Question

The molar absorptivity for aqueous solutions of phenol at 211 nm is 6.17x103L/mol/cm. Calculate the linear range of phenol concentrations if the transmittance is to be less than 85% and greater than 7% when the measurements are made in 1cm cell.

Answer 1

Explanation:

The given data is as follows.

          $\lambda$ = 211 nm,          $\sum = 6.17 \times 10^{3} mol/L/cm$

           l = 1 cm,             7% < Transmittance < 85%

Suppose the aqueous solution follows Lambert-Beer's law. Therefore,

                   Absorbance = $-log \frac{\text{Percentage transmittance}}{100}$

Hence, for 7% transmittance the value of absorbance will be as follows.

                  Absorbance = $-log \frac{7}{100}$

                           $A_{1}$ = 1.155

For 85% transmittance the value of absorbance will be as follows.

                 Absorbance = $-log \frac{85}{100}$

                           $A_{2}$ = 0.07058

According to Lambert-Beer's law.

                  A = $\sum \times l \times C$

where,       A = absorbance

                 $\sum$ = molar extinction coefficient

                 C = concentration

Therefore, concentration for 7% absorbance is as follows.

                    $A_{1} = \sum \times l \times C_{1}$

                  $C_{1}$ = $\frac{1.155}{6.17 \times 10^{3} \times 1}$

                                  = $0.187 \times 10^{-3} mol/L$

                                  = 0.187 mmol/L

Concentration for 85% absorbance is as follows.

                   $A_{2} = \sum \times l \times C_{2}$

                  $C_{2}$ = $\frac{0.07058}{6.17 \times 10^{3} \times 1}$

                                  = $0.01144 \times 10^{-3} mol/L$

                                  = 0.01144 mmol/L

Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.