基础化学梁逸增PPT课件

1、 本章教学内容 4.1 缓冲溶液与缓冲原理缓冲溶液与缓冲原理 4.2 缓冲溶液的缓冲溶液的pH值计算值计算 4.3 缓冲容量和缓冲范围缓冲容量和缓冲范围 4.4 缓冲溶液的配置缓冲溶液的配置 4.5 缓冲溶液在医学上的意义缓冲溶液在医学上的意义第1页/共46页内容提要内容提要:1.缓冲溶液的组成及缓冲机制 5. 血液中的缓冲系2.缓冲溶液pH的计算3.缓冲容量和缓冲范围及其影响因素4. 缓冲溶液的配制第2页/共46页重点:难点:1. 缓冲溶液缓冲溶液pH值的计算值的计算2. 缓冲容量的概念缓冲容量的概念3. 缓冲溶液的选择与配制方法。缓冲溶液的选择与配制方法。缓冲容量的概念缓冲容量的概念 第3

2、页/共46页 solutions that resist change in hydronium ion, H+, and the hydroxide ion,OH-, concentration (and consequently pH) upon addition of small amounts of acid or base, or upon dilution. 4.1.1 The Essential Feature of a Buffer Buffer Solution:4.1 缓冲溶液与缓冲原理缓冲溶液与缓冲原理第4页/共46页 Buffers are mixtures conta

3、ining a common ion: either weak acids and their conjugate bases or weak bases and their conjugate acid. Two common buffers: ammonium- ammonia, carbonate-bicarbonate NH4+(aq) H+(aq) + NH3 (aq) CO32- (aq) + H2O (l) H CO3-(aq) + OH- (aq)第5页/共46页共轭酸共轭碱HAcNH4ClH2PO4-NaAcNH3H2OHPO42-抗酸成分缓冲系抗碱成分缓冲溶液组成示意图第6

4、页/共46页Buffer systemConjugate acidConjugate basepKa( at 25)HAcNaAcHAcAc-4.76H2CO3 NaHCO3H2CO3HCO3-6.35H3PO4 NaH2PO4H3PO4H2PO4-2.16TrisHCl TrisTrisH+Tris7.85H2C8H4O4 KHC8H4O4H2C8H4O4HC8H4O4-2.89NH4Cl NH3NH4+NH39.25CH3NH3+Cl- CH3NH2CH3NH3+CH3NH210.63NaH2PO4 Na2HPO4H2PO4-HPO42-7.21Buffer systems that ar

5、e useful at various pH values第7页/共46页Tris: Tris(Hydroxymethy)methanamin三羟甲基氨基甲烷三羟甲基氨基甲烷 CCH2OHCH2OHHOH2CNH2第8页/共46页 下列情况均需pH一定的缓冲溶液： 大多数为酶所控制的生化反应 微生物的培养 组织切片 细胞染色 药物调剂、研制等第9页/共46页Buffer with equal concentrations of conjugate base and acidOH-H3O+Buffer after addition of H3O+H2O + CH3COOH H3O+ + CH3C

6、OO-Buffer after addition of OH-CH3COOH + OH- H2O + CH3COO-4.1.2 How a Buffer Works第10页/共46页HAc + H2O H3O+ + AcH+ +Shift left+OHH2OShift rightAnti-acidAnti-baseanti-acid mechanismanti-base mechanism第11页/共46页The amounts of weak acid and weak base in the buffer must be significantly larger than the amo

7、unts of H3O+ or OH- that will be added, otherwise the pH cannot remain approximately constant. Thus addition of limited amounts of a strong acid or base is counteracted by the species present in the buffer solution, and the pH changes very little. No solution can keep the pH approximately constant i

8、f you add larger amounts of either acid or base that are present in the original buffer. 第12页/共46页For a HB-NaB buffer system,HB + H2OH3O+ + BNaB Na+ + BKa =H+BHBH+ = KaHBBApply log on both sides of above equation,pH = pKa + lg BHBThe Henderson-Hasselbalch Equation4.2 缓冲溶液缓冲溶液pH值的计算值的计算第13页/共46页pH =

9、pKa + lg BHB= pKa + lgconjugate baseconjugate acidpKa ：the log of Ka of the conjugate acidB、HB：equilibrium concentrationB / HB：buffer ratioB+HB： total concentrationHB = cHB cHB(dissociated) B = cNaB + cHB(dissociated) cHB cNaBThe Henderson-Hasselbalch Equation第14页/共46页pH = pKa + lgcBcHB= pKa + lgnB

