Question

 

Required molar concentration of potassium ferricyanide to yield A420 = 20

- given The extinction coefficient of potassium ferricyanide at 420 nm is 1040 M-1cm-1, and F.W = 329.26

Also how do I calculate dilution value to get A420=0.1 from dilutin A420=20 and H2O

 

 

Answer 1

Ans. #1. Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                                    A = Absorbance

                                    e = molar extinction coefficient (M^-1cm^-1)

                                    L = path length (in cm)

                                    C = concentration

Given,

            Absorbance = 0.20

            e = 1040 M^-1 cm^-1

Putting the values in equation 1-

            0.20 = 1040 M^-1 cm^-1 x C x 1.0 cm

            Or, C = 0.20 / (1040 M^-1) = 1.9231 x 10^-4 M

Therefore, [ K₃[Fe(CN)₆] ] in solution = 1.9231 x 10^-4 M

#. Yes, you can C1V1 = C2V2 to prepare dilutions with the condition –

            C1 : C2 = Abs1 : Abs2        - that is, the ratio two concentrations is proportional to ratio of their respective absorbance.

# Note in equation that, concertation is proportional to absorbance. SO, C1:C2 = Abs1 : Abs2.

For example, [ K₃[Fe(CN)₆] ] = 1.9231 x 10^-4 M - Abs = 0.20

That is, if you want a solution with abs = 0.10 (i.e. half the absorbance of above solution), you need to halve the concertation , too.

So,

1.9231 x 10^-4 M / 2 = (Abs1) / 2

Or, 1.9231 x 10^-4 M / 2 = (0.20) / 2

Or, 9.6155 x 10^-5 M = 0.10

That is, [ K₃[Fe(CN)₆] ] = 9.6155 x 10^-5 M would give an absorbance of 0.10 unit.