The UV-Vis absorption spectrum of the mixture of Ru(bpy)^2+_3 and Ir(bpy)^2+_3 w

Question

The UV-Vis absorption spectrum of the mixture of Ru(bpy)^2+_3 and Ir(bpy)^2+_3 was measured to identify the unknown concentrations of Ru(bpy)^2+_3 and Ir(bpy)^ 2+_3. Absorbance at 350 nm and 545 nm were used for this experiment, and the molar extinction coefficients (epsilon) were estimated as shown in the table below. Calculate the concentration of each species in the mixture, using the following data: the total absorbance values at 350 nm and 545 nm are 0.954 and 0.672, respectively, and the light path length of a cuvette is 1 cm.

Explanation / Answer

Ans. Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                       A = Absorbance

                       e = molar absorptivity at specified wavelength (M-1cm-1)

                        L = path length (in cm)

                        C = Molar concentration of the solute

Let the [Ru(bpy)32+] = R molar       , and [Ir(bpy)32+] = C molar in the given sample.

It’s assumed that the unit of molar absorptivity is in terms of M-1cm-1 and path length is 1.0 com in all cases.

#1. At 350 nm,

Total absorbance of the mixture = Abs of Ru(bpy)32+ + Abs of Ir(bpy)32+

Or, 0.954 = (600 M-1cm-1) x R M x 1.0 cm + (100 M-1cm-1) x C M x 1.0 cm

Or, 0.954 = 600R + 100C

Hence, 600R + 100C = 0.954                    - equation 1

#2. At 545 nm,

Total absorbance of the mixture = Abs of Ru(bpy)32+ + Abs of Ir(bpy)32+

Or, 0.672 = (58 M-1cm-1) x R M x 1.0 cm + (325 M-1cm-1) x C M x 1.0 cm

Or, 0.672 = 58R + 325C

Hence, 58R + 325C = 0.672                      - equation 2

#3. Comparing (equation 1 x 58) – (equation 2 x 600)-

            34800 R          + 5800 C                    = 55.332

       (-) 34800 R        + 195000 C                = 403.0

                                    - 189200 C     = -347.868

                                    Or, C = 347.868/ 189200 = 1.839 x 10-3

Therefore, [Ir(bpy)32+] in the sample = C M = 1.839 x 10-3 M

Putting the values of C in equation 1-

            600R + 100C = 0.954

            Or, 600R = 0.954 – 100 x (1.839 x 10-3) = 0.7701

            Or, R = 0.7701 / 600 = 1.284 x 10-3

Hence, [Ru(bpy)32+] = R M = 1.284 x 10-3 M

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