# How to without calculations understand the liquid-liquid extraction efficiency?

I'm asking myself this question for a very long time and I haven't found a simple an good explanation yet for the fact that it's better (if you want to clean them) to wash your hands two times with one millilitre of soap rather than one time with two millilitres.

If you are an organic chemist you will probably more ask yourself,

Why is it better to extract a solute in a solvent $\ce{S1}$ (non-miscible with a solvent $\ce{S2}$) doing two extractions using $\ce{10 mL}$ of $\ce{S2}$ rather than one extraction using $\ce{20 mL}$ of $\ce{S2}$?

This is what I learnt at school and what I always used in organic chemistry lab sessions.

As mention above I am looking for a simple and good explanation, that is, without using any calculation. I am looking for a reasoning a bit like Feynman. I tried to resonate using dots but every time I fail and it leads me to the same result, that is, there is no better way to extract a solute...

EDIT : I put the below parts in very small because I think some people misunderstood my question. The question is just above, even though there is not interogation point.

For those who wonder if that is true or not, and as I wasn't convinced at all, I made some calculations before to be "sure" about this fact.

So let's start and imagine you want to extract iodine from water using benzene. In the following, the subscripts $aq$ will stand for the aqueous phase and $org$ will stand for the organic phase. $C_T$ will refer to the total concentration. $A$ will refer to the molecule we want to extract.

Let's call $D$ the partition coefficient between the organic and the aqueous phase, $$D = \frac{C_{T, org}(A)}{C_{T, aq}(A)} = \frac{C_{T, org}}{C_{T, aq}}\tag 1$$

The quantity $Q$ of the solute $A$ in both phases, is determined by, $$\begin{cases} Q_{T, org} = C_{T, org} \times V_{org} \\ Q_{T, aq} = C_{T, aq} \times V_{aq} \end{cases} \tag 2$$

At the equilibrum, for the first extraction, a mass balance gives, $$C_0 \times V_{aq} = C_{1, aq} \times V_{aq} + C_{1, org} \times V_{org} \tag3$$

leading to, $$C_{1, aq} = C_{0, aq} \times \frac{1}{1+Du} \tag 4$$ where $u = V_{org}/V_{aq}$.

Then after $n$ extractions, $$C_{n, aq} = C_{0, aq} \times \left( \frac{1}{1+Du}\right)^n \tag 5$$

For the case of the extraction of $\ce{I2}$ from water to benzene, I used the solubility values found on the French Wikipedia page to calculate the partition coefficient (I am not taking in consideration the activity coefficients as I don't know them). It leads to $D = 373$ (assuming my calculations are correct)

The values I used are, $C_0 = 1.300 \times 10^{-3} \text{ mol/L}$, two extractions with $u = 0.2$ compared to one with $u = 0.4$.

For the two extractions, $C_{2, aq} = 2.27 \times 10^{-7} \text{ mol/L}$

For one extraction, $C_{1, aq} = 8.66 \times 10^{-6} \text{ mol/L}$

We clearly see the difference which is huge, even though these calculations are approximative.

• Basically, it all boils down to $\left(1+{x\over2}\right)^2>1+x$. – Ivan Neretin Sep 16 '18 at 3:07
• @IvanNeretin could you elaborate, please? I don't understand what your $x$ refers to. And if it refers to the concentration of the solute I don't understand your reasoning that boils it down to this inequality. – ParaH2 Sep 16 '18 at 3:14
• My $x$ stands for your $D\cdot u$. If you split the solution in halves, then you make your $x$ twice smaller, but your $n$ twice bigger. – Ivan Neretin Sep 17 '18 at 9:58
• @IvanNeretin Oh ok., but my question is about finding a logic reasoning leading to this fact. My calculations are just for non-convinced people. Because even though my calculations are correct this fact it absolutely not logic for me. Imagine you want to explain this to a non-scientific person. – ParaH2 Sep 17 '18 at 15:59
• That's why I made it a comment and not an answer. – Ivan Neretin Sep 17 '18 at 16:23