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% 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 determine the number of extractions required.
Let's calculate the amount of θ that remains in the aqueous solution after each extraction.
After the first extraction of 10.0 mL of ethyl acetate, the remaining aqueous solution is 50.0 mL - 10.0 mL = 40.0 mL.
The distribution coefficient tells us that θ is 5 times more soluble in the ethyl acetate layer, so we can calculate the amount of θ that remains in the aqueous solution after each extraction using the equation:
Amount of θ in aqueous solution = Initial amount of θ / (Distribution coefficient)n,
Where n is the number of extractions performed.
For the first extraction, the amount of θ remaining in the aqueous solution is:
Amount of θ in aqueous solution after first extraction = Initial amount of θ / (Distribution coefficient)^1
= Initial amount of θ / 5^1
= Initial amount of θ / 5,
Now, let's calculate the amount of θ remaining in the aqueous solution after the second extraction:
Amount of θ in aqueous solution after second extraction = Initial amount of θ / (Distribution coefficient)^2
= Initial amount of θ / 5^2
= Initial amount of θ / 25,
We can see a pattern here. After each extraction, the amount of θ remaining in the aqueous solution is divided by the distribution coefficient.
Using this pattern, we can calculate the amount of θ remaining in the aqueous solution after each extraction as follows:
After the third extraction: Initial amount of θ / 5^3
After the fourth extraction: Initial amount of θ / 5^4
After the fifth extraction: Initial amount of θ / 5^5
To extract at least 95.0% of θ, we want to have at most 5.0% of θ remaining in the aqueous solution after all the extractions. In other words, we want to have the amount of θ remaining in the aqueous solution to be less than or equal to 0.05 times the initial amount of θ.
Let's set up the inequality:
Initial amount of θ / 5^n ≤ 0.05 * Initial amount of θ,
We can cancel out the common factor of Initial amount of θ:
1 / 5^n ≤ 0.05,
To solve this inequality, we can take the logarithm (base 5) of both sides:
log5(1 / 5^n) ≤ log5(0.05),
Simplifying the left side:
- n ≤ log5(0.05),
Using a logarithmic property:
- n ≤ log5(5^-2),
Simplifying the right side:
- n ≤ -2,
We can multiply both sides by -1 and flip the inequality sign:
n ≥ 2.
Therefore, at least 2 extractions are required to extract at least 95.0% of θ from a 50.0 mL aqueous solution with a distribution coefficient of 5.0.