Question

Before working this pre-lab exercise read the theory, experimental precedures and sample calculations outlined in the laboratory manual. 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 0.310 and at 520 nm was 0.770, 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

The absorbance of the mixture of HMR and MR^- at 425 nm = 0.310

                                                                     at 520 nm = 0.770

According to beer law A = k c          where k = beer law constant which is given by the slope of the beer law plot.

We know that the total absorbance in a mixture A = A₁ + A₂

where A₁ = absorption due to component HMR

    A₂ = absorption due to component MR^-

A₁ = k₁c₁   k₁ = beer law constant of component HMR         A₂ = k₂c₂ , k₂ = beer law constant of component MR^-

                 c₁ = concentration of componet HMR                          c₂ = concentration of component MR^-

At wavelength 425 nm

A = 0.310 , k₁ and k₂ are 0.072 and 0.306 respectively

A₁ = k₁c₁ = 0.072 c₁

A₂ = k₂c₂ = 0.306 c₂

therefore 0.310 = 0.072 c₁ + 0.306 c₂ (or)     0.072 c₁ + 0.306 c₂ = 0.310   ----------------- (1)

similarly at wavelength 525 nm

A = 0.770 , k₁ and k₂ are 0.760 and 0.036 respectively

A₁ = k₁c₁ = 0.760 c₁

A₂ = k₂c₂ = 0.036 c₂

therefore 0.770 = 0.760 c₁ + 0.036 c₂ (or) 0.760 c₁ + 0.036 c₂ = 0.770 -------------------- (2)

solving (1) and (2)

0.072 c₁ + 0.306 c₂ - 0.310   = 0   ----------------- (1)

0.760 c₁ + 0.036 c₂ - 0.770   = 0    ----------------- (2)

c₁ / [ -0.306 * 0.770 + 0.310*0.036] = c₂ / [-0.310*0.760 + 0.072*0.770] = 1 / [0.072 * 0.036 - 0.306 * 0.760]

c₁ / [ -0.23562 + 0.01116] = c₂ / [ - 0.2356 + 0.05544] = 1 / [ 0.002592 - 0.23256]

c₁ / -0.22446 = c₂ / -0.18016 = 1 / -0.23

[MR^- ] / [ HMR] = c₂ / c₁ = 0.18016 / 0.22446

                       = 0.80

Thus the ratio of relative concentrations of MR^- and HMR in solution is 0.80