# Required molar concentration o

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

Ans. #1. Beer-Lambert’s Law, A =e C L           – equation1,

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 x10-4 M

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

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

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

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

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

That is, if you want a solution with abs = 0.10 (i.e. half theabsorbance 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, [K[Fe(CN)]] = 9.6155 x 10-5 M would give anabsorbance of 0.10 unit.

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