Question

Suppose that a student collects the following spectrophotometric data to determine the relative concentrations of HMR and MR in solution: The abosorbance at 425 nm was measured to be 0.170 and at 520 nm was 0.690, the slopes corresponding to standard Beer's Law plots were as follows:

αIA = 0.072

αIB = 0.306

αIIA = 0.760

αIIB = 0.036

Using this data, calculate the ratio between the relative concentrations of MR- and HMR in this solution.

Answer 1

We know that according to Beer's law

Absrobance = e.c.l

e = Extinction coeffecient

c = Concentration

l = length

Now absrobance of mixture = Absrobance of HMR + Absrobance of MR

Now at two wavelengths the two abrobances are different

The slope mentioned is the molar absroptivities for two wavelength absroption v/s concentration

Generally the length of path is = 1cm [So we assume the same here also]

A1 = 0.170 = 0.072 X [MR] + 0.306 [HMR]

[MR] = (0.170 - 0.306[HMR] ) / 0.072 .........(1)

A2 = 0.690 = 0.76 [MR] + 0.036 [HMR] .........(2)

Let us put (1) in (2)

0.690 = 0.76 X (0.170 - 0.306[HMR] ) / 0.072 + 0.036[HMR]

0.690 = 10.55 (0.170 - 0.306[HMR] ) + 0.036 [HMR]

0.690 = 1.79 - 3.23 [HMR] + 0.036[HMR]

1.1 = 3.194[HMR]

[HMR] = 0.345 .............(3)

Put (3) in (1)

[MR] = (0.170 - 0.306[HMR] ) / 0.072

[MR] = 0.170 - 0.306 X 0.345 / 0.072

[MR] = 0.895

So ratio of [MR] / [HMR] = 0.895 / 0.345 = 2.59 : 1