基础化学答案第04

1、 章后习题解答TOP习题什么是缓冲溶液？试以血液中的HCO-HCO-缓冲系为例，说明缓冲作用的原理及其在医学上TOC o 1-5 h z233的重要意义。答能抵抗少量外来强酸、强碱而保持其pH基本不变的溶液称为缓冲溶液。血液中溶解的CO2与HCO-组成缓冲系。正常人体HCO-/CO(aq)为20/1,pH=7.40。若pH7.35,发生酸中毒,pH7.45,332发生碱中毒。当酸性代谢产物增加时，抗酸成分HCO-与HO+结合，增加的HCO可通过加快呼吸以C033232的形式呼出；消耗的HCO-则由肾减少对其的排泄而得以补充；当碱性代谢产物增加时，OH-与HO+生33成HO，促使抗碱成分HCO离

2、解以补充消耗的H0+。同理，减少的HC0及增加的HCO-可通过肺和肾来233233调控。血液中的HC0-HCO-缓冲系与其他缓冲系共同作用，维持pH为7.357.45的正常范围。233什么是缓冲容量？影响缓冲溶量的主要因素有哪些？总浓度均为O.lOmolL-i的HAc-NaAc和HCO-HCO-缓冲系的缓冲容量相同吗？233解缓冲容量是衡量缓冲溶液缓冲能力大小的尺度，表示单位体积缓冲溶液pH发生一定变化时，所能抵抗的外加一元强酸或一元强碱的物质的量。影响缓冲容量的主要因素是缓冲系的总浓度和缓冲比：缓冲比一定时，总浓度越大缓冲容量越大；总浓度一定时，缓冲比越接近于1缓冲容量越大。总浓度及缓冲比相

3、同的HAc-NaAc和HCO-HCO-缓冲系的缓冲容量相同。TOC o 1-5 h z233下列化学组合中，哪些可用来配制缓冲溶液？(1)HCl+NHH0(2)HCl+Tris(3)HCl+NaOH2(4)NaHPO+NaPO(5)HPO+NaOH(6)NaCl+NaAc243434解可用来配制缓冲溶液的是：(1)HCl+NHH0、(2)HCl+Tris、(4)NaHPO+NaPO和HPO32243434+NaOH将0.30molL-!吡啶(CHN，pK=8.77)和0.10molHCl溶液等体积混合，混合液是否为缓冲55b溶液？求此混合溶液的pH。解CHN与HCl反应生成CHNH+Cl-(卩

4、比啶盐酸盐)，混合溶液为0.10molL-iCHN和0.050molL-i555555CHNH+Cl-缓冲溶液，pK=14.00-&77=5.2355aPH=PKa+lgC(CCHHNHl)=也+lg000=E55将10.0gNaCO和10.0gNaHCO溶于水制备250mL缓冲溶液，求溶液的pH。n(HCO-)=3100g84.0g-mol-i=0.119mol=0.094molpH=pK+lgan(CO2-)n(HCO-)3=10-33+lg0.094mol0.119mol=10.236.求pH=3.90，总浓度为0.400molL-i的HCOOH（甲酸）-HCOONa（甲酸钠）缓冲溶液中

5、，甲酸和甲酸钠的物质的量浓度(HCOOH的pK=3.75)a解设c(HCOONa)=xmolL-i,则c(HCOOH)=0.400molL-i-xmolL-ipH=3.75+lgxmol-L-1(0.400-x)mol-L-1=3.90解得c(HCOO-)=xmolL-1=0.234molL-ic(HC00H)=(0.400-0.234)molL-i=0.166molL-i7.向100mL某缓冲溶液中加入0.20gNaOH固体，所得缓冲溶液的pH为5.60.。已知原缓冲溶液共轭酸HB的pK=5.30,c(HB)=0.25molL-1，求原缓冲溶液的pH。an(NaOH)0.20g/40gmol

6、-11000mL100mL1L-0.050molL-i加入NaOH后,pH=5.30+lgB-+0.050molL-1(0.25-0.050)molL-1=5.60解得B-=0.35molL-i原溶液0.35molL-1pH=5.30+lg=5.450.25molL-1&阿司匹林（乙酰水杨酸、以HAsp表示）以游离酸（未解离的）形式从胃中吸收，若病人服用解酸药,调整胃容物的pH为2.95，然后口服阿司匹林0.65g。假设阿司匹林立即溶解，且胃容物的pH不变,问病人可以从胃中立即吸收的阿司匹林为多少克（乙酰水杨酸的Mr=180.2、pK=3.48）?a解pH=pKa+lg需=3.48+lg册=2

