Question

Calculate the absorbance of solutions having twice the percent transmittance of those in Problem 13-2.

Short answer

a) The absorbance for the percent transmittance of 5.38% is 1.57.

b) The absorbance for the percent transmittance of 0.246 is 0.609.

c) The absorbance for the percent transmittance of 19.7% is 0.705.

d) The absorbance for the percent transmittance of 11.9% is 0.924.

e) The absorbance for the percent transmittance of 0.062 is 1.207.

f) The absorbance for the percent transmittance of 7.9% is 1.102.

Step-by-step solution

Given information

Convert the following transmittance data to absorbances:

(a) 5.38% (d) 23.8%

(b) 0.492 (e) 0.124

(c) 39.4% (f) 15.8%

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

A = -log T

Where,

• A is the absorbance of the solution.

• T is the transmittance of the solution.

Rearrange the above expression for the percent transmittance.

$A=-\log \left(\frac{\%T}{100}\right)$..................(1)

Here, A is the absorbance of the solution and T is the transmittance of the solution.

Part (a) Step 1: Explanation

Substitute the value in equation 1:

$$\begin{gathered} A=-\log \left(\frac{\%T}{100}\right) \\ A=-\log \left(\frac{2.69}{100}\right) \\ =-\log (0.0538) \\ =1.57 \end{gathered}$$

The absorbance for the percent transmittance of 5.38% is 1.57.

Part (b) Step 1: Explanation

Substitute the value in equation 1:

$$\begin{gathered} A=-\log T \\ =-\log (0.246) \\ =0.609 \end{gathered}$$

The absorbance for the percent transmittance of 0.246 is 0.609.

Part (c) Step 1: Explanation

Substitute the value in equation 1:

$$\begin{gathered} A=-\log \left(\frac{\%T}{100}\right) \\ A=-\log \left(\frac{19.7}{100}\right) \\ =-\log (0.197) \\ =0.705 \end{gathered}$$

The absorbance for the percent transmittance of 19.7% is 0.705.

Part (d) Step 1: Explanation

Substitute the value in equation 1:

$$\begin{gathered} A=-\log \left(\frac{\%T}{100}\right) \\ A=-\log \left(\frac{11.9}{100}\right) \\ =-\log (0.119) \\ =0.924 \end{gathered}$$

The absorbance for the percent transmittance of 11.9% is 0.924.

Part (e) Step 1: Explanation

Substitute the value in equation 1:

$$\begin{gathered} A=-\log T \\ =-\log (0.062) \\ =1.207 \end{gathered}$$

The absorbance for the percent transmittance of 0.062 is 1.207.

Part (f) Step 1: Explanation

Substitute the value in equation 1:

$$\begin{gathered} A=-\log \left(\frac{\%T}{100}\right) \\ A=-\log \left(\frac{7.9}{100}\right) \\ =-\log (0.079) \\ =1.102 \end{gathered}$$

The absorbance for the percent transmittance of 7.9% is 1.102.