase+H+

共轭酸共轭碱proton

AcidandBaseAcid:aprotondonor.Base:aprotonacceptor.Amphiprotic:aspeciescapableofactingasbothanacidandabase.

HA

A+H+

Half-reactionHFF-+H+H2PO4-HPO42-+H+H6Y2+H5Y++H+NH4+NH3+H+Generalexpression:

例:HF在水中的离解反应

半反应:

HFF-+H+

半反应:H++H2OH3O+

总反应:HF+H2OF-+H3O+

简写:HFF-+H+

酸碱反应的实质是质子转移Essenceofacid-basereaction——transferofproton水的自递AciddissociationBasedissociationAutoprotolysisAcid-basereactionTypesofequilibriumconstantsKwAutoprotolysisconstantforwaterKaDissociationconstantforanacidKbDissociationconstantforabaseKspSolubilityproductnFormationconstantKdNameandsymbolofequilibriumconstantp-functionspx=-lgx酸碱反应的平衡常数AciddissociationBasedissociationConjugateacid-basepairHA—A-

EquilibriumConstantAutoprotolysis1.Autoprotolysis

(质子自递反应)H2O+H2OH3O++OH-(25°C)Equilibriumconstant:

KW＝aH3O+

×

aOH-＝1.00×10-14（25℃）KW:Autoprotolysisconstantoractivityproductofwater水的质子自递常数或活度积HA+H2OA-+H3O+A-+H2OHA

+OH-

2.Dissociationofmonoproticweakacid/baseaH

+aA

-Ka=aHAaHA

aOH

-Kb=aA

-HalfreactionDissociationconstantRelationshipofKaandKb:aH

+aA

-KaKb

==KwaHAaHA

aOH

-aA

-

pKa+pKb=pKw=14.00Dissociationofpolyproticacid/base多元酸碱pKb1+pKa3=14.00pKb2+pKa2=14.00pKb3+pKa1=14.00H3PO4

H2PO4-HPO42-PO43-Kb2Kb1Kb3Ka1Ka2Ka3Kbi

=KwKa(n-i+1)

Reactionconstant,KtH++OH-

H2O

H++Ac-

HAcOH-+HAcH2O

+Ac-3.Acid-basereaction(Titrationreaction)1Kt==1014.00Kw1Kt==KaKbKw1Kt==KbKaKwTitrationreaction活度与浓度ai=gi

ci活度：在化学反应中表现出来的有效浓度，通常用a表示溶液无限稀时:g=1中性分子:g

=1溶剂活度:a

=1Debye-Hückel公式：（稀溶液I<0.1mol/L）I：离子强度,I=1/2∑ciZi2,zi：离子电荷,B:常数,(=0.00328@25℃),与温度、介电常数有关,å：离子体积参数(pm)-lggi=0.512zi2I1+BåI-lggi=0.512zi2I活度常数K◦

——与温度有关反应：HA＋BHB++A-平衡常数aHB

+

aA

-K◦=aBaHA浓度常数Kc——与温度和离子强度有关[HB+][A-]Kc==[B][HA]aHB

+

aA

-=aBaHAgB

gHA-gHB+

gA-K◦gHB+

gA-Material(Mass)Balance

(物料平衡)各物种的平衡浓度之和等于其分析浓度ChargeBalance

(电荷平衡)溶液中正离子所带正电荷的总数等于负离子所带负电荷的总数(电中性原则)ProtonBalance

(质子平衡)溶液中酸失去质子数目等于碱得到质子数目Equilibriuminsolution——MBE;CBE;PBE

MassBalanceEquationMBE

物料平衡式

Thetotalamountofaspeciesaddedtoasolutionmustequalthesumoftheamountofeachofitspossibleformspresentinsolution.各物种的平衡浓度之和等于其分析浓度Mixtureof210-3mol/LZnCl2and0.2mol/LNH3[Cl-]=410-3mol/L[Zn2+]+[Zn(NH3)2+]

+[Zn(NH3)22+]

+[Zn(NH3)32+]

+[Zn(NH3)42+]

=210-3mol/L[NH3]+[Zn(NH3)2+]

+2[Zn(NH3)22+]

+3[Zn(NH3)32+]

+4[Zn(NH3)42+]

=0.2mol/L[Na+]+[H+]=[OH-]+[HC2O4-]+2[C2O42-]Thetotalconcentrationofpositivechargeinasolutionmustequalthetotalconcentrationofnegativecharge.