7、.95=0.295n(HAsp)依题意n(Asp-)+n(HAsp)=:=0.0036mol180g-mol-i解得n(HAsp)=0.0028mol可吸收阿司匹林的质量二0.0028molX180.2gmol-i=0.50g在500mL0.20molL-iCHCOOH(丙酸，用HPr表示)溶液中加入NaOH1.8g,求所得溶液的近似pH和校正后的精确pH。已知CHCOOH的pK=4.87，忽略加入NaOH引起的体积变化。25a解求近似pHpH=4.78pK+lg=4.87+Ig呢/他罰-1an(HPr)0.500Lx0.20mol-L-i1.8g/40g-mol-i求精确pH,丙酸钠是强电解

8、质0.045mol(0.500LX12+0.045mlX12)0.500L=0.09molL-i心0.1molL-i当Z=0,1=010时，校正因数lg駕=-0-11pH=pKa+lgn(Pr-)+lg丫(Pr-)=4.78+(-0.11)=4.67n(HPr)y(HPr)某医学研究中，制作动物组织切片时需pH约为7.0的磷酸盐缓冲液作为固定液。该固定液的配方是：将29gNaHPO12HO和2.6gNaHPO2H0分别溶解后稀释至1L。若校正因数242242lg丫(HPO4-)=-0.53，计算该缓冲溶液的精确pH。y(HPO-)24解c(NaHPO)=29g=0.081molL-i24358

9、.0g-mol-1x1Lc(NaHP0)=2:6g=0.017molL-i24156g-mol-1x1LTOC o 1-5 h zpH=pK+丫(HPO4-)+lgc(HP2-)=7.21+(-.0.53)+lg.081nolL-1=7.36a2y(HPO-)c(HPO-)0.017nol-L-12424将0.10molL-iHAc溶液和0.10molL-iNaOH溶液以3:1的体积比混合，求此缓冲溶液的pH及缓冲容量。解HAc溶液和NaOH溶液的体积分别为3V和V，c(HAc)=(0.10X3V-0.10XV)molL-i/(3V+V)=0.050molL-ic(Ac-)=0.10molL-

10、iXV/(3V+V)=0.025molL-ipH=4.75+lg=4.450.025mol-L10.050mol-L-1B=2.303x=0.038mol-L-i(0.050+0.025)molL-i0.050molL-ix0.025molL-i某生物化学实验中需用巴比妥缓冲溶液，巴比妥（CHN0）为二元有机酸（用HBar表示，pK812232al=7.43）。今称取巴比妥18.4g,先加蒸馏水配成100mL溶液，在pH计监控下，加入6.00molL-iNaOH溶液4.17mL，并使溶液最后体积为1000mL。求此缓冲溶液的pH和缓冲容量。（已知巴比妥的Mr=184gmol-i)解HBar与N

11、aOH的反应为2HBar(aq)+NaOH(aq)=NaHBar(aq)+HO(l)反应生成的NaHBar的物质的量n(NaHBar)=c(NaOH)V(NaOH)=6.OmolL-iX4.17mL=25mmol，剩余HBar的物质的量为2n(HBar)=n(HBar)-n(NaOH)=一18.4g一X1000-25mmol=75mmol余22184g-mol-ipH=pK+lg=7.43+lg25mml=6.95an(HBar)75mmol2B=2.303X=0.043molL-i(75mmol/i000mL)x(25mmol/i000mL)(75+25)mmol/i000mL)分别加NaO

12、H溶液或HCl溶液于柠檬酸氢钠（缩写Na2HCit）溶液中。写出可能配制的缓冲溶液的抗酸成分、抗碱成分和各缓冲系的理论有效缓冲范围。如果上述三种溶液的物质的量浓度相同，它们以何种体积比混合，才能使所配制的缓冲溶液有最大缓冲容量？（已知HCit的pK=3.13、pK=4.76、3a1a2pK=6.40）a3解.溶液组成缓冲系抗酸成分抗碱成分有效缓冲范围B最大时最大体积比NaHCit+HCl2HCit-HCit2-2HCit2-HCit-23.765.762:1NaHCit+HCl2HCit-HCit-32HCit-2HCit32.134.132:3NaHCit+NaOHHCit2-Cit3-Ci

