第四讲分析化学中化学平衡课件.ppt

1、第四章第四章 分析化学中的化学平衡分析化学中的化学平衡 Chemical Equilibrium in Analytical Chemistry 滴定分析中化学平衡滴定分析中化学平衡四大平衡体系：四大平衡体系：酸碱平衡酸碱平衡配位平衡配位平衡氧化还原平衡氧化还原平衡沉淀平衡沉淀平衡四种滴定分析法：四种滴定分析法：酸碱酸碱滴定法滴定法配位配位滴定法滴定法氧化还原氧化还原滴定法滴定法沉淀沉淀滴定法滴定法4A Chemical equilibrium in solution 滴定分析中化学平衡滴定分析中化学平衡4B Distribution fraction 平衡浓度及分布分数平衡浓度及分布分数4C

2、 Calculation of proton concentration 酸碱溶液的酸碱溶液的H+浓度计算浓度计算4D Buffer solution 缓冲溶液缓冲溶液4A Chemical equilibrium in solutionModern theory of acids and basesBrsted-Lowry theory质子酸碱理论质子酸碱理论Conjugate acid Conjugate base + H+ 共轭酸共轭酸 共轭碱共轭碱 proton Acid and BaseAcid: a proton donor.Base: a proton acceptor.Amph

3、iprotic: a species capable of acting as both an acid and a base. HA A + H+ Half-reaction HF F - + H+ H2PO4- HPO42- + H+ H6Y2+ H5Y+ + H+ NH4+ NH3 + H+General expression: 例例: HF在水中的离解反应在水中的离解反应 半反应半反应: HF F- + H+ 半反应半反应: H+ + H2O H3O+ 总反应总反应: HF + H2O F- + H3O+ 简写简写: HF F- + H+ 酸碱反应的实质是质子转移酸碱反应的实质是质子转

4、移Essence of acid-base reactiontransfer of protonAcHHAcAcOHOHHAcOHOHHAcHHAc3232简化：OHNHOHNHOHHOHNHHNH423243水的自递水的自递OHHOHOHOHOHOHOHHOHOHHOH2322232简化：OHNHNHHClOHNHOHNHClOHOHHCl243243332Acid dissociationBase dissociationAutoprotolysisAcid-base reactionTypes of equilibrium constantsKwAutoprotolysis consta

5、nt for waterKaDissociation constant for an acidKbDissociation constant for a baseKspSolubility product nFormation constantKdName and symbol of equilibrium constantnp-functions px=-lgx酸碱反应的平衡常数酸碱反应的平衡常数HAAHAHHAaKAOHHA-2HAOHOHAbKAcid dissociationBase dissociation)C25(1000. 1OHHOHo14OHHw2KConjugate aci

6、d-base pair HAA-wwOHHAOHHAHAAHpppKKKKKKbaba Equilibrium Constant Autoprotolysis1. Autoprotolysis (质子自递反应质子自递反应)H2O + H2O H3O+ + OH- (25C)Equilibrium constant: KWaH3O+ aOH-1.0010-14（25）KW: Autoprotolysis constant or activity product of water 水的质子自递常数或活度积水的质子自递常数或活度积HA + H2O A- + H3O+ A- + H2O HA + OH

7、- 2. Dissociation of monoprotic weak acid/baseaH + aA -Ka=aHAaHA aOH -Kb=aA -Half reactionDissociation constantRelationship of Ka and Kb:aH + aA -Ka Kb = = KwaHAaHA aOH -aA - pKa + pKb = pKw= 14.00Dissociation of polyprotic acid/base 多元酸碱多元酸碱pKb1 + pKa3 = 14.00pKb2 + pKa2 = 14.00pKb3 + pKa1= 14.00 H