(Solutionelectroneutrality

电中性原则)。

ChargeBalanceEquationCBE

电荷平衡式Na2C2O4solutionNaClsolution[Na+]+[H+]=[OH-]+[Cl-]ProtonBalanceEquationPBE质子平衡式

HowtoobtainPBE：Theamountofprotonsreleasedfromacidsmustequaltothatacceptedbybases.溶液中酸失去质子数目等于碱得到质子数目(1)Findoutthereferenceprotonlevels

(参考水准),orzerolevelofprotons

(零水准)whichshouldbethepredominantspeciesinthesolutionandinvolvedintheprotontransfer.先选零水准(大量存在,参与质子转移的物质)，一般选取投料组分及H2O(2)Puttheproductsviathereferenceprotonlevelsacceptingorreleasingprotonsonbothsidesoftheequation.

将零水准得质子产物写在等式一边,失质子产物写在等式另一边(3)Theconcentrationofeachproductmustmultiplytheamountoftranferedprotons.

浓度项前乘上得失质子数

Example:（1）H2OReferenceprotonlevels

：H2O（2）HAcsolutionPBE：Referenceprotonlevels

：H2O、HAcPBE（3）H3PO4

solutionPBEReferenceprotonlevels

：H2O、H3PO4（4）Na2HPO4solution（5）NH4Ac(6)Mixtureofstrongacid/baseandweakacid/base.HCl+HAcReferenceprotonlevels：H2O、HAcPBE:[H+]-cHCl=[OH-]+[Ac-]NaOH+NH3Referenceprotonlevels：H2O、NH3PBE:[H+]+[NH4+]=[OH-]-cNaOH

(7)Conjugatedsystem

cbmol/LNaAc

andcamol/LHAcPBEPBExxReferenceprotonlevels

：H2O、HAcReferenceprotonlevels

：H2O、Ac-CharacteristicofPBE(1)

ReferenceprotonlevelsneverappearinPBE(2)OnlythespeciesinvolvedintheprotontransferappearinPBE.Na2HPO4Exercises：Na2HPO4

solution[H+]+[H2PO4-]+2[H3PO4]=[OH-]+[PO43-]Referenceprotonlevels

：H2O、HPO42-Na(NH4)HPO4[H+]+[H2PO4-]+2[H3PO4]=[OH-]+[NH3]+[PO43-]Na2CO3[H+]+[HCO3-]+2[H2CO3]=[OH-]Referenceprotonlevels

：H2O、CO32-Aratioexpressingtheconcentrationofonecomponenttotheapparentconcentrationofsolute.溶液中某酸碱组分的平衡浓度占其分析浓度的分数.“δ”establishesthecorrelationbetweenequilubriumconcentrationandapparantconcentration.

“δ”将平衡浓度与分析浓度联系起来

[HA]＝δHA

c

HA,[A-]=δA-

c

HA

MonoproticweakacidPolyproticacid/base4BDistributionfraction

分布分数酸度对弱酸(碱)形体分布的影响酸度和酸的浓度酸度：溶液中H＋的平衡浓度或活度，通常用pH表示

pH=-lg[H+]酸的浓度：酸的分析浓度，包含未解离的和已解离的 酸的浓度对一元弱酸：cHA＝[HA]+[A-][HAc][HAc]Ka[HAc]+[H+]=1.MonoproticweakacidHAcAc-

H+

+

cHAc

=

[HAc]+[Ac-][HAc]δHAc

=cHAc[H+]=[H+]

