基础化学梁逸增PPT学习教案

1、会计学1基础化学梁逸增基础化学梁逸增 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5. 血液中的缓冲系血液中的缓冲系2.缓冲溶液缓冲溶液pH的计算的计算3.缓冲容量和缓冲范围及其影响因素缓冲容量和缓冲范围及其影响因素4. 缓冲溶液的配制缓冲溶液的配制第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。

2、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. 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页第5页/共46页共轭酸共轭酸共轭碱共轭碱HAcNH4ClH2PO4-NaA

3、cNH3H2OHPO42-抗抗酸酸成分成分缓冲系缓冲系抗抗碱碱成分成分缓冲溶液组成示意图缓冲溶液组成示意图第6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.

4、63NaH2PO4 Na2HPO4H2PO4-HPO42-7.21Buffer systems that are 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 下列情况均需下列情况均需pH一定的缓冲溶液：一定的缓冲溶液： 大多数为酶所控制的生化反应大多数为酶所控制的生化反应 微生物的培养微生物的培养 组织切片组织切片 细胞染色细胞染色 药物调剂、研制等药物调剂、研制等第9页/共46页Buffer with equa

5、l concentrations of conjugate base and acidOH-H3O+Buffer after addition of H3O+H2O + CH3COOH H3O+ + CH3COO-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 mechanis

6、m第11页/共46页The amounts of weak acid and weak base in the buffer must be significantly larger than the amounts 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 t

7、he buffer solution, and the pH changes very little. No solution can keep the pH approximately constant if 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 o

8、f 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 = 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 = cN

9、aB + cHB(dissociated) cHB cNaBThe Henderson-Hasselbalch Equation第14页/共46页pH = pKa + lgcBcHB= pKa + lgnB / 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-Hasselba

10、lch Equation Fight dilutionHBBVVlg pKapH-第15页/共46页 HBBlg p lg HBBlg p HBBlg paalg p pH-,HBB-aHBB_aHBBa_aKKKK如果用活度代替浓度，如果用活度代替浓度，校正因数校正因数第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

11、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) 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. 4lgHAApKpHaCalculat

12、ing 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 buffer solution50. 452504 . 052502 . 0lg74. 4)()()()(lg11mmolmLLmolmmolmLLmolOHnHAnOHnAnpKpHaCalculating the pH of a Buffer Solution-1Sample Problem 4-1SOLUTION第19页/共46页7.21 1.0mmol

13、1.0mmollg7.21 POHHPOlg p pH42-24a2K反应反应2： NaH2PO4 + NaOH Na2HPO4 + H2O反应前 2.0mmol 1.0mmol 反应后 1.0mmol 1.0mmol反应反应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 amount of strong acid or base n

14、eeded 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 (or b) stands for moles of stro

15、ng 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 of , the greater the capacity

16、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 decrease minFor a given buffer

17、 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，因为， = 2.303 HBB / c总即：即： max= 0.576ctotal缓冲比等于缓冲比等于1时，时，HB=B-=1/2 c总所以，所以，第26页/共46页When the buffer

18、 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 buffer capacity. When

19、 c(B-)/c(HB)=1, the buffer has the highest capacity.Conclusion: 第27页/共46页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 have 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 de

20、crease 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 buffer concentration (0.05molL-10.2molL-1. 渗透压 )4. Mix the solution and adjust the pH.General steps for a buffer preparation:4.

21、4 缓冲溶液的配置缓冲溶液的配置第30页/共46页第31页/共46页(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 limestone-rich soils. How many grams of Na2CO3 must be added to 1.5L of freshly prepared 0.20M NaHCO3

22、 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-1PROBLEMSample 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 con

23、c. 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 = (1.5L)(0.094mols/L)= 0.14 = 15 g Na2CO30.14 moles 105.99gmolSOLUTION第34页/共46页PROBLEMPreparing a buffe

24、r-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 addition of water. How many liters of buffer can the technician prepare at most ? What are the conc

25、entrations 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, while only a part of the HCl can be used. Let the volume of HCl used be x L, so, the total volume o

26、f 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 +0.5VEquil-(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, 需要H3PO4 和NaOH 各多少毫升？第38页/共46页The pH of the bloo