This equation you've mentioned in your answer to my recent question regarding kd, f=[1+(KD*Vo/Va)]^-(n) ,is this a general equation we can use directly?

Question

This is the first time I've come across with this..

Answer 1

Sorry for any inconveniences caused due to the mistake done by me by posting "@DrBob222" in the name section except for the subject question.

Answer 2

Not to worry. Yes, it is a general equation. If you were to do 4 extractions you would need to go through the Kc = [(mg x org layer/volume)/(mg x aq layer/volume)] four times. After the first time you would take the amount in the aq layer, extract that a second time, take what's left in the aq layer and extract a third time etc etc. This equation I gave you does it all in 1 step by giving you the fraction left in just 1 math equation. By the way, Im headed for bed (it's almost 1 A.M. here) so I won't be available until later tomorrow. This problem wasn't as tough as the others you've posted. :-)

Answer 3

Its okay there's no rush.I was working out some questions we were given.

I also agree with you.This question is all about substituting values to a equation.Yet I have doubts on this one :-(

But that formula is not taught in our curriculum and I'm sure the lesson was over.
So do I have to apply values 4 times??
Maybe I should derive this formula in a general method and apply the values.

Answer 4

I think you are talking about two different applications. The formula not taught in your school is for the purpose of calculating Kd ONLY (or if one knows Kd, it will give you the fraction of solute remaining in the aqueous phase after n extractions). It allows you to go through one calculation instead of four calculations to obtain the solute in the aq phase after four extractions. Of course you may use the conventional formula four times but that's the long way around. There may be a way to use the two sets of data described in the problem (1 where 10 mg remains and the other where 5 mg remains) to arrive at the Kd but I didn't work it out BECAUSE I had access to the formula where we could use the fraction of 5 mg/10 mg. My best guess is that if you set up the regular formula with its two unknowns using the 10 mg remaining and divide by the reular formula with its two unknowns using the 5 mg remaining, solve them simultaneously, you will get that formula you haven't seen before; however, I think that would be a bear of a problem. I have used the fraction equation a number of times and I know it works so I'm not inclined to want to derive it.

Answer 5

Thank you.I'll try that