+Ka

δHAc[HAc]+[Ac-][HAc]=[Ac-][Ac-]δAc-

==cHAc[HAc]+[Ac-][H+]

+Ka

Ka

=δAc-δHA＋δA-＝1Characteristicofδδ

isafunctionofpH

andpKa，independentoftheapparentconcentrationofacidc[H+]=[H+]

+Ka

δHA[H+]

+Ka

Ka

=δA-

ExampleCalculateδHAc

andδAc-ofHAcsolutionatpH4.00

and8.00,respectively.Solution:Ka,HAc=1.75×10-5

AtpH4.00

AtpH8.00

δHAc=5.7×10-4,δAc-

≈1.0[H+]δHAc

==0.85[H+]

+Ka

KaδAc-

==0.15[H+]

+Ka

pHδHAδA-pKa-2.00.990.01*pKa-1.30.950.05pKa-1.00.910.09*pKa0.500.50pKa+1.00.090.91*pKa+1.30.050.95pKa+2.00.010.99δHAandδA-atdifferentpH*pKa-1.30.950.05*pKa0.500.50*pKa+1.30.050.95对于给定弱酸，

δ

对pH作图→分布分数图DistributionplotofHAc（pKa=4.76）3.466.06pKa±1.3pHHAcAc-4.76Predominantrangeδ优势区域图DistributionplotofHF（PKA=3.17）HF

F-pKa3.17pH1.00.50.0δ024681012pH3.17HFF-PredominantrangeDistributionplotofHCN（pKA=9.31）pKa9.31HCN

CN-pH024681012pH1.00.50.0δ9.31HCNCN-PredominantrangeHA的分布分数图（pKa）分布分数图的特征

两条分布分数曲线相交于（pka，0.5）pH<pKa时，溶液中以HA为主

pH>pKa时，溶液中以A-为主分布分数－多元弱酸二元弱酸H2AH2AH++HA-

H++A2-cH2CO3=[H2CO3]+[HCO3-]+[CO32-][H2A]==δH2AcH2AδA2-δHA-def[HA-]==cH2Adef[A2-]==cH2Adef物料平衡酸碱解离平衡H2AH++HA-

H++A2-cH2CO3=[H2CO3]+[HCO3-]+[CO32-][H2A]==δH2AcH2AδA2-δHA-[HA-]==cH2A[A2-]==cH2A[H+]2[H+]2+[H+]Ka1+Ka1

Ka2===[H+]2+[H+]Ka1+Ka1

Ka2[H+]2+[H+]Ka1+Ka1

Ka2[H+]Ka1

Ka1

Ka2Polyproticweakacid:HnAHnAH++Hn-1A-

……H++HA(n+1)-

H++An-[H+]n=δ0[H+]n+[H+]n-1Ka1+…+Ka1

Ka2..Kan[H+]n-1

Ka1

=δ1[H+]n+[H+]n-1Ka1+…+Ka1

Ka2..Kan…=δn[H+]n+[H+]n-1Ka1+…+Ka1

Ka2..KanKa1

Ka2..Kan…分布分数定义物料平衡酸碱解离平衡DistributionplotofH2CO31.00.0024681012pHδH2CO3HCO3-CO32-H2CO3

HCO3-

CO32-6.38pKa110.25pKa2△pKa=3.87pHPredominantrangeDistributionplotofH3PO4H3PO4

H2PO4-

HPO42-

PO43-2.16pKa5.057.21pKa5.1112.32pKa1pKa2pKa31.00.0024681012pHδH3PO4H2PO4-HPO42-PO43-PredominantrangeConclusionsondistributionfraction

δ

isafunctionofpHandpKa，independentoftheapparentconcentrationofacidc.

