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1、第四章 分析化学中的化学平衡 Chemical Equilibrium in Analytical Chemistry,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 equilibrium in solution 滴定分析中化学平衡4B Distribution fraction 平衡浓度及分布分数4C Calculation of proton concentration 酸碱溶液的H+浓度计算4D Buffer solution 缓冲溶液,4A Chemical equilibr
2、ium in solution,Modern theory of acids and bases,Brsted-Lowry theory质子酸碱理论,Conjugate acid Conjugate base + H+ 共轭酸 共轭碱 proton,Acid and Base,Acid: a proton donor.Base: a proton acceptor.Amphiprotic: a species capable of acting as both an acid and a base.,HA A + H+ Half-reaction,HF F - + H+ H2PO4- HPO4
3、2- + H+ H6Y2+ H5Y+ + H+ NH4+ NH3 + H+,General expression:,例: HF在水中的离解反应 半反应: HF F- + H+ 半反应: H+ + H2O H3O+ 总反应: HF + H2O F- + H3O+ 简写: HF F- + H+,酸碱反应的实质是质子转移,Essence of acid-base reactiontransfer of proton,水的自递,Acid dissociation,Base dissociation,Autoprotolysis,Acid-base reaction,Types of equilibri
4、um constants,Name and symbol of equilibrium constant,p-functions px=-lgx,酸碱反应的平衡常数,Acid dissociation,Base dissociation,Conjugate acid-base pair HAA-,Equilibrium Constant,Autoprotolysis,1. Autoprotolysis (质子自递反应),H2O + H2O H3O+ + OH- (25C),Equilibrium constant: KWaH3O+ aOH-1.0010-14（25）,KW: Autoproto
5、lysis constant or activity product of water 水的质子自递常数或活度积,HA + H2O A- + H3O+ A- + H2O HA + OH-,2. Dissociation of monoprotic weak acid/base,Half reaction,Dissociation constant,Relationship of Ka and Kb:,pKa + pKb = pKw= 14.00,Dissociation of polyprotic acid/base 多元酸碱,pKb1 + pKa3 = 14.00pKb2 + pKa2 =
6、14.00pKb3 + pKa1= 14.00,Reaction constant, Kt,H+ + OH- H2O H+ + Ac- HAcOH- + HAc H2O + Ac-,3.Acid-base reaction (Titration reaction),Titration reaction,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 =1,Debye-Hckel公式：（稀溶液I0.1 mol/L）,I：离子强度, I=1/2ciZi2, zi：离子电荷,B: 常数, (=0.00
7、328 25), 与温度、介电常数有关,：离子体积参数(pm),活度常数 K 与温度有关,反应：HAB HB+ +A-,平衡常数,浓度常数 Kc 与温度和离子强度有关,HB+A-,Kc = =,BHA,aHB + aA -,=,aBaHA,gB 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,Charge Balance (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),Proton Balance (质子平衡)溶液中酸失去质子数目等于碱得到质子数目,Equili
8、brium in solutionMBE;CBE;PBE,Mass Balance Equation MBE 物料平衡式,The total amount of a species added to a solution must equal the sum of the amount of each of its possible forms present in solution.各物种的平衡浓度之和等于其分析浓度,Mixture of 2 10-3 mol/L ZnCl2 and 0.2 mol/L NH3,Cl- = 4 10-3 mol/L,Zn2+ +Zn(NH3) 2+ +Zn(
9、NH3)22+ +Zn(NH3)32+ +Zn(NH3)42+ = 2 10-3 mol/L,NH3 +Zn(NH3) 2+ +2Zn(NH3)22+ +3Zn(NH3)32+ +4Zn(NH3)42+ = 0.2 mol/L,Na+ + H+ = OH- + HC2O4- + 2C2O42-,The total concentration of positive charge in a solution must equal the total concentration of negative charge. (Solution electroneutrality 电中性原则)。,Char
