第十一章 吸光光度法

1、第十一章 吸光光度法习题 11-1 朗伯-比尔定律的物理意义是什么？答：溶液的吸光度A与液层厚度b成正比，与溶液浓度c成正比，即A = kbc习题11-2摩尔吸光系数的物理意义是什么？它与哪些因素有关？答：摩尔吸光系数数值上等于吸光物质浓度为l.OmoLL-i,液层厚度为1.0cm时溶液的吸 光度。它与吸光物质的性质有关。习题 11-3 将下列透光度换算成吸光度 10% 60%100%解： A= -lgT T=10%A = -lg10% = 1 T = 60%A = -lg60% = 0.22 T=100%A = -lg100% = 0习题11-4某试液用2.0 cm的比色皿测量时，T = 60%，若改用1.0 cm或3.0 cm比色皿，T及A等于多少?解： T1 = 60%A1 = 0.22,T2= 10-A= 10-0.11 = 78%T3 = 10-A = 10-0.33 = 47%若改用1.0 cm比色皿A2 = 0.11若改用3.0 cm比色皿A3 = 0.33习题11-5 5.溶液，在久=525 nm处用3.0 cm吸收皿测得吸光度!= 0.3364max 计算吸光系数a和

2、摩尔吸光系数；1.94 x10-4m=0.12%若仪器透光度绝对误差AT = 0.4%，计算浓度的相对误差，c解： =A/bc = 0.336/(5.0x10-5 x3.0) = 2.2 x103 Lmol-1cm-1c =5.0x 10-5x 158.04 = 7.9x 10-3a = 0.336 / (7.9 x 10-3 x 3.0) = 14 Lg-1cm-1 T = 10-0.336 = 0.461Ac0.434 x 0.4%1c0.461xlg0.461习题11-6某钢样含镍约0.12%，用丁二酮肟比色法( =1.3x104)进行测定。试样溶解后， 显色、定容至100.0 mL。取部分试液于波长470 nm处用1.0 cm比色皿进行测量，如希望此 时测量误差最小，应称取试样多少克？解：c =0434 = 3.3 x10-5mol L-11x1.3x104m = 3.3 x 10-5 x 0.100 x 58.69 = 1.94 x 10-4gm= 0.16 g习题11-7 5.00x10-5molL-1的KMnO4溶液在520 nm波长处用2.0 cm比色皿测得吸光度A =

3、0.224。称取钢样1.00 g溶于酸后，将其中的Mn氧化成MnO -,定容100.00 mL后，在 4 上述相同条件下测得吸光度为 0.314。求钢样中锰的含量。0.224=2.24 x 103L - mol -1 - cm-1解：2x5.00x10-5A 0.314c( x) = 7.01 x 10 -5mol - L-1 - b 2.24 x 10-5 x 2e(Mn) =7.01x 10-5 x 0.100 x 54.9/1.00 = 3.9x10-4习题11-8准确称取0.536 g NH4Fe(SO4)212H2O，溶于水后定容500.00 mL,再取不同体积溶液在50.0 mL比色管内用邻二氮菲显色，定容后在510 nm处测得吸光度如下：V(Fe2+) /mL01.002.003.004.005.00A00.120.250.380.510.63取1.00 mL未知含Fe2+溶液稀释到100.00 mL，再取稀释液5.00 mL，在50.0 mL比色管 内用同样方法显色定容后测得吸光度A = 0.47。求未知溶液中Fe2+的浓度。解:0.536 g NH4Fe(SO4)21

