基本化學(xué)答案解析第04

****章后習(xí)題解答[TOP]習(xí)題1.什么是緩沖溶液？試以血液中的H2CO3-HCO一緩沖系為例，說明緩沖作用的原理及其在醫(yī)學(xué)上3的重要意義。答能抵抗少量外來強酸、強堿而保持其pH基本不變的溶液稱為緩沖溶液。血液中溶解的CO2與HCO-組成緩沖系。正常人體[HCO-]/[CO2(aq)]為20/1，pH=7.40。若pH<7.35，發(fā)生酸中毒，3 3pH>7.45，發(fā)生堿中毒。當(dāng)酸性代謝產(chǎn)物增加時，抗酸成分hco-與H3O+結(jié)合，增加的H2CO3可通過3加快呼吸以CO2的形式呼出；消耗的HCO-則由腎減少對其的排泄而得以補充;當(dāng)堿性代謝產(chǎn)物增加時,3[OH-]與H3O+生成H2O，促使抗堿成分H2CO3離解以補充消耗的H3O+。同理，減少的H2CO3及增加的HCO-可通過肺和腎來調(diào)控。血液中的H2cO3-HCO-緩沖系與其他緩沖系共同作用，維持pH為3 37.35?7.45的正常范圍。2.什么是緩沖容量？影響緩沖溶量的主要因素有哪些?總濃度均為0.10molL1的HAc-NaAc和H2CO3-HCO-緩沖系的緩沖容量相同嗎？3解緩沖容量是衡量緩沖溶液緩沖能力大小的尺度，表示單位體積緩沖溶液pH發(fā)生一定變化時，所能抵抗的外加一元強酸或一元強堿的物質(zhì)的量。影響緩沖容量的主要因素是緩沖系的總濃度和緩沖比：緩沖比一定時，總濃度越大緩沖容量越大；總濃度一定時，緩沖比越接近于1緩沖容量越大?？倽舛燃熬彌_比相同的HAc-NaAc和H2CO3-hco-緩沖系的緩沖容量相同。33.下列化學(xué)組合中，哪些可用來配制緩沖溶液？(1)HCl+NH3-H2O (2)HCl+Tris(3)HCl+NaOH⑷Na2HpO4+Na3PO4 ⑸H3PO4+NaOH ⑹NaCI+NaAc解可用來配制緩沖溶液的是：(1)HCl+NH3-H2O、(2)HCl+Tris、(4)Na2HpO4+Na3PO4和(5)H3PO4+NaOH

4.將0.30molLi吡啶(C5H5N,pKb=8.77)和0.10molLiHCl溶液等體積混合混合液是否為緩沖溶液？求此混合溶液的pH。解C5H5N與HCl反應(yīng)生成C5H5NH+Cl-(吡啶鹽酸鹽)，混合溶液為0.10molLiC5H5N和0.050molL1C5H5NH+Cl-緩沖溶液，pKa=14.00-8.77=5.23pH=pK+lgc(C5H5N)=5.23+lg010=5.33ac(CHNH+) 0.055 55.將10.0gNa2CO3和10.0gNaHCO3溶于水制備250mL緩沖溶液，求溶液的pH。n(HCO-)=n(HCO-)=3n(CO2-)=310.0g84.0g-mol10.0g—=0.119mol-i=0.094mol口n□n(CO2-)1-0.094molpH=pK+lg 3—=10-33+lg =10.23an(HCO-) 0.119mol36.求pH=3.90，總濃度為0.400molLi的HCOOH(甲酸)-HCOONa(甲酸鈉)緩沖溶液中，甲酸和甲酸鈉的物質(zhì)的量濃度(HCOOH的pK=3.75)a解設(shè)c(HCOONa)=xmolLi,則c(HCOOH)=0.400molLi-xmolLiTTcrL、 xmol-L-1 cccpH=3.75+lg =3.90(0.400-x)mol-L-1解得c(HCOO-)=xmolLi=0.234molLi解得c(HCOOH)=(0.400-0.234)molLi=0.i66molLi7.向100mL某緩沖溶液中加入0.20gNaOH固體，所得緩沖溶液的pH為5.60.。已知原緩沖溶液共軛酸HB的pKa=5.30,c(HB)=0.25molLi,求原緩沖溶液的pH。n(NaOH)=0.20g/40g.mol-1x1000mL=0.050molLi1L100mL1L加入NaOH后，pH=5.30+lg[B-]+。050molL1=5.60

