15.用固体NH4Cl和NaOH溶液来配制1L总浓度为0.125mol·L-1,pH=9.00的缓冲溶液,问需NH4Cl多少克?
求需1.00mol·L-1的NaOH溶液的体积(mL)。
解设需NH4Cl的质量为xg
pKa(NH4+)=14.00-4.75=9.25
得1.00mol·L-1×V(NaOH)=0.562[x/53.5g·mol-1-1.00mol·L-1×V(NaOH)]
又[1.00mol·L-1×VNaOH)+0.562(x/53.5g·mol-1-1.00mol·L-1×V(NaOH))]/1L
=0.125mol·L-1
解得x=6.69,V(NaOH)=0.045L
即:
需NH4Cl6.69g,NaOH溶液0.045L。
16.用0.020mol·L-1H3PO4溶液和0.020mol·L-1NaOH溶液配制100mLpH=7.40的生理缓冲溶液,求需H3PO4溶液和NaOH溶液的体积(mL)。
解设第一步反应需H3PO4和NaOH溶液体积各为xmL
⑴H3PO4(aq)+NaOH(aq)=NaH2PO4(aq)+H2O(l)
xmLH3PO4与xmLNaOH完全反应,生成NaH2PO40.020mol·L-1×xmL=0.020xmmol
⑵第二步反应:
设生成的NaH2PO4再部分与NaOHymL反应,生成Na2HPO4,其与剩余NaH2PO4组成缓冲溶液
NaH2PO4(aq)+NaOH(aq)=Na2HPO4(aq)+H2O(l)
起始量mmol+0.020x+0.020y
变化量mmol-0.020y-0.020y+0.020y
平衡量mmol0.020(x-y)0+0.020y
=1.55
依题意又有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℃时,近似pH为7.40的生理缓冲溶液,计算在Tris和Tris·HCl浓度均为0.050mol·L-1的溶液l00mL中,需加入0.050mol·L-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Tris·HCl在37℃时的pKa=7.85,忽略离子强度的影响。
)
解⑴
V(HCl)=47.6mL
⑵设加入NaClxg,血浆渗透浓度为300mmol·L-1
=0.018mol·L-1
=0.050mol·L-1
(0.018+2×0.050)mol·L-1+
0.300mol·L-1
x=0.79,即需加入NaCl0.79g
18.正常人体血浆中,[
]=24.0mmol·L-1、[CO2(aq)]=1.20mmol·L-1。
若某人因腹泻使血浆中[
]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H2CO3的pKa1ˊ=6.10。
解pH=pKa1′
pH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。
Exercises
1.Howdotheacidandbasecomponentsofabufferfunction?
Whyaretheytypicallyaconjugateacid-basepair?
SolutionAbuffersolutionconsistsofaconjugateacid-basepair.Theconjugatebasecanconsumetheaddedstrongacid,andtheconjugateacidcanconsumetheaddedstrongbase,tomaintainpH。
Theconjugateacid-basepairsofweakelectrolytespresentinthesamesolutionatequilibrium.
2.WhenH3O+isaddedtoabuffer,doesthepHremainconstantordoesitchangeslightly?
Explain.
SolutionThepHofabufferdependsonthepKaoftheconjugateacidandthebuffercomponentratio.WhenH3O+isaddedtoabuffer,thebuffercomponentratiochangesslightly,sothepHchangesslightly.
3.AcertainsolutioncontainsdissolvedHClandNaCl.Whycan’tthissolutionactasabuffer?
SolutionThissolutioncan’tactasabuffer.HClisnotpresentinsolutioninmolecularform.Therefore,thereisnoreservoirofmoleculesthatcanreactwithaddedOH-ions.LikewisetheCl-doesnotexhibitbasebehaviorinwater,soitcannotreactwithanyH3O+addedtothesolution.
4.Whatistherelationshipbetweenbufferrangeandbuffer-componentratio?
SolutionThepHofabufferdependsonthebuffercomponentratio.When[B-]/[HB]=1,pH=pKa,thebufferismosteffective.Thefurtherthebuffer-componentratioisfrom1,thelesseffectivethebufferingactionis.Practically,ifthe[B-]/[HB]ratioisgreaterthan10orlessthan0.1,thebufferispoor.ThebufferhasaeffectiverangeofpH=pKa±1.