13、t3-HCit2-5.407.402:12现有（1）0.10molL-1NaOH溶液，（2）0.10molL-1NH溶液，（3）0.10molL-1NaHPO溶液各32450mL,欲配制pH=7.00的溶液，问需分别加入0.10molL“HCl溶液多少mL?配成的三种溶液有无缓冲作用？哪一种缓冲能力最好？解HCl与NaOH完全反应需HCl溶液50mL。(2)HCl(aq)+NHHO(aq)=NHCl(aq)+HO(l)3242NH+的pK=14.004.75=9.25,4a70025+10OmolL_ix50mL-O.lOmolL_ixV(HC1)+gO.lOmolL-ixV(HC1)解得V(

14、HCl)=50mLHCl(aq)+NaHPO(aq)=NaHPO(aq)+NaCl(aq)2424HPO的pK=7.21，34a2r心rciiO.lOmolL-ix50mL-O.lOmolL-ixV(HC1)7.00=7.21+lgO.lOmolL-ixV(HCl)解得V(HCl)=31mL第一种混合溶液无缓冲作用；第二种pHpK-1，无缓冲能力；第三种缓冲作用较强。a用固体NHCl和NaOH溶液来配制1L总浓度为0.125molL-1,pH=9.00的缓冲溶液，问需NHCl4多少克？求需1.00molL-1的NaOH溶液的体积(mL)。解设需NHCl的质量为xg4pK(NH+)=14.00-

15、4.75=9.25a49.OO=9.25+lg叫LTxV(NaOH)x/53.5gmol-i-l.OOmolL-ixV(NaOH)得1.00molL-iXV(NaOH)=0.562x/53.5gmol-i-1.00molL-iXV(NaOH)又1.00molL-1XVNaOH)+0.562(x/53.5gmol-1-1.00molL-1XV(NaOH)/1L=0.125molL-1解得x=6.69,V(NaOH)=0.045L即：需NHCl6.69g,NaOH溶液0.045L。4用0.020molL-1HPO溶液和0.020molL-1NaOH溶液配制100mLpH=7.40的生理缓冲溶液，T

16、OC o 1-5 h z34求需HPO溶液和NaOH溶液的体积(mL)34解设第一步反应需HPO和NaOH溶液体积各为xmL34HPO(aq)+NaOH(aq)=NaHPO(aq)+HO(l)34242xmLHPO与xmLNaOH完全反应，生成NaHPO0.020molL-1XxmL=0.020 xmmol3424(2)第二步反应：设生成的NaHPO再部分与NaOHymL反应，生成NaHPO，其与剩余NaHPO组成242424缓冲溶液NaHPO(aq)+NaOH(aq)=NaHPO(aq)+HO(l)24242起始量mmol+0.020 x+0.020y变化量mmol-0.020y-0.020

17、y+0.020y平衡量mmol0.020(x-y)0+0.020y0.020ymmol7.40=7.21+lg0.020(x-y)mmol=1.55依题意又有2x+y=100解得x=38.4,y=23.2即需HPO溶液38.4mL，NaOH溶液(38.4+23.2)mL=61.6mL。34今欲配制37C时，近似pH为7.40的生理缓冲溶液，计算在Tris和TrisHCl浓度均为0.050molL-1的溶液100mL中，需加入0.050molL-1HCl溶液的体积(mL)。在此溶液中需加入固体NaCl多少克，才能配成与血浆等渗的溶液？(已知TrisHCl在37C时的pK=7.85,忽略离子强度的

18、影响。)a解(1)7.40=7.85+lgn(Tris)n(TrisHC1)0.0500molL-ix100mL-0.0500molL-ixV(HC1)_.+g0.0500molL-ix100mL+0.0500molL-ixV(HCl).355=100-V(HCl)100+V(HCl)V(HCl)=47.6mL设加入NaClxg,血浆渗透浓度为300mmolL-1c(Tris)=0.050molL-1x100mL一0.050molL-1x47.6mL147.6mL=0.018molL-1c(TrisHCl)=0.050molL-1x100mL+0.050molL-1x47.6mL_=0.050

19、molL-i147.6mL(0.018+2X0.050)molL-i+=0.300molL-i58.5gmol-1x147.6mLx=0.79，即需加入NaCl0.79g18.正常人体血浆中，HCO-=24.0mmolL-i、CO(aq)=1.20mmolL-i。若某人因腹泻使血32浆中HCO-减少到为原来的90%，试求此人血浆的pH，并判断是否会引起酸中毒。已知HCO的323pK/=6.10。al解pH=pKa1+lgHCO3=6.10+lgCO(aq)1.2mmolL12pH虽接近7.35，但由于血液中还有其他缓冲系的协同作用，不会引起酸中毒。ExercisesHowdotheacidan