(0.25-0.050)molL

解得[B解得[B-]=0.35molLi原溶液pH=5.30+lg0.35mol.LT=5.450.25mol-L-i.阿司匹林(乙酰水楊酸、以HAsp表示)以游離酸(未解離的)形式從胃中吸收，若病人服用解酸藥，調(diào)整胃容物的pH為2.95，然后口服阿司匹林0.65g。假設(shè)阿司匹林立即溶解，且胃容物的pH不變，問病人可以從胃中立即吸收的阿司匹林為多少克(乙酰水楊酸的Mr=180.2、pK=3.48)？a解 pH=pK+lgn(Asp-)=3.48+lgn(Asp-)=2.95an(HAsp) n(HAsp)n(Asp-)=0.295n(HAsp)依題意 n(Asp-)+n(HAsp)=一065g—=0.0036mol180g-mol-1解得 n(HAsp)=0.0028mol可吸收阿司匹林的質(zhì)量=0.0028molx180.2g-mol-i=0.50g.在500mL0.20molLiC2H5COOH(丙酸，用HPr表示)溶液中加入NaOH1.8g，求所得溶液的近似pH和校正后的精確pH。已知C2H5COOH的pKa=4.87，忽略加入NaOH引起的體積變化。解⑴求近似pHpH=pKa+lgn(Pr-)=487+lgL8g/40g.mol-1二4.78n(HPr) 0.500Lx0.20mol-L-1-1.8g/40g-mol-1⑵求精確pH，丙酸鈉是強電解質(zhì)I=1Ec.z.2=1(0.045molx12+0.045molx12)=0.09molL1~0.1molU2ii2 0.500L 0.500L當(dāng)Z=0,1=0.10時，校正因數(shù)lg迎。=-011Y(HB),pH=pKa+lg鑫+lg東"78+(一0.11)"67.某醫(yī)學(xué)研究中，制作動物組織切片時需pH約為7.0的磷酸鹽緩沖液作為固定液。該固定液的配方是：將29gNa2HpO4-12H2O和2.6gNaH2P。4叫0分別溶解后稀釋至1L。若校正因數(shù)

1gy30;9=-0.53，計算該緩沖溶液的精確pH。y(HPO-)2 4解c(Na2HPO4)= 29g =0.081molLi358.0g-moltx1Lc(NaH2PO4)= 2.6g =0.017molLi156g-mol-ix1LpH=pK2+y(hpo2-)+1gc(HpO2-)=7.21+(..0.53)+1g0.081mo1L1=7.36ay(HPO-)c(HPO-) 0.017mo1.L-12 4 2 4.將0.10molLiHAc溶液和0.10molLiNaOH溶液以3:1的體積比混合求此緩沖溶液的pH及緩沖容量。解HAc溶液和NaOH溶液的體積分別為3V和V,c(HAc)=(0.10x3V-0.10xV)mo1L1/(3V+V)=0.050molUc(Ac-)=0.10molUxV/(3V+V)=0.025molUpH=4.75+lg0.025moiIpH=4.75+lg0.050mol-L-10.050moi?L-1x0.025moi?L-1B=2.303x =0.038moi?L-1(0.050+0.025)mol-L-1.某生物化學(xué)實驗中需用巴比妥緩沖溶液，巴比妥（C8Hl2N2O3）為二元有機酸（用H2Bar表示，pKa1=7.43）。今稱取巴比妥18.4g，先加蒸餾水配成100mL溶液，在pH計監(jiān)控下，加入6.00molLNaOH溶液4.17mL,并使溶液最后體積為1000mL。求此緩沖溶液的pH和緩沖容量。（已知巴比妥的Mr=184g?mol.1)解H2Bar與NaOH的反應(yīng)為H2Bar(aq)+NaOH(aq)=NaHBar(aq)+H2O(l)反應(yīng)生成的NaHBar的物質(zhì)的量n(NaHBar)=c(NaOH)V(NaOH)=6.0mol-L-1x4.17mL=25mmol,剩余H2Bar的物質(zhì)的量為n余(H2Bar)=n(H2Bar)-n(NaOH)= 區(qū)4gx1000-25mmol=75mmol八 184g-mol-1

