Question

Beer lambert law. Simple Q

Can some one find the molar absorbitiy "e" for the  1,1'-diethyl 2,2'-cyanine iodide. ( SEE DATA BELLOW AND CALCULATIONS) and the theoretical value and explain where there is a diffrence between them. Thank you so much.

3) Compare the experimental and literature values of ε (dye I only). Do you notice any discrepancies between these values?

Answer 1

Answer :

According to Beer-Lambert law,

A =  $\epsilon$​​​​​​. C.L

Where,

A = absorbance of the solution (unit less)

$\epsilon =$ molar absorptivity or molar extinction Co-efficient.

C = Concentration of the solution ( mol /L)  

L = length of the solution i.e. Path length of the cuvette in cm

And given that L =1.94 nm = 1.94 × 10^-7 cm  [ from the given calculation]

So if we plot A vs C the the graph will be a straight line with a positive slope of  $\epsilon. L$ and pass through the origin.

Experimentally we have also found the straight line for the A vs C curve which is given in Figure 2 in the question.

Now from the experimental cure [ fig 2] we have to find the slope of the straight line.

For this purpose we have to chose two points in the straight line and for this two point the respective value in the two axis is to be determined.

The slope then be calculated by the below formula,

Slope = (C₂ - C₁) /( A₂ - A₁​​​​​​)

Where,

C₂ and A₂ are the concentration and Absorption respectively corresponding to the 2^nd point

C₁ and A₁ are concentration and Absorption respectively corresponding to the 1^st point.

Let for the 1st point,

A₁ =0.3 ( from the Absorption axis of figure 2)

Corresponding Concentration,

C₁ = 4×10^-6 mol L^-1 ( from the Concentration axis of fugure 2)

Similarly considered a 2nd point,

A₂ =0.6 [ from Absorption axis ]

Corresponding concentration,

C₂ = 8×10^-6 mol L^-1 [ from concentration axis]

So,

Slope = (0.6 - 0.3) /{( 8×10^-6 mol L^-1) - (4×10^-6 mol L^-1)}

Or, Slope = {0.3 /(4×10^-6)} mol^-1 L

Or, Slope = 75×10³ L mol^-1

But according to Beer-Lambert law​​​​​​,

The slope is = $\epsilon. L$​​​​​​

So ,

$\epsilon. L$ = 75×10³ L mol^-1

Or,  $\epsilon$. (1.94×10^-7 cm) = (75×10³ L mol^-1 )

Or, $\epsilon =$ (75×10³ L mol^-1 )/ (1.94×10^-7 cm)

Or, $\epsilon =$ 3.86 ×10¹¹ L mol^-1 cm^-1​​​​​​

(Answered)