What is the molar absorptivity

Question

Ferrous ion reacts with an unknown ligand X to form a bright red complex with λmax = 512 nm.

Fe2+(aq) + 3X(aq) ↔ [FeX3]2+(aq)

Standard solutions of [FeX3]2+(aq) of varying concentrations were prepared, and their

absorbances at λmax were obtained using cuvettes of path length of 2.00 cm. Data are shown

below.

Table 3. Absorbance measurements of [FeX3]2+ standard solutions

Concentration of [FeX3]2+(aq), M	Absorbance
0.00	0.005
1.00 x 10-5	0.170
2.00 x 10-5	0.334
3.00 x 10-5	0.501
4.00 x 10-5	0.665
5.00 x 10-5	0 .831

a. Give the equation of the line and R2 value of the calibration curve from the data above.

b. What is the molar absorptivity of [FeX3]2+ in proper units?

c. An unknown solution of the Fe-complex has an absorbance of 0.555. What is the concentration of [FeX3]2+ in the unknown solution?

Answer 1

a) You will plot concentration (x axes) vs absorbance (y axes) and hopefully get a straight line of the form

y = mx + b. If you use a general plotting program, it will give you the equation and R2 value.

b) The molar absorptivity is obtained from the Beer Lambert law: A = ecl where....

A = absorptivity

e = molar absorptivity / molar extinction

c = concentration in moles/liter (M)

l = path length in cm = 2

e = A / c x l

For example:

using data for 1x10^-5 M standard, after subtracting blank of 0.005 ...

e = 0.165 / (1x10^-5 M)(2 cm)

e = 8250 M^-1cm^-1

using data for 5x10^-5 M standard, after subtracting blank of 0.005

e = 0.826 / (5x10^-5) (2 cm)

e = 8260 M^-1cm^-1

Or you can find e from the equation of the straight line obtained in (a) above. Just substitute 1 M for the concentration (x) and solve for y in the y = mx + b equation.

c) Plug 0.555 in for y in the y = mx + b equation generated in (a) above and solve for x. From a cursory review of the data in the table, the value of x should be between 3x10^-5 and 4x10^-5 M. Estimating from the Beer Lamber law, and using a value of 8260 for the molar absorptivity, we can estimate it as follows...

0.555 = 8260 (c) (2)

c = 3.36x10^-5 M