10、ge Balance Equation CBE 电荷平衡式,Na+ + H+ = OH- + Cl-,Proton Balance Equation PBE 质子平衡式,How to obtain PBE：,The amount of protons released from acids must equal to that accepted by bases.溶液中酸失去质子数目等于碱得到质子数目,(1) Find out the reference proton levels (参考水准), or zero level of protons (零水准) which should be t
11、he predominant species in the solution and involved in the proton transfer. 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) Put the products via the reference proton levels accepting or releasing protons on both sides of the equation. 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) The concentration of each product must mul
12、tiply the amount of tranfered protons. 浓度项前乘上得失质子数,Example:（1） H2O,Reference proton levels ：H2O,（2） HAc solution,PBE：,Reference proton levels ：H2O、HAc,PBE,（3） H3PO4 solution,PBE,Reference proton levels ：H2O、H3PO4,（4） Na2HPO4 solution,（5） NH4Ac,(6) Mixture of strong acid/base and weak acid/base.HCl +
13、 HAc Reference proton levels ：H2O、HAc PBE: H+ - cHCl = OH- + Ac-NaOH + NH3 Reference proton levels ：H2O、NH3PBE: H+ + NH4+= OH- - cNaOH,(7) Conjugated system cb mol/L NaAc and ca mol/L HAc,PBE,PBE,Reference proton levels ：H2O、HAc,Reference proton levels ：H2O、Ac-,Characteristic of PBE,(1) Reference pr
14、oton levels never appear in PBE(2) Only the species involved in the proton transfer appear in PBE.,Exercises：Na2HPO4 solution,H+ + H2PO4- +2H3PO4 = OH- +PO43-,Reference proton levels ：H2O、HPO42-,Na(NH4)HPO4,H+ + H2PO4- +2H3PO4 = OH- +NH3 + PO43-,Na2CO3,H+ + HCO3- + 2H2CO3 = OH-,Reference proton leve
15、ls ：H2O、CO32-,A ratio expressing the concentration of one component to the apparent concentration of solute. 溶液中某酸碱组分的平衡浓度占其分析浓度的分数.,“” establishes the correlation between equilubrium concentration and apparant concentration. “” 将平衡浓度与分析浓度联系起来 HA HA c HA , A-= A- c HA,4B Distribution fraction 分布分数,酸
16、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：cHAHA+A-,1. Monoprotic weak acid,HAc Ac-, H+ +,cHAc = HAc+Ac-,HA A -1,Characteristic of , is a function of pH and pKa，independent of the apparent concentration of acid c,=,A-,Example CalculateHAc andAc- of H
17、Ac solution at pH4.00 and 8.00, respectively.,Solution: Ka, HAc=1.7510-5 At pH4.00,At pH8.00 HAc = 5.710-4, Ac- 1.0, HA andA- at different pH,对于给定弱酸， 对pH作图分布分数图,Distribution plot of HAc（pKa=4.76）,Predominant range,优势区域图,Distribution plot of HF（PKA=3.17）,Predominant range,Distribution plot of HCN（pKA
18、=9.31）,pKa9.31,Predominant range,HA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）,HpKa时，溶液中以A-为主,分布分数多元弱酸,二元弱酸H2A,H2AH+HA- H+A2-,c H2CO3=H2CO3+HCO3-+CO32-,H2AH+HA- H+A2-,cH2CO3=H2CO3+HCO3-+CO32-,Polyprotic weak acid: HnA,HnAH+Hn-1A- H+HA(n+1)- H+An-,H+n,=,0,H+n + H+n-1Ka1 +Ka1 Ka2.Kan,H+n-1 Ka1,=,1,H+n +
