分析化學(xué)中的化學(xué)平衡

第四章分析化學(xué)中的化學(xué)平衡

ChemicalEquilibriuminAnalyticalChemistry

滴定分析中化學(xué)平衡四大平衡體系：酸堿平衡配位平衡氧化還原平衡沉淀平衡四種滴定分析法：酸堿滴定法配位滴定法氧化還原滴定法沉淀滴定法4A

Chemical

equilibriuminsolution滴定分析中化學(xué)平衡4B

Distributionfractionδ

平衡濃度及分布分?jǐn)?shù)4C

Calculationofprotonconcentration酸堿溶液的H+濃度計(jì)算4D

Buffersolution緩沖溶液4A

ChemicalequilibriuminsolutionModerntheoryofacidsandbasesBr?sted-Lowrytheory——質(zhì)子酸堿理論Conjugateacid

Conjugatebase+H+

共軛酸共軛堿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在水中的離解反應(yīng)

半反應(yīng):

HFF-+H+

半反應(yīng):H++H2OH3O+

總反應(yīng):HF+H2OF-+H3O+

簡(jiǎn)寫(xiě):HFF-+H+

酸堿反應(yīng)的實(shí)質(zhì)是質(zhì)子轉(zhuǎn)移Essenceofacid-basereaction——transferofproton水的自遞AciddissociationBasedissociationAutoprotolysisAcid-basereactionTypesofequilibriumconstantsKwAutoprotolysisconstantforwaterKaDissociationconstantforanacidKbDissociationconstantforabaseKspSolubilityproductnFormationconstantKdNameandsymbolofequilibriumconstantp-functionspx=-lgx酸堿反應(yīng)的平衡常數(shù)AciddissociationBasedissociationConjugateacid-basepairHA—A-

EquilibriumConstantAutoprotolysis1.Autoprotolysis

(質(zhì)子自遞反應(yīng))H2O+H2OH3O++OH-(25°C)Equilibriumconstant:

KW＝aH3O+

×

aOH-＝1.00×10-14（25℃）KW:Autoprotolysisconstantoractivityproductofwater水的質(zhì)子自遞常數(shù)或活度積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多元酸堿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活度與濃度ai=gi

ci活度：在化學(xué)反應(yīng)中表現(xiàn)出來(lái)的有效濃度，通常用a表示溶液無(wú)限稀時(shí):g=1中性分子:g

=1溶劑活度:a

=1Debye-Hückel公式：（稀溶液I<0.1mol/L）I：離子強(qiáng)度,I=1/2∑ciZi2,zi：離子電荷,B:常數(shù),(=0.00328@25℃),與溫度、介電常數(shù)有關(guān),?：離子體積參數(shù)(pm)-lggi=0.512zi2I1+B?I-lggi=0.512zi2I活度常數(shù)K?

——與溫度有關(guān)反應(yīng)：HA＋BHB++A-平衡常數(shù)aHB

+

aA

-K?=aBaHA濃度常數(shù)Kc——與溫度和離子強(qiáng)度有關(guān)[HB+][A-]Kc==[B][HA]aHB

+

aA

-=aBaHAgB

gHA-gHB+

gA-K?gHB+

gA-Material(Mass)Balance

(物料平衡)各物種的平衡濃度之和等于其分析濃度ChargeBalance

(電荷平衡)溶液中正離子所帶正電荷的總數(shù)等于負(fù)離子所帶負(fù)電荷的總數(shù)(電中性原則)ProtonBalance

(質(zhì)子平衡)溶液中酸失去質(zhì)子數(shù)目等于堿得到質(zhì)子數(shù)目Equilibriuminsolution——MBE;CBE;PBE

MassBalanceEquationMBE

物料平衡式

Thetotalamountofaspeciesaddedtoasolutionmustequalthesumoftheamountofeachofitspossibleformspresentinsolution.各物種的平衡濃度之和等于其分析濃度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

電中性原則)。

ChargeBalanceEquationCBE

電荷平衡式Na2C2O4solutionNaClsolution[Na+]+[H+]=[OH-]+[Cl-]ProtonBalanceEquationPBE質(zhì)子平衡式

HowtoobtainPBE：Theamountofprotonsreleasedfromacidsmustequaltothatacceptedbybases.溶液中酸失去質(zhì)子數(shù)目等于堿得到質(zhì)子數(shù)目(1)Findoutthereferenceprotonlevels

(參考水準(zhǔn)),orzerolevelofprotons

(零水準(zhǔn))whichshouldbethepredominantspeciesinthesolutionandinvolvedintheprotontransfer.先選零水準(zhǔn)(大量存在,參與質(zhì)子轉(zhuǎn)移的物質(zhì))，一般選取投料組分及H2O(2)Puttheproductsviathereferenceprotonlevelsacceptingorreleasingprotonsonbothsidesoftheequation.

