Analytical Chem istry - DePauw University

590 Analytical Chemistry 2.01.0moles metal = moles ligand0.8absorbance0.60.4metal in excessligand in excess0.2Figure 10.38 Continuous variations plotfor Example 10.8. The photo shows thesolutions used in gathering the data. Eachsolution is displayed directly below its correspondingpoint on the continuous variationsplot.0.00.0 0.2 0.4 0.6 0.8 1.0X LPractice Exercise 10.8So l u t i o nA plot of absorbance versus the mole fraction of ligand is shown in Figure10.38. To find the maximum absorbance, we extrapolate the two linearportions of the plot. The two lines intersect at a mole fraction of ligand of0.75. Solving for y givesXL075 .y = = = 31 − X 1 − 0.75LThe formula for the metal–ligand complex is Fe(o-phenanthroline) 3 2+ .Use the continuous variations data in the following table to determine the formula for the complexbetween Fe 2+ and SCN – . The data for this problem is adapted from Meloun, M.; Havel, J.; Högfeldt,E. Computation of Solution Equilibria, Ellis Horwood: Chichester, England, 1988, p. 236.X L absorbance X L absorbance X L absorbance X L absorbance0.0200 0.068 0.2951 0.670 0.5811 0.790 0.8923 0.3250.0870 0.262 0.3887 0.767 0.6860 0.701 0.9787 0.0710.1792 0.471 0.4964 0.807 0.7885 0.540Click here to review your answer to this exercise.