The distribution coefficient, k = concentration in hexanes

Answered question

2022-04-25

The distribution coefficient, k = concentration in hexanes / concentration in water, between hexanes and water for salute A is 7.5. How much hexanes would be required to remove 97.5% of A in a single extraction from a solution of 10 g of A in 100 mL of water?

Answer & Explanation

star233

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Skilled2023-04-29Added 403 answers

Let x be the amount of hexanes required to remove 97.5% of A from the solution.
Initially, the concentration of A in water is 10 g/100 mL = 0.1 g/mL. If 97.5% of A is removed from the solution, the remaining concentration of A in water will be 0.025 g/mL.
Let cH and cW be the concentrations of A in hexanes and water, respectively, after the extraction. Then we have:
k=cHcW=7.5
Since the amount of A is conserved in the extraction, we have:
cH·VH+cW·VW=cW·(VH+VW)=0.025 g/mL·100 mL=2.5 g
where VH and VW are the volumes of hexanes and water used in the extraction, respectively. We want to find VH, which is equal to x since we are assuming a single extraction.
Substituting cH=k·cW=7.5·0.1 g/mL=0.75 g/mL, we get:
0.75 g/mL·x+0.1 g/mL·(100 mLx)=2.5 g
Solving for x, we get:
x=25 mL
Therefore, 25 mL of hexanes is required to remove 97.5% of A in a single extraction.

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