Analytical Chemistry 3 rd EXAM. June 4, 2013

Transcription

1 Analytical Chemistry 3 rd EXAM. June 4, 2013 I. Suppose that we place a mixture of substances A and B in equal concentrations in the first box in column 2 (figure 1). Assume that substance A dissolves 75 % in phase 1 (stationary) and 25 % in phase 2 (mobile), while substance B dissolves 25 % in phase 1 and 75 % in phase 2. When solvent (eluent) is added, they move down the column (Figure 2). (20pt) Figure 1 Figure 2 a) Fill in the match boxes (No.1 ~No.5) in Figure 2 for both stationary and mobile phases. (15pt) b) Plot the distribution of A and B (fraction in mobile phase vs. box number) along the column of Figure 2 after five additions of solvent. (5pt)

2 a) b)

3 II. Explain or describe the followings: (20pt) a) Principle of compound electrode (carbon dioxide gas -sensing electrode) (5pt) b) Disadvantages of ion-selective electrodes (5pt) c) The formation constant (K f ) is usually defined in terms of the species Y 4- reacting with the metal ion although Y 4- is the only one of the seven forms of EDTA (HY 3-, H 2 Y 2 2-, etc.) (5pt) d) In the EDTA titration, the auxiliary ligand must be kept below a certain level. (5pt) a) The CO 2 gas-sensing electrode consists of an ordinary glass ph electrode surrounded by a thin layer of electrolyte solution enclosed in a semipermeable membrane made of rubber, Teflon, or polyethylene. A Ag AgCl reference electrode is immersed in the electrolyte solution. When CO 2 diffuses through the semipermeable membrane, it lowers the ph in the electrolyte. The response of the glass electrode to the change in ph is a measure of the CO 2 concentration outside the electrode. b) 1) They respond to the activity of analytes. We usually want concentrations, not activities 2) They respond to the activity of uncomplexed ions. Therefore, ligands must be absent or masked. 3) Precision is rarely better than 1%. * 1 mv error 4% error in monovalent ion activity. It doubles for divalent ions and tripled for trivalent ions. 4) They can be fouled by organic solutes like proteins which lead to sluggish, drifting response. 5) They are fragile and have limited shelf-life. c) Given defining K f in terms of makes it more convenient., K f is not dependent on ph. Thus,

4 d) A ligand should bind the metal of interest strongly enough to prevent metal hydroxide from precipitating but weakly enough to give up the metal when EDTA is added. If too much is present, ligands won t give up metals when EDTA is added and if there are too few ligands, metal hydroxide will precipitate. Thus, the auxiliary ligand must be kept below a certain level. III. A Gaussian bandshape results when the partition coefficient, K is independent of the concentration of solute on the column in chromatography. In real columns, K changes as the concentration of solute increases, and band shapes are skewed. Silica surfaces of columns and stationary phase particles have hydroxyl groups that form hydrogen bonds with polar solutes. (20pt) a) What kind of skewed bandshape do you expect? Draw the bandshape you may obtain and explain why you get such a skewed bandshape. (10pt) b) The hydrogen bonding between hydroxyl groups and polar solutes could be suppressed by blocking the hydroxyl groups. Describe a blocking method. (10pt) a)

5 Due to hydrogen bonding between hydroxyl groups (on silica surfaces of columns and stationary phase particles) and polar solutes, small quantities of solute are retained more strongly than large quantities. This leads to a long tail of gradually decreasing concentration after the peak as shown in boxed isotherm. b) Silanization can reduce tailing by blocking the hydroxyl groups with nonpolar trimethylsilyl groups as shown below: IV. (20pt) a) Calcium ion was titrated with EDTA at ph 11, using Calmagite as indicator (Table 11-3). Which is the principal species of Calmagite at ph 11? What color was observed before the equivalence point? After the equivalence point? (10pt) b) Pyrocatechol violet (Table 11-3) is to be used as a metal ion indicator in an EDTA titration. The procedure is as follows: (10pt) 1. Add a known excess of EDTA to the unknown metal ion. 2. Adjust the ph with a suitable buffer. 3. Back-titrate the excess chelate with standard Al 3+. From the following available buffers, select the best buffer, and then state what color change will be observed at the end point. Explain your answer. (a) ph 6-7 (b) ph 7-8 (c) ph 8-9 (d) ph 9-10

6 a) - Which is the principal species of Calmagite at ph 11? HIn 2- Route 1: compare pka values or Route 2: calculate ratio of species. [H 2 In - ] [HIn 2- ] + [H + ] [HIn 2- ] [In 3- ] + [H + ] Therefore, HIn 2- is the principal species. ph pka log HIn H In HIn H In ph pka log In HIn In HIn What color was observed before the equivalence point? wine red - After the equivalence point? blue

7 b) Color change for each buffers (a) ph 6-7: yellow blue (b) ph 7-8: yellow~violet blue (c) ph 8-9: violet blue (d) ph 9-10: violet blue Buffer (a) ph 6-7 will be the best among buffers from (a) through (d) because the color change from yellow to blue is easier to observe than that from violet to blue in other buffers. V. The following figure shows how the ion-selective electrode works. (20pt) (a) Why is the electrical potential (E outer or E inner ) created across the membrane? (8pt)

8 (b) Derive the following equation for the electric potential difference across phase boundary between membrane and analyte. (6pt) A o : activity of C + in the outer aqueous solution A m : activity of C + in the membrane (c) Derive the following equation for the potential difference between the outer and inner solution. (6pt) A a) Due to phase separation by membrane, ions R - (hydrophobic) and A - (hydrophilic) cannot enter aqueous and organic phases respectively, whereas C + can freely diffuse across the interface. Inside the membrane, C + complexes with L forming LC which develops equilibrium with a small amount of free C +. As soon as a few C + ions diffuse from the membrane into the aqueous phase, there is a buildup of positive charge in the water immediately adjacent to the membrane. This charge separation creates an electric potential difference (E outer ) across the membrane. b) When C + diffuses from a region of activity A m in the membrane to a region of activity A o in the outer solution, the free energy change is given as As discussed in a) charge imbalance creates an electric potential difference (E outer ). The free energy difference for C + in the two phases is nfe where F is the Faraday constant and n is the charge of the ion. At equilibrium, the net change in free energy for diffusion of C + across the membrane boundary must be 0. So we get

9 nfe 0 Solve for E outer and obtain c) Potential difference between the outer and the inner solutions is: Constant Constant Constant Since the first, third and forth terms on the right hand side of above equation are constant, combining those yields A