Chapter 11, Problem 11A.1ST

Interpretation Introduction

Interpretation:

Molar absorption coefficient and absorbance of the aqueous solution containing amino acid tyrosine has to be calculated where transmittance is 0.14 and cell length is $\text{5}\text{.0 mm}$. Also transmittance through a cell with a length of 1.0 mm has to be determined.

Concept Introduction:

• According to Beer–Lambert law, intensity of radiation absorbed at a particular frequency depends on number of species absorbed (N), molar concentration [J] and the path length, L.

  $\begin{array}{l}{\text{I = I}}_{\text{0}}{10}^{-\epsilon \left[J\right]L}\\ \\ where,\\ \epsilon \text{ = }Molarabsorptioncoefficient\\ \text{I = }Intensityofincidentradiation\\ {\text{I}}_0\text{ = }Intensityoftransmittedradiation\end{array}$

• Equation for transmittance, T can be given as,

  $\begin{array}{c}\text{ T= }\frac{\text{I}}{{\text{I}}_{\text{0}}}\\ \\ T\text{ = }{10}^{-\epsilon \left[J\right]L}\\ \\ where,\\ \text{I = }Intensityofincidentradiation\\ {\text{I}}_0\text{ = }Intensityoftransmittedradiation\\ \text{N = }Numberofspeciesabsorbed\\ \left[J\right]\text{ = }Molarconcentration\\ L\text{ = }Pathlength\end{array}$

• Equation for absorbance, A can be given as,

  $\begin{array}{c}\text{A = }-\log \text{ T}\\ \\ T\text{ = }\epsilon \left[J\right]L\\ \\ where,\\ \text{I = }Intensityofincidentradiation\\ {\text{I}}_0\text{ = }Intensityoftransmittedradiation\\ \text{N = }Numberofspeciesabsorbed\\ \left[J\right]\text{ = }Molarconcentration\\ L\text{ = }Pathlength\end{array}$

Explanation of Solution

Given information is shown below,

$\begin{array}{l}\text{Molar concentration, }\left[J\right]\text{= 0}\text{.10 mmol}{\text{.dm}}^{\text{-3}}\\ \text{Transmittance, T = 0}\text{.14}\\ \text{Wavelength = 240 nm }\end{array}$

• Molar absorption coefficient is calculated by the given formula.

  $\begin{array}{l}T\text{ = }{10}^{-\epsilon \left[J\right]L}\\ \\ \epsilon \text{ = }-\frac{10\left(T\right)}{\left[J\right]L}\end{array}$

Substitute the given values as shown,

  $\begin{array}{c}\epsilon \text{ = }-\frac{10\left(T\right)}{\left[J\right]L}\\ \\ \text{ = }-\frac{10\left(0.14\right)}{\left(\text{ 0}\text{.10 mmol}{\text{.dm}}^{\text{-3}}\right)\left(\text{5}\text{.0 mm}\right)}\\ \\ =\text{ 1}{\text{.70 dm}}^3.{\text{mmol}}^{-1}.m{m}^{-1}\\ \\ \text{}\text{= 1}\text{.70}\times {\text{10}}^4{\text{ dm}}^3.{\text{mol}}^{-1}.c{m}^{-1}\end{array}$

Molar absorption coefficient is $1.70\times 10^{4}d{m}^3.mo{l}^{-1}.c{m}^{-1}$.

• Absorbance of the solution is determined as given,

  $\begin{array}{c}\text{A = }-\log \text{ T}\\ \\ \text{ = }-\log \left(0.14\right)\\ \\ =0.85\end{array}$

Absorbance of the solution is $0.85$

• Absorbance of radiation through a $1.0\text{ mm}$ cell length is calculated as follows,

  $\begin{array}{c}A\text{ = }\epsilon \left[J\right]L\\ \\ \text{= }\left(\text{1}{\text{.70 dm}}^3.{\text{mmol}}^{-1}.m{m}^{-1}\right)\left(\text{0}\text{.10 mmol}{\text{.dm}}^{\text{-3}}\right)\left(1.0\text{ mm}\right)\\ \\ \text{= 0}\text{.17}\end{array}$

Transmittance of radiation through a $1.0\text{ mm}$ cell length is calculated as follows,

$\begin{array}{c}\text{A = }-\log \text{ T}\\ \\ \text{T = }{10}^{-\text{A}}\\ \\ \text{ = }{10}^{\left(-0.17\right)}\\ \\ =0.68\end{array}$

Transmittance of radiation through a $1.0\text{ mm}$ cell length is $0.68$.