第四章分析化学习题参考答案

1第四章习题参考答案8解：（1）32110100.50.203.17%2989.0VV1=66mL（2）32210100.50.205.60%10005.1VV2=57mL（1）32310100.50.209.98%9684.1VV3=56mL11解：CaO+2HCl=CaCl2+H2O依CaOHClABCaOHClMCMCbaT33/102110所以0.005000=08.5610213HClCCHCl=0.1783mol·L-11783.0000.12000.0000.1VV=0.1217L=121.7mL12解：依m=32)(21CONaHClMCVV=20mL时，m=g11.099.105102010.0213称量误差Er=%18.011.00002.0V=25mL时，m=g13.099.105102510.0213称量误差Er=%15.013.00002.0213解：C=105545.01000.099.105/5877.0/LmolVMmNaCO3+2HCl=NaCl2+H2O+CO2依HCln：1:232CONanCHCl×0.02196:(0.02000×0.05545)=2:1解得：CHCl=0.1010mol•L-115解：MgCO3+2HCl=MgCl2+H2O+CO2NaOH+HCl=NaCl+H2O30.33NaOHC=36.40CHClNaOHC=1.2CHCl依HCln：1:23MgCOn所以(48.48×CHCl-3.83×1.2CHCl)×10-3:32.84850.1=2:1CHCl=1.000mol•L-1NaOHC=1.2CHCl=1.2×1.000=1.200mol•L-116解：(1)1000.05000.018.294/709.14/VMmCmol•L-1（2）6Fe2++Cr2O72-+14H+=6Fe3++2Cr3++7H2O依TB/A=a/b310ABMCFeOCrKT/722=6722OCKrC×eFM×10-3=6×0.1000×55.845×10-3=0.03351g·mL-1因Cr2O72-≌6Fe2+≌3Fe2O332722/OFeOCrKT=3722OCKrC×32OFeM×10-3=3×0.1000×159.69×10-3=0.04791g·mL-1或：32722/OFeOCrKT=FeOCrKT/722·eeFOFMM232=0.03351×845.55269.159=0.04791g·mL-1317解：1200.01000.146.36/00.1004374.03HClCmol•L-1(1)NaOH+HCl=NaCl+H2O依TB/A=a/b310ABMCNaOHHClNaOHHClMCT3/101=1×0.1200×10-3×40.00=0.004800g·mL-1(2)CaO+2HCl=CaCl2+H2OCaOHClCaOHClMCT3/1021=21×0.1200×10-3×56.08=0.003365g·mL-119解：CaCO3+2HCl=CaCl2+H2O+CO2HCln：1:23CaCOn(0.5100×50.00-0.4900×25.00)×10-3:1:209.100mm=0.6631g=m/ms=0.6631/1.000=66.31%21解：依题意=V%2NaOH+H2C2O4=Na2C2O4+2H2O解法一：1:2:422OCNaNaOHnn0.1018×V×10-3:1:204.90102Vmm=0.4583g解法二：=%1021/3BsABBsABBsAVmMVCmMVbCamm故gVMVCmBABBs4583.021004.901018.0102113解法三：004583.01004.901018.0213/422OCHNaOHTg·mL-1=%422422/VmVTmmsOCHNaOHsOCHms=422/OCHNaOHT×100=0.004583×100=0.4583g