δ1+δ2+…+δn=1[H+]n=δ0[H+]n+[H+]n-1Ka1+…+Ka1

Ka2..Kan[H+]n-1

Ka1

=δ1[H+]n+[H+]n-1Ka1+…+Ka1

Ka2..Kan…=δn[H+]n+[H+]n-1Ka1+…+Ka1

Ka2..KanKa1

Ka2..Kan…δ

仅是pH和pKa的函数，与酸的分析浓度c无关4C

Calculationoftheconcentrationofproton1.Strongacid/base2.Monoproticweakacid/baseHA

Polyproticweakacid/baseH2A,H3A3.Amphiproticcompound

HA-4.Mixtureofacidandbasestrong+weakweak+weak5.Conjugateweakacid/weakbasepair

HA+A-1强酸碱溶液强酸(HCl):强碱(NaOH):cHCl=10-5.0

and

10-8.0mol·L-1,pH=?质子条件:[H+]+cNaOH=[OH-]最简式:[OH-]=cNaOH质子条件:[H+]=cHCl+[OH-]最简式:[H+]=cHCl公式推导？Kwisaconstant,soExample

Calculate[H+]and[OH-]in0.200mol/LNaOHYoucan’talwaysassumethattheamountofH3O+

orOH-contributedfromwaterisnegligible.

Example

DeterminethepHofa10-8mol/LHClsolution.

IfyouassumeallH3O+comesfromHCl,youwouldcalculatethepHas8.0.

Thatwouldmeanthatyouaddedanacidandmadeabasicsolution–wrong!!FormularCondition10-6.0-10-8.0[H+]=ca>10-6.0[OH-]=cb<10-8.02弱酸(碱)溶液展开则得一元三次方程,数学处理麻烦!

一元弱酸(HA)

质子条件式:[H+]=[A-]+[OH-]

平衡关系式精确表达式:[H+]＝Ka[HA]+Kw[H+]=+[H+]Ka[HA][H+]Kw[H+]

+Ka

ca[H+][HA]=若:

Kaca>10Kw,忽略Kw(即忽略水的酸性)[HA]=ca-[A-]=ca-([H+]-[OH-])≈ca-[H+]

近似计算式:展开得一元二次方程:[H+]2+Ka[H+]-caKa=0，求解即可[H+]＝Ka[HA]+Kw精确表达式：[H+]＝Ka(ca-[H+])

IfKac＜10Kw,andc/Ka

>100

thenignoretheeffectof

dissociatedacidon[HA],

[HA]≈c

Simplified:[H+]＝Ka[HA]+Kw[H+]＝Kac

+KwSimplified:

Furthermore,ifc/Ka≥

100,thenc

－

[H+]≈

c[H+]＝Kac(1)Kac≥10Kw:

(3)c/Ka

≥100

；Kac≤10Kw:(2)Kac≥10Kw；ca/Ka

≥100:[H+]＝Ka[HA]+KwConclusion：[H+]＝Ka(c

-[H+])[H+]＝Kac

+Kw(Thesimplest)[H+]＝Kac例计算0.20mol·L-1Cl2CHCOOH的pH.(pKa=1.26)如不考虑酸的离解(用最简式:pH=0.98),

则Er=29%解:

Kac=10-1.26×0.20=10-1.96>>10Kwc/Ka

=0.20/10-1.26=100.56

＜100故近似式:解一元二次方程:[H+]=10-1.09则pH=1.09[H+]＝Ka(ca-[H+])2CalculatethepHof0.1mol/Lmonochloroaceticacid

(一氯乙酸).(Ka=1.410-3)Solution:3CalculatethepHof1.010-4mol/LHCN.(Ka=6.210-10)Solution:pH=1.961CalculatethepHof0.10mol/LHAc.(pKa=4.76)Solution:Exercises处理方式与一元弱酸类似用Kb

代替Ka，[OH-]代替[H+]一元弱酸的公式可直接用于一元弱碱的计算直接求出：[OH-],

再求[H+]

pH=14-pOH一元弱碱(B-)质子条件式:[OH-]=[H+]+[HB]

