00:01
Hello everyone, so let's do this parallel about solvent extraction, but before i get started, i recommend that you do the question yourself and come back to see if you've got a better right or not.
00:13
So hopefully you've done the question yourself and i'll also work in it together.
00:16
So we're given a substance and we're asked if we're extracting a substance from water into ether, is it more effective to do one extraction with 300 milliliters of ether or three extraction with 100 milliliters? well, multiple extractions are always more effective than single instructions.
00:36
And this is because when we're extracting a substance, the equilibrium that is established between the two layers, the ratio of the concentration of the substances in each layer will be constant for a particular biphasic system.
00:50
So here in this case, it would be water ether, and we obviously represent that with k, which is a partition coefficient.
00:58
So i'll write down what i'm, sometimes what i basically said.
01:02
So multiple single extractions.
01:15
So when we're extracting a substance, our ratio of the constant for the biphasic system, which is for the water, ether system represented with the partition coefficient so that is k equals malaria of ether phase right because there is the water ether system the bi -placistence system so we have the ether phase polarity and we have the malady of a case which is obviously water so the k value in this case is experimental but can be estimated using the solubility data.
02:57
So if we have a substance in the question that has a solubility of, so we have its substance, the solubility of 1 grams over 1 10 -mill i mentioned in the question of ether and 1 1 gram.
03:31
50ml of water, then we can obviously estimate our k.
03:38
So we can do k equal 1 over g.
03:43
Then i get 1g over 10 a will divided by 1g over 50 mil.
03:57
And this would give us a k of 5.
04:00
Our partition coefficient.
04:02
Now if there's five grams of a substance in 300 millilit of water, then x is the amount extracted in the ether layer.
04:11
So let's say we have five grams of substance in 300 and low water, we make six of the amount extracted in the ether layer, so we can form a mathematical equation from that, we would get 5 grams minus x.
04:54
It's 5 grams of the sustenage minus x, which is the amount of extracted in the interior.
05:01
So we could do our k as x divided by 300 and divide that and divide that by 5 minus x over 200 of water and we can equal 5 because we got that as our k before and then we can rearrange to find our x and we would we x is equal to 4 .16 grams so that's our x that's the amount extracted in the ether there now if we're using three times uh so this this is the situation if there was multiple extractions we were to use sorry my bad this we get x equals 4 .16 grams if we're doing only one extraction so this whole case where we did 300 emmals of water this is only one single extraction and we get x as 4 .16 and if we do that is our x right so if we do the if we move on to the situation that we're using 300 ml of ether for the extraction.
06:24
So let's move on.
06:25
So let's label this as situation one, a single extraction.
06:34
And if we move on to the situation two, which is multiple extractions.
06:41
We have three times a hundred ether.
06:50
Then we're going to have a different set of equations.
06:53
So we're going to have three.
06:54
Different k equations so we have k equal it's labeled it as x1 over 300mm of ether and divide by 5 minus x over 300m water and we made that equal of 5 right because that's our partition and then we get that x1 is equal to when we rearrange 3 .1 .25 square else.
07:28
Okay and that means that we have 1 .875 rounds of water.
07:37
Okay now we do a second extraction.
07:43
We have k -2uple x2 over under the mill ether and 5 minus x it's not only x it's x1 you know why because this is extraction from the previous we're using the previous think of it as three bowls we're using the first bowl to do the second extraction so we're going to have to have to follow it in a sequence and we're going to use 5 minus x1 we're going to use x1 from our previous calculation so we have 5 minus x1 over j100 and then we can equal 5 and we get that our x2 is equal to 1 .172 grams, right? and that means that these we have 0 .7 grams of water.
08:44
Because we had 1 .875 grams of water in the first extraction.
08:50
So we're going to do 1 .875 minus 1 .172 grams and that should leave us with 0 .7 grams of water.
08:58
I know that a lot of students make a mistake of thinking that when we're doing the second extraction, we are going to be using a new set of equipment, fresh equipment, but no, when we're doing the second extraction, we're using the previous extraction to the next one.
09:14
So like when we're doing this third extraction here, we're going to have to use the second extraction data to follow our steps.
09:21
So we have xp over 100 in the third.
09:27
We divide that by 0 .7 grams...