pH=pKa+lgn(HBar-)=7.43+lg25mmol=6.95n(HBar) 75mmol20=2303x(75mmol/1000mL)x(25mmol/1000mL)二0043moL(75+25)mmol/1000mL).分別加NaOH溶液或HCl溶液于檸檬酸氫鈉(縮寫Na2HCit)溶液中。寫出可能配制的緩沖溶液的抗酸成分、抗堿成分和各緩沖系的理論有效緩沖范圍。如果上述三種溶液的物質(zhì)的量濃度相同，它們以何種體積比混合，才能使所配制的緩沖溶液有最大緩沖容量？(已知H3Cit的pKa1=3.13、pKa2=4.76、pKa3=6.40)'解.有效緩沖 0最大時體溶液組成緩沖系抗酸成分抗堿成分 1r范圍積比Na2HCit+HH2CitNa2HCit+HH2Cit--HCit3.76?5.7HCit2-H2Cit-2:12-Cl2-2:3Na2HCit+H H3Cit-H2Cit 2.13?4.12:3Cl - 2 3 35.40?7.405.40?7.40HCit2--Cit3- Cit3-aOH.現(xiàn)有(1)0.10molL1NaOH溶液，(2)0.10molL1NH3溶液，(3)0.10molL1Na2HpO4溶液各50mL,欲配制pH=7.00的溶液，問需分別加入0.10molL1HCl溶液多少mL?配成的三種溶液有無緩沖作用？哪一種緩沖能力最好？解⑴HCl與NaOH完全反應(yīng)需HCl溶液50mL。⑵HCl(aq)+NH3-H2O(aq)=NH4Cl(aq)+H2O(l)NH4+的pKa=14.00-4.75=9.25,

7.00=9.25+1gO.IOmol-L-ix50mL—0.10mo1.L-7.00=9.25+1gO.IOmol-L-ixV(HC1)解得 V(HCl)=50mL⑶HCl(aq)+Na2HPO4(aq)=NaH2PO4(aq)+NaCl(aq)H3PO4的pKa2=7.21,7.00=7.21+1gO.IOmol-L-ixV(HC1)O.IOmol-L-i7.00=7.21+1gO.IOmol-L-ixV(HC1)解得V(HCl)=31mL第一種混合溶液無緩沖作用；第二種pH<pKa-1，無緩沖能力；第三種緩沖作用較強。.用固體NH4a和NaOH溶液來配制1L總濃度為0.125molU，pH=9.00的緩沖溶液，問需NH4C1多少克？求需1.00molL1的NaOH溶液的體積(mL)解設(shè)需NH4C1的質(zhì)量為xgpKa(NH4+)=14.00-4.75=9.259.OO=9.25+lgi.OOmol-L-9.OO=9.25+lgx/53.5g?mol-i-i.OOmol-L-ixV(NaOH)得1.00molL1xV(NaOH)=0.562[x/53.5g-mol-1-1.00molL1xV(NaOH)]又[1.00molL1xVNaOH)+0.562(x/53.5g-mol-1-1.00molL1xV(NaOH))]/1L=0.125molL1解得x=6.69,V(NaOH)=0.045L即：需NH4C16.69g,NaOH溶液0.045L。.用0.020molL1H3PO4溶液和0.020molL1NaOH溶液配制100mLpH=7.40的生理緩沖溶液，求需H3PO4溶液和NaOH溶液的體積(mL)解設(shè)第一步反應(yīng)需H3PO4和NaOH溶液體積各為xmL⑴ H3PO4(aq)+NaOH(aq)=NaH2PO4(aq)+H2O(l)

