How to set up the core PK calculations
Five quantities describe how the body handles a drug that follows first-order kinetics: the elimination rate constant k, half-life, volume of distribution, clearance, and the concentration a dose produces. They are connected by a small set of equations, and most exam questions give you two of the quantities and ask for a third. The work is picking the right equation, and the trap is confusing which quantity controls which dose.
t½ = 0.693 ÷ k • Vd = dose ÷ C₀ • CL = k × Vd • loading dose = C_target × Vd ÷ (S × F) • k = ln(C₁ ÷ C₂) ÷ (t₂ − t₁)
Half-life and k
A drug has k = 0.1 hr⁻¹. Its half-life:
t½ = 0.693 ÷ 0.1 = 6.93 hr
The 0.693 is ln 2, and it appears because a half-life is the time for the concentration to fall by a factor of two. The same equation solves for k when the half-life is given: k = 0.693 ÷ t½. Steady state is reached after about 4 to 5 half-lives of repeated dosing, and about 94% of a drug is gone 4 half-lives after the last dose, so this one number answers a lot of "how long until" questions.
Volume of distribution
A 500 mg IV bolus produces an initial concentration of 12.5 mg/L.
Vd = 500 ÷ 12.5 = 40 L
Vd is the volume the dose would have to be dissolved in to give the measured concentration. It is not a real space. A Vd larger than total body water (about 42 L in a 70 kg adult) means the drug is mostly in tissue, not plasma. Keep the units straight: mg divided by mg/L gives L.
Clearance
For the same drug, with k = 0.1 hr⁻¹ and Vd = 40 L:
CL = 0.1 × 40 = 4 L/hr
Clearance is the volume of plasma cleared of drug per unit time. It's the quantity that determines the maintenance dose, because at steady state the dosing rate has to equal the elimination rate, and elimination rate is CL × concentration.
Loading dose
Target concentration is 15 mg/L and Vd is 40 L. IV, so S and F are both 1.
loading dose = 15 × 40 = 600 mg
Loading dose depends on Vd, not on clearance or half-life. You are filling the volume up to the target concentration, and how fast the drug leaves afterward has nothing to do with how much it takes to fill it. Maintenance dose depends on clearance, not on Vd. Exams swap these constantly, and the fix is to ask which physical thing the dose is doing: filling a volume (Vd) or replacing what's been cleared (CL).
S is the salt factor, the fraction of the salt form that is active drug, and F is bioavailability. Both are 1 for an IV dose of the active moiety, which is why the practice problems here say to assume them. If a question gives S = 0.8 or F = 0.5, the loading dose is divided by them, which makes it larger.
k from two concentrations
Levels are 30 mg/L at hour 0 and 5 mg/L at hour 12.
ratio = 30 ÷ 5 = 6
ln(6) = 1.7918
k = 1.7918 ÷ 12 = 0.1493 hr⁻¹
t½ = 0.693 ÷ 0.1493 = 4.64 hr
The natural log, not log base 10. It's the "ln" key on the calculator. Put the higher concentration on top so the log comes out positive. And carry k to four decimals; it's a small number that everything downstream gets divided by, so a k of 0.15 instead of 0.1493 shifts the half-life by a tenth of an hour.
Where students lose the points
- Vd for maintenance, CL for loading. Backwards. Loading fills the volume; maintenance replaces clearance.
- log instead of ln. ln(6) is 1.79; log(6) is 0.78.
- Concentrations in the wrong order. ln(C₁ ÷ C₂) with C₁ the earlier, higher level.
- Half-life used where k is needed. Convert with 0.693 first. The equations above all use k.
- mg/L versus mcg/mL. They're the same number (1 mg/L = 1 mcg/mL). ng/mL is a thousand times smaller.
- Truncating k. Four decimals minimum.