8、3PO4 H2PO4- HPO42- PO43-Kb2Kb1Kb3Ka1Ka2Ka3Kbi = KwKa(n-i+1) Reaction constant, KtH+ + OH- H2O H+ + Ac- HAcOH- + HAc H2O + Ac-3.Acid-base reaction (Titration reaction)1Kt = =10 14.00Kw1Kt = =KaKbKw1Kt = =KbKaKwTitration reaction活度与浓度活度与浓度ai = g gi ci活度：在化学反应中表现出来的有效浓度，活度：在化学反应中表现出来的有效浓度， 通常用通常用a表示表示

9、溶液无限稀时溶液无限稀时: g g =1中性分子中性分子: g g =1溶剂活度溶剂活度: a =1Debye-Hckel公式：公式：（稀溶液（稀溶液I0.1 mol/L）I：离子强度：离子强度, I=1/2ciZi2, zi：离子电荷：离子电荷,B: 常数常数, (=0.00328 25), 与温度、介电常数有关与温度、介电常数有关,：离子体积参数：离子体积参数(pm)-lgg gi=0.512zi2 I1+B I-lgg gi=0.512zi2 I活度常数活度常数 K 与温度有与温度有关关反应：反应：HAB HB+ +A-a(H )(Ac )Kgggg 平衡常数平衡常数aHB + aA -

10、K =aBaHA浓度常数浓度常数 Kc 与温度和离子强度有关与温度和离子强度有关HB+A-Kc = =BHAaHB + aA -=aBaHAg gB g gHA-g gH HB+ g gA-K K g gH HB+ g gA-Material (Mass) Balance (物料平衡物料平衡)各物种的平衡浓度之和等于其分析浓度各物种的平衡浓度之和等于其分析浓度Charge Balance (电荷平衡电荷平衡)溶液中正离子所带正电荷的总数等于负离子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所带负电荷的总数(电中性原则电中性原则)Proton Balance (质子平衡质子平衡)溶液中

11、酸失去质子数目等于碱得到质子数目溶液中酸失去质子数目等于碱得到质子数目Equilibrium in solutionMBE;CBE;PBE Mass Balance Equation MBE 物料平衡式物料平衡式 3NaHCOccCOHCOCOH,NaMBE23332：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各物种的平衡浓度之和等于其分析

12、浓度Mixture of 2 10-3 mol/L ZnCl2 and 0.2 mol/L NH3Cl- = 4 10-3 mol/L Zn2+ +Zn(NH3) 2+ +Zn(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/LNa+ + H+ = OH- + HC2O4- + 2C2O42- The total concentration of positive charge in a solution must eq

13、ual the total concentration of negative charge. (Solution electroneutrality 电中性原则电中性原则)。 Charge Balance Equation CBE 电荷平衡式电荷平衡式Na2C2O4 solutionNaCl solutionNa+ + H+ = OH- + Cl- Proton Balance Equation PBE 质子平衡式质子平衡式 How to obtain PBE：The amount of protons released from acids must equal to that accep

14、ted by bases.溶液中酸失去质子数目等于碱得到质子数目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 (参考水准参考水准), or zero level of protons (零水准零水准) which should be the predominant species in the solution and involved in the proton transfer. 先选零水准先选零水准 (大量存在大量存在,参与质子转移的物质参与质子转移的物质)，一般选取，一般选取投料组分投料组分及及H2O(2) Pu

15、t the products via the reference proton levels accepting or releasing protons on both sides of the equation. 将零水准得质将零水准得质子产物写在等式一边子产物写在等式一边,失质子产物写在等式另一边失质子产物写在等式另一边(3) The concentration of each product must multiply the amount of tranfered protons. 浓度项前浓度项前乘上得失质子数乘上得失质子数 Example:（1） H2OReference pro

16、ton levels ：H2O（2） HAc solutionPBE：AcOHHReference proton levels ：H2O、HAcPBEOHHOHOH3Simplified（3） H3PO4 solutionPBEPO 3HPO2POHOHH342442Reference proton levels ：H2O、H3PO4（4） Na2HPO4 solutionPOOHPOHPOH 2H344243（5） NH4AcNHOHHAcH3(6) Mixture of strong acid/base and weak acid/base.vHCl + HAc Reference pro

