无机及分析化学答案.docx

无机及分析化学答案.docx根据下列酸碱的解离常数,选取适当的酸度是共觇碱来配置PH==,其共轨酸,碱的浓度比应是多少?HAC,NH3•H20,H2C2O4,NaHCOs,H3PO4,NaAC,NaHPO4,C6H5NH2,NH4cl解:选择HAC和NaAC来配置PH=,共轨酸碱的浓度比可由以下来计算检表得:Ka°(hao—1•8X10° Kb(ac-)=Kw/Ka=10-14/('5)=-10由PH=14-Pkb°+lg((C(AC-))/(C(HAC)))=14-lg(-10)+lg((C(ac-))/(Cam))解得:(C(AC-))/(C(HAC))=4/7=•H20和NH4cl来配制PH=,共轨酸碱的浓度比如下Kb°(NH3•H20)^PH=14-Kb(NH3•H2O)+lg(C(NH3・H20)/C(NH4cl))10=14-lg()+1g(C(NH3•H2O)/C(NH4cl))解得:C(NH3・H2O)/c(XH4cl)—5:1欲配制250mlPH=,问在125ml,<?解:由题意得可设还应加入xml,-L-1的HAC溶液检表得Ka(HAC)==Pkae(HAC)+lg((C(AC->)/(C(HAC)))5二-lg()+lg((C(Ac-))/(C(hac)))解得:(C<AC-))/(C(IIAC))=(AC-)=•L_1=;n(uAc)=6xX10~/(+xX10-3)/(6xX10_3/(+xX10-3))=/(6xX10-3)解得:x=:(3)•L?•!?NH4cl溶液混合解:由题意的检表可得Kb(NH3)-5 ; K/(NH;)=Kw/Kb(NH3)°=10H7(-5)=-10n(0H-)= n(NH「)(NH:rH20)=/(+)=•L_1C(NH「)=/(+)=•L1PH—Pka°(NH4)+]g(C(NH3・H20)/C(NH4cl))=_lg()+0=(5)••L_1NaAC溶液混合混合前:n(H+)=(AC-)=:C(HAC)/(+)mol•L_1=•!/C(AC')=/(+)mol>L05mol・I?PH二Ph0(HAC)+lg((C(AC-))/(C(HAc))) 检表得:PkJ(HAC)-5=-lg(-5)+lg()(9)<(H3POi)=・5=(0H)=°(HP042_)=~13混合后:n(Na3P04)=(Na2HPO4)=:n(Na3PO4)/n(Na2HPO4)=2PH=IC(HPO产)+lg2二-lg(-13)+•L_1NH3(aq)•L_1Hcl溶液,,不允许再加水,最多能配制多少升缓冲溶液,其中c(nh3),C(nh4+)各为多少?解:由表可知:Kb(NI13)°="5则PH二14-Kb°(NH3•H2O)+lg(C(NH3-H20)/C(NH4cl))=9即:PH二14一4・74+lg(C(nh3・H20)/C(nh4ci))二9得]g(C(NH3・H20)/c(NH4cl))=—(NH3・H20)/C(NH4cl)二0・55设xLNH3•H20yLHC1则有C(NH3•H20)=n(•H20)/(x-y)/(x+y)mol•L1C(NH4+)=n(Hcl)/V总二0・5y/(x+y)mol•L_1即C(NHb-H20)/C(NH「)==