將零水準(zhǔn)得質(zhì)子產(chǎn)物寫(xiě)在等式一邊,失質(zhì)子產(chǎn)物寫(xiě)在等式另一邊(3)Theconcentrationofeachproductmustmultiplytheamountoftranferedprotons.

濃度項(xiàng)前乘上得失質(zhì)子數(shù)

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.溶液中某酸堿組分的平衡濃度占其分析濃度的分?jǐn)?shù).“δ”establishesthecorrelationbetweenequilubriumconcentrationandapparantconcentration.

“δ”將平衡濃度與分析濃度聯(lián)系起來(lái)

[HA]＝δHA

c

HA,[A-]=δA-

c

HA

MonoproticweakacidPolyproticacid/base4BDistributionfraction

分布分?jǐn)?shù)酸度對(duì)弱酸(堿)形體分布的影響酸度和酸的濃度酸度：溶液中H＋的平衡濃度或活度，通常用pH表示

pH=-lg[H+]酸的濃度：酸的分析濃度，包含未解離的和已解離的 酸的濃度對(duì)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對(duì)于給定弱酸，

δ

對(duì)pH作圖→分布分?jǐn)?shù)圖DistributionplotofHAc（pKa=4.76）3.466.06pKa±1.3pHHAcAc-4.76Predominantrangeδ優(yōu)勢(shì)區(qū)域圖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?jǐn)?shù)圖（pKa）分布分?jǐn)?shù)圖的特征

兩條分布分?jǐn)?shù)曲線相交于（pka，0.5）pH<pKa時(shí)，溶液中以HA為主

pH>pKa時(shí)，溶液中以A-為主分布分?jǐn)?shù)－多元弱酸二元弱酸H2AH2AH++HA-

H++A2-cH2CO3=[H2CO3]+[HCO3-]+[CO32-][H2A]==δH2AcH2AδA2-δHA-def[HA-]==cH2Adef[A2-]==cH2Adef物料平衡酸堿解離平衡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?jǐn)?shù)定義物料平衡酸堿解離平衡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

僅是pH和pKa的函數(shù)，與酸的分析濃度c無(wú)關(guān)4C

Calculationoftheconcentrationofproton1.Strongacid/base2.Monoproticweakacid/baseHA

Polyproticweakacid/baseH2A,H3A3.Amphiproticcompound

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

HA+A-1強(qiáng)酸堿溶液強(qiáng)酸(HCl):強(qiáng)堿(NaOH):cHCl=10-5.0

and

10-8.0mol·L-1,pH=?質(zhì)子條件:[H+]+cNaOH=[OH-]最簡(jiǎn)式:[OH-]=cNaOH質(zhì)子條件:[H+]=cHCl+[OH-]最簡(jiǎn)式:[H+]=cHCl公式推導(dǎo)？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(堿)溶液展開(kāi)則得一元三次方程,數(shù)學(xué)處理麻煩!

一元弱酸(HA)

質(zhì)子條件式:[H+]=[A-]+[OH-]

平衡關(guān)系式精確表達(dá)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+]

近似計(jì)算式:展開(kāi)得一元二次方程:[H+]2+Ka[H+]-caKa=0，求解即可[H+]＝Ka[HA]+Kw精確表達(dá)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例計(jì)算0.20mol·L-1Cl2CHCOOH的pH.(pKa=1.26)如不考慮酸的離解(用最簡(jiǎn)式:pH=0.98),

則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則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處理方式與一元弱酸類似用Kb

代替Ka，[OH-]代替[H+]一元弱酸的公式可直接用于一元弱堿的計(jì)算直接求出：[OH-],

再求[H+]

pH=14-pOH一元弱堿(B-)質(zhì)子條件式:[OH-]=[H+]+[HB]

代入平衡關(guān)系式[B-]Kb[OH-][OH-][OH-]Kw=+精確表達(dá)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最簡(jiǎn)式:

多元弱酸溶液二元弱酸(H2A)質(zhì)子條件:

[H+]=[HA-]+2[A2-]+[OH-]2Ka2[H+]=Ka1[H2A](1+)+Kw[H+]酸堿平衡關(guān)系KwKa1[H2A][H+]=++2Ka1Ka2[H2A][H+][H+]2[H+]≤0.05,可略

近似式:以下與一元酸的計(jì)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](忽略二級(jí)及以后各步離解)Ka1c>10Kw，<0.05

c/Ka1

≥100

[H+]＝Ka1c2Ka2[H+]=Ka1[H2B](1+)[H+][H+]=Ka1[H2B]Ka2[H+]Ka2Ka1c≈計(jì)算飽和H2CO3溶液的pH值(0.040mol/L)