10、/ VnHB / VpH= pKa + lg nBnHBIf the concentrations of conjugate acid and base used are equal, i.e. cB = cHB .pH = pKa + lg cB VB cHB VHB= pKa + lgVBVHBthree different types of of Henderson-Hasselbalch Equation Fight dilutionHBBVVlg pKapH-第15页/共46页 HBBlg p lg HBBlg p HBBlg paalg p pH-,HBB-aHBB_aHBBa_a

11、KKKK如果用活度代替浓度，校正因数第16页/共46页Calculating the pH of a Buffer Solution-1PROBLEMSample Problem 4-1A buffer is prepared by mixing equal volumes of 0.2 molL-1 NaAc and 0.4 molL-1 HAc. What is the pH of the final solution? The pKa of HAc is 4.74. What will the pH be after the addition of 0.005mol of NaOH(s)

12、 to 500 mL of the buffer solution described above?第17页/共46页SOLUTIONBefore the addition of 0.005mol of NaOH(s) to 500 mL of the buffer solution44. 42 . 01 . 0lg74. 4lgHAApKpHaCalculating the pH of a Buffer Solution-1Sample Problem 4-1第18页/共46页After the addition of 0.005mol of NaOH(s) to 500 mL of the

13、 buffer solution50. 452504 . 052502 . 0lg74. 4)()()()(lg11mmolmLLmolmmolmLLmolOHnHAnOHnAnpKpHaCalculating the pH of a Buffer Solution-1Sample Problem 4-1SOLUTION第19页/共46页例例4-2： 将将0.10 molL-1 的的 H3PO4溶液溶液20 mL与与 0.10 molL-1 的的 NaOH溶液溶液30 mL混合，混合溶液是缓冲混合，混合溶液是缓冲溶液吗溶液吗? 求混合溶液的求混合溶液的pH。已知：。已知：pKa1=2.16,

14、pKa2=7.21, pKa3=12.32。7.21 1.0mmol1.0mmollg7.21 POHHPOlg p pH42-24a2K反应2： NaH2PO4 + NaOH Na2HPO4 + H2O反应前 2.0mmol 1.0mmol 反应后 1.0mmol 1.0mmol反应1： H3PO4 + NaOH NaH2PO4 + H2O反应前 200.10mmol 300.10mmol 反应后 1.0mmol 2.0mmol解:第20页/共46页4.3.1 The Concepts of Buffer CapacityBuffer capacity is defined as the a

15、mount of strong acid or base needed to change the pH of one liter of buffer by 1 unit.Buffer capacity is the ability to resist pH change.Or, more specifically,dpH d)b(aVn4.3 缓冲容量和缓冲范围缓冲容量和缓冲范围第21页/共46页dpH d)b(aVnwhere is the buffer capacity and has units of moles per liter per pH (molL-1pH-1); dna (

16、or b) stands for moles of strong acid or strong base which are added to a buffer solution to cause the change in pH, dpH.第22页/共46页The following can be derived from above one： = 2.303 HBB / ctotalunit：mol L1 pH1The magnitude of indicates the relative strength of buffer capacity. The larger the value

17、of , the greater the capacity of the buffer to resist changes in pH.第23页/共46页4.3.2 Factors Affecting Buffer CapacityBuffer capacity depends on two factors:Relative one: buffer ratio, B / HB.Absolute one: total conc. of buffer, B + HB第24页/共46页When Ctotal if fixed： cBcHB=11(max)cBcHB=101 mincBcHB=110

18、decrease minFor a given buffer pair, the more the buffer ratio approach 1, the stronger the capacity ; when the buffer ratio equals 1, the capacity reaches its maximum. decrease max= 0.576ctotal第25页/共46页总总总总cccc576. 02/12/1303. 2因为， = 2.303 HBB / c总即：即： max= 0.576ctotal缓冲比等于1时，HB=B-=1/2 c总所以，第26页/共4

19、6页When the buffer ratio, c(B-)/c(HB), is fixed, the more concentrated the components of a buffer, the greater the buffer capacity.The pH of a buffer is distinct from its buffer capacity.When the total concentration of buffer is fixed, the more the ratio of c(B-)/c(HB) approaches 1, the more the buff

20、er capacity. When c(B-)/c(HB)=1, the buffer has the highest capacity.Conclusion: 第27页/共46页（1）HCl（2）0.1 molL-1 HAc+NaOH（3）0.2 molL-1 HAc+NaOH（4）0.05 molL-1 KH2PO4+NaOH（5）0.05 molL-1 H2BO3+NaOH（6）NaOH缓冲容量与缓冲容量与pH的关系的关系第28页/共46页Buffer range the pH range over which the buffer acts effectively.Buffers ha

