基礎(chǔ)化學(xué)梁逸增PPT學(xué)習(xí)教案

1、會(huì)計(jì)學(xué)1基礎(chǔ)化學(xué)梁逸增基礎(chǔ)化學(xué)梁逸增 本章教學(xué)內(nèi)容 4.1 緩沖溶液與緩沖原理緩沖溶液與緩沖原理 4.2 緩沖溶液的緩沖溶液的pH值計(jì)算值計(jì)算 4.3 緩沖容量和緩沖范圍緩沖容量和緩沖范圍 4.4 緩沖溶液的配置緩沖溶液的配置 4.5 緩沖溶液在醫(yī)學(xué)上的意義緩沖溶液在醫(yī)學(xué)上的意義第1頁(yè)/共46頁(yè)5. 血液中的緩沖系血液中的緩沖系2.緩沖溶液緩沖溶液pH的計(jì)算的計(jì)算3.緩沖容量和緩沖范圍及其影響因素緩沖容量和緩沖范圍及其影響因素4. 緩沖溶液的配制緩沖溶液的配制第2頁(yè)/共46頁(yè)重點(diǎn):難點(diǎn):1. 緩沖溶液緩沖溶液pH值的計(jì)算值的計(jì)算2. 緩沖容量的概念緩沖容量的概念3. 緩沖溶液的選擇與配制方法。

2、緩沖溶液的選擇與配制方法。緩沖容量的概念緩沖容量的概念 第3頁(yè)/共46頁(yè) 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 緩沖溶液與緩沖原理緩沖溶液與緩沖原理第4頁(yè)/共46頁(yè)第5頁(yè)/共46頁(yè)共軛酸共軛酸共軛堿共軛堿HAcNH4ClH2PO4-NaA

3、cNH3H2OHPO42-抗抗酸酸成分成分緩沖系緩沖系抗抗堿堿成分成分緩沖溶液組成示意圖緩沖溶液組成示意圖第6頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)Tris: Tris(Hydroxymethy)methanamin三羥甲基氨基甲烷三羥甲基氨基甲烷 CCH2OHCH2OHHOH2CNH2第8頁(yè)/共46頁(yè) 下列情況均需下列情況均需pH一定的緩沖溶液：一定的緩沖溶液： 大多數(shù)為酶所控制的生化反應(yīng)大多數(shù)為酶所控制的生化反應(yīng) 微生物的培養(yǎng)微生物的培養(yǎng) 組織切片組織切片 細(xì)胞染色細(xì)胞染色 藥物調(diào)劑、研制等藥物調(diào)劑、研制等第9頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)HAc + H2O H3O+ + AcH+ +Shift left+OHH2OShift rightAnti-acidAnti-baseanti-acid mechanismanti-base mechanis

6、m第11頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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 緩沖溶液緩沖溶液pH值的計(jì)算值的計(jì)算第13頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè) HBBlg p lg HBBlg p HBBlg paalg p pH-,HBB-aHBB_aHBBa_aKKKK如果用活度代替濃度，如果用活度代替濃度，校正因數(shù)校正因數(shù)第16頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)7.21 1.0mmol

13、1.0mmollg7.21 POHHPOlg p pH42-24a2K反應(yīng)反應(yīng)2： NaH2PO4 + NaOH Na2HPO4 + H2O反應(yīng)前 2.0mmol 1.0mmol 反應(yīng)后 1.0mmol 1.0mmol反應(yīng)反應(yīng)1： H3PO4 + NaOH NaH2PO4 + H2O反應(yīng)前 200.10mmol 300.10mmol 反應(yīng)后 1.0mmol 2.0mmol解:第20頁(yè)/共46頁(yè)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 緩沖容量和緩沖范圍緩沖容量和緩沖范圍第21頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)總總總總cccc576. 02/12/1303. 2因?yàn)?，因?yàn)椋?= 2.303 HBB / c總即：即： max= 0.576ctotal緩沖比等于緩沖比等于1時(shí)，時(shí)，HB=B-=1/2 c總所以，所以，第26頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)緩沖容量與緩沖容量與pH的關(guān)系的關(guān)系第28頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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. 滲透壓 )4. Mix the solution and adjust the pH.General steps for a buffer preparation:4.

21、4 緩沖溶液的配置緩沖溶液的配置第30頁(yè)/共46頁(yè)第31頁(yè)/共46頁(yè)(HAc)(AclgppHaVVK240100lg.xxmL )100(mL lg76. 400. 5xx第32頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè))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頁(yè)/共46頁(yè)P(yáng)ROBLEMPreparing 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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)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頁(yè)/共46頁(yè)用0.030molL-1H3PO4溶液和同濃度的NaOH溶液配制pH=7.40的生理緩沖溶液200mL, 需要H3PO4 和NaOH 各多少毫升？第38頁(yè)/共46頁(yè)The pH of the bloo