A) 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?
B) 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?
C) 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.
D) 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.
I have no idea how to approach these problems. Any help is much appreciated!
I am Dr. Henary and I am very mad after seeing this. I will use a thicker accent for future lectures because of you. Have a good day.
I understand that the 2.7 of problem #1 is the distribution coefficient(all of these are equilibrium type problems) but I don't know the system you are using for "solvent 1" and "solvent 2". Am I correct that Korganicphase/aqueousphase = 2.7 or is it the reciprocal of that.
Here is the formula to use for #1.
fn = [1+K*(Vo/Va)]-n
fn = fraction remaining in the aq phase after n extractions.
K = distribution constant organic/aqeuous
Vo = volume of the organic phase
Va = volume of the aqueous phase
n = # extractions.
For the other problems, use
Ko/a = (concn org layer)/(concn aq layer)
A) To determine how much compound can be extracted by three 10-mL portions of benzene, we need to calculate the amount of compound that will be transferred from water to benzene in each extraction.
Let's start by calculating the amount of compound dissolved in water. We know that 1.0g of the compound was dissolved in 100mL of water. Therefore, the concentration of the compound in water is:
Concentration (C1) = 1.0g / 100mL = 0.01g/mL
Now, let's calculate the amount of compound that will be transferred from water to benzene in each extraction. The distribution coefficient (Kp) is defined as the concentration of the compound in solvent 2 divided by the concentration of the compound in solvent 1. In this case:
Kp = Concentration in benzene (C2) / Concentration in water (C1)
Given Kp = 2.7, we can rearrange the equation to solve for C2:
C2 = Kp * C1 = 2.7 * 0.01g/mL = 0.027g/mL
Now, let's calculate the amount of compound that can be extracted by a 10-mL portion of benzene. The volume of benzene used in each extraction is 10mL, and the concentration of the compound in benzene is 0.027g/mL. Therefore, the amount of compound extracted in each portion is:
Amount extracted = Concentration in benzene * Volume of benzene = 0.027g/mL * 10mL = 0.27g
Since there are three 10-mL portions of benzene, the total amount of compound that can be extracted is:
Total amount extracted = Amount extracted per portion * Number of portions = 0.27g * 3 = 0.81g
Therefore, 0.81g of the compound can be extracted by three 10-mL portions of benzene.
B) To determine the number of extractions required to recover at least 90% of the compound, we need to consider the value of Kp and the equal volumes of the solvents.
Given the value of Kp = 0.5, we can interpret this as the compound being more soluble in solvent 1 (water) compared to solvent 2 (pentane).
When using equal volumes of the two solvents, it means that the compound will distribute equally between the two solvents. At each extraction, approximately half of the compound will be transferred to the pentane layer, and the other half will remain in the water layer.
To recover at least 90% of the compound, we need to remove at least 90% of the compound from the water layer. Since only half of the compound is transferred to the pentane layer in each extraction, the fraction remaining in the water layer after each extraction is 0.5.
Let's calculate the fraction remaining in the water layer after each extraction:
After 1 extraction: Remainder = 0.5
After 2 extractions: Remainder = 0.5 * 0.5 = 0.25
After 3 extractions: Remainder = 0.5 * 0.5 * 0.5 = 0.125
After n extractions: Remainder = 0.5^n
Now, we can set up an inequality to find the minimum number of extractions required to have a remainder less than or equal to 0.1 (90% recovery):
0.5^n ≤ 0.1
Taking the logarithm of both sides, we get:
n * log(0.5) ≤ log(0.1)
Solving for n, we find:
n ≥ log(0.1) / log(0.5)
Using a calculator, we can determine the value of n:
n ≥ 3.3219
Therefore, we need at least 4 extractions to recover at least 90% of the compound.
C) To calculate the percentage of the compound that can be removed from liquid phase 1 by using one to four extractions with liquid phase 2, we need to consider the value of Kp and the volume ratio of the two phases.
Given Kp = 2 and the volume of phase 2 is 50% that of phase 1, it means that the compound is more soluble in phase 2.