21、ve a usable range within 1 pH unit of the pKa of its acid component.4.3.3 Buffer Range 1cBcHB=1(max)cBcHB=101 mincBcHB=110 decrease min decreasepH = pKa 1buffer effective range第29页/共46页1. Choose the conjugate acid-base pair.2. Calculate the ratio of buffer component concentrations.3. Determine the b

22、uffer concentration (0.05molL-10.2molL-1. 渗透压 )4. Mix the solution and adjust the pH.General steps for a buffer preparation:4.4 缓冲溶液的配置缓冲溶液的配置第30页/共46页例例4-3： 如何配制pH=5.00的缓冲溶液100mL? 解：解： (1) 选择缓冲系（表4-1） 由于HAc的pKa= 4.76，接近所配缓冲溶液 pH=5.00，所以可选用HAc-Ac-缓冲系。 (2) 确定总浓度，计算所需缓冲系的量 一般具备中等缓冲能力 (0.050.2molL-1)即可

23、， 并考虑计算方便，选用0.10molL-1的HAc和 0.10molL-1 NaAc溶液。应用式(4-2)可得第31页/共46页设 V(NaAc) = x mL则 ： V(NaAc) = x mL = 64mLV(HAc)= 100mL - 64mL= 36mL(HAc)(AclgppHaVVK240100lg.xxmL )100(mL lg76. 400. 5xx第32页/共46页An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on lim

24、estone-rich soils. How many grams of Na2CO3 must be added to 1.5L of freshly prepared 0.20M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11.PLANWe know the Ka and the conjugate acid-base pair. Convert pH to H3O+, find the number of moles of carbonate and convert to mass.Preparing a buffer-1PROBL

25、EMSample Problem 4-4第33页/共46页)c(HCO)c(COlg pKapH-3-23We know the pH of the buffer is 10.0. The conc. of NaHCO3 is 0.20M, Using Henderson-Hasselbalch Equation we can find out the conc. of CO32- in the buffer. Ka of HCO3- is 4.7x10-11.2 . 0)(log11107 . 4log0 .1023COcCO32- = 0.094Mmoles of Na2CO3 = (1.

26、5L)(0.094mols/L)= 0.14 = 15 g Na2CO30.14 moles 105.99gmolSOLUTION第34页/共46页PROBLEMPreparing a buffer-2Sample Problem 4- 5There is 2 liter of 0.50molL-1NH3H2O and 2 liter of 0.50molL-1HCl (hydrochloric acid) in a laboratory. A technician wants to use them to prepare a buffer with pH=9.00 without the a

27、ddition of water. How many liters of buffer can the technician prepare at most ? What are the concentrations of NH3H2O and NH4+ in the buffer? The pkb(NH3H2O)=4.74.第35页/共46页SOLUTIONTo prepare the buffer with the volume as more as possible, 2 liter of 0.500molL-1 NH3H2O must be utilized completely, w

28、hile only a part of the HCl can be used. Let the volume of HCl used be x L, so, the total volume of the buffer prepared is (2.00+x )L. After the neutralization,Preparing a buffer-2Sample Problem 4- 5第36页/共46页NH3 (aq) + HCl (aq) NH4+ (aq) + Cl-(l)Initial(mol) 1.0 0.5V 0change(mol) -0.5V -0.5V +0.5VEq

29、uil-(mol) 1.0- 0.5V 0 0.5V0 . 95 . 05 . 00 . 1lg0 .14lgVVpKHAApKpHbaV = 1.3L , so the biggest volume of buffer is 3.3L.13106. 03 . 335. 03 . 35 . 00 . 1)(LmolVNHc14200. 03 . 365. 03 . 35 . 0)(LmolVNHcPreparing a buffer-2Sample Problem 4- 5第37页/共46页用0.030molL-1H3PO4溶液和同浓度的NaOH溶液配制pH=7.40的生理缓冲溶液200mL,

30、 需要H3PO4 和NaOH 各多少毫升？第38页/共46页The pH of the blood in a healthy individual remains remarkably constant at 7.35 to 7.45. This is because the blood contains a number of buffers that protect against pH change due to the presence of acidic or basic metabolites. From a physiological viewpoint, a change of

31、 0.3 pH unit is extreme. Acid metabolites are ordinarily produced in greater quantities than basic metabolites, and carbon dioxide is the principal one. 4.5 缓冲溶液在医学上的意义缓冲溶液在医学上的意义第39页/共46页 体液中存在多种生理缓冲系，使体液的体液中存在多种生理缓冲系，使体液的pH保持保持基基 本稳定本稳定 例如：血液的pH保持在7.357.45之间 血浆中: H2CO3 -HCO3- 、H2PO4-HPO42- 、HnP-Hn-1P- 红细胞中： H2b-Hb-(血红蛋白血红蛋白)、H2bO2-HbO2-(氧合血红蛋白氧