Differential Extraction and Choosing a Method

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Some useful compound is dissolved in water, and you want to pull it out using ether. You have exactly one hundred millilitres of ether. Should you pour it all in at once, or use it in two portions of fifty millilitres each? Same ether, same total volume, so the same result, right? Hold on to your answer. By the end of this video, you will see that one choice clearly wins. Questions on extraction and on choosing a purification method do appear in JEE, so this is worth your time. So, let's begin.

0:37Differential (solvent) extraction: the idea

So, here is the situation. We have an organic compound, dissolved in water. We want the compound back, and we want to leave the water behind. Okay? So we add an organic solvent, such as ether, dichloromethane, or chloroform. The word immiscible on screen means the solvent does not mix with water, so it forms its own separate layer. Right? Now we shake the two layers together in a separating funnel. The compound is more soluble in the organic solvent, so it passes largely into the organic layer. Now tell me, how do we finally get the compound out of that layer? Yes, exactly! We separate the organic layer from the water, and then we evaporate the solvent. The compound is left behind. That's it, that is differential extraction, also called solvent extraction. Got it? So, let me ask you this: which two properties must the extracting solvent have? Very good! First, it must be immiscible with water, otherwise there is no separate layer to take away. Second, it must dissolve the compound better than water does, otherwise the compound simply stays in the water. Clear?

1:49The distribution law

Now, how much of the compound really moves across? For a solute shared between the two layers, the distribution law gives us this ratio. Here, c org, with the small word org written below the c, is the concentration in the organic layer. And c water is the concentration in the water layer. We write the ratio, c org over c water, as K D, a capital K with a small D below it, okay? So tell me this. If K D equals four, which layer has the higher concentration of solute? The organic layer, correct, four times the concentration in the water! Now the course turns this into a formula. V w, with a small w below the V, is the volume of water, and V is the volume of solvent used each time. After one extraction, the fraction left in the water is V w over, V w plus K D times V, with me? Now suppose we take the water layer and extract it again with fresh solvent, and again, n times in all. Each time, the same fraction of whatever is still there stays behind. So what happens to the fraction left? It gets multiplied again and again! So we put the fraction inside big round brackets, and raise the whole bracket to the power n, the small n at the top right. And note this carefully: this formula gives the fraction left in the water, not the fraction extracted.

3:16Choosing a method: read the clue

Right? Alright, now the second half of today's title, choosing a method. In these questions, the mixture itself gives you a clue, and the arrow on screen points from the clue to the method. First clue: an organic compound in its aqueous solution, which means dissolved in water. That is differential extraction, exactly what we just learnt. Okay? Now try this one from the course. Aniline is in its reaction mixture with water and some tarry, non-volatile matter. Which method would you choose? Steam distillation, very good! Aniline is steam-volatile and immiscible with water, and the tar is not volatile at all. Right? Next, two liquids whose boiling points are far apart, like chloroform and aniline. Distillation does the job. With me? If two solids differ in solubility in hot water, like benzoic acid and naphthalene, we use crystallisation. Clear? And if one part sublimes, like naphthalene with common salt, or iodine with sand, we use sublimation. So always read the clue first, and then choose. Got it?

4:25Example 1: one extraction with 100 mL ether, KD = 4

Now, Example one. We have a solute with K D equal to four, between ether and water. One hundred millilitres of water containing it is extracted once, with one hundred millilitres of ether. What fraction is extracted? This is the first choice from our puzzle, all the ether at once. Pause the video and try it first. Done? Okay, let's check. First, the data. V w, the volume of water, is one hundred millilitres. The ether volume, V, is also one hundred millilitres. K D is four, and the number of extractions, n, is one. Okay? Now the formula. On top, one hundred. Below, one hundred plus four times one hundred, which makes five hundred. So one hundred over five hundred gives zero point two. So the answer is zero point two, right? No, no, be careful here, zero point two is the fraction left in the water, not the fraction extracted. Extracted is one minus zero point two, which is zero point eight, that is eighty per cent. So all the ether at once pulls out eighty per cent. Remember this number. Clear?

5:40Example 2: two 50 mL portions beat one 100 mL portion

Now Example two, the second choice from our puzzle. Same solute, with K D equal to four, and the same one hundred millilitres of water. But this time we extract twice, with fifty millilitres of ether each time. What fraction is extracted now? Pause the video and try it first. Done? Before we calculate, let me ask you something. The total ether is still one hundred millilitres. So the answer must be eighty per cent again, right? Let's not guess, let's calculate. Now V is fifty millilitres, so K D times V is four times fifty, which is two hundred. And the number of extractions, n, is two, because we extract twice. Okay? Fraction left: one hundred over three hundred gives one third, and we square it, because there are two extractions. One third, squared, is one ninth, about zero point one one. The wavy equals sign on screen means approximately. Right? So extracted is one minus one ninth, which is eight ninths, about eighty eight point nine per cent. That is more than eighty per cent, with the same ether! So our puzzle is solved: the two portions clearly win. With me? And the course pushes this one step further. Three portions of thirty three point three millilitres each, the same total again, leave only about zero point zero seven nine in the water. That is ninety two point one per cent extracted. Clear? So here is the lesson: many small extractions are more efficient than one large one with the same total volume. This idea can appear in statement-type questions, so remember it. Got it?

7:23Quick revision: n extractions formula, choosing a method

So, quick revision. Differential extraction uses a solvent that does not mix with water, and dissolves the compound better than water. The fraction left after n extractions is V w over, V w plus K D times V, all to the power n. Many small extractions beat one large one of the same total volume. And to choose a method, read the clue in the mixture first. Now go and try the three practice problems on this page. You will find them easy. All the best!