Question

For a project, the radical scavenging capacities of plant extracts, you use a DPPH assay. A 1.21x10^-4 M solution of the stable free radical 2,2-diphenyl-1-picrylhydrazyl DPPH has a molar absorbance of 0.690 when measured in a 0.500 cm cell at wavelength of 517 nm.

calculate:
a. the molar abs for dpph at this wavelength

b. the abs when concentration is expressed in ppm.

c. molar concentration of dpph in solution with abs of 0.929 when in a 1.00cm cell at wavelength of 517nm.

d. abs of solution that as twice transmittance of solution in c.

Answer 1

We shall we the Lambert - Bear's law to solve this question Matite where A a beobanu l=1cm, & molar extinction wefficient te wreentration aiven, Asia 0.690; c 1.21 x 10-4 (M) 1.21% 10-u r., & = 5400. UZ M-1 cm - a. Henu molar absorbance at this wavelength is 5702.49 Mum-l som b. c = 1.21 x 16" (M): 121 x 10-6 (M) - 121 bpm. But, the molar extinction officient is independant of the concentration of the absorption species. Hence, even if it is presented in ppm, absorbance is still 0.690. (Any c. we have to 5302.47 M 1 cm 1 A = 0.929 ..CA0.929 Mb M = 1.63*105 MO t 5702 un Thus; concentration is 1.63 x 10-5 M. (Am 63x10-5 (m) d. for bolution in a concentration is 1:6 3 x 1 Now: a beosbance and y. Transmittance are related as ; A-609(1.1) For A 0.429 0.92922 - 109 14.1) of log(4.) = 2-0.929 = 11071 ory (),1), 11:72 for the given solution ind, 1 T = 2X 11.97 = 23:55 Absorbance of this solution - 2 - log (23.55 ) - 0.628 i concentration = 0.628 M. l.1ox 10-4M. 5702.49 Henu, concentration is mox 164 (M) (Aus) Норя + a, b} ш. 2)