acid – base：酸–基地

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1、Acid–BaseEXAM#1(a)WhatisthepHofa2.0molarsolutionofaceticacid?Kaaceticacid=1.8x10-5(b)Abuffersolutionispreparedbyadding0.10literof2.0molaraceticacidsolutionto0.1literofa1.0molarNaOHsolution.Whatisthehydrogenionconcentrationofthebuffersolution?(c)Supposethat0.10literof0.50molarHClisaddedto0

2、.040literofthebufferpreparedin(b).Whatisthehydrogenionconcentrationoftheresultingsolution?Answer:(a)CH3COOH(aq)<=>H+(aq)+CH3COO-(aq)Ka=1.8x10-5=[H+][CH3COO-][CH3COOH][H+]=[CH3COO-]=X[CH3COOH]=2.0-X,X<<2.0,thereforeXisnegligiblehere.1.8x10-5=X2/2.0X=6.0x10-3=[H+];pH=-log[H+]=2.22(b)0.1Lx2.

3、0mol/L=0.20molCH3COOH0.1Lx1.0mol/L=0.10molNaOH=0.10molOH-CH3COOH(aq)+OH-(aq)<=>CH3COO-(aq)+H2OI0.20moles0.10moles0C-0.10moles-0.10moles+0.10molesE0.10moles00.10moles(0.10moles/0.2liters=0.5Msolutionsofbothbuffercomponents).the0.10molofhydroxideneutralizes0.10molCH3COOHwith0.10molremaining

4、withaconcentrationof0.10mol/0.20L=0.5M.ofbothCH3COOHandCH3COO-.Howeveryoucansolvethiseasilybecause½waytotheequivalencepoint,pH=pKa.-log(1.8x10-5)=pHof4.74(c)0.10litersofHClx0.50M=0.05molesofHCl.Themolesforthebuffercomponentsinpart(b)wouldbe0.50Mx0.040liters=0.02molesCH3COO-+HCl<=>CH3COOH+

5、Cl-I0.020moles0.050moles0.020molesC-0.020moles-0.020moles+0.020molesE0moles0.03moles0molesTotalvolume=0.10=0.040liters=0.140liters0.030moles/0.140litersofsolution=0.214M[H+]=0.214MEXAM#2H3PO2,H3PO3,andH3PO4aremonoprotic,diproticandtriproticacids,respectively,andtheyareaboutequalinacidstre

6、ngth.HClO2,HClO3,andHClO4areallmonoproticacids,butHClO2isaweakeracidthanHClO3whichisweakerthanHClO4.Accountfor:(a)Thefactthatthemoleculesofthethreephosphorusacidscanprovidedifferentnumbersofprotons.(b)Thefactthatthethreechlorineacidsdifferinstrengths.Answer:(a)thestructureforthethreeacids

7、areasfollows:PHOOHOHOPHOHOHHPOHOHOThehydrogenatom(s)bondeddirectlytothephosphorusatomarenotacidicinaqueoussolution;onlythosehydrogenatomsbondedtotheoxygenatomscanbereleasedasprotons.(b)Theacidstrengthissuccessivelygreaterasthenumberofoxygenatomsincreasesbecausetheve