17、ton levels ：H2O、HAc PBE: H+ - cHCl = OH- + Ac-vNaOH + NH3 Reference proton levels ：H2O、NH3PBE: H+ + NH4+= OH- - cNaOH (7) Conjugated system cb mol/L NaAc and ca mol/L HAc PBEbc- AcOHH-PBEOH-HAcHacxxReference proton levels ：H2O、HAcReference proton levels ：H2O、Ac-Characteristic of PBE(1) Reference pro

18、ton levels never appear in PBE(2) Only the species involved in the proton transfer appear in PBE. Na2HPO4POOHPOHPOH2H344243Exercises：Na2HPO4 solution H+ + H2PO4- +2H3PO4 = OH- +PO43-Reference proton levels ：H2O、HPO42-Na(NH4)HPO4H+ + H2PO4- +2H3PO4 = OH- +NH3 + PO43-Na2CO3H+ + HCO3- + 2H2CO3 = OH-Ref

19、erence proton levels ：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将平衡浓度与分析浓度联系起来 HA H

20、A c HA , A-= A- c HA Monoprotic weak acidPolyprotic acid/base4B Distribution fraction 分布分数分布分数酸度对弱酸酸度对弱酸(碱碱)形体分布的影响形体分布的影响酸度和酸的浓度酸度和酸的浓度酸度：溶液中酸度：溶液中H的平衡浓度或活度，通常用的平衡浓度或活度，通常用pH表示表示 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：酸的分析浓度，包含未解离的和已解离的 酸的浓度酸的浓度 对一元弱酸：对一元弱酸：cHAHA+A-HAcHAcKaHAc+H+=1. Monoprotic weak

21、acidHAc Ac- H+ + cHAc = HAc+Ac-HAcHAc =cHAcH+=H+ + Ka HAcHAc+Ac-HAc= Ac- Ac-Ac- = =cHAcHAc+Ac-H+ + Ka Ka =Ac-HA A -1Characteristic of is a function of pH and pKa，independent of the apparent concentration of acid cH+=H+ + Ka HAH+ + Ka Ka =A- Example CalculateHAc andAc- of HAc solution at pH4.00 and 8

22、.00, respectively.Solution: Ka, HAc=1.7510-5 At pH4.00 At pH8.00 HAc = 5.710-4, Ac- 1.0H+HAc = = 0.85H+ + Ka Ka Ac- = = 0.15H+ + Ka pH HAA- pKa- 2.00.990.01*pKa- 1.30.950.05 pKa- 1.00.910.09*pKa0.500.50 pKa+ 1.00.090.91*pKa+ 1.30.050.95 pKa+ 2.00.010.99 HA andA- at different pH*pKa- 1.30.950.05*pKa0

23、.500.50*pKa+ 1.30.050.95对于给定弱酸，对于给定弱酸， 对对pH作图作图分布分数图分布分数图0 2 4 6 8 10120.00.51.04.76pHDistribution plot of HAc（pKa=4.76）3.46 6.06pKa1.3pHHAcAc-4.76Predominant range优势区域图优势区域图Distribution plot of HF（PKA=3.17）HF F-pKa3.17pH1.00.50.00 2 4 6 8 10 12 pH 3.17HFF-Predominant rangeDistribution plot of HCN（p

24、KA=9.31）pKa9.31HCN CN-pH0 2 4 6 8 10 12 pH1.00.50.09.31HCNCN-Predominant rangeHA的分布分数图（的分布分数图（pKa）分布分数图的特征分布分数图的特征0.00.51.0pkapH 两条分布分数曲线相交于两条分布分数曲线相交于（pka，0.5） pHpKa时，溶液中以时，溶液中以A-为主为主分布分数多元弱酸分布分数多元弱酸二元弱酸二元弱酸H2AH2AH+HA- H+A2-c H2CO3=H2CO3+HCO3-+CO32-H2A=H2Ac H2AA2-HA-defHA-=c H2AdefA2-=c H2Adef物料平衡物

