A 50 M solution of chromophore A had an absorbance of 0.861 at 405 nm and 0.21 a

Question

A 50 M solution of chromophore A had an absorbance of 0.861 at 405 nm and 0.21 at 610 nm. Chromophore B (50 M) had an absorbance of 0.352 at 405 mn and 1.26 at 610 nm. An unknown mixture of chromophores A and B yielded absorbances of 0.586 At 405 nm and 0.462 at 610 nm. Assume that the two chromophores do not react with each other, that is the absorbances of the chromophores are additive, and that all measurements were in a 1 cm path length. Calculate the concentration of each chromophore in the mixture.

Explanation / Answer

given that 5*10-5 Solution of A has an absorbance of = 0.861 nm at 405nm

from A= e1bC

0.861= e*1*C ( e1= extinction coefficient)

0.861= e*50*10-6

e1= 0.861/(50*10-6)=17220 for chormophere A

at 620nm e1' = 0.21/(50*10-6)=4200 Chromophere A

for chromophere B e2= 0.352/(50*10-6)=7040 at 405 nm and   e2'= 0.462/(50*10-6)=9240

Unknown sample A at 405nm = 0.586

= e1*1*C1+e2*1*c2 = 17220*c1+7040c2 =0.586 (c1 and c2 are concentration of A and B in the sample) (1)

at 610 nm A = 0.462,    4200c1+9240c2= 0.462 (2)

eq.1 becomes c1+0.41c2= 0.586/17220 = 3.4*10-5 (1A) (divide the equation by 17220)

Eq.2 becomes c1+2.2c2= 0.00011   (2A) (divide by 4200)

Eq.2A- Eq.1A gives 1.79c2= 0.000076

c2= 4.245*10-5M and c1= 0.00011-2.2*4.245*10-5=1.659*10-5

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