代入平衡关系式[B-]Kb[OH-][OH-][OH-]Kw=+精确表达式:[OH-]

=Kb[B-]

+

Kw(1)Kbc>10Kw:(2)c/Kb

>100:(3)Kbc>10Kw,c/Kb

>100:[OH-]=Kb(cb-[OH-])[OH-]=Kbcb+Kw[H+]=KaKwcb[OH-]=Kbcb最简式:

多元弱酸溶液二元弱酸(H2A)质子条件:

[H+]=[HA-]+2[A2-]+[OH-]2Ka2[H+]=Ka1[H2A](1+)+Kw[H+]酸碱平衡关系KwKa1[H2A][H+]=++2Ka1Ka2[H2A][H+][H+]2[H+]≤0.05,可略

近似式:以下与一元酸的计算方法相同Ka1ca>10Kw2Ka2[H+]=Ka1[H2A](1+)+Kw[H+]2Ka2[H+]=Ka1[H2A](1+)+Kw[H+]2Ka2[H+]2Ka2[H+][H+]=Ka1[H2A](忽略二级及以后各步离解)Ka1c>10Kw，<0.05

c/Ka1

≥100

[H+]＝Ka1c2Ka2[H+]=Ka1[H2B](1+)[H+][H+]=Ka1[H2B]Ka2[H+]Ka2Ka1c≈计算饱和H2CO3溶液的pH值(0.040mol/L)

强酸(HCl)+弱酸(HA)质子条件:[H+]=cHCl+[A-]+[OH-](近似式)忽略弱酸的离解:[H+]≈cHCl

(最简式)3.混合酸碱体系Kw[H+]=cHCl++KacaKa+[H+][H+]酸碱平衡关系

强碱(NaOH)+弱碱(B-)质子条件:[H+]+[HB]+cNaOH=[OH-]忽略弱碱的离解:[OH-]≈c(NaOH)(最简式)Kw[OH-]=cHCl++KbcbKb+[OH-][OH-]两弱酸(HA+HB)溶液质子条件:

[H+]=[A-]+[B-]+[OH-][HA]≈cHA[HB]≈cHB酸碱平衡关系KwKHA[HA][H+]=++KHB[HB][H+][H+][H+][H+]＝KHAcHA[H+]＝KHAcHA+KHBcHBKHAcHA>>KHBcHB弱酸+弱碱(HA+B-)溶液质子条件:

[H+]+[HB]=[A-]+[OH-][HA]≈cHA[HB]≈cHB酸碱平衡关系Kw[H+][HB][H+]+=+KHA[HA]KHB[H+][H+][H+]＝KHAKHBcHA/cB4.AmphiproticcompoundsAmphiproticcompoundsbehaveasacidsinthepresenceofbasicsolutesandbasesinthepresenceofacidicsolutes.在溶液中既起酸(给质子)、又起碱（得质子）的作用**PolyacidsaltNa2HPO4,NaH2PO4,

Weak-acidweak-baseSaltNH4Ac

Aminoacid质子条件:[H+]+[H2A]=[A2

-]+[OH-]精确表达式:酸碱平衡关系式

酸式盐NaHAKw[H+][HA-][H+]+=+Ka2[HA-]Ka1[H+][H+][H+]＝Ka1(Ka2[HA-]+Kw)Ka1+[HA-]若:Ka1>>Ka2,[HA-]≈c(ΔpKa≥3.2)近似计算式:如果c>10Ka1,则“Ka1”可略,得最简式:[H+]=Ka1Ka2

c

Ka1+c[H+]=Ka1(Ka2

c＋Kw)Ka1+c若Ka2c>10Kw

则Kw可忽略[H+]＝Ka1Ka2精确式：Ka1>>Ka2,[HA-]≈cKa2c>10Kwc>10Ka1pH=1/2(pKa1+pKa2)[H+]＝Ka1(Ka2c+Kw)

