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Offline sunflowerzz

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Distribution Coefficient (Kd) Question
« on: May 18, 2013, 05:27:11 PM »
I need a little help with these questions...not sure if I'm going in the right direction:


1. Calculate the distribution coefficient of a solution of 15.1g caffeine dissolved in 25mL of water. The solution is extracted once with 50mL of chloroform. The solubility of caffeine in water is 1.5g/100mL at 25C. The solubility of caffeine in chloroform is 14.0g/100mL at 25C

I know Kd = (solubility of a compound - organic phase)/(solubility of a compound - aqueous phase)
So in other words, Kd = concentration of caffeine in chloroform/concentration of caffeine in water

What is throwing me off are the solubilities that were given - how do they fit into the formula for Kd?

This is what I did: (15.1g/25mL)/(1.5g/100mL)

--> however, not sure how to fit chloroform into the equation?


2. A 50mL aliquots of a 0.3000M of compound A is extracted with 50.0mL of diethyl ether. Titration of the extracted aqueous solution required 23.50mL of 0.5000M NaOH. Calculate the distribution coefficient of compound A.

Not sure if this is right but this is what I did:
Kd = (0.5000M * 23.5mL)/(0.3000M * 50mL) = 0.783

Is this right?


3. A 7.3g sample of propanoic acid is dissolved in 300mL of water. How much 0.1N NaOH is required to neutralize a 50mL aliquot of propanoic acid?

I used C1V1 = C2V2
So the concentration of propanoic acid is:
moles = 7.3g/80.06g/mol = 0.0912moles
C1 = 0.0912mol/0.3L = 0.304 M
V1 = 50mL or 0.05L aliquot
C2 = 0.1N = 0.1M of NaOH
Therefore, V2 = (0.304M * 0.05L)/0.1M = 0.152L or 152mL - this seems like a very large volume - not sure if this is right?

Offline Borek

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Re: Distribution Coefficient (Kd) Question
« Reply #1 on: May 19, 2013, 03:07:04 AM »
1. Calculate the distribution coefficient of a solution of 15.1g caffeine dissolved in 25mL of water. The solution is extracted once with 50mL of chloroform. The solubility of caffeine in water is 1.5g/100mL at 25C. The solubility of caffeine in chloroform is 14.0g/100mL at 25C

Ignore first two phrases, they are irrelevant to the question.

Quote
2. A 50mL aliquots of a 0.3000M of compound A is extracted with 50.0mL of diethyl ether. Titration of the extracted aqueous solution required 23.50mL of 0.5000M NaOH. Calculate the distribution coefficient of compound A.

Not sure if this is right but this is what I did:
Kd = (0.5000M * 23.5mL)/(0.3000M * 50mL) = 0.783

Is this right?

No, for two reasons - one is your fault, the other is fault of whoever wrote the question. You are not told how the substance A reacts with NaOH (1:1? 1:2? something else?) so technically it is impossible to answer the problem. And even assuming - as you did, probably not even knowing you were assuming anything - they react 1:1, you still did wrong. Try to systematically calculate concentrations in both phases remembering, that the total amount of substance doesn't change.

Quote
3. A 7.3g sample of propanoic acid is dissolved in 300mL of water. How much 0.1N NaOH is required to neutralize a 50mL aliquot of propanoic acid?

I used C1V1 = C2V2

Right and wrong. It will give the correct answer, but do you know why?

Quote
So the concentration of propanoic acid is:
moles = 7.3g/80.06g/mol = 0.0912moles

Check the molar mass of the propanoic acid.

And yes, the final volume looks rather large - but you are in the right ballpark.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sunflowerzz

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Re: Distribution Coefficient (Kd) Question
« Reply #2 on: May 19, 2013, 12:05:37 PM »
1. Calculate the distribution coefficient of a solution of 15.1g caffeine dissolved in 25mL of water. The solution is extracted once with 50mL of chloroform. The solubility of caffeine in water is 1.5g/100mL at 25C. The solubility of caffeine in chloroform is 14.0g/100mL at 25C

Ignore first two phrases, they are irrelevant to the question.

Quote
2. A 50mL aliquots of a 0.3000M of compound A is extracted with 50.0mL of diethyl ether. Titration of the extracted aqueous solution required 23.50mL of 0.5000M NaOH. Calculate the distribution coefficient of compound A.

Not sure if this is right but this is what I did:
Kd = (0.5000M * 23.5mL)/(0.3000M * 50mL) = 0.783

Is this right?

No, for two reasons - one is your fault, the other is fault of whoever wrote the question. You are not told how the substance A reacts with NaOH (1:1? 1:2? something else?) so technically it is impossible to answer the problem. And even assuming - as you did, probably not even knowing you were assuming anything - they react 1:1, you still did wrong. Try to systematically calculate concentrations in both phases remembering, that the total amount of substance doesn't change.

