Drug Partitioning and Extractions (Cut off for Exam 1)

Question 1

Distribution Coefficient

Answer 1

• If excess liquid or solid is added to a mixture of immiscible liquids, it will distribute itself between the 2 phases so it becomes saturated
• If substance is added in an amount insufficient to saturate the solutions, then it will distribute between the 2 layers in a definite concentration ratio

Question 2

Partitioning - Semi-Permeable Membrane

Answer 2

• Solvent system is the same on both sides

- ONLY unionized moiety can pass thru membrane

Question 3

Distribution Coefficient - Weak Acids

Answer 3

D1/2 = [1 + (Ka/[H3O+]1)]/[1 + (Ka/[H3O+]2)]

Question 4

Distribution Coefficient - Weak Base

Answer 4

D1/2 = [1 + ([H3O+]1/Ka)]/[1 + ([H3O+]2/Ka)]

Question 5

Extractions

Answer 5

• Partitioning of solute from 1 phase to another

- Commonly used method to separate/isolate rugs/metabolites from biological samples

Question 6

Selection Factors of Extracting Solvent

Answer 6

• Must be immiscible with water
• High affinity for drug
• Cost, toxicity, flammability

Question 7

Extraction Equation

Answer 7

w1 = w * [KV1/(KV1 + V2)]

• w = weight (g) of solute in given volume
• V2 = volume of extracting solvent
• K = distribution coefficient
• w1 = how many grams remaining

Question 8

Repeated Extractions

Answer 8

wn = w * [KV1/(KV1 + V2)]^n