Ka1+c[H+]＝Ka1Ka2cKa1+c[H+]＝Ka1Ka2[H+]＝Ka1(Ka2[HA-]+Kw)Ka1+[HA-]弱酸弱碱盐NH4Ac质子条件式:

[H+]+[HAc]=[NH3]+[OH-]Ka’c>10Kwc>

10Ka[H+]＝Ka(Ka’c+Kw)

Ka1+c酸碱平衡关系[NH4+]≈[Ac-]≈c[H+]=

KaKa’cKa+c[H+]＝KaKa’Ka’NH4+KaHAc例计算0.0010mol/LCH2ClCOONH4

溶液的pHCH2ClCOOH:Ka=1.4×10-3NH3:Kb=1.8×10-4

Ka’=5.6×10-10

Ka’c≥

10Kw,c<10KapH=6.24[H+]＝Ka1Ka2cKa1+c氨基酸H2N-R-COOHPBE:

[H+]+[+H3N-R-COOH]=[H2N-R-COO-]+[OH-]Ka2c>10Kwc/Ka1>10[H+]＝Ka1Ka2cKa1+c[H+]＝Ka1(Ka2c+Kw)

Ka1+c酸碱平衡关系[H+]＝Ka1Ka2

综合考虑、分清主次、合理取舍、近似计算酸碱溶液[H+]的计算总结质子条件物料平衡电荷平衡

酸碱平衡关系[H+]的精确表达式近似处理[H+]的近似计算式和最简式Abuffersolutionisasolutionofaconjugateacid/basepairthatresistschangesinpH.

4DBuffersolutionsAconjugateweakacid/weakbasepair:camol/LHBandcbmol/LB-H+OH-SmallpHshift缓冲溶液：能减缓强酸强碱的加入或稀释而引起的pH变化

PlotsofpHvs.mLofwateraddedto(a)0.500mLof0.100mol·L-1HCland(b)0.500mLofasolution0.100mol·L-1inbothHAcandNaAc.ChangeofpHwiththedilutionofsolutionHClNaAc-HAc[H+]=Ka=Ka[HA][A-]ca-[H+]+[OH-]cb+[H+]-[OH-][H+]=Ka

[HA][A-]camol/LHA+cbmol/LNaAPBE：[HA]=ca+[OH-]-[H+][A-]=cb+[H+]-[OH-]物料平衡:[HA]+[A-]=ca+cb电荷平衡:[H+]+[Na+]=[OH-]+[A-][HA]=ca-[H+]+[OH-][A-]=ca+cb-[HA]=cb+[H+]-[OH-]+)4D-1Calculationofbuffersolution'spHpH<6(酸性),略去[OH-]pH=pKa+lg

cacb[H+]=Kaca-[H+]+[OH-]cb+[H+]-[OH-][H+]=Kaca-[H+]cb+[H+][H+]=Kaca+[OH-]cb-[OH-]pH>8(碱性),略去[H+]若ca

≥20[H+];

cb

≥20[H+],

或ca

≥20[OH-];

cb

≥20[OH-],最简式[H+]=Ka

cacb计算方法：(1)先按最简式计算[OH-]或[H+]。(2)再计算[HA]或[A-],看其是否可以忽略.如果不能忽略,再按近似式计算。通常情况下，由共轭酸碱对组成的缓冲溶液可以用最简式直接计算pH例(1)0.10mol/LNH4Cl–0.20mol/LNH3

先按最简式:

(2)0.080mol/L二氯乙酸–0.12mol/L二氯乙酸钠先用最简式求得[H+]＝0.037mol/L∵ca>>[OH+],cb>>[OH-]∴结果合理pH=9.56应用近似式:解一元二次方程，[H+]=10-1.65mol·L-1,

pH=1.65pH=pKa+lg=9.56cacb[H+]=Kaca-[H+]cb+[H+]Example

Howtoprepare200mLpH2.0bufferwithNH2CH2COOHandHCl?(MNH2CH2COOH=75.01;pKa1=2.35;c