基礎(chǔ)化學(xué)-梁逸增-第四章課件教學(xué)文案

1、基礎(chǔ)化學(xué)-梁逸增-第四章課件 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n Buffers are mixtures containing a common ion: either weak acids and their conjugate bases or weak bases and their conjugate acid. n Two common buffers: ammonium-

2、ammonia, carbonate-bicarbonate NH4+(aq) H+(aq) + NH3 (aq) CO32- (aq) + H2O (l) H CO3-(aq) + OH- (aq)共軛酸共軛酸共軛堿共軛堿HAcNH4ClH2PO4-NaAcNH3H2OHPO42-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Buffer systemConjugate acidConjugate basepKa( at 25)HAcNaAcHAcAc-4.76H2CO3 NaHCO3H2CO3HCO3-6.35H3PO4 NaH2PO4H3PO4H2PO4

3、-2.16TrisHCl TrisTrisH+Tris7.85H2C8H4O4 KHC8H4O4H2C8H4O4HC8H4O4-2.89NH4Cl NH3NH4+NH39.25CH3NH3+Cl- CH3NH2CH3NH3+CH3NH210.63NaH2PO4 Na2HPO4H2PO4-HPO42-7.21Buffer systems that are useful at various pH valuesTris: Tris(Hydroxymethy)methanamin三羥甲基氨基甲烷三羥甲基氨基甲烷 CCH2OHCH2OHHOH2CNH2 下列情況均需下列情況均需pH一定的緩沖溶液：一定

4、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Buffer with equal 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 WorksHAc + H2O H3O+ +

5、AcH+ +Shift left+OHH2OShift rightAnti-acidAnti-baseanti-acid mechanismanti-base mechanismThe 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 limit

6、ed 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 if you add larger amounts of either acid or base that are present in the original buffer. For a HB-NaB buffer system,H

7、B + H2OH3O+ + BNaB Na+ + BKa =H+BHBH+ = KaHBBApply log on both sides of 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pH = pKa + lg BHB= pKa + lgconjugate baseconjugate acidpKa ：the log of Ka of the conjugate acidB、HB：equilibrium concentrationB / HB：buffer r

8、atioB+HB： total concentrationHB = cHB cHB(dissociated) B = cNaB + cHB(dissociated) cHB cNaBThe Henderson-Hasselbalch EquationpH = 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 + l

9、gVBVHBthree different types of of Henderson-Hasselbalch Equation Fight dilutionHBBVVlg pKapH- 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，校正校正因數(shù)因數(shù)Calculating the pH of a Buffer Solution-1PROBLEMSample Problem 4-1A buffer is prepared by mixing equal volumes of 0.2 mol

10、L-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) to 500 mL of the buffer solution described above?SOLUTIONBefore the addition of 0.005mol of NaOH(s) to 500 mL of the buffer solution44. 42 . 01 . 0

11、lg74. 4lgHAApKpHaCalculating the pH of a Buffer Solution-1Sample Problem 4-1After 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例例4-2

12、： 將將0.10 molL-1 的的 H3PO4溶液溶液20 mL與與 0.10 molL-1 的的 NaOH溶液溶液30 mL混合，混合溶液是緩沖溶液嗎混合，混合溶液是緩沖溶液嗎? 求混合求混合溶液的溶液的pH。已知：。已知：pKa1=2.16, pKa2=7.21, pKa3=12.32。7.21 1.0mmol1.0mmollg7.21 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 + H2

13、O反應(yīng)前 200.10mmol 300.10mmol 反應(yīng)后 1.0mmol 2.0mmol解:4.3.1 The Concepts of Buffer CapacityBuffer capacity is defined as the amount 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 緩沖容量和緩

14、沖范圍緩沖容量和緩沖范圍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 strong acid or strong base which are added to a buffer solution to cause the change in pH, dpH.The following can be derived from above one： = 2.303 HBB / ctotaluni

15、t：mol L1 pH1The magnitude of indicates the relative strength of buffer capacity. The larger the value of , the greater the capacity of the buffer to resist changes in pH.4.3.2 Factors Affecting Buffer CapacityBuffer capacity depends on two factors:Relative one: buffer ratio, B / HB.Absolute one: tot

16、al conc. of buffer, B + HBWhen Ctotal if fixed： cBcHB=11(max)cBcHB=101 mincBcHB=110 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總總總總cccc576. 02/12/130

17、3. 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，時，HB=B-=1/2 c總所以，所以，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 buf

18、fer is fixed, the more the ratio of c(B-)/c(HB) approaches 1, the more the buffer capacity. When c(B-)/c(HB)=1, the buffer has the highest capacity.Conclusion: （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緩沖容量與緩沖容量與pH的關(guān)系的關(guān)系Buffer range t

19、he 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 decrease min decreasepH = pKa 1buffer effective range1. Choose the conjugate acid-base pair.Calculate the ratio of b

20、uffer component concentrations. 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:4.4 緩沖溶液的配置緩沖溶液的配置例例4-3： 如何配制如何配制pH=5.00的緩沖溶液的緩沖溶液100mL? 解：解： (1) 選擇緩沖系（表選擇緩沖系（表4-1） 由于由于HAc的的pKa= 4.76，接近所配緩沖溶液，接近所配緩沖溶液 pH=5.00

21、，所以可選用，所以可選用HAc-Ac-緩沖系。緩沖系。 (2) 確定總濃度，計(jì)算所需緩沖系的量確定總濃度，計(jì)算所需緩沖系的量 一般具備中等緩沖能力一般具備中等緩沖能力 (0.050.2molL-1)即可，即可， 并考慮計(jì)算方便，選用并考慮計(jì)算方便，選用0.10molL-1的的HAc和和 0.10molL-1 NaAc溶液。應(yīng)用式溶液。應(yīng)用式(4-2)可得可得設(shè) V(NaAc) = x mL則 ： V(NaAc) = x mL = 64mLV(HAc)= 100mL - 64mL= 36mL(HAc)(AclgppHaVVK240100lg.xxmL )100(mL lg76. 400. 5xx

22、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 to make the buffer? Ka of HCO3- is 4.7x10-11.PLANWe know the Ka and the conjugate acid-base pai

23、r. Convert pH to H3O+, find the number of moles of carbonate and convert to mass.Preparing a buffer-1PROBLEMSample Problem 4-4)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 buf

24、fer. 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 105.99gmolSOLUTIONPROBLEMPreparing a buffer-2Sample Problem 4- 5There is 2 liter of 0.50molL-1NH3H2O and 2 liter of 0.50molL-1HCl (hydrochloric acid) in a l

25、aboratory. 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 concentrations of NH3H2O and NH4+ in the buffer? The pkb(NH3H2O)=4.74.SOLUTIONTo prepare the buffer with the volume as

26、 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 of the buffer prepared is (2.00+x )L. After the neutralization,Preparing a buffer-2Sample Problem 4- 5NH3 (aq) + HCl (aq) NH

27、4+ (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 Pro

28、blem 4- 5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？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 meta

29、bolites. From a physiological viewpoint, a change of 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 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 體液中存在多種生理緩沖系，使體液的體液中存在多種生理緩沖系，使體液的pH保持基保持基 本穩(wěn)定本穩(wěn)定 例如：血液的pH保持在7.357.45之間 血漿中: H2CO3 -HCO3- 、H2PO4-HPO42- 、HnP-Hn-1P- 紅細(xì)胞中： H2b-Hb-(血紅蛋白血紅