无机分析化学答案

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1、第二章定量分析概论（答案）1・有关反应和滴定反应MnO+2Na2O2+H2O=MnO42+2OH+4Na+3MnO42+4H+=2MnO4+MnO2I+2H2OMnO2+2Fe2++4H+=2Fe3+Mn2++2H2OMnO4+5Fe2++8H+=Mn2++5Fe3++4H2OCr2O3+6Fe2++14H+=2Cr3++6Fe3+4-17H2O解：3MnO2〜3MnO4厶〜MnO?〜2Fe"MnO4'〜5F03MnO=B/2]X[o.1000X10.00—5x0.01000x0.24]x70.94=003]32.000x1000•MnO%=3.13%ICT2O3〜2CrO4

2、2-CrO72-6Fe2+3Cl*2O3==0.0144[l/6]x[0.1000x50.00—5x0.010x10.40—5x0.5880]xl52.02.000x1000Cr2O3%=1.44%2•解：n(Fc)=5n(MnO4-)C(MnO4尸x1000=0.2000(moI/L)5Xjj.ojn{H[CqO^}=5n(MnO4)/4C(H3[C2O4]2=5X0.40X0.2000/4=0.10(moI/L)n(OH)/3=n(H3[C2O4⑵K(NaOH)={[0.1X10.00]/0.1}X3=30.00(ml)(5Fe2++MnO4+8H+=5Fe3++Mn2+

3、+4H2O5H3(C204F2+4Mn04+17H=4Mnc++20C02f+I6H2O)3.解:Na2B4O7•10H20〜2HCI〜2NaOHn(Na2B407•10H20)二(riHCI总—^HCI余)/2Va/Vb二Cb/Ca二1/1.0352lIlNa2B4O7・10H20/MNa2B4O7・101120=VaCa-VbCa/l.0352X0.5000/381.4=[30.00XCa-5.50XCa/l・035]/1000Ca=[2X0.5/0.3814]/[30.00-5.50/1.035]=2.622/24.69=0.1062[mol/L]Cb=Ca/1.035

4、=0.1026[mol/L]3.解Sb2S3+302=Sb203+3S02S02+2Fe3++2H2O=2Fe2++SO42+4H+6Fe2++Cr2O72+14H+=2Cr3++6Fe3++7H2O2Sb〜SbzSs〜3SO2〜6Fe*〜Cr2O72-3Sb2S3=CK2CrO7XVK2CrO7XMsb2S3/m样=0.02000X25.20X0.33968/0.3000=0.57073Sb=2XCK2CrO7XVK2CrO7XAg/m样=2X0.02000X25.20X121.8/[0.3000X1000]=0.40924.解：Mn02+C2042_+4H+=Mn2++2

5、C02+2H202Mn04_+5C2042'+l6H=2Mn2++10C02+8H20MnOz〜GO广；500广〜2MnO「戚嘶2。4_5CrMnQ.VrMn0xMMnO2oMnO尸2加样<（）5<、---x0.01964xl6.50xW3x86.94二（1342丿0.3000二0.8465or84.65%6•解:Ca[C103]2~12Fe3+—2Cr2O72-T0C6iqq—x[0.1000x26.00—0.02000x6xl0.00]x——oCa[C103]2=12皿0.2000=0.1207or12.07%第三章误差及数据处理1•⑴尢=£Xi/n=[20.48+20

6、.55+20.58+20.60+20.53+20.50]/6=20.546_J=[l/6]Jxi-Xi=0.037/=!dr=[d/X]x100%=0.18%=0.046%[2]e=X-u=20.54-20.46=-0.08e^=[X-p]/u=[20.54-20.46]/20.46=-0.39%二37.45+37.20+37.30+37.254Erf37303735二xl00%=0.13%37.35n2.[1]X=-Yxi=nX1+X2+X3+X44x-X[2]d=^=-ni==0.075(%)0.15+0.10+0.00+0.05X=^xl00%=0.20%n_[3

7、]S=工(疋一乂)2旦0.11%n-1=*二罕竺=0.055%JnJ43•当X=67.04时，卩二—=67.04—66.62=2000.21当X=65.82时，U二匚以聞-66.62=3600.21查口表P[65.87〜67.04]二0.4999+0.4773=97.72%出现在此区间可能的个数为(1-0.9772)xioo=2.28«2个4.心g」iif"」00.2查U表

8、〃

9、=3的面积为0.4987分析结果小于11.6g/t概率P二0・5-0.4987=0.13%5・F二F°+0.0291gCx二E