[mass]

A 50ml aliquot of a 0.1 M solution of a neutral drug was extracted with 2 x 5 ml ethyl acetate. the drug concentration in the combined organic layers was found to be 0.37 M.
Calculate the partition coefficient for this drug under these conditions

How can  I work this out, I have tried the multiple extraction forumla.

[sjb]

A start then.

How much drug do you have to start with?

If the partition cofficient between ethyl acetate and your unknown other solvent (I assume water, as you have mentioned "combined aqueous" for the rest) is x:1, how much would you have in each phase after one extraction?

[ARGOS++]

Dear Mass;

What you call the “Multiple Extraction Formula”?

As I told you already for first time when you asked this Question: 

"Partition Coefficient”:

You have only to combine both extractions into one Formula.

But please tell me what the “Multiple Extraction Formula” is for you.
Maybe that’s a Base we can start from to build the Formula.

Good Luck!
                    ARGOS⁺⁺

[mass]

Well according to my notes, the multiple extraction forumla is =

wa/wt = (A)/ (PS+A)n


wa = mass of solute in aqueous phase after n extractions
wt = initial mass of solute used
a = volume of aqueous phase
s = volume of organic phase
p = parititon coffeificne of solute
n= number of extractions

If anyone can please show me their working, I would appreciate it as the exam is nearing and I still don't get how to do it.

[ARGOS++]

Dear Mass;

Yes!,  -  Your Formula is correct, if I assume that your n means in real an Exponent.

But your Formula is very inconvenient to rearrange for your p (what most and I call k_N)!

Let’s do the Begin of the Formula together with Nernst:
(Then you know also where the Formula is coming from.)

    k_N =  c_TOP / c_BOTTOM = c_T / c_B = (m_T / v_T) / (m_B / v_B)  = m_T / m_B * v_B / v_T.

For simplification, because the volume ratio is all time constant we set:
         v_T / v_B =  V_RN.

And so we get    k_N = m_T / m_B  * 1/V_RN.

It’s the easiest to observe all the time the lower Phase, because that mass you can easy calculate from your given data (= total  –  total extracted).

That means for m_B, if we use at the same time the Equation: 
            m_START = m_S =   m_T + m_B:

            m_B =  m_T / (k_N * V_RN)  =  (m_S – m_B)  / (k_N * V_RN).

But now it’s your turn:
Rearrange the Formula to get all m_B to the left side!
Hint: If you, after you have rearranged for m_B, substitute the whole nominator, containing only constants, by q_N, then you will get a formula that holds not only for a single Extraction.
It holds for n = 1, 2, 3, 4, ….. z
Use the correct one for your situation and substitute q_N back.
Now it’s also easy to resolve for k_N !

Try it and show your work in case you got in troubles, and then I can try to help you.
(I have already all done!)

This time I hope to have been of more help to you.


Good Luck!
                    ARGOS⁺⁺

[mass]

hi argos, you have totally left me confused with those equations, and what's kn. Is that the US method?

[Borek]

There is no such thing as US or non US method, chemistry is one. kn is just a partition coefficient.

[ARGOS++]

Dear Mass;

If it seems sooooo difficult to you: 

Use your Letter p instead of k_N!

(We can later translate it back to k_N!)

It's only important that you try!!

Good Luck!
                    ARGOS⁺⁺

[mass]

Dear Mass;

If it seems sooooo difficult to you: 

Use your Letter p instead of k_N!

(We can later translate it back to k_N!)

It's only important that you try!!

Good Luck!
                    ARGOS⁺⁺

It's not so much that, its the equation itself.

[ARGOS++]

Dear Mass;

Your last Posting is not of any help  — to give you any help !!
You have at least to tell where you run into troubles!

Only expanding and rearranging your Formula for you is never a Solution, even if you would like!

On what Step you have troubles to follow? 
I don’t see any place where I can make smaller Steps.

Have you troubles to solve the last Equation for the mass of the Bottom Layer (m_B)^*)?:
            m_B =  (m_S – m_B)  / (k_N * V_RN).

Is it easier for you if we write it your way?:
            m_B =  (m_S – m_B)  / (P * V_RN).

^*) Think of the Bottom Layer as the Layer that stays “stationary”!
    (In your case the aqueous Layer.)

Good Luck!
                    ARGOS⁺⁺

[sjb]

If you're having difficulties following ARGOS++'s approach, let's break it down a bit further.

I assume you can work out how much substance you have originally.

If the partition coefficient, p, of your substance was 1.000, in a mixture of 50 ml EtOAc and 50 ml H₂O how much is there in each layer?

How about if there was only 5 ml EtOAc, and still 50 ml water?

S