CHemistry lab

Question

A beam of light is reduced to one-fourth of its initial intensity on passing through a 10 cm thick cell containing a 0.05 M (molar = moles per liter) solution. Calculate the extinction coefficient for the solute.

Answer 1

Hello Nelson

 

To solve this question, we need to use the Beer's Law equation, which is:

 

A = ε l c

 

where A is the absorbance, ε is the molar extinction coefficient, l is the length (or thickness) of the cell, and c is the concentration.

 

From your question, we know the values of l (10 cm) and c (0.05 M).

 

In UV spectroscopy, we can measure the intensity of light absorbed by a solution either in units of absorbance or %-transmittance.  Absorbance and %-Transmittance are inversely proportional to each other.  For example, 100% T equals a value of 0 (zero) absorbance.  If %-Transmittance is known, we can calculate A using the following equation:

 

A = 2 - log(%-T)

 

The question states that the intensity of light was reduced by 1/4.  If we assume that intensity of the starting light = 100%, then 1/4 of 100% = 25%.  This is the value of %-T.  Now use the above equation to solve for A.

 

A = 2 - log(25%) = 2 - log(25) = 2 - 1.40 = 0.60   (FYI - Absorbance has no units!)

 

Now rearrange the Beer's Law equation to solve for ε.

 

ε = A / lc  =  0.60 / (10 cm x 0.05 M) = 1.2 M^-1 cm^-1