The partition coefficient (Kp) for caffeine in dichloromethane is 5.2 You have an aqueous solution of caffeine (80 mg) in 30 mL of water. Calculate the amount (in mg) of caffeine that can be extracted with 1 portion of 30 mL of dichloromethane (i.e. how much caffeine will be in dichloromethane). Write your answer in the form (xx.x mg)

Introduction to Chemical Engineering Thermodynamics
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ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
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### Partition Coefficient Calculation: Caffeine Extraction

**Problem Statement:**

The partition coefficient (K\(_p\)) for caffeine in dichloromethane is 5.2.

You have an aqueous solution of caffeine (80 mg) in 30 mL of water. Calculate the amount (in mg) of caffeine that can be extracted with 1 portion of 30 mL of dichloromethane (i.e., how much caffeine will be in dichloromethane). Write your answer in the form (xx.x mg).

**Solution Explanation:**

To solve this problem, we can use the partition coefficient \( K_p \), which is the ratio of the concentration of solute in the organic phase (dichloromethane) to the concentration of solute in the aqueous phase (water).

The partition coefficient formula is given by:
\[ K_p = \frac{[Caffeine]_{organic}}{[Caffeine]_{aqueous}} \]

Given data:
- \( K_p \) for caffeine in dichloromethane: 5.2
- Initial caffeine amount \( C_{total} \): 80 mg
- Volume of water \( V_{water} \): 30 mL
- Volume of dichloromethane \( V_{DCM} \): 30 mL

Let \( x \) be the amount of caffeine in dichloromethane after extraction.

According to the partition coefficient:
\[ K_p = \frac{\frac{x}{V_{DCM}}}{\frac{C_{total} - x}{V_{water}}} \]

Substituting the known values:
\[ 5.2 = \frac{\frac{x}{30}}{\frac{80 - x}{30}} \]

Solving for \( x \):
\[ 5.2 = \frac{x}{80 - x} \]
\[ 5.2 (80 - x) = x \]
\[ 416 - 5.2x = x \]
\[ 416 = 6.2x \]
\[ x = \frac{416}{6.2} \]
\[ x \approx 67.1 \text{ mg} \]

Thus, the amount of caffeine that can be extracted with 30 mL of dichloromethane is approximately \( \mathbf{67.1 \text{ mg}} \).
Transcribed Image Text:### Partition Coefficient Calculation: Caffeine Extraction **Problem Statement:** The partition coefficient (K\(_p\)) for caffeine in dichloromethane is 5.2. You have an aqueous solution of caffeine (80 mg) in 30 mL of water. Calculate the amount (in mg) of caffeine that can be extracted with 1 portion of 30 mL of dichloromethane (i.e., how much caffeine will be in dichloromethane). Write your answer in the form (xx.x mg). **Solution Explanation:** To solve this problem, we can use the partition coefficient \( K_p \), which is the ratio of the concentration of solute in the organic phase (dichloromethane) to the concentration of solute in the aqueous phase (water). The partition coefficient formula is given by: \[ K_p = \frac{[Caffeine]_{organic}}{[Caffeine]_{aqueous}} \] Given data: - \( K_p \) for caffeine in dichloromethane: 5.2 - Initial caffeine amount \( C_{total} \): 80 mg - Volume of water \( V_{water} \): 30 mL - Volume of dichloromethane \( V_{DCM} \): 30 mL Let \( x \) be the amount of caffeine in dichloromethane after extraction. According to the partition coefficient: \[ K_p = \frac{\frac{x}{V_{DCM}}}{\frac{C_{total} - x}{V_{water}}} \] Substituting the known values: \[ 5.2 = \frac{\frac{x}{30}}{\frac{80 - x}{30}} \] Solving for \( x \): \[ 5.2 = \frac{x}{80 - x} \] \[ 5.2 (80 - x) = x \] \[ 416 - 5.2x = x \] \[ 416 = 6.2x \] \[ x = \frac{416}{6.2} \] \[ x \approx 67.1 \text{ mg} \] Thus, the amount of caffeine that can be extracted with 30 mL of dichloromethane is approximately \( \mathbf{67.1 \text{ mg}} \).
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