19、 H+n-1Ka1 +Ka1 Ka2.Kan,=,n,H+n + H+n-1Ka1 +Ka1 Ka2.Kan,Ka1 Ka2.Kan,分布分数定义物料平衡酸碱解离平衡,Distribution plot of H2CO3,pH,Predominant range,Distribution plot of H3PO4,Predominant range,Conclusions on distribution fraction, is a function of pH and pKa，independent of the apparent concentration of acid c. 1+2+
20、 +n = 1,H+n,=,0,H+n + H+n-1Ka1 +Ka1 Ka2.Kan,H+n-1 Ka1,=,1,H+n + H+n-1Ka1 +Ka1 Ka2.Kan,=,n,H+n + H+n-1Ka1 +Ka1 Ka2.Kan,Ka1 Ka2.Kan, 仅是pH和pKa 的函数，与酸的分析浓度c无关,4C Calculation of the concentration of proton,1. Strong acid/base2. Monoprotic weak acid/base HA Polyprotic weak acid/base H2A, H3A3. Amphiprotic
21、 compound HA-4. Mixture of acid and base strong+weak weak+weak5. Conjugate weak acid/weak base pair HA+A-,1 强酸碱溶液,强酸(HCl):强碱(NaOH):,cHCl=10-5.0and10-8.0 molL-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？,Kw is a constant, so,Example Calculate H+ and OH- in
22、0.200 mol/L NaOH,You cant always assume that the amount of H3O+ or OH- contributed from water is negligible.Example Determine the pH of a 10-8 mol/L HCl solution.If you assume all H3O+ comes from HCl, you would calculate the pH as 8.0.That would mean that you added an acid and made a basic solution
23、wrong!,2 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,若: Kaca10Kw , 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+KaH+-caKa=0，求解即可,精确表达式：,If Kac10Kw, and c/Ka 100 then ignore the effect of dissociated acid on HA, HA c Simplified:,Simplified:,Furthermore, if c/Ka 100,
24、 then c H+ c,(1) Kac10Kw :,(3) c/Ka 100 ； Kac 10Kw :,(2) Kac10Kw ；ca/Ka 100 :,Conclusion：,(The simplest),例 计算0.20molL-1 Cl2CHCOOH 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 =10-1.260.20=10-1.9610Kw c/Ka = 0.20 / 10-1.26 =100.56 100,2 Calculate the pH of 0.1 mol/L monochloroacetic acid (一
25、氯乙酸). (Ka = 1.410-3),Solution:,3 Calculate the pH of 1.010-4 mol/L HCN. (Ka = 6.210-10),Solution:,pH=1.96,1 Calculate the pH of 0.10 mol/L HAc. (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,(1) Kbc 10Kw
26、:,(2) c/Kb 100 :,(3) Kbc 10Kw, c/Kb 100 :,最简式:,多元弱酸溶液,二元弱酸(H2A)质子条件:,H+ = HA- + 2A2- + OH-,酸碱平衡关系,0.05, 可略 近似式:,以下与一元酸的计算方法相同,Ka1ca 10Kw,(忽略二级及以后各步离解),Ka1c 10Kw，,0.05,c/Ka1 100,计算饱和H2CO3溶液的pH值(0.040 mol/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+ c HCl (最简式),3. 混合酸碱体系,酸碱平衡关系,强碱(
27、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),两弱酸(HA+HB)溶液,质子条件: H+ = A- + B- + OH-,HA cHA HBcHB,酸碱平衡关系,KHAcHAKHBcHB,弱酸+弱碱(HA+B-)溶液,质子条件: H+ + HB = A- + OH-,HA cHA HBcHB,酸碱平衡关系,4. Amphiprotic compounds,Amphiprotic compounds behave as acids in the presence of basic solutes and
28、bases in the presence of acidic solutes.在溶液中既起酸(给质子)、又起碱（得质子）的作用,*Polyacid salt Na2HPO4, NaH2PO4, Weak-acid weak-base Salt NH4Ac Amino acid,质子条件: H+H2A=A2 -+OH-,精确表达式:,酸碱平衡关系式,酸式盐 NaHA,若: Ka1Ka2, HA-c (pKa3.2),近似计算式:,如果 c 10Ka1, 则“Ka1”可略,得 最简式:,若Ka2c 10Kw 则 Kw可忽略,精确式：,Ka1Ka2, HA-c,Ka2c 10Kw,c 10 Ka1