xmLH3PO4與xmLNaOH完全反應(yīng)，生成NaH2Po40.020molLixxmL=0.020xmmol⑵第二步反應(yīng)：設(shè)生成的NaH2P04再部分與NaOHymL反應(yīng)，生成Na2HpO4,其與剩余NaH2PO4組成緩沖溶液NaH2PO4(aq)+NaOH(aq)=Na2HpO4(aq)+H2O(l)起始量mmol+0.020X +0.020y變化量mmol-0.020y -0.020y +0.020y平衡量mmol0.020(x-y) 0 +0.020y0.020ymmol7.40=7.21+1g 0.020(X-y)mmo1y=1.55X-y依題意又有 2x+y=100解得x=38.4,y=23.2即需H3PO4溶液38.4mL,NaOH溶液(38.4+23.2)mL=61.6mL。17.今欲配制37℃時，近似pH為7.40的生理緩沖溶液，計算在Tris和Tris-HCl濃度均為0.050molLi的溶液l00mL中，需加入0.050molL1HCl溶液的體積(mL)。在此溶液中需加入固體NaCl多少克，才能配成與血漿等滲的溶液？（已知Tris-HCl在37℃時的pKa=7.85,忽略離子強度的影響。）解⑴7.40=7.40=7.85+1gn(Tris)n(Tris?HC1)_ 10.0500molLx100mL-0.0500molLxV(HC1)一.+g0.0500molLx100mL+0.0500mol.L-1xV(HCl)0.355=100-V(HCl)100+V(HCl)V(HCl)=47.6mL⑵設(shè)加入NaClxg，血漿滲透濃度為300mmolL1

c(Tris)==0.018molLi0.050mol-L-ix100mL—0.050mol-L-c(Tris)==0.018molLi147.6mLc(Tris-HQ)==0.050molLi0.050mol-L-ix100mL+0.050mol-c(Tris-HQ)==0.050molLi147.6mL(0.018+2x0.050)molLi+ 2器x1000mL1-1=0.300molLi58.5g-mol-1x147.6mLx=0.79，即需加入NaCI0.79g18.正常人體血漿中，［hco-］=24.0mmoI.L-i、［CO2（aq）］=1.20mmolLi。若某人因腹瀉使血3漿中［HCd］減少到為原來的90%，試求此人血漿的pH，并判斷是否會引起酸中毒。已知H2cO3的PKai3'=6.10。PH=PKPH=PKai'+lg[HCO-][CO2(叫)]=6.10+lg24mmol-L-1x0.90 =7.361.2mmol-L-1pH雖接近7.35，但由于血液中還有其他緩沖系的協(xié)同作用，不會引起酸中毒。ExercisesHowdotheacidandbasecomponentsofabufferfunction?Whyaretheytypicallyaconjugateacid-basepair?SolutionAbuffersolutionconsistsofaconjugateacid-basepair.Theconjugatebasecanconsumetheaddedstrongacid,andtheconjugateacidcanconsumetheaddedstrongbase,tomaintainpH。Theconjugateacid-basepairsofweakelectrolytespresentinthesamesolutionatequilibrium.WhenH3O+isaddedtoabuffer,doesthepHremainconstantordoesitchangeslightly?Explain.SolutionThepHofabufferdependsonthepKaoftheconjugateacidandthebuffercomponentratio.WhenH3O+isaddedtoabuffer,thebuffercomponentratiochangesslightly,sothepHchangesslightly.

