Given a drug with a half-life of 2 days and a noon concentration today of 10 μg/mL, what is the expected concentration at noon tomorrow?

Given a drug with a half-life of 2 days and a noon concentration today of 10 μg/mL, what is the expected concentration at noon tomorrow?

• A. 7 μg/mL
• B. 7.5 μg/mL
• C. 5 μg/mL
• D. 3.5 μg/mL

Answer: (A)

Drug concentrations decay exponentially when elimination follows first-order kinetics, and the half-life tells you how long it takes for the concentration to drop to half. After a time t, the remaining amount is C0 multiplied by 2^(−t/T1/2). Here the half-life is 2 days, and the time elapsed is 1 day (noon today to noon tomorrow), so t/T1/2 = 0.5. That gives a factor of 2^(−0.5) = 1/√2 ≈ 0.707. Multiplying the starting concentration 10 μg/mL by 0.707 yields about 7.07 μg/mL, so the expected concentration at noon tomorrow is roughly 7 μg/mL. This aligns with exponential decay: not a full halving yet, just a partial drop.

Drug concentrations decay exponentially when elimination follows first-order kinetics, and the half-life tells you how long it takes for the concentration to drop to half. After a time t, the remaining amount is C0 multiplied by 2^(−t/T1/2). Here the half-life is 2 days, and the time elapsed is 1 day (noon today to noon tomorrow), so t/T1/2 = 0.5. That gives a factor of 2^(−0.5) = 1/√2 ≈ 0.707. Multiplying the starting concentration 10 μg/mL by 0.707 yields about 7.07 μg/mL, so the expected concentration at noon tomorrow is roughly 7 μg/mL. This aligns with exponential decay: not a full halving yet, just a partial drop.