Finding the Average Epsilon and Molarity of Unknown Solution

Question

Stock Solution: 8.8×10^-5 M

Solution #1: 1.809×10^-5 M

Solution #2: 1.046×10^-5 M

You measure the absorbance of each solution (at λ_max = 524 nm) and obtain the following results:

Absorbance:

Solution
Stock solution	3.28
Solution #1	0.669
Solution #2	0.384

Determine the average epsilon for FD&C Red 40.

=______M^-1cm^-1

You are given an unknown solution of FD&C Red 40. You measure the absorbance and find it to be 0.570 at 524 nm.

Determine the concentration of the unknown solution in terms of molarity.

Answer 1

Beer Lambert Law:

A = ecl

A = absorbance

e = molar extinction coefficient

c = concentration (M)

l = path length = 1 cm (assumed)

Stock:

3.28 = e x 8.8x10^-5 x 1

e = 4.318x10⁴

Solution 1:

0.669 = e x 1.809x10-5 x 1

e = 3.648x10⁴

Solution 2:

0.384 = e x 1.046x10^-5 x 1

e = 3.671x10⁴

MEAN = 3.879x10⁴ M^-1cm^-1

Concentration of unknown:

A = ecl

0.570 = (3.879x10⁴)(c)(1)

c = 1.469x10^-5 M