25、料平衡酸碱解离平衡酸碱解离平衡H2AH+HA- H+A2-cH2CO3=H2CO3+HCO3-+CO32-H2A=H2Ac H2AA2-HA-HA-=c H2AA2-=c H2AH+2H+2 + H+Ka1 + Ka1 Ka2=H+2 + H+Ka1 + Ka1 Ka2H+2 + H+Ka1 + Ka1 Ka2H+ Ka1 Ka1 Ka2Polyprotic weak acid: HnAHnAH+Hn-1A- H+HA(n+1)- H+An-H+n=0H+n + H+n-1Ka1 +Ka1 Ka2.KanH+n-1 Ka1 =1H+n + H+n-1Ka1 +Ka1 Ka2.Kan=nH+n

26、+ H+n-1Ka1 +Ka1 Ka2.KanKa1 Ka2.Kan分布分数定义分布分数定义物料平衡物料平衡酸碱解离平衡酸碱解离平衡Distribution plot of H2CO31.00.00 2 4 6 8 10 12 pHH2CO3HCO3-CO32-H2CO3 HCO3- CO32-6.38pKa110.25pKa2pKa = 3.87pHPredominant rangeDistribution plot of H3PO4 H3PO4 H2PO4- HPO42- PO43-2.16 pKa 5.05 7.21 pKa 5.11 12.32 pKa1 pKa2 pKa31.00.0

27、0 2 4 6 8 10 12 pHH3PO4H2PO4-HPO42-PO43-Predominant rangeConclusions on distribution fraction is a function of pH and pKa，independent of the apparent concentration of acid c. 1+2+ +n = 1H+n=0H+n + H+n-1Ka1 +Ka1 Ka2.KanH+n-1 Ka1 =1H+n + H+n-1Ka1 +Ka1 Ka2.Kan=nH+n + H+n-1Ka1 +Ka1 Ka2.KanKa1 Ka2.Kan 仅是

28、仅是pH和和pKa 的函数，与酸的分析浓度的函数，与酸的分析浓度c无关无关 4C Calculation of the concentration of proton1. Strong acid/base2. Monoprotic weak acid/base HA Polyprotic weak acid/base H2A, H3A3. Amphiprotic compound HA-4. Mixture of acid and base strong+weak weak+weak5. Conjugate weak acid/weak base pair HA+A-1 强酸碱溶液强酸碱溶液强

29、酸强酸(HCl):强碱强碱(NaOH): cHCl=10-5.0and10-8.0 molL-1, pH=?质子条件质子条件: H+ + cNaOH = OH-最简式最简式: OH- = cNaOH质子条件质子条件: H+ = cHCl + OH-最简式最简式: H+ = cHCl242wKCCH公式推导？公式推导？Kw is a constant, soOHOHKw3141014143100.52.010OHOHwK3 .13100 . 5lgOHlgpH143Example Calculate H+ and OH- in 0.200 mol/L NaOHYou cant always as

30、sume 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 wrong!242KccHwaaFormularCondition10

31、-6.0 - 10-8.0H+=ca 10-6.0OH-= cb10Kw , 忽略忽略Kw (即忽略水的酸性即忽略水的酸性) HA=ca-A-=ca-(H+-OH-) ca-H+ 近似计算式近似计算式:展开得一元二次方程展开得一元二次方程: H+2+KaH+-caKa=0，求解即可，求解即可H+ KaHA + Kw精确表达式：精确表达式：H+ Ka (ca - H+) If Kac10Kw, and c/Ka 100 then ignore the effect of dissociated acid on HA, HA c Simplified:H+ KaHA + KwH+ Kac + Kw

