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

Step-by-step 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 = (0.5)/(200)`

`8x + y = (0.5)/(200)l`

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

Then;

`8x + y = (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 = (0.5)/(400 \ l)`

since l = 1

`x + 20 y = (0.5)/(400 \ ) --- (2)`

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

`8x + y - 8x - 160 y = (0.5)/(200) - (0.5 * 8)/(400) \\ \\ \implies - 159 y = (0.5)/(200) ( 1 - (8)/(2)) \\ \\ -159 y = (-0.5 * 3)/(200) \\ \\ 159 \ y = 0.0075 \\ \\ y = (0.0075)/(159) \\ \\ y = 0.00004716 \\ \\ y = 4.7 * 10^(-5 ) \ M`