This is part of a "short" problem set that I just don't understand. My work below. 1.70 x 10-5 M ML was found to generate a detector response of 426 mV at 532 nm. When a non-absorbing species was placed into the cuvette, the detector response measured was 765 mV at 532 nm. Calculate molar absorption.

Question

kkutie7 · Answer

My idea:
$$A=\epsilon bC=-log\frac{P}{P_{0}}$$
$$-log\frac{426 mV}{765 mV}=\epsilon (1cm)(1.70*10^{-5}M)$$
$$\epsilon =\frac{-log\frac{426mV}{765mV}}{(1cm)(1.70*10^{-5}M)}$$

kkutie7 · Answer

Hey @dan815 am I thinking about this right?

kkutie7 · Answer

@imqwerty

kkutie7 · Answer

@Kainui

kainui · Answer

Yeah this is correct, I was just checking cause it's been a while. Since power is related to current and voltage by P=IE the fact that you were given the voltage (E) in milivolts is the same as if you were given the power (P) since the current is a constant throughout the experiment I believe.

Specifically this substitution is safe: (if this is your concern)

$$\log \frac{P}{P_0} = \log \frac{IE}{IE_0} = \log \frac{E}{E_0}$$

kkutie7 · Answer

I got a very large number and I was worried about it
$$14955.99M^{-1}cm^{-1}$$

kkutie7 · Answer

After this I have to find this:
If the uncomplexed metal (Mn+) also absorbs weakly at 532 nm (with  = 575 M-1 cm-1 ) what is the concentration of ML in a solution containing ML + 0.0916 mM Mn+ if the detector response for the solution is 169 mV?

kainui · Answer

Hmm I haven't done much with molar absorptivity in a long while so I can't be very helpful I gotta go review. I think you're right though, that value did seem pretty high since it did have a pretty high transmittance, it doesn't make sense that it would have high absorptivity.

kkutie7 · Answer

All I know is that it has to be complicated because my Professor doesn't like simple and I'm confused

kainui · Answer

Gimme a little time, I should probably know this cause I have chemistry degree so I'm gonna read some stuff but I might get distracted on side tangents and take an hour or so haha

kkutie7 · Answer

As Long as you can help be by tomorrow I'll be very grateful

kkutie7 · Answer

@TheProfessor35 please help me

kkutie7 · Answer

@Whitemonsterbunny17 do you know how to do this?