Compound θ has a distribution coefficient of 5.0 when extracted from water with ethyl acetate, with θ being more soluble in the top layer. How many 10.0 mL ethyl acetate extractions would be required to extract at least 95.0% (do not round up) of θ from a 50.0 mL aqueous solution?
To extract at least 95.0% of θ from a 50.0 mL aqueous solution, we need to calculate the number of extractions required.
First, let's calculate the volume of θ in the aqueous solution:
Volume of θ in the aqueous solution = 50.0 mL * (1 - 0.95) = 2.5 mL
Now, let's calculate the volume of θ in the ethyl acetate layer after each extraction:
Volume of θ in the ethyl acetate layer after each extraction = 10.0 mL * 5.0 = 50.0 mL
We need to perform extractions until the volume of θ in the ethyl acetate layer is greater than or equal to the volume of θ in the aqueous solution:
Number of extractions = (2.5 mL / 50.0 mL) + 1 = 0.05 + 1 = 1.05
Since we cannot perform a fraction of an extraction, we need to round up the number of extractions required.
Therefore, we need to perform 2 extractions (rounded up) to extract at least 95.0% of θ from a 50.0 mL aqueous solution.
why is it not 5 extractions?
I apologize for the confusion caused by my previous response. To extract at least 95% of θ from a 50.0 mL aqueous solution, the distribution coefficient and the number of extractions are not directly proportional.
Let's reconsider the problem.
First, let's calculate the volume of θ in the aqueous solution:
Volume of θ in the aqueous solution = 50.0 mL * (1 - 0.95) = 2.5 mL
Now, let's consider each extraction:
In the first extraction:
Volume of θ in the ethyl acetate layer = 10.0 mL * 5.0 = 50.0 mL
After the first extraction, the volume of θ in the aqueous solution is reduced to:
2.5 mL - 50.0 mL = -47.5 mL
Since we cannot have a negative volume, we consider that the remaining θ in the aqueous solution is negligible. Therefore, only one extraction is sufficient to extract more than 95% of θ from the 50.0 mL aqueous solution.
Therefore, 1 extraction is required to extract at least 95% of θ from a 50.0 mL aqueous solution.
when answered earlier you had said five extraction were needed? what changed?
I apologize for the confusion in my previous responses. I made an error in my calculations.
To extract at least 95.0% of θ from a 50.0 mL aqueous solution, we need to consider the distribution coefficient and the number of extractions.
The distribution coefficient is given as 5.0, which means that θ is more soluble in the ethyl acetate layer.
To determine the number of extractions required, we can calculate the remaining volume of θ in the aqueous solution after each extraction.
After the first extraction, the volume of θ in the aqueous solution is reduced to:
50.0 mL - 10.0 mL = 40.0 mL
After the second extraction:
40.0 mL - 10.0 mL = 30.0 mL
After the third extraction:
30.0 mL - 10.0 mL = 20.0 mL
After the fourth extraction:
20.0 mL - 10.0 mL = 10.0 mL
After the fifth extraction:
10.0 mL - 10.0 mL = 0.0 mL (negligible remaining volume)
Therefore, to extract at least 95.0% of θ from a 50.0 mL aqueous solution, 5 extractions are required.
To determine the number of 10.0 mL ethyl acetate extractions required to extract at least 95.0% of θ from a 50.0 mL aqueous solution, we need to calculate the distribution ratio (D) and then use it to calculate the number of extractions.
The distribution ratio (D) is defined as the ratio of the concentration of θ in the top layer (ethyl acetate) to its concentration in the bottom layer (water). In this case, the distribution coefficient is given as 5.0, which is the same as the distribution ratio (D).
D = [θ]top layer / [θ]bottom layer
We can rearrange this equation to solve for [θ]top layer:
[θ]top layer = D * [θ]bottom layer
Now, let's assume that after x extractions, we have extracted a fraction of θ, f, from the aqueous solution. The remaining fraction of θ in the aqueous solution is (1 - f).
After each extraction, the volume of the aqueous solution becomes 50.0 mL - (10.0 mL * x) and the volume of θ extracted into the ethyl acetate layer becomes (10.0 mL * x).
The fraction of θ extracted after each extraction is given by the distribution ratio:
f = D * (10.0 mL * x) / (50.0 mL - (10.0 mL * x))
To calculate the number of extractions required to extract at least 95.0% (0.95) of θ from the aqueous solution, we need to solve for x in the equation:
0.95 = D * (10.0 mL * x) / (50.0 mL - (10.0 mL * x))
Let's solve this equation:
0.95 * (50.0 mL - (10.0 mL * x)) = D * (10.0 mL * x)
47.5 mL - 9.5 mL * x = 5.0 * x
14.5 mL = 14.5 * x
x = 1
Hence, at least one 10.0 mL ethyl acetate extraction is required to extract at least 95.0% of θ from the 50.0 mL aqueous solution.