強(qiáng)酸(HCl)+弱酸(HA)質(zhì)子條件:[H+]=cHCl+[A-]+[OH-](近似式)忽略弱酸的離解:[H+]≈cHCl

(最簡(jiǎn)式)3.混合酸堿體系Kw[H+]=cHCl++KacaKa+[H+][H+]酸堿平衡關(guān)系

強(qiáng)堿(NaOH)+弱堿(B-)質(zhì)子條件:[H+]+[HB]+cNaOH=[OH-]忽略弱堿的離解:[OH-]≈c(NaOH)(最簡(jiǎn)式)Kw[OH-]=cHCl++KbcbKb+[OH-][OH-]兩弱酸(HA+HB)溶液質(zhì)子條件:

[H+]=[A-]+[B-]+[OH-][HA]≈cHA[HB]≈cHB酸堿平衡關(guān)系KwKHA[HA][H+]=++KHB[HB][H+][H+][H+][H+]＝KHAcHA[H+]＝KHAcHA+KHBcHBKHAcHA>>KHBcHB弱酸+弱堿(HA+B-)溶液質(zhì)子條件:

[H+]+[HB]=[A-]+[OH-][HA]≈cHA[HB]≈cHB酸堿平衡關(guān)系Kw[H+][HB][H+]+=+KHA[HA]KHB[H+][H+][H+]＝KHAKHBcHA/cB4.AmphiproticcompoundsAmphiproticcompoundsbehaveasacidsinthepresenceofbasicsolutesandbasesinthepresenceofacidicsolutes.在溶液中既起酸(給質(zhì)子)、又起堿（得質(zhì)子）的作用**PolyacidsaltNa2HPO4,NaH2PO4,

Weak-acidweak-baseSaltNH4Ac

Aminoacid質(zhì)子條件:[H+]+[H2A]=[A2

-]+[OH-]精確表達(dá)式:酸堿平衡關(guān)系式

酸式鹽NaHAKw[H+][HA-][H+]+=+Ka2[HA-]Ka1[H+][H+][H+]＝Ka1(Ka2[HA-]+Kw)Ka1+[HA-]若:Ka1>>Ka2,[HA-]≈c(ΔpKa≥3.2)近似計(jì)算式:如果c>10Ka1,則“Ka1”可略,得最簡(jiǎn)式:[H+]=Ka1Ka2

c

Ka1+c[H+]=Ka1(Ka2

c＋Kw)Ka1+c若Ka2c>10Kw

則Kw可忽略[H+]＝Ka1Ka2精確式：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-]弱酸弱堿鹽NH4Ac質(zhì)子條件式:

[H+]+[HAc]=[NH3]+[OH-]Ka’c>10Kwc>

10Ka[H+]＝Ka(Ka’c+Kw)

Ka1+c酸堿平衡關(guān)系[NH4+]≈[Ac-]≈c[H+]=

KaKa’cKa+c[H+]＝KaKa’Ka’NH4+KaHAc例計(jì)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酸堿平衡關(guān)系[H+]＝Ka1Ka2

綜合考慮、分清主次、合理取舍、近似計(jì)算酸堿溶液[H+]的計(jì)算總結(jié)質(zhì)子條件物料平衡電荷平衡

酸堿平衡關(guān)系[H+]的精確表達(dá)式近似處理[H+]的近似計(jì)算式和最簡(jiǎn)式Abuffersolutionisasolutionofaconjugateacid/basepairthatresistschangesinpH.

4DBuffersolutionsAconjugateweakacid/weakbasepair:camol/LHBandcbmol/LB-H+OH-SmallpHshift緩沖溶液：能減緩強(qiáng)酸強(qiáng)堿的加入或稀釋而引起的pH變化

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電荷平衡:[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(堿性),略去[H+]若ca

≥20[H+];

cb

≥20[H+],

或ca

≥20[OH-];

cb

≥20[OH-],最簡(jiǎn)式[H+]=Ka

cacb計(jì)算方法：(1)先按最簡(jiǎn)式計(jì)算[OH-]或[H+]。(2)再計(jì)算[HA]或[A-],看其是否可以忽略.如果不能忽略,再按近似式計(jì)算。通常情況下，由共軛酸堿對(duì)組成的緩沖溶液可以用最簡(jiǎn)式直接計(jì)算pH例(1)0.10mol/LNH4Cl–0.20mol/LNH3

先按最簡(jiǎn)式:

(2)0.080mol/L二氯乙酸–0.12mol/L二氯乙酸鈉先用最簡(jiǎn)式求得[H+]＝0.037mol/L∵ca>>[OH+],cb>>[OH-]∴結(jié)果合理pH=9.56應(yīng)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