[LHM]

The distribution coefficient, K_D, for an organic compound between water and methylene chloride is 3.40. An aqueous solution of the organic compound contains 0.500 g per 100 mL and is extracted with 50.0 mL of methylene chloride. What percentage of the organic compound originally in the water is extracted?

I've never really done any problems involving distribution coefficients before except for this one time during a chromatography lab, and all I remember is that it has to do something with the ratio of separation or something like that? Since I really don't even know where to start, how do you do this?

[rabolisk]

The distribution coefficient is just an equilibrium constant for [A]_aq  [A]_org.

See if you can get started there.

[LHM]

So is it maybe something along the lines of:
3.40=(x g org/50.0 mL org)/(0.500 g org/100 mL H₂O)?
If this is right, then x=0.85 g.

But then how do you know how much of the organic compound was originally in the water? 0.85 g + 0.5 g = 0.90 g? I don't think that's right though, because 0.5/0.90*100=55.6%, which isn't even close to any of the answer choices.

[rabolisk]

I think you misread the question. Originally, before extraction, there was 0.500 g of organic compound in the water. However, you should still have gotten the right answer. Your biggest mistake was 0.85 + 0.5 = 0.90...

There are a couple ways of solving this. One is to treat it like any other equilibrium reaction.

The second way is to use an explicit formula for the mole fraction (or percentage) of the solute remaining in the aqueous phase. Turns out this value (the mole fraction, which we will call q) is independent of the original concentration (Note that this is why it would not have mattered how you comprehended the problem). You could solve the problem even if you didn't know that there was 0.500g to begin with.

First see if you can solve this problem like any equilibrium reaction. Then try to find the formula for q, where q = mol_aq / total mol.

[LHM]

Sorry, but I have a few more questions.

I got the right answer by using 0.85/1.35*100=63.0%. But how can there initially be 0.500 g of organic compound in the water, and then 0.85 g is extracted?

And as for the second way, I don't see how to get a formula?