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Given the molar absorptivity for a species X of 1600 M-1cm-1 at a wavelength of 270 nm, and 400 M-1cm-1 at a wavelength of 540 nm. For species Y, the molar absorptivity at 270 nm is 200 M-1cm-1 and 8000 M-1cm-1 at 540 nm. For a certain mixture, the absorbance at 270 nm is 0.5 and, at 540 nm, is also 0.5. What is the concentration of species Y in the mixture

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Answer:

Explanation:

From the given information:

At wavelength = 270 nm

[tex]\varepsilon x_1 = 1600 \ m^{-1} \ cm^{-1} \\ \\ \varepsilon y_1 = 200 \ m^{-1} \ cm^{-1}[/tex]

At 270 nm

Suppose x is said to be the solution for the concentration of x and y to be the solution for the concentration of y;

Then:

[tex]\varepsilon x_1 \ l + \varepsilon y_1 \ l= 0.5 \\ \\ A = A_1 + A_2[/tex]

[tex]1600 xl + 200 yl= 0.5[/tex]

Divide both sides by 200

[tex]8xl + yl = \dfrac{0.5}{200}[/tex]

[tex]8x + y = \dfrac{0.5}{200}l[/tex]

Use l = 1cm (i.e the standard length)

Then;

[tex]8x + y = \dfrac{0.5}{200} ---- (1)[/tex]

For 540 nm:

[tex]\varepsilon x_2 x \ l + \varepsilon y_2 y \ l= 0.5 \\ \\ 40 xl + 800 yl = 0.5[/tex]

[tex]x + 20 y = \dfrac{0.5}{400 \ l}[/tex]

since l = 1

[tex]x + 20 y = \dfrac{0.5}{400 \ } --- (2)[/tex]

Equating both (1) and (2) together, we have:

[tex]8x + y - 8x - 160 y = \dfrac{0.5}{200} - \dfrac{0.5 \times 8}{400} \\ \\ \implies - 159 y = \dfrac{0.5}{200} ( 1 - \dfrac{8}{2}) \\ \\ -159 y = \dfrac{-0.5 \times 3}{200} \\ \\ 159 \ y = 0.0075 \\ \\ y = \dfrac{0.0075}{159} \\ \\ y = 0.00004716 \\ \\ y = 4.7 \times 10^{-5 } \ M[/tex]

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