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### Topic: extraction  (Read 9033 times)

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#### machozitatu

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« on: November 19, 2009, 04:51:08 PM »
the distribution coefficient of a substance (unknown)between CCl4 and water is 5.00. if 30mL of .100M of this unknown is extracted 5 times with 20mL water each time, what fraction of this unknown is in the water after the 5th extraction

#### sjb

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« Reply #1 on: November 19, 2009, 05:13:31 PM »
Probably easiest to do this one step at a time.

How much is there in the aqueous after 1 extraction?

#### machozitatu

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« Reply #2 on: November 20, 2009, 08:17:31 AM »
I am just learning how to do this so I have no idea how to set it up. I have the formula but I don't know what goes where.....can you help me set it up?

#### sjb

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« Reply #3 on: November 21, 2009, 07:33:04 AM »
OK, what formulae do you have that may be of use?

#### machozitatu

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« Reply #4 on: November 21, 2009, 06:12:42 PM »
the formula is distribution coefficient is equal to the concentration of solvent 1 divided by solvent 2.
I do not really think that this equation applies to the problem. What I was thinking of doing was multiplying the volume (of the solute) given by the coefficient...so that I can get the ratio. Meaning that the solute much more soluble in solvent 1 than water (solvent 2). I would then use this ratio and multiply it by the mL of water used for extraction to determine how much is taken by the water. I would do this for each of the extractions performed
If there is another way to do this. please let me know.

#### sjb

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« Reply #5 on: November 22, 2009, 05:36:16 AM »
the formula is distribution coefficient is equal to the concentration of solvent 1 divided by solvent 2.

Concentration *in* solvent 1, divided by concentration in solvent 2, right.

So initially you have 3 mmols of product in the CCl4.

After one extraction you have 3-x mmols in the CCl4, and x mmols in the aqueous, such that the ratio of their concentrations is equal to your distribution coefficient, so how much after one extraction?

#### machozitatu

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« Reply #6 on: November 22, 2009, 01:07:02 PM »
So I would set it up as 3mmol-x/x is equal to 5 and solve for x then subtract that value from the original 3mmols to get the initial mmol in the second extraction. I would then do the same thing for all of the extractions basically.

#### sjb

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« Reply #7 on: November 22, 2009, 01:23:42 PM »
Not quite. Don't forget that you have different volumes of aqueous and of organic phase here.

#### machozitatu

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« Reply #8 on: November 22, 2009, 03:52:32 PM »
I don't know what to do. I thought we were putting in 20mL of water each time so initially we were starting with none. and for each extraction we start with no water.

#### sjb

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« Reply #9 on: November 22, 2009, 04:23:13 PM »
Remember that your Kd = [org]/[aq], so you can rewrite the concentrations in terms of moles and volume to give..?

#### machozitatu

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« Reply #10 on: November 22, 2009, 05:42:18 PM »
so for the first extraction it would be that .5mmols was extracted therefore the concentration in the water is .025M and the mL taken from the CCl4 is 5ml (leaving 25ml).
how does this play a factor in the calculation. I just don't get how to apply the question that you ask me...do i still follow the same formula of using 3-x/x = 5 or is that just a step to follow to get to the real equation

#### sjb

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« Reply #11 on: November 23, 2009, 01:06:43 PM »
Initially, if there are x mmol of compound in the aqueous, there are 30 - x mmol in the organic. The concentration of the compound in the aqueous is x/20e-3, and in the organic is (30-x)/30e-3. So Kd = 5 = (30-x)/30e-3/x/20e-3. So how many mmol are there in the aqueous after one extraction?

#### machozitatu

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« Reply #12 on: November 23, 2009, 03:15:44 PM »
I calculated it out and got 3.5 for X. so that means that 3.5ml of the solute is extracted initially or is that 3.5 mmol extracted. so for the next extraction I would start with 26.5ml and do the same thing (since 30-3.5 is 26.5).

#### sjb

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« Reply #13 on: November 23, 2009, 03:16:45 PM »
Initially, if there are x mmol of compound in the aqueous, there are 30 - x mmol in the organic. The concentration of the compound in the aqueous is x/20e-3, and in the organic is (30-x)/30e-3. So Kd = 5 = (30-x)/30e-3/x/20e-3. So how many mmol are there in the aqueous after one extraction?

Sorry, this should of course be 3 - x mmol in the organic, and the resulting final equation 5 = (3-x)/30e-3/x/20e-3

#### sjb

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« Reply #14 on: November 23, 2009, 05:04:07 PM »
I calculated it out and got 3.5 for X. so that means that 3.5ml of the solute is extracted initially or is that 3.5 mmol extracted. so for the next extraction I would start with 26.5ml and do the same thing (since 30-3.5 is 26.5).

If the answer you got was 3.5, then that would be 3.5 mmol extracted, and so 26.5 mmol left in the organic to plug into a second iteration of the equation. Note that I messed the formula up, which is obvious here as 3.5 mmol is greater than your original 3 mmol.