5.Choosespecificacid-baseconjugatepairsofsuitableforpreparethefollowingbeffers(UseTable4-1forKaofacidorKbofbase):
(a)pH≈4.0;(b)pH≈7.0;(c)[H3O+]≈1.0×10-9mol·L-1;
Solution(a)HAcandAc-(b)
and
(c)
andNH3
6.Choosethefactorsthatdeterminethecapacityofabufferfromamongthefollowingandexplainyourchoices.
(a)Conjugateacid-basepair(b)pHofthebuffer(c)Bufferranger
(d)Concentrationofbuffer-componentreservoirs
(e)Buffer-componentratio(f)pKaoftheacidcomponent
SolutionChoose(d)and(e).Buffercapacitydependsonboththeconcentrationofthereservoirsandthebuffer-componentratio.Themoreconcentratedthecomponentsofabuffer,thegreaterthebuffercapacity.Whenthecomponentratioisclosetoone,abufferismosteffective.
7.WouldthepHincreaseordecrease,andwoulditdosotoalargerorsmallextent,ineachofthefollowingcases:
(a)Add5dropsof0.1mol·L-1NaOHto100mLof0.5mol·L-1acetatebuffer
(b)Add5dropsof0.1mol·L-1HClto100mLof0.5mol·L-1acetatebuffer
(c)Add5dropsof0.1mol·L-1NaOHto100mLof0.5mol·L-1HCl
(d)Add5dropsof0.1mol·L-1NaOHtodistilledwater
Solution(a)ThepHincreasestoasmallextent;(b)ThepHdecreasestoasmallextent;(c)ThepHincreasestoasmallextent;(d)ThepHincreasestoalargerextent.
8.Whichofthefollowingsolutionswillshowbufferproperties?
(a)100mLof0.25mol·L-1NaC3H5O3+150mLof0.25mol·L-1HCl
(b)100mLof0.25mol·L-1NaC3H5O3+50mLof0.25mol·L-1HCl
(c)100mLof0.25mol·L-1NaC3H5O3+50mLof0.25mol·L-1NaOH
(d)100mLof0.25mol·L-1C3H5O3H+50mLof0.25mol·L-1NaOH
Solution(b)and(d)
9.AchemistneedsapH10.5buffer.ShouldsheuseCH3NH2andHClorNH3andHCltoprepareit?
Why?
Whatisthedisadvantageofchoosingtheotherbase?
SolutionThepKaofCH3NH2·HClis10.65.ThepKaof
is9.25.ThepKaoftheformerismorecloseto10.5.AbufferismoreeffectivewhenthepHisclosetopKa.SheshouldchooseCH3NH2.Theotherisnotagoodchoice.
10.Anartificialfruitcontains11.0goftartaricacidH2C4H4O6,and20.0gitssalt,potassiumhydrogentartrate,perliter.WhatisthepHofthebeverage?
Ka1=1.0×10-3
Solution
11.Whatarethe[H3O+]andthepHofabenzoatebufferthatconsistsof0.33mol·L-1C6H5COOHand0.28mol·L-1C6H5COONa?
Kaofbenzoicacid=6.3×10-5。
Solution
[H+]=7.41×10-5mol·L-1
12.Whatmassofsodiumacetate(NaC2H3O2·3H2O,Mr=136.1g·mol-1)andwhatvolumeofconcentratedaceticacid(17.45mol·L-1)shouldbeusedtoprepare500mLofabuffersolutionatpH=5.00thatis0.150mol·L-1overall?
Solutionc(CH3COO-)+c(CH3COOH)=0.150mol·L-1
n(CH3COO-)+n(CH3COOH)=0.150mol·L-1×500mL×10-3L·mL-1=0.0750mol
n(CH3COOH)=0.0270mol,n(CH3COO-)=0.0480mol
Massofsodiumacetate=0.0480mol×136.1g·mol-1=6.53g
13.NormalarterialbloodhasanaveragepHof7.40.Phosphateionsformoneofthekeybufferingsystemsintheblood.Findthebuffer-componentratioofaKH2PO4/Na2HPO4solutionwiththispHpKa2ˊof
=6.80
Solution