For a single IV bolus of 100 mg with Vd = 20 L and Cl = 4 L/h, compute k, t1/2, and the approximate steady-state peak concentration after dosing every 12 h with accumulation.

• A. k = 0.1 h^-1; t1/2 ≈ 6.93 h; Css,max ≈ 2.0 mg/L
• B. k = 0.25 h^-1; t1/2 ≈ 2.77 h; Css,max ≈ 3.0 mg/L
• C. k = 0.3 h^-1; t1/2 ≈ 2.31 h; Css,max ≈ 4.0 mg/L
• D. k = 0.2 h^-1; t1/2 ≈ 3.47 h; Css,max ≈ 5.5 mg/L

Answer: (D)

First-order elimination and accumulation with repeated IV bolus dosing are being tested here. The rate constant for elimination is k = Cl/Vd. With Cl = 4 L/h and Vd = 20 L, k = 4/20 = 0.2 h^-1, giving a half-life t1/2 = 0.693/k ≈ 3.47 h.

For repeated dosing every 12 hours, there is accumulation because not all drug is cleared before the next dose. After a single dose, the concentration jump is D/Vd = 100 mg / 20 L = 5 mg/L. At steady state, the peak concentration after a dose is this amount multiplied by the accumulation factor R, where R = 1 / (1 − e^(−kτ)) and τ is the dosing interval (12 h). Compute e^(−kτ) = e^(−0.2×12) = e^(−2.4) ≈ 0.0907, so R ≈ 1 / (1 − 0.0907) ≈ 1.10. Therefore Css,max ≈ 5 mg/L × 1.10 ≈ 5.5 mg/L.

So you get k ≈ 0.2 h^-1, t1/2 ≈ 3.47 h, and Css,max ≈ 5.5 mg/L, which matches the stated answer.