AcertainsolutioncontainsdissolvedHClandNaCl.Whycan'tthissolutionactasabuffer?SolutionThissolutioncan'tactasabuffer.HClisnotpresentinsolutioninmolecularform.Therefore,thereisnoreservoirofmoleculesthatcanreactwithaddedOH-ions.LikewisetheCl-doesnotexhibitbasebehaviorinwater,soitcannotreactwithanyH3O+addedtothesolution.Whatistherelationshipbetweenbufferrangeandbuffer-componentratio?SolutionThepHofabufferdependsonthebuffercomponentratio.When[B-]/[HB]=1,pH=pKa,thebufferismosteffective.Thefurtherthebuffer-componentratioisfrom1,thelesseffectivethebufferingactionis.Practically,ifthe[B-]/[HB]ratioisgreaterthan10orlessthan0.1,thebufferispoor.ThebufferhasaeffectiverangeofpH=pKa±1.Choosespecificacid-baseconjugatepairsofsuitableforpreparethefollowingbeffers(UseTable4-1forKaofacidorKbofbase):(a)pH~4.0;(b)pH~7.0;(c)[H3O+]~1.0x10-9molL1;Solution(a)HAcandAc- (b)hpo.andhpo2.(c)nh+andNH32 4 4 4Choosethefactorsthatdeterminethecapacityofabufferfromamongthefollowingandexplainyourchoices.Conjugateacid-basepair(b)pHofthebuffer(c)BufferrangerConcentrationofbuffer-componentreservoirsBuffer-componentratio(f)pKaoftheacidcomponentSolutionChoose(d)and(e).Buffercapacitydependsonboththeconcentrationofthereservoirsandthebuffer-componentratio.Themoreconcentratedthecomponentsofabuffer,

thegreaterthebuffercapacity.Whenthecomponentratioisclosetoone,abufferismosteffective.WouldthepHincreaseordecrease,andwoulditdosotoalargerorsmallextent,ineachofthefollowingcases:Add5dropsof0.1molLiNaOHto100mLof0.5molLiacetatebufferAdd5dropsof0.1molLiHClto100mLof0.5molLiacetatebufferAdd5dropsof0.1molLiNaOHto100mLof0.5molLiHClAdd5dropsof0.1molLiNaOHtodistilledwaterSolution(a)ThepHincreasestoasmallextent;(b)ThepHdecreasestoasmallextent;(c)ThepHincreasestoasmallextent;(d)ThepHincreasestoalargerextent.Whichofthefollowingsolutionswillshowbufferproperties?100mLof0.25molLiNaC3H5O3+150mLof0.25molLiHCl100mLof0.25molLiNaC3H5O3+50mLof0.25molLiHCl100mLof0.25molLiNaC3H5O3+50mLof0.25molLiNaOH100mLof0.25molLiC3H503H+50mLof0.25molLiNaOHSolution(b)and(d)AchemistneedsapH10.5buffer.ShouldsheuseCH3NH2andHClorNH3andHCltoprepareit?Why?Whatisthedisadvantageofchoosingtheotherbase?SolutionThepKofCHNH?HClis10.65.ThepKofnh+is925ThepKoftheformerisa 32 a 4 amorecloseto10.5.AbufferismoreeffectivewhenthepHisclosetopKmorecloseto10.5.AbufferismoreeffectivewhenthepHisclosetopKa.SheshouldchooseCH3NH2.Theotherisnotagoodchoice.Anartificialfruitcontains11.0goftartaricacidH2C4H4O6,and20.0gitssalt,potassiumhydrogentartrate,perliter.WhatisthepHofthebeverage?Ka1=1.0x10-3Solutionnn(KHCHO)+Solutionnn(KHCHO)+1g 4-4-6--3.00+lgn(HCHO)2 4 4 620.0g188.2g-mol-i11.0g150.1g-mol-i-3.00.+