20、dbasecomponentsofabufferfunction?Whyaretheytypicallyaconjugateacid-basepair?SolutionAbuffersolutionconsistsofaconjugateacid-basepair.Theconjugatebasecanconsumetheaddedstrongacid,andtheconjugateacidcanconsumetheaddedstrongbase,tomaintainpH。Theconjugateacid-basepairsofweakelectrolytespresentinthesames

21、olutionatequilibrium.WhenH3O+isaddedtoabuffer，doesthepHremainconstantordoesitchangeslightly？Explain.SolutionThepHofabufferdependsonthepKaoftheconjugateacidandthebuffercomponentratio.WhenH3O+isaddedtoabuffer，thebuffercomponentratiochangesslightly，sothepHchangesslightly.Acertainsolutioncontainsdissolv

22、edHC1andNaCl.Whycantthissolutionactasabuffer?SolutionThissolutioncantactasabuffer.HClisnotpresentinsolutioninmolecularform.Therefore,thereisnoreservoirofmoleculesthatcanreactwithaddedOH-ions.LikewisetheCl-doesnotexhibitbasebehaviorinwater,soitcannotreactwithanyH3O+addedtothesolution.Whatistherelatio

23、nshipbetweenbufferrangeandbuffer-componentratio?SolutionThepHofabufferdependsonthebuffercomponentratio.WhenB-/HB=1,pH=pKa,thebufferismosteffective.Thefurtherthebuffer-componentratioisfrom1,thelesseffectivethebufferingactionis.Practically,iftheB-/HBratioisgreaterthan10orlessthan0.1,thebufferispoor.Th

24、ebufferhasaeffectiverangeofpH=pKa1.Choosespecificacid-baseconjugatepairsofsuitableforpreparethefollowingbeffer(UseTable4-1forKofacidorKofbase):ab(a)pH-4.0；(b)pH-7.0；(c)H3O+1.0 x10-9molL-1；Solution(a)HAcandAc-(b)HPO-andHPO2(c)NH+andNH33Choosethefactorsthatdeterminethecapacityofabufferfromamongthefoll

25、owingandexplainyourchoices.(a)Conjugateacid-basepair(b)pHofthebuffer(c)BufferrangerConcentrationofbuffer-componentreservoirsBuffer-componentratiopKaoftheacidcomponentSolutionChoose(d)and(e).Buffercapacitydependsonboththeconcentrationofthereservoirsandthebuffer-componentratio.Themoreconcentratedtheco

26、mponentsofabuffer,thegreaterthebuffercapacity.Whenthecomponentratioisclosetoone,abufferismosteffective.WouldthepHincreaseordecrease,andwoulditdosotoalargerorsmallextent,ineachofthefollowingcases:Add5dropsof0.1molLiNaOHto100mLof0.5molL-iacetatebufferAdd5dropsof0.1molLiHClto100mLof0.5molL-iacetatebuff

27、erAdd5dropsof0.1molLiNaOHto100mLof0.5molL-iHClAdd5dropsof0.1molL-iNaOHtodistilledwaterSolution(a)ThepHincreasestoasmallextent;(b)ThepHdecreasestoasmallextent;(c)ThepHincreasestoasmallextent;(d)ThepHincreasestoalargerextent.Whichofthefollowingsolutionswillshowbufferproperties?100mLof0.25molLiNaC3H5O3

28、+150mLof0.25molLiHCl100mLof0.25molLiNaC3H5O3+50mLof0.25molL-iHCl100mLof0.25molLiNaC3H5O3+50mLof0.25molLiNaOH100mLof0.25molLiC3H5O3H+50mLof0.25molLiNaOHSolution(b)and(d)AchemistneedsapH10.5buffer.ShouldsheuseC耳NH2andHClorNH3andHCltoprepareit?Why?Whatisthedisadvantageofchoosingtheotherbase?is9.25.Thep

29、KaoftheformerismorecloseSolutionThepKaofCH3NH2HClis10.65.ThepKaofNHto10.5.AbufferismoreeffectivewhenthepHisclosetopKa.SheshouldchooseCH3NH2.Theotherisnotagoodchoice.Anartificialfruitcontains11.0goftartaricacidH2C4H4O6,and20.0gitssalt,potassiumhydrogentartrate,perliter.WhatisthepHofthebeverage?Ka1=1.0 x10-3SolutionpH二pKai十lgn(KHCHO)4_46n(HCHO)2446=3.00+lg2