Question

A Standard solution X has a concentration of 2.10 x 10-2 M, with an absorbance of 0.0621. The absorbance of a solution of X with an unknown concentration was measured to be 0.0314. All solutions were placed a container with 0.500 cm pathlength. (10pts)

 a. What is the molar absorptivity constant? Show and EXPLAIN your solution.

 b. What is the concentration of the unknown? Show and EXPLAIN your solution.  c. What is the transmittance of the standard solution? Show and EXPLAIN your solution. 

d. What is the transmittance of the unknown solution? Show and EXPLAIN your solution. 

e. What will be the absorbance of the standard solution if a 1.50 cm pathlength was used? Show and EXPLAIN your solution. 

 

Answer 1

The equations used here :

A = εCl , where A = Absorbance, ε = Molar absorptivity constant, C = concentration (in Molarity) and l = path length

Also, if I0 = intensity of incident light, It= Intensity of transmitted light, then:

Also, Absorbance (A) = , and ,

log It

Percent Transmittance (%T) = [Trasmittance is generally expressed in percentage]

It 1. X 100%

Combining absorbance and %T :

t %T 100 lo

So,

100 A = log %T

Or, A = 2 – log(%T) [ Using the relations : log (a/b) = log(a) – log(b) and log(10x) =x, so log(100) = log(102) = 2]

9.

(a)

Take the standard solution. Equation : A = εCl

Here, A = 0.0621 , C = 2.10 x 10-2 M , l = 0.500 cm

So, 0.0621 = (ε)( 2.10 x 10-2 M)( 0.500 cm)

Solving, ε =5.914 M-1 cm-1

(b)

The value of ε is constant for a given solution. So, ε for unknown = ε of standard =5.914 M-1 cm-1

A = 0.0314 , l = 0.500 cm

Putting the values in A = εCl :

0.0314= (5.914 M-1 cm-1)(C)( 0.500 cm)

Solving, C = 0.0106 M = 1.06 x 10-2 M

(c)

Here, formula is : A = 2-log(%T)

For standard, A = 0.0621

So, 0.0621 = 2-log(%T)

Or, %T = 86.67 %

(d)

A = 2-log(%T)

For unknown, A = 0.0314

So, 0.0314 = 2-log(%T)

Or, %T = 93.02%

(e)

A = εCl

ε =5.914 M-1 cm-1 , C = 2.10 x 10-2 M , l = 1.50 cm

A = (5.914 M-1 cm-1)(2.10x 10-2 M)(1.50 cm)

Or, A =0.186