分析化学课件第十章__紫外—可见分光光度法习题解答

1第十章紫外—可见分光光度法习题解答13、解：33%111012.100.1104962.0557.0ClAEcm43%111065.21012.11023610cmEM14、解：被测样品溶液的浓度为C样)100/(1000.110010000.20.1000500.03mlgC样被测样品中维生素C的浓度为C测)100/(1084.910839.900.1560551.044%11mlglEACcm测样品中维生素C的百分质量分数为ω%4.981001000.110839.910034样测CC15、解：依据Lambert-Beer定律：TEClAlg则22211121lglglCElCETT当溶液的组成及浓度一定时，则11222121lglgTllTCCEE当%6000.211Tcml时当时cml00.12，则111.060.0lg00.200.1lg2T%5.772T同理：%5.46333.060.0lg00.200.3lg00.3333TTcml%0.36444.060.0lg00.200.4lg00.4444TTcml17、解：某碱的苦味酸盐（B.A）的百分吸光系数为24100.110481.2598.03%11ClAEcm依据8302411010210104%11%11cmcmEMEM盐2BOH+HABA+H2O碱的分子量为M=M盐-M酸+MH20=830-229+18=61918、解：依据误差要求，当%5.0T，吸光度测量范围为0.2~0.7时，浓度的测量相对误差较小，此时对应的浓度为适宜浓度。当吸光度A=0.4343时，浓度的测量相对误差最小，此时对应的浓度为最佳浓度。依题意6.540107.147.31410104%11MEcm当A=0.2时，)100/(1070.300.15412.04%11mlgEACcm当A=0.7时，)100/(1029.100.15417.03%11mlgEACcm故适宜浓度范围为3.70×10-4(g/100ml)~1.29×10-3(g/100ml)当A=0.4343时，最佳浓度为)100/(1003.800.15414343.04%11mlgEACcm19、解：M+＋X-=MXMX的稳定常数]][[][XMMXK依题意得：在第一种溶液中，M＋全部转化为MX，则CMX=0.000500ml/L1600000500.0800.0.MXMXMXMXCACA在第二种溶液中，有部分M＋转化MX。反应达平衡时LmollAMX/000400.000.11600640.0.][)/(000100.0000400.0000500.0][LmolM)/(0246.00004.00250.0][LmolX1630246.0000100.0000400.0]][[][XMMXK320、解：（a）A=-lgT=-lg0.716=0.145(b)50.41548333.00101ACl(c)548333.001010100.36736.7%ClT21、解：（1）对照品)100/(1000.21001000100.2033mlgC对20700.11000.2414.03%11lCAEcm对对（2）原料药)100/(1000.21001000100.2033mlgC样)100/(10932.100.1207400.03%11mlglEACcm测%6.961001000.210932.110033样测原料CC（3）注射液)100/(0.1010000.1000.1mlmlC样)100/(502.2)100/(002502.000.1207581.0%11mlmgmlglEACcm测)/(250.0)100/(00.10)100/(50.2mlmgmlmlmlmgCC样测注射液22、解：依题意得：BABAAAAAAA238238238282282282因BBAA238282故lCEEAAAAAAAAA.).(2382822382822382824)100/(364.0100000.1)270720(278.0442.0).(238282238282mlmglEEAACAAA23、解：()[]HAaAAAKHAAAHA=0.002,AA-=1.024,A=0.430,[H+]=1.0×10-4450.0020.430()[]1.0107.21100.4301.024HAaAAAKHAApKa=－lg7.21×10-5=4.1424、改错：有一浓度为2.00×10－3mol/L的有色溶液。解：依据Lambert-Beer定律A=ECl=-lgT，相同物质在相同波长处，吸光系数相等。使用相同的吸收池，则l相等。此时2121CCAA)/(1066.41000.2300.0200.0lglg331121122LmolCATCAAC25、解：当透光率测量误差%5.0T时，吸光度A=0.4343时，浓度的测量误差最小。此时)/(1041.200.1108.14343.054LmollAC)(0135.0005.0105085.551041.2%5.05035gCMm26、解；1200120001001010%11MECM)100/(000400.00.1000.10000.22500500.0mlgC样%2.79100000400.0000317.0100样测CC)100/(000317.000.112000417.0lg.lg%11%11mlglETEACcmcmM测