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.

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

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

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.

3) To determine the number of extractions required to recover at least 90% of the compound, you need to consider the distribution coefficient (Kp) and the efficiency of each extraction.

The distribution coefficient (Kp) is the ratio of the concentration of the compound in the organic phase to its concentration in the aqueous phase. In this case, Kp is 0.5, which means the compound prefers to stay in the organic phase.

Each extraction can remove a certain fraction of the compound from the aqueous phase. To calculate this fraction, you can use the equation:

Fraction extracted = 1 - (1/Kp)^n

Where 'n' is the number of extractions.

In this case, you want to recover at least 90% of the compound, so we can set the fraction extracted equal to or greater than 0.9:

0.9 <= 1 - (1/0.5)^n

Solving for 'n', you can use logarithms:

n >= log(1 - 0.9) / log(1/0.5)

To get a whole number for 'n', you can round up to the nearest integer. So, you will need at least 'n' extractions, rounded up to the nearest whole number, to recover at least 90% of the compound.

4) To calculate the percentage of the compound that can be removed from liquid phase 1 using one to four extractions with liquid phase 2, you need to consider the distribution coefficient (Kp) and the efficiency of each extraction, as well as the volume ratio between the two phases.

Assuming Kp = 2, the concentration of the compound in the organic phase will be twice the concentration in the aqueous phase. Let's denote the volume of liquid phase 1 as V1, and the volume of liquid phase 2 as V2 (which is 50% of V1 based on the given information).

For one extraction, the amount of compound extracted is given by:

Amount extracted = [Initial concentration in phase 1 - Final concentration in phase 1] * V2

The initial concentration in phase 1 is just the concentration of the compound in the liquid phase. The final concentration in phase 1 can be calculated using the distribution coefficient (assuming complete extraction), and multiplying by the initial concentration in phase 1.

In general, for 'n' extractions, the amount of compound extracted is given by:

Amount extracted = [Initial concentration in phase 1 - Final concentration in phase 1 after 'n' extractions] * V2

To calculate the percentage of the compound removed, you can divide the amount extracted by the initial amount in phase 1 and multiply by 100.

Repeat this calculation for 'n' ranging from one to four to get the percentage of the compound that can be removed from liquid phase 1 using one to four extractions with liquid phase 2.

5) To increase the partition coefficient (Kp) of a slightly polar organic compound between diethyl ether and water, you can adjust the polarity of the solvents.

A simple method to increase the partition coefficient is to add a salt to the aqueous phase. By adding a salt, you increase the ionic strength of the aqueous phase, which can decrease the solubility of the compound in water and increase its tendency to partition into the organic phase. This results in a higher partition coefficient.

The presence of ions from the salt in the aqueous phase can also provide a stronger attractive force for the polar compounds, encouraging them to stay in the organic phase.

It's important to note that the choice of salt to use may depend on the specific compound and solvents being used. It's recommended to consult the literature or seek expert advice for the most appropriate salt to use in a particular situation.