分析化學(xué)題庫和答案(精心整理)-分析化學(xué)經(jīng)典題庫

分析化學(xué)題庫和答案分析化學(xué)題庫和答案(精心整理)-分析化學(xué)經(jīng)典題庫#5NH：[H5NH：[H+][H：]+Ka10-8.0010-8.00+10-9.26=10-0.023丁 [Mg2+][NH4+][PO43-]二KspS?C?5 ?10-4.42,S=KNH3 NH: 叩: K : 10-12.7 wS二sp二二10-3.78C?5 -10-4.42 \0.2X10-0.023X10-4.42NH3 NH+34s=1.7X10-4(molL-i)6計(jì)算下列換算因數(shù)：測(cè)定物 稱量物(1)FeO Fe2O3⑵Ni 丁二酮肟鎳(Ni(C4H8N202)2)(3)Al2O3 Al(C9H6ON)3解：換算因數(shù)f一被測(cè)組分式量稱量形式量2M 2x71.85(1)F=—r(Fe0^=——=0.8999M159.69r(Fe2O3)A5869⑵F= r(Ni)——=5869=0.2018M 290.9r[Ni(C4H8N2O2)2](3)MF(3)MF= r(A/_0_)2 23Mr[Al(C9H6ON)3]101.96=0.11102x459.447黃鐵礦中硫的質(zhì)量分?jǐn)?shù)約為36%，用重量法測(cè)定硫，欲得0.80g左右的BaSO4沉淀，問應(yīng)稱取質(zhì)量為若干克？解：設(shè)應(yīng)稱取試樣的質(zhì)量為m試樣g,則M0.80x——嗎」Mr(BaSC)4)-X100%=36%m試樣0.80x32.06m試樣233.436%0.80x32.06m試樣233.436%x100%=0.31(g)8用重量法測(cè)定NH4)2SO4-FeSO4-6H2O的純度，若天平稱量誤差為0.2巾8，為了使灼燒后Fe2O3的稱量誤差不大于0.1%,應(yīng)最少稱取試樣多少克？

M解：m=F?m= Fe2O3 xm解：Fe2O3 試樣 2M 試樣(NH4)2SO4-FeSO4-6H2O相對(duì)誤差二絕對(duì)誤差

稱量值x100%0.2x10-3x100%m試樣m試樣m相對(duì)誤差二絕對(duì)誤差

稱量值x100%0.2x10-3x100%m試樣m試樣mFe2O30.2x10-3M 皿5) xm2M 試樣r[(NH4)2SO4-FeSO4-6H2O]0.2x10-3M0.1%x r(ie7O3^ 2Mr[(NH4)2SO4-FeSO4.6H2O]0.2x10-30.1%x1602x392x100%W0.1%x100%x100%=1(g)9稱取1.0000g化學(xué)純MgSOj7H2O試樣，將鎂沉淀為MgNH4P04灼燒成Mg2P2。7,得0.4857g;若將硫酸根沉淀為BaS04,灼燒后得1.0224g,試問該試樣是否符合已知的化學(xué)式？原因何在？M=18.02,M=222.6,r(H2M=18.02,M=222.6,r(H2O) r(Mg2P2O7)r(MgSO4-7H2O) r(MgSO4)M=233.4r(BaSO4)解：純MgS04-7H20中含MgS04理論值為48.84%,含H2051.16%,試樣中MgS04質(zhì)量為:2M r(Mg2M r(MgSO4)Mx0.4875=2x120.4222.6x0.4875=0.5274(g)r(Mg2P2O7)Mr(MgSOMr(MgSO4)x1.0224=Mr(BaSO4)120.4233.4x1.0224=0.5274(g)0.5274w= x100%=52.74%MgSO4 1.0000w=1-52.74=47.26%H2O由計(jì)算結(jié)果可知，MgS04-7H20試樣不符合已知的化學(xué)式，原因是失去部分結(jié)晶水。10稱取含NaCl和NaBr的試樣（其中還有不與Ag+發(fā)生反應(yīng)的其它組分）0.4000g,溶解后，用0.1043mol-L-1AgN03標(biāo)準(zhǔn)溶液滴定，用去21.09mL。另取同樣質(zhì)量的試樣溶解后加過量的AgNO3溶液，得到的沉淀經(jīng)過濾洗滌烘干后，得沉淀重0.4020g。計(jì)算試樣中NaCl和NaBr的質(zhì)量分?jǐn)?shù)。已知：Mr(NaCl)=58.44, Mr(NaBr)A02.9，Mr(AgCl)A?3，Ms)=187.8解：設(shè)試樣中NaCl的質(zhì)量為xg,NaBr的質(zhì)量為yg0.1043x21.09x10-3= +58.44102.90.4020二業(yè)+358.44 102.9將①和②聯(lián)立求解，得 x=0.0147g，y=0.2005gwNaClwBrCl00147.x100%=3.68%0.40000,2005x100%=50.13%0.4000s=sx100%=905-x100%=0.09%r2 _ 55.242%V=V,溶液的組成是CO2-；12 31+106.13-1+1011.15-2+1015.00-3+1017.75-4+1019.37-5+1019.84-6―1—=2.8X10-15=10-14.553.5x10146 =P[F-]5 =106.13-1-14.55=3.8x10-io(AlF)2+ 1 Al3+5 =p[F-]25 =1011.15-2-14.55=4.0x10-6(AlF2)+ 2**5 Al3+4取某含Cl-廢水樣100mL,加入20.00