chemistry -drbob222

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I'm not sure if my 1 and 2 are correct, and I don't know how to do 3, 4, and 5. Any help is appreciated!

1) You wish to extract an organic compound from an aqueous phase into an organic layer. to minimizing the number of transfer steps, would it be better to use an organic solvent that is heavier or lighter than water?
my response:
I would use an organic solvent that is heavier, because you don't need to pour the upper aqueous layer out of the separatory funnel after each extraction.
2) A compound distributes between water (solvent 1) and benzene (solvent 2) with Kp = 2.7. If 1.0g of the compound were dissolved in 100mL of water, how much compound could be extracted by THREE 10-mL portions of benzene?
work:
Kp = [x/ (100)]/ [(1.0-x)/ 10] =2.7
Kp = [x/ (100)]/ [(1/28-x)/ 10] =2.7
Kp = [x/ (100)]/ [([1/28]^2-x)/ 10] =2.7
x=(1/28)^3 g

3) If the value of Kp is 0.5 for the distribution of a compound between pentane (solvent 2) and water (solvent 1), and equal volumes of the two solvents were used, how many extractions of the aqueous layer will be required to recover at least 90% of the compound?

4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2. Assume that Kp = 2 and the volume of phase 2 equals to 50% that of phase 1.

5) A slightly polar organic compound distributes between diethyl ether and water with a partition coefficient equal to 3 (in favor of the ether). What simple method can be used to increase the partition coefficient? Explain.

  • chemistry -drbob222 (correction) -

    work for 2: (typo, correction below)
    Kp = [(1.0-x)/ 10]/[x/ (100)] =2.7
    Kp = [(1/28-x)/ 10]/[x/ (100)] =2.7
    Kp = [([1/28]^2-x)/ 10]/[x/ (100)] =2.7
    x=(1/28)^3 g

  • chemistry -drbob222 (just 5) -

    I think I got the rest of the problem. The only one I need help on is number 5.

  • chemistry -drbob222 -

    This is really out of my field but I think you make a difference by adding a soluble salt, such as NaCl, to the aqueous phase. That makes the polar solute, which is already more favorably inclined to go with the ether, less able to fill the "spaces" in the water solvent. Check my think. Look in your text to see if you can find anything.

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