4、2H2O,溶于水后定容 500.00 mL,溶液浓度为 c(NH4Fe(SO4)212H2O) =1.07x10-3 gmL-1,亦为 c(Fe2+) =1.24x10-4 gmL-10NH4Fe(SO4)212H2O 系列溶液的吸光度 A 对V(Fe2+)作图得工作曲线，在工作曲线上查得A(x ) = 0.47处V(x) = 3.70 mL。故未知溶液 中Fe2+的浓度c(Fe2+) =1.24x10-4x3.70x100.00 /5.00x1.00=9.18x10-3 g-mL-1习题11-9普通光度法分别测定0.50 x10-4, 1.0x10-4 moLLZn2*标液和试液的吸光度A为0.600， 1.200， 0.800。 若以0.50 x10-4molL-1Zn2+标准溶液作参比溶液，调节T100%，用示差法测定第二 标液和试液的吸光度各为多少？ 两种方法中标液和试液的透光度各为多少？ 示差法与普通光度法比较，标尺扩展了多少倍？ 根据中所得有关数据，用示差法计算试液中Zn的含量(mgL-1) 解： A = AA = sbAc = KAcrA 严 1.200 0.600 =

5、0.600rs2A = 0.800 0.600 = 0.200rx 普通法：T = 10-1.200 = 6.31%s2T = 10-0.800 = 15.8%xT = 10-0.600 = 25.1%s1示差法： T =100%s1T = 10-0.600 = 25.1%rs2T = 10-0.200 = 63.1%rx 扩展 4 倍 AA( s) = Ac(s)AA( x) Ac( x)0.600 = 1.0 xl0 0.5 xl00.200 c(x)0.5 x10-4c( x)- 0.5 x 10-4= 0.17 x 10-4c (x) = 0.67 x 10-4 molL-1c(Zn) = 0.67 x 10-4 x 65.39 = 4.4 x 10-3 gL-1 = 4.4 mgL-1习题11-10用分光光度法测定含有两种配合物x和y的溶液的吸光度(b =1.0 cm),获得下列数据：溶液浓度 c/molL-1吸光度 A1吸光度 A2久=285 nm久=365 nmx5.0x10-40.0530.430y1.0x10-30.9500.050x+y未知0.6400.370计算未

6、知液中 x 和 y 的浓度。解：久=285 nm时0.053 =285x5.0x 10-4 x 1.0x0.950 = S285 x 1.0 x 10-3 x 1.0y* 285 = 1.1x 102L - mol -1 - cm -1x* 285 = 9.5 x 102L - mol -1 - cm-1y久=365 nm时0.430 =*365x5.0x10-4 x1.0x0.050 =*365x1.0x10-3 x1.0y* 365 = 8.6 x IO2L - mol-1 - cm-1x* 365 = 0.050 x 103L - mol-1 - cm-1y285 nm 0.640 =1.1 x 102c(x)+ 9.5x 102c(y)365 nm 0.370 = 8.6x 102c(x)+ 0.050x 103c(y)c(x) = 3.9 x 10-4molL-1c(y) = 6.3 x 10-4molL-1习题 11-11 A solution containing iron (as the thiocyanate complex) was observed to tran

7、smit 74.2% of the incident light with X=510 nm compared to an appropriate bland. What is the absorbance of this solution? What is the transmittance of a solution of iron with four times as concentrated?解： A1 = -lgT = -lg74.2% = 0.130 A2 = 4 A1 = 0.520习题 11-12 Zinc(II) and the ligand L form a product cation that absorbs strongly at 600 nm. As long as the concentration of L excess that of zinc (II) by a factor of 5, the absorbance of the solution is only lined on the cation concentration. Neither zinc (II) nor L absorbs at 600 nm. A solution that is 1.60X 10-6 molL-1 in zinc (II) 1.00 mol L-1 in L has an absorbance of 0.164 in a 1.00cm cell at 600 nm. Calculate the transmittance of this solution. the transmittance of this solution in a 3 cm cell. the molar absorbance of the complex at 600 nm.解： T1= 10-A= 10-0.164 = 68.5% A = 3A = 0.49221T = 10-A= 10-0.492 = 32.2%2 = A/bc = 0.164/(1.0x1.60x10-6) = 1.03x105 Lmol-1cm-1

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