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Topic: Distribution ratio etc.  (Read 6870 times)

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Offline Nagenyleveable

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Distribution ratio etc.
« on: August 10, 2010, 06:22:54 AM »
Hi, I would really appreciate any help with this question. I am studying for repeat exams and although i have attempted the question, and think i have it right, the last part confuses me.

Q. One liter of aqueous sample containing benzoic acid with a Ph of 6.2 is to be analysed. The acid dissociation constant of benzoic acid in water at 25c is known to be 6.3x10-5 M-1 and the distribution coefficient of the molecular form of benzoic acid in diethyl ether and water is 720.

(i) Calculate the distribution ratios of benzoic acid in diethyl and water at a pH of 6.2. If the Ph value of the sample solution is adjusted to 2.2 what will the value of the distribution ratio be?

This is exactly as the question is asked. Using this eq: D= K_D/(1+K_a/([H^+])) I get 7.14 and 712.88.

(ii) What will be the extraction efficiencies when 50ml of diethyl ether is used to extract the sample solution at Ph values of 6.2 and 2.2 respectively?

For this question I used the following formula: E(%)=D/(D+x/y)  x 100%

and got 26% and 97%

(iii) Which Ph value, 2.2 or 6.2, is optimal for the extraction of benzoic acid by an anion exchange extraction from an aqueous sample? Determine the % extraction efficiency at Ph value selected.
 
This is the question which makes me doubt the rest of what i have done as I thought this should be the same as the answer to part (ii).

Thank you for all your help. ???

Offline Borek

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Re: Distribution ratio etc.
« Reply #1 on: August 10, 2010, 09:43:08 AM »
I have not checked your earlier work - shame on me. I can just confirm that numbers presented make sense - as expected you got higher distribution coefficient and higher extraction efficiency for lower pH.

You are right that question is poorly worded and asks to calculate exactly the same thing for the second time. No need for that, just copy the answer you got earlier.
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Offline Nagenyleveable

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Re: Distribution ratio etc.
« Reply #2 on: August 10, 2010, 10:13:19 AM »
Thank you so much for your help. My stress level has lowered! ;)

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