Post a New Question


posted by .

Suppose that a compound Z had a partition coefficient of 8 in ether-water. If 10g of z was dissolved in 100mL of water, and if the water solution was extracted with two 100 mL portions of ether, answer the following questions

a) what is the total amount of z contained in 200mL of ether extract after the two extractions?

b) if instead of using two 100 mL portions of ether a single extraction had been carried out using one 200 mL portion of ether what weight of Z would be contained in the ether solution under these conditions?

  • chemistry -

    K = 8 = (z/100)/(10/100)
    Solve for z for the first extraction, subtract from 10 to determine the amount left in the aqueous phase, then repeat the extraction.
    b) is done the same way.

  • chemistry -

    I think you're wrong bob.

  • chemistry -

    Yeah, Bob is wrong.
    What you would actually do for the following:

    a) Do the first extraction:
    K=(z/?mL ether)/((10g-z)/?mL water)
    8=(z/100mL ether)/((10g-z)/100mL water)

    Rearrange to make it easier:
    8=(z/(10g-z))*(100mL water/100mL ether)

    Now solve for z and it should come out to about 8.9g. (Simple algebra shouldn't be too hard)

    Then do the same thing for the second extraction, but instead of (10g-z), replace it with (1.1g-z) because that's the amount of Z that remains in water so you're using ether again to extract the remaining z.

    8=(z/100mL ether)/((1.1g-z)/100mL water)
    8=(z/(1.1g-z))*(100mL water/100mL ether)
    and then solve for the second z value.

    Finally, add both Z's from both extractions and you should get your total Z extracted.

    b) Use the same equation from the first part:
    K=(z/?mL ether)/((10g-z)/?mL water)
    8=(z/200mL ether)/((10g-z)/100mL water)
    8=(z/(1.1g-z))*(100mL water/200mL ether)
    then solve for z only one time because you're using all 200mL ether to extract in one go, so when you get your z, that will be the answer.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question