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

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 pathlength. Calculate the concentration of each chromophore in the mixture

Explanation / Answer

A= absorptivity*C*l

absorptivity (A) = 17220 and 4200

absorptivity (B) = 11720 and 25200

for the mixture,

A = cAabsA + cBabsB

0.586 = 17220cA + 11720cB at 405 nm

0.462 = 4200cA + 25200cB at 610 nm

solving for cA and cB using the two equations yields,

concentration of A = 39.91 uM and that of B = 11.68 uM.

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