Question

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

Answer 1

Answer:

Explanation:

From the given information:

At wavelength = 270 nm

$\varepsilon x_1 = 1600 \ m^{-1} \ cm^{-1} \\ \\ \varepsilon y_1 = 200 \ m^{-1} \ cm^{-1}$

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:

$\varepsilon x_1 \ l + \varepsilon y_1 \ l= 0.5 \\ \\ A = A_1 + A_2$

$1600 xl + 200 yl= 0.5$

Divide both sides by 200

$8xl + yl = \dfrac{0.5}{200}$

$8x + y = \dfrac{0.5}{200}l$

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

Then;

$8x + y = \dfrac{0.5}{200} ---- (1)$

For 540 nm:

$\varepsilon x_2 x \ l + \varepsilon y_2 y \ l= 0.5 \\ \\ 40 xl + 800 yl = 0.5$

$x + 20 y = \dfrac{0.5}{400 \ l}$

since l = 1

$x + 20 y = \dfrac{0.5}{400 \ } --- (2)$

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

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