A drug that follows first-order kinetics has a half-life of 6 hours. A patient receives a continuous IV infusion starting at t=0. What is the approximate time required to reach 90% of steady-state plasma concentration?

• A 12 hours (2 half-lives)
• B 20 hours (~3.3 half-lives) ✓
• C 18 hours (3 half-lives)
• D 30 hours (5 half-lives)

Explanation

During constant-rate IV infusion with first-order elimination, the fraction of steady-state concentration reached at any time t is (1 − e^−(0.693t/t½)). Reaching 90% of steady state requires: 1 − e^(−0.693t/6) = 0.9, so e^(−0.693t/6) = 0.1, meaning −0.693t/6 = ln(0.1) = −2.303, giving t = 2.303 × 6/0.693 = 13.86/0.693 ≈ 20 hours (approximately 3.32 half-lives). As a clinical rule of thumb, 90% of Css is reached in 3.3 half-lives; 95% requires ~4.3 half-lives and 99% requires ~6.6 half-lives.

Reference: KD Tripathi, Essentials of Medical Pharmacology, 8th ed.

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