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The distribution co-efficient of butanoic acid (pKa = 4.8) between water and two organic solvents are given below; Benzene D= 3...Asked by James
The distribution co-efficient of butanoic acid (pKa = 4.8) between water and two organic solvents are given below;
Benzene D= 3.0
Octanol D= 6.2
a) which organic solvent is expected to extract the acid most efficiently from the water? explain your answer.
So you expect that Benzene, as it has a lower K, and therefore a higher concentration in the aqeous phase, to be more efficient?
b) When extracting butanoic acid into benzene is it more effective to do one extraction with 300mL or three 100mL extractions? Support your answer with calculations.
The equation as I recall is,
q^n = (Vaq/Vaq+K*Vorg)^n,
where n =number of extractions.
Benzene D= 3.0
Octanol D= 6.2
a) which organic solvent is expected to extract the acid most efficiently from the water? explain your answer.
So you expect that Benzene, as it has a lower K, and therefore a higher concentration in the aqeous phase, to be more efficient?
b) When extracting butanoic acid into benzene is it more effective to do one extraction with 300mL or three 100mL extractions? Support your answer with calculations.
The equation as I recall is,
q^n = (Vaq/Vaq+K*Vorg)^n,
where n =number of extractions.
Answers
Answered by
DrBob222
You didn't define enough for me. Here is what I have where f<sub>n</sub> is the fraction left of the solute REMAINING in the water layer after n extractions.
f<sub>n</sub> = [1+K*(Vo/Va)]<sup>-n</sup>
Of course, you can calculate it three times from a single extract, too.
f<sub>n</sub> = [1+K*(Vo/Va)]<sup>-n</sup>
Of course, you can calculate it three times from a single extract, too.
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