32、Simplified: Furthermore, if c/Ka 100, then c H+ cH+ Ka c(1) Kac10Kw : (3) c/Ka 100 ； Kac 10Kw : (2) Kac10Kw ；ca/Ka 100 :H+ KaHA + KwConclusion：H+ Ka (c - H+)H+ Kac + Kw(The simplest)H+ Kac例例 计算计算0.20molL-1 Cl2CHCOOH 的的pH.(pKa=1.26)如不考虑酸的离解如不考虑酸的离解(用最简式用最简式:pH=0.98), 则则 Er=29%解解: Kac =10-1.260.20=10-

33、1.9610Kw c/Ka = 0.20 / 10-1.26 =100.56 100故近似式故近似式: 解一元二次方程解一元二次方程: H+=10-1.09 则则pH=1.09H+ Ka (ca - H+)2 Calculate the pH of 0.1 mol/L monochloroacetic acid (一氯乙酸一氯乙酸). (Ka = 1.4 10-3),10071/aaKc,10waaKcK)H(HAHaaacKKSolution:0HH2aaacKK3 Calculate the pH of 1.0 10-4 mol/L HCN. (Ka = 6.2 10-10)Solutio

34、n:,100/aaKc,10waaKcKwaaKcKHpH=1.961 Calculate the pH of 0.10 mol/L HAc. (pKa = 4.76)Solution:,10waaKcK,100/aaKc88. 2)00. 176. 42/1)p(p21pH（aacKExercises处理方式与处理方式与一元弱酸一元弱酸类似类似用用Kb 代替代替Ka，OH-代替代替H+一元弱酸的公式可直接用于一元弱碱的计算一元弱酸的公式可直接用于一元弱碱的计算直接求出直接求出：OH-, 再求再求H+ pH=14-pOH一元弱碱一元弱碱(B-)质子条件式质子条件式: OH-= H+ + HB

35、代入代入平衡关系式平衡关系式B- KbOH-OH- OH-Kw= +精确表达式精确表达式: OH- =KbB- + Kw(1) Kbc 10Kw : (2) c/Kb 100 : (3) Kbc 10Kw, c/Kb 100 :OH-=Kb (cb-OH-) OH-=Kb cb + KwH+= KaKwcbOH-= Kbcb最简式最简式: 多元弱酸溶液多元弱酸溶液二元弱酸二元弱酸(H2A)质子条件质子条件: H+ = HA- + 2A2- + OH-2Ka2H+= Ka1H2A (1+ ) + KwH+酸碱平衡关系酸碱平衡关系KwKa1H2AH+= + +2Ka1Ka2H2AH+H+2H+ 0

36、.05, 可略可略 近似式近似式: 以下与一元酸的计算方法相同以下与一元酸的计算方法相同Ka1ca 10Kw2Ka2H+= Ka1H2A (1+ ) + KwH+2Ka2H+= Ka1H2A (1+ ) + KwH+2Ka2H+2Ka2H+H+= Ka1H2A(忽略二级及以后各步离解忽略二级及以后各步离解)Ka1c 10Kw， KHBcHB弱酸弱酸+弱碱弱碱(HA+B-)溶液溶液质子条件质子条件: H+ + HB = A- + OH-HA cHA HBcHB酸碱平衡关系酸碱平衡关系KwH+HBH+ = +KHAHAKHBH+H+H+ KHAKHBcHA/cB4. Amphiprotic com

37、poundsAmphiprotic compounds behave as acids in the presence of basic solutes and bases in the presence of acidic solutes.在溶液中既起酸在溶液中既起酸(给质子给质子)、又起碱（得质子）的作用、又起碱（得质子）的作用*Polyacid salt Na2HPO4, NaH2PO4, Weak-acid weak-base Salt NH4Ac Amino acid质子条件质子条件: H+H2A=A2 -+OH-精确表达式精确表达式:酸碱平衡关系式酸碱平衡关系式 酸式盐酸式盐 Na

