Compound A has a molar absorptivity of 21.5 L mol-1 cm-1 at...

Compound A has a molar absorptivity of 21.5 L mol-1 cm-1 at 300 nm and molar absorptivity of 0.5 L mol-1 cm-1 at 600 nm. Compound B has a molar absorptivity of 5.3 L mol-1 cm-1 at 300 nm and a molar absorptivity of 11.5 L mol-1 cm-1 at 600 nm. In a 1 cm pathlength cell, a solution containing both Compound A and Compound B had an absorbance of 0.287 at 300 nm and an absorbance of 0.576 at 600 nm. What are the molar concentrations of Compound A and Compound B in the solution? [Compound A] = 0.015 mol L-1, [Compound B] = 0.010 mol L-1 [Compound A] = 0.010 mol L-1, [Compound B] = 0.015 mol L-1 [Compound A] = 0.001 mol L-1, [Compound B] = 0.050 mol L-1 [Compound A] = 0.05 mol L-1, [Compound B] = 0.001 mol L-1 [Compound A] = 0.050 mol L-1, [Compound B] = 0.005 mol L-1

Transcript

00:01 Okay, so this is a kind of complicated problem.

00:04 We can make some assumptions.

00:07 So let's see, a compound, we can assume the concentration of the compound a is x, and the concentration of the compound b is y.

00:19 And of course, it says both compound b and a has absorption, right, as 300 nanometer and 600 nanometer.

00:28 So let's first look at 300 nanometer.

00:31 Absorption.

00:33 So in this case, we can calculate in the 300 nanometer, the absorbance for the compound a equals to the concentration of the a, right? the coefficient times the length of the pass and times the concentration of the compound a.

00:59 So that's equal to, here it tells you that's equal to about 20, 1 .5 and l is 1 times x and absorption for the b in 300 nanometer same that's 5 .3 times y and tells you right in this mixture the total absorption as 300 nanometer is 0 .287 so tells you absorption a plus absorption b equals to about .287.

01:50 So that's actually about 21 .5x plus 5 .3y y equals to .287.

02:07 Another case is 600 nanometers.

02:15 Dissorbents of the a equals to same beer's law, but the coefficient here is .5.

02:25 So it's .5x.

02:26 And for b it's 11 .5 so same become 11 .5 y so at this case the absorbance a plus absorbens b equals to in this case it says it's 0 .5 .76 that's equal to 0 .5x plus 11 .5y.

02:59 So we just need to combine these two expressions one two.

03:06 So what you get here is 21 .5x plus 5 .3y equals to .287.

03:19 And 0 .5x plus 11 .5y equals to .576.

03:29 So what we can do here is actually to, for equivalent 1 and equation 2, try to use equivalent 2 times a equivalent 2 times.

03:43 About 21 .5 over 0 .5.

03:57 So that's 2 times 43.

04:10 So that becomes what? that becomes 21 .5x plus 494 .5y equals to 24 .768...

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