《分析化学》阅读材料 06
摘自 Analytical Chemistry (FECS)
 Molecular systems can be identifies by their characteristic energy term schemes consisting of discrete
electronic, vibrational and rotational states. At room temperature the substances are mainly in their electronic
and vibrational ground states（电子和振动基态）. Upon interaction with the appropriate type of electromagnetic
radiation, characteristic electronic, vibrational and rotational transitions can be induced in the sample. Thses
excited states（激发态）usually decay to their original ground states within 10-8s, either by emitting the
previously absorbed radiation in all directions with the same or a lower frequency, or by “radiationless”
relaxation（无辐射松弛）.
 Molecular spectra（分子光谱）can be obtained in the absorption or emission mode from samples in the gaseous,
liquid or solid state. They are images of the interactions mentioned above and contain analytical information
about the sample.
 Most UV-VIS spectra are obtained by measuring the intensity of the absorption of monochromatic radiation
across a range of wavelengths passing through a solution in a cuvette（液池）. The practical wavelength region
extends from 190-400 nm (UV range) and from 400-780 nm (VIS range).
 In a typical experiment, a light beam of intensity I0 strikes a sample consisting of a quartz or glass cell
containing a solution. After passing through the cell, the light beam has a reduced intensity I due to reflection
losses at the cell windows, absorption in the sample and, eventually, by scattering at dispersed particles. Only
the absorption losses are caused by the dissolved sample. The run is repeated using an identical cell containing
only solvent to compensate for reflection and scattering losses, and the transmittance（透射率）T is calculated
using the following equation, T=I / I0 Isolution / Isolvent . The presentation of transmittance T, as a function of
wavelength , is the required spectrum of the sample.
 The intensity of an absorption band, i.e., the absorbance, is proportional to the number of absorbing species in
the illuminated part of the cell. Absorbance, A, is defined by the equation, A = –logT =log(I0/I) and is
proportional to the cell thickness, b [cm], the concentration of the solution c [mol/L]; and a substance-specific
proportionality constant  called the molar absorptivity, [L.mol-1 cm-1]. A =  b c , the Beer’s Law.
 For a given system, a linear relationship exists between A and the sample concentration, but usually only for
dilute solution (c  0.1 mol/L).
 The distribution of a solute S, equilibrated between an aqueous phase and an organic solvent may be described
by an equilibrium equation: Saq
Sorg .There is no need to choose an aqueous phase as one of the two
phase, and we can describe distributions between any pair of immiscible solvents, liquid 1 and liquid 2, with
the appropriate solutes, S1 and S2, respectively: S1
S2. Such systems are described by an equilbrium
constant: KD = [S]org / [S]aq or [S]2 / [S]1. Where KD is called the partition coefficient（分配系数）.
 In the context of practical chemical procedures, we have to go a step further beyond ideal thermodynamic
behavior. For example, in the extraction of an acid HA from an aqueous into an organic phase, we dealing with
more than one chemical entity, namely, the dissociated and undissociated acid forms. We can define a partition
coefficient only to the ratio of the undissociated acid KD = [HA]org / [HA]aq, which dose not tell the whole story,
because, in the aqueous phase the acid may coexist in the dissociated form, and in the organic phase, higher
ion pair products may be formed. In general total solute concentrations are determined by analytical
investigations, and the information gained is used to explain the distribution of the species between the two
phases, including speciation. The ratio of the total concentrations of the solute is a practical means of dealing
with distribution equilibrium situations, and is called the distribution ratio（分配比）Dc:
Dc 
[tatal concentration of all forms of HA]org
[tatal concentration of all forms of HA] aq
Problem
1. A solution contains 3.0 mg of iron per liter. The iron is converted into a complex with 1.10-phenanthroline（邻
二氮菲）
； and the absorbance of the solution in a 2.0 cm cell is 1.20. The molecular weight of the complex is
596. Calculate (a) the absorptivity（吸收系数）and (b)the molar absorptivity（摩尔吸收系数）of the ferrous
complex.
2. 100 mL of a solution which is 0.100 mol/L in the weak acid HA is extracted with 25.0 mL of ether. After the
extraction a 25.0 aliquot of the aqueous phase required 20.0 mL of 0.0500 mol/L NaOH for titration.
Calculate the distribution ratio of HA organic to aqueous.
3. A sample solution contains Cd2+ and Zn2+.A 25.00 mL aliquot is titration with 0.01000 mol/L EDTA at pH
5.0, requiring 32.50 mL; another 25.00 mL portion adds potassium iodide in excess, then it is passed through
a anion-exchange column（阴离子交换柱）in the basic form. If 20.20 mL of 0.0100 mol/L EDTA is required to
titrate the effluent（流出液）at pH 5.0. Calculate the concentrations of Cd2+ and Zn2+ in the sample solution.