38、HAKwH+HA-H+ = +Ka2HA-Ka1H+H+H+ Ka1(Ka2HA-+Kw)Ka1+HA-若若: Ka1Ka2, HA-c (pKa3.2)近似计算式近似计算式:如果如果 c 10Ka1, 则则“Ka1”可略可略,得得 最简式最简式:H+=Ka1Ka2 c Ka1+ cH+=Ka1(Ka2 cKw)Ka1+ c若若Ka2c 10Kw 则则 Kw可忽略可忽略H+ Ka1Ka2精确式：精确式：Ka1Ka2, HA-cKa2c 10Kw c 10 Ka1pH = 1/2(pKa1 + pKa2)H+ Ka1(Ka2c+Kw) Ka1+cH+ Ka1Ka2c Ka1+ cH+ Ka1Ka

39、2H+ Ka1(Ka2HA-+Kw)Ka1+HA-弱酸弱碱盐弱酸弱碱盐 NH4Ac质子条件式质子条件式: H+ + HAc = NH3 + OH-Kac 10Kw c 10 KaH+ Ka(Kac+Kw) Ka1+c酸碱平衡关系酸碱平衡关系NH4+ Ac-cH+= KaKac Ka+ cH+ KaKaKa NH4+Ka HAc例例 计算计算 0.0010 mol/L CH2ClCOONH4 溶液的溶液的pH CH2ClCOOH: Ka=1.410-3NH3: Kb=1.810-4 Ka=5.610-10 Kac 10Kw , c 10Kw c/Ka1 10H+ Ka1Ka2c Ka1+ cH+

40、 Ka1(Ka2c+Kw) Ka1+c酸碱平衡关系酸碱平衡关系H+ Ka1Ka2 综合考虑、分清主次、合理取舍、近似计算综合考虑、分清主次、合理取舍、近似计算酸碱溶液酸碱溶液H+的计算总结的计算总结质子条件质子条件物料平衡物料平衡电荷平衡电荷平衡 酸碱平衡关系酸碱平衡关系 H+的精确表达式的精确表达式近似处理近似处理H+的近似计算式和最简式的近似计算式和最简式nA buffer solution is a solution of a conjugate acid/base pair that resists changes in pH. 4D Buffer solutionsA conjuga

41、te weak acid/weak base pair: ca mol/L HB and cb mol/L B-BHHBH+BHHBOH-Small pH shiftHBB- + H+缓冲溶液：缓冲溶液：能减缓强酸强碱的加入或稀释而引能减缓强酸强碱的加入或稀释而引起的起的pH变化变化 0 1 2 3 4 5 6 7 0 10 20 30 40 50 mL of Water pH a) 0 1 2 3 4 5 6 7 0 10 20 30 40 50 mL of Water pH b) Plots of pH vs. mL of water added to (a ) 0.500 mL of 0

42、.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 solutionHClNaAc-HAcH+= Ka =KaHAA-ca-H+OH-cb+H+-OH-H+= Ka HAA-camol/L HA+ cbmol/L NaAPBE：HA=ca+OH-H+ A- = cb+H+-OH-物料平衡物料平衡: HA + A- = ca + cb电荷平衡电荷平衡: H+ + Na+ = OH- + A- HA = ca - H+ +

43、 OH- A- = ca + cb-HA = cb + H +-OH-+)4D-1 Calculation of buffer solutions pHpH 8 (碱性碱性),略去略去H+若若ca 20H+; cb 20H+, 或或ca 20OH-; cb 20OH-,最简式最简式H+= Ka ca cb计算方法：计算方法：(1) 先按最简式计算先按最简式计算OH-或或H+。(2) 再计算再计算HA或或A-,看其是否可以忽略看其是否可以忽略.如果不如果不能忽略能忽略,再按近似式计算。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