29、,pH = 1/2(pKa1 + pKa2),弱酸弱碱盐 NH4Ac,质子条件式: H+ + HAc = NH3 + OH-,Kac 10Kw,c 10 Ka,酸碱平衡关系NH4+ Ac-c,Ka NH4+Ka HAc,例 计算 0.0010 mol/L CH2ClCOONH4 溶液的pH,CH2ClCOOH: Ka=1.410-3,NH3: Kb=1.810-4 Ka=5.610-10,Kac 10Kw , c10Ka,pH = 6.24,氨基酸 H2N-R-COOH,PBE: H+ + +H3N-R-COOH = H2N-R-COO- + OH-,Ka2c 10Kw,c/Ka1 10,酸碱
30、平衡关系,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,A buffer solution is a solution of a conjugate acid/base pair that resists changes in pH.,4D Buffer solutions,A conjugate weak acid/weak base pair: ca mol/L HB and cb mol/L B-,H+,OH-,Small pH shift,缓冲溶液：能减缓强酸强碱的加入或稀释
31、而引起的pH变化,Plots of pH vs. mL of water added to (a ) 0.500 mL of 0.100 molL-1 HCl and (b) 0.500 mL of a solution 0.100 molL-1 in both HAc and NaAc.,Change of pH with the dilution of solution,HCl,NaAc-HAc,camol/L HA+ cbmol/L NaA,PBE：HA=ca+OH-H+ A- = cb+H+-OH-,4D-1 Calculation of buffer solutions pH,H 6
32、 (酸性),略去OH-,H 8 (碱性),略去H+,若ca 20H+; cb 20H+, 或ca 20OH-; cb 20OH-,最简式,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) 0.10 mol/L NH4Cl 0.20 mol/L NH3 先按最简式: (2) 0.080 mol/L二氯乙酸 0.12mol/L二氯乙酸钠 先用最简式求得 H+0.037 mol/L,caOH+, cbOH- 结果合理 pH=9.56,应用近似式:,解一元
33、二次方程，H+=10-1.65 molL-1 , pH=1.65,pH = pKa + lg =9.56,ca,cb,Example How to prepare 200 mL pH2.0 buffer with NH2CH2COOH and HCl? (MNH2CH2COOH=75.01; pKa1=2.35; cNH2CH2COOH=0.1 molL-1; cHCl=1.0molL-1),Solution：,NH2CH2COOH + HCl NH3+CH2COOH + Cl,If cNH3+CH2COOH = x,NH2CH2COOH 1.50g; 1.0 molL-1 HCl 13.8m
34、L; to 200 mL solution,Multiprotic weak acid-based buffers display many different pH. A diprotic weak acid can be used to prepare buffers at two pHs. Buffers of malonic acid (pKa1 = 2.85 and pKa2 = 5.70) can be prepared:,The effect of dilutionThe effect of added acids and bases,4D-2 Properties of buf
35、fer solution,缓冲容量,缓冲容量：衡量缓冲溶液缓冲能力大小，用表示 dc/dpH,加合性： = H+ OH- HA =2.3H+2.3OH-+2.3HAAcHA 对于pH在pKa1范围内的HA =2.3HAAcHA,HAA的缓冲体系 有极大值 pHpKa时， 即HA=A 极大0.58cHA,4D-3. Buffer capacity,Factors: The absolute concentration of the weak acid and the weak base. Their relative proportions of the weak acid and the we
36、ak base.,标准缓冲溶液,校准酸度计,常用缓冲溶液,常用缓冲溶液,缓冲溶液的选择原则,不干扰测定 例如：EDTA滴定Pb2+,不用HAc-Ac-有较大的缓冲能力，足够的缓冲容量 较大浓度 (0.011molL-1); pHpKa 即cacb11 HAc NaAc : pKa=4.76 (45.5) NH4OHNH3: pKb=4.75 (810 ) (CH2)6N4 (CH2)6N4H+: pKb=8.87 （4.56）,Preparation of buffersA buffer solution of any desired pH can be prepared by combing calculated quantities of a suitable conjugate acid/base pair.Prepared buffers by making up a solution of approximately the desired pH and then adjust by adding acid or conjugate base until the required pH is indicated by a pH meter.,