Quote
3. A 7.3g sample of propanoic acid is dissolved in 300mL of water. How much 0.1N NaOH is required to neutralize a 50mL aliquot of propanoic acid?

I used C1V1 = C2V2

Right and wrong. It will give the correct answer, but do you know why?

Quote
So the concentration of propanoic acid is:
moles = 7.3g/80.06g/mol = 0.0912moles

Check the molar mass of the propanoic acid.

And yes, the final volume looks rather large - but you are in the right ballpark.

#1 So is it Kd = (14.0/100)/(1.5/100) = 9.3?
- why do you ignore the first two phases? Is that to throw you off or something? Does that mean that Kd doesn't change no matter the volumes/weights of caffeine or chloroform?


#2 So I don't assume they react 1:1? How do you know this?
You find the concentrations of both and then Kd = concentration of the organic phase/concentration of the aqueous phase?

So this is what I did: I found the moles of compound A = 0.015 (before extraction) and moles of NaOH = 0.01175 (after extraction with ether)

So 0.015 - 0.01175 = 0.00325 mol (ether phase)

Kd = conc'n of ether phase/conc'n of aqueous phase
= (0.00325 mol/0.05L)/(0.01175 mol/0.05L) = 0.277


#3 My bad! The molar mass of propanoic acid is 74g/mol so moles = 0.0986
So the final volume is 164 mL (0.164L) - is this right?

- why is C1V1 = C2V2 right and wrong?

Offline Borek

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Re: Distribution Coefficient (Kd) Question
« Reply #3 on: May 19, 2013, 12:52:46 PM »
#1 So is it Kd = (14.0/100)/(1.5/100) = 9.3?
- why do you ignore the first two phases? Is that to throw you off or something? Does that mean that Kd doesn't change no matter the volumes/weights of caffeine or chloroform?

Kd is ratio of concentrations, and as such doesn't depend on amount of solvents. Yes, these additional information were just muddying the water.

Quote
#2 So I don't assume they react 1:1? How do you know this?

Let me answer with the question: I have a substance B on mind. Tell me in what molar ratio it reacts with NaOH.

Quote
You find the concentrations of both and then Kd = concentration of the organic phase/concentration of the aqueous phase?

Yes.

Quote
So this is what I did: I found the moles of compound A = 0.015 (before extraction) and moles of NaOH = 0.01175 (after extraction with ether)

So 0.015 - 0.01175 = 0.00325 mol (ether phase)

Kd = conc'n of ether phase/conc'n of aqueous phase
= (0.00325 mol/0.05L)/(0.01175 mol/0.05L) = 0.277

Logic looks right, I have not checked the numbers.

Quote
#3 My bad! The molar mass of propanoic acid is 74g/mol so moles = 0.0986
So the final volume is 164 mL (0.164L) - is this right?

That's what I got.

Quote
why is C1V1 = C2V2 right and wrong?

It works ONLY when the substances react in 1:1 molar ratio. Here they do, but I know many students have tendency to use this equation for every titration, regardless of the stoichiometry.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sunflowerzz

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Re: Distribution Coefficient (Kd) Question
« Reply #4 on: May 19, 2013, 01:21:05 PM »
#1 So is it Kd = (14.0/100)/(1.5/100) = 9.3?
- why do you ignore the first two phases? Is that to throw you off or something? Does that mean that Kd doesn't change no matter the volumes/weights of caffeine or chloroform?

Kd is ratio of concentrations, and as such doesn't depend on amount of solvents. Yes, these additional information were just muddying the water.

Quote
#2 So I don't assume they react 1:1? How do you know this?

Let me answer with the question: I have a substance B on mind. Tell me in what molar ratio it reacts with NaOH.

Quote
You find the concentrations of both and then Kd = concentration of the organic phase/concentration of the aqueous phase?

Yes.

Quote
So this is what I did: I found the moles of compound A = 0.015 (before extraction) and moles of NaOH = 0.01175 (after extraction with ether)

So 0.015 - 0.01175 = 0.00325 mol (ether phase)

Kd = conc'n of ether phase/conc'n of aqueous phase
= (0.00325 mol/0.05L)/(0.01175 mol/0.05L) = 0.277

Logic looks right, I have not checked the numbers.

Quote
#3 My bad! The molar mass of propanoic acid is 74g/mol so moles = 0.0986
So the final volume is 164 mL (0.164L) - is this right?

That's what I got.

Quote
why is C1V1 = C2V2 right and wrong?

It works ONLY when the substances react in 1:1 molar ratio. Here they do, but I know many students have tendency to use this equation for every titration, regardless of the stoichiometry.



Thank you so much! I understand Kd a little better thanks to your comments!

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