Let's calculate the fraction of the compound that remains in phase 1 after each extraction:
After 1 extraction: Remainder = 1 - (1 / Kp) = 1 - (1 / 2) = 0.5
After 2 extractions: Remainder = 0.5 - (0.5 / Kp) = 0.5 - (0.5 / 2) = 0.25
After 3 extractions: Remainder = 0.25 - (0.25 / Kp) = 0.25 - (0.25 / 2) = 0.125
After 4 extractions: Remainder = 0.125 - (0.125 / Kp) = 0.125 - (0.125 / 2) = 0.0625
Now, let's calculate the percentage of the compound that can be removed from phase 1 after each extraction:
After 1 extraction: Percentage removed = (1 - Remainder) * 100 = (1 - 0.5) * 100 = 50%
After 2 extractions: Percentage removed = (1 - Remainder) * 100 = (1 - 0.25) * 100 = 75%
After 3 extractions: Percentage removed = (1 - Remainder) * 100 = (1 - 0.125) * 100 = 87.5%
After 4 extractions: Percentage removed = (1 - Remainder) * 100 = (1 - 0.0625) * 100 = 93.75%
Therefore, by using one to four extractions with liquid phase 2, the compound can be removed from liquid phase 1 by approximately 50%, 75%, 87.5%, and 93.75%, respectively.
D) To increase the partition coefficient of a compound between diethyl ether and water, we can use the salting-out method. This method involves adding a salt (such as sodium chloride) to the water phase.
By adding salt to the water phase, it increases the ionic strength of the solution. This, in turn, decreases the solubility of the organic compound in the water phase and increases its solubility in the diethyl ether phase.
The partition coefficient is influenced by the relative solubilities of the compound in the two phases. By decreasing the solubility of the compound in water and increasing its solubility in diethyl ether, the partition coefficient will increase.
The increased partition coefficient allows for a more efficient extraction of the compound from the water phase to the diethyl ether phase.
A) To determine how much compound could be extracted by three 10-mL portions of benzene, we need to consider the distribution of the compound between water and benzene.
Given:
Kp = 2.7 (distribution coefficient)
Initial mass of compound = 1.0g
Volume of water = 100mL
Volume of each portion of benzene = 10mL
Number of portions of benzene = 3
First, let's calculate the amount of compound that remains in the water after the extraction with benzene. We can use the formula:
Amount in water = Initial amount - Amount extracted
Amount in water = (Mass of compound) - (Mass of compound extracted by one portion of benzene)
To calculate the mass of the compound extracted by one portion of benzene, we use the distribution coefficient (Kp):
Mass extracted by one portion of benzene = (Kp * Volume of benzene) * (Amount in water)
Now, we can calculate the amount of compound extracted by each portion of benzene:
Amount extracted by one portion of benzene = Mass extracted by one portion of benzene / Volume of benzene
Finally, to find the total amount of compound extracted by three portions of benzene:
Total amount extracted = Amount extracted by one portion of benzene * Number of portions of benzene
B) The question asks how many extractions of the aqueous layer will be required to recover at least 90% of the compound, given the distribution coefficient (Kp) between pentane and water.
Given:
Kp = 0.5
Volume of pentane = Volume of water
To determine the number of extractions required, we need to find the fraction of the compound remaining in the aqueous layer after each extraction. This can be calculated using:
Fraction remaining in water = (1 - Fraction extracted) = (1 - 1/Kp)
After each extraction, the fraction remaining in the aqueous layer decreases. We need to continue the extraction until the fraction remaining in water is less than or equal to 0.10 (90% recovery).
So, we can set up an equation to solve for the number of extractions:
(1 - 1/Kp)^n ≤ 0.10
Solving for n (number of extractions) will give us the answer.
C) To calculate the percentage of a compound that can be removed from liquid phase 1 using one to four extractions with liquid phase 2, given a distribution coefficient (Kp) and volumes of the two phases, we can use the equation:
Percentage removed = (1 - (1 - 1/Kp)^n) * 100
For each value of n (number of extractions), calculate the percentage removed using the given formula. Repeat the calculation for n = 1, 2, 3, 4, and compute the percentage removed for each case.
D) To increase the partition coefficient (Kp) for a slightly polar organic compound between diethyl ether and water, there are a few methods we can consider:
1. Adjust pH: Changing the pH of the aqueous phase can significantly influence the partition coefficient. By adjusting the pH to favor ionization of the compound, it can enhance its solubility in either the organic or aqueous phase.
2. Modify solvent polarity: Adding salts or other polar solvents to the water phase can increase the ionic strength and affect the solubility of the compound. Using cosolvents or modifying the content of organic solvents can also alter the solvent polarity and impact partitioning.
3. Change temperature: Temperature can play a role in solubility and distribution between different phases. By optimizing the temperature conditions, it is possible to increase the partition coefficient.
4. Utilize solid-phase extraction: Solid-phase extraction techniques using specialized resins or columns can selectively adsorb the compound from the aqueous phase, thereby increasing the effective partition coefficient.
These methods offer different strategies for manipulating the system to increase the partition coefficient and enhance the extraction efficiency. The choice of the appropriate method depends on the specific compound and desired outcome.