44、算以用最简式直接计算pH例例(1) 0.10 mol/L NH4Cl 0.20 mol/L NH3 先按最简式先按最简式: (2) 0.080 mol/L二氯乙酸二氯乙酸 0.12mol/L二氯乙酸钠二氯乙酸钠 先用最简式求得先用最简式求得 H+0.037 mol/LcaOH+, cbOH- 结果合理结果合理 pH=9.56应用近似式应用近似式: :解一元二次方程，解一元二次方程，H+=10-1.65 molL-1 , pH=1.65 pH = pKa + lg =9.56cacbH+= Kaca-H+ cb+H+Example How to prepare 200 mL pH2.0 buff

45、er with (MNH2CH2COOH=75.01; pKa1=2.35; cNH2CH2COOH=0.1 molL-1; cHCl=1.0molL-1)Solution：)(50. 107.7510200100. 03gmxx1 . 0lg35. 200. 21069. 0Lmolx)(8 .130 . 1200069. 0mLVHClIf c= x1.50g; 1.0 molL-1 HCl 13.8mL; to 200 mL solutionnMultiprotic weak acid-based buffers display many different pH. nA diproti

46、c weak acid can be used to prepare buffers at two pHs. nBuffers of malonic acid (pKa1 = 2.85 and pKa2 = 5.70) can be prepared:nThe effect of dilutionnThe effect of added acids and bases4D-2 Properties of buffer solution缓冲容量缓冲容量缓冲容量：缓冲容量：衡量缓冲溶液缓冲能力大小，用衡量缓冲溶液缓冲能力大小，用 表示表示 dc/dpH 加合性：加合性： = H+ OH- HA =

47、2.3H+2.3OH-+2.3HAAcHA 对于对于pH在在pKa 1范围内的范围内的HA =2.3HAAcHA HAA的缓冲体系的缓冲体系 有极大值有极大值 pHpKa时，时， 即即HA=A 极大极大0.58cHA4D-3. Buffer capacityFactors: The absolute concentration of the weak acid and the weak base. Their relative proportions of the weak acid and the weak base.OH- HOHHAHAHbaaaccKKbaaccKH标准缓冲溶液标准缓冲

48、溶液校准酸度计校准酸度计常用缓冲溶液常用缓冲溶液pH 标标 准准 溶溶 液液pH（25）饱和酒石酸氢钾饱和酒石酸氢钾 (0.034molL-1)3.560.050 molL-1 邻苯二甲酸氢钾邻苯二甲酸氢钾4.010.025 molL-1 KH2PO4 0.025 molL-1 Na2HPO46.860.010 molL-1硼砂硼砂 9.18常用缓冲溶液常用缓冲溶液8.5109.25NH4+NH38.5109.24硼砂硼砂(H3BO3+H2BO3-)7.598.21三羟甲基甲胺三羟甲基甲胺+HCl6.587.21H2PO4-+HPO42-4.56.05.13六次甲基四胺六次甲基四胺+HCl45

49、.54.76HAc+NaAc34.53.77甲酸甲酸+NaOH23.52.86氯乙酸氯乙酸+NaOH1.53.02.35氨基乙酸氨基乙酸+HCl缓冲范围缓冲范围pKa缓冲溶液缓冲溶液缓冲溶液的选择原则缓冲溶液的选择原则 不干扰测定不干扰测定 例如：例如：EDTA滴定滴定Pb2+,不用不用HAc-Ac-有较大的缓冲能力，足够的缓冲容量有较大的缓冲能力，足够的缓冲容量 较大浓度较大浓度 (0.011molL-1); pHpKa 即即ca cb1 1 HAc NaAc : pKa=4.76 (45.5) NH4OHNH3: pKb=4.75 (810 ) (CH2)6N4 (CH2)6N4H+: p

50、Kb=8.87 （4.56）nPreparation of buffersvA buffer solution of any desired pH can be prepared by combing calculated quantities of a suitable conjugate acid/base pair.vPrepared buffers by making up a solution of approximately the desired pH and then adjust by adding acid or conjugate base until the requi