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 determine the number of extractions required to extract at least 95.0% of compound θ, we need to consider the solubility of the compound and the volume of the solvent used.
Given:
- Distribution coefficient (K) = 5.0
- Volume of aqueous solution (Vaq) = 50.0 mL
- Volume of each extraction (Vext) = 10.0 mL
The distribution coefficient (K) is defined as the ratio of the concentration of the compound in the organic phase to the concentration of the compound in the aqueous phase. In this case, K = 5 means that compound θ is 5 times more soluble in ethyl acetate than in water.
Let's calculate the cumulative extraction efficiency (CE) after each extraction using the equation:
CE = (Vext / Vaq) * (K / (1 + (K * (Vext / Vaq))))
For each extraction, the concentration of compound θ in the aqueous phase decreases and the concentration in the organic phase increases. The cumulative extraction efficiency accounts for the previous extractions as well.
Let's calculate the cumulative extraction efficiency after each extraction:
1st extraction:
CE1 = (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE1 ≈ 0.0909
2nd extraction:
CE2 = CE1 + (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE2 ≈ 0.1714
3rd extraction:
CE3 = CE2 + (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE3 ≈ 0.2540
4th extraction:
CE4 = CE3 + (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE4 ≈ 0.3396
5th extraction:
CE5 = CE4 + (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE5 ≈ 0.4272
We can see that after the 5th extraction, the cumulative extraction efficiency is approximately 0.4272, which is less than 95.0%.
Therefore, it would take at least 5 extractions to extract less than 95.0% of compound θ from the 50.0 mL aqueous solution.
why not 4 extractions?
I apologize for the mistake in my previous response. Let's re-calculate the cumulative extraction efficiency for 4 extractions:
1st extraction:
CE1 = (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE1 ≈ 0.0909
2nd extraction:
CE2 = CE1 + (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE2 ≈ 0.1714
3rd extraction:
CE3 = CE2 + (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE3 ≈ 0.2540
4th extraction:
CE4 = CE3 + (10.0 mL / 50.0 mL) * (5.0 / (1 + (5.0 * (10.0 mL / 50.0 mL))))
CE4 ≈ 0.3396
We can see that after the 4th extraction, the cumulative extraction efficiency is approximately 0.3396, which is less than 95.0%.
Therefore, it would take at least 5 extractions to extract at least 95.0% of compound θ from the 50.0 mL aqueous solution.
To determine the number of ethyl acetate extractions required to extract at least 95.0% of compound θ from a 50.0 mL aqueous solution, we need to consider the distribution coefficient and the volumes involved.
First, let's calculate the total volume of the ethyl acetate needed:
50.0 mL (aqueous solution) + X mL (ethyl acetate extractions) = Total volume
Since each extraction uses 10.0 mL of ethyl acetate, the equation becomes:
50.0 mL + (10.0 mL × X) = Total volume
Next, let's calculate the total amount of θ extracted after X extractions:
Total θ extracted = (5.0 × X) / (1 + 5.0 × X) × 100%
We want to extract at least 95.0% of θ, so we can set up the following equation:
Total θ extracted ≥ 95.0%
(5.0 × X) / (1 + 5.0 × X) × 100% ≥ 95.0%
Now we can solve this equation to find the value of X:
(5.0 × X) / (1 + 5.0 × X) ≥ 0.95
Cross-multiplying the equation:
5.0 × X ≥ 0.95 × (1 + 5.0 × X)
5.0 × X ≥ 0.95 + 4.75 × X
0.25 × X ≥ 0.95
X ≥ 0.95 / 0.25
X ≥ 3.8
Therefore, you would need at least 3.8, or rounded up to 4, 10.0 mL ethyl acetate extractions to extract at least 95.0% of compound θ from a 50.0 mL aqueous solution.