Clinical pharmacokinetics worksheet with answers

Free printable Clinical pharmacokinetics worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
t/t½C/Css (%)100%597%

Practice exercise (made-up values, not clinical guidance). A patient starts regular doses of phenobarbital, which has a half-life of 80 h. Using the rule that steady state is reached after about five half-lives, after how many hours tss is steady state reached?

t½ = 80 h
tss = h
2.Easy
110100t (h)CC₀t½ = 2 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, lidocaine has an elimination half-life of 2 h. Find the elimination rate constant k in h⁻¹.

t½ = 2 h
k = h⁻¹
3.Medium
110t (h)C (mg/L)C₀ = 8 mg/Lt½ = 2 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 210 mg of tobramycin. The level back-extrapolated to time zero is 8 mg/L and the half-life is 2 h. Find the clearance CL in L/h.

D = 210 mgC₀ = 8 mg/Lt½ = 2 h
CL = L/h
4.Medium
VCss​ = 2 mg/LR = ? mg/minCL = 850 mL/min

Practice exercise (made-up values, not clinical guidance). A patient has a lidocaine clearance of 850 mL/min. What infusion rate R in mg/min keeps the steady-state level at 2 mg/L?

CL = 850 mL/minCss = 2 mg/L
R = mg/min
5.Hard
VCss,avg​ = 10 mg/LD = ? mg · q8hCL = 0.03 L/kg/h89 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). An 89 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.03 L/kg/h. What dose D in mg gives an average steady-state level of 10 mg/L?

CL = 0.03 L/kg/hτ = 8 hS = 0.8F = 189 kgCss,avg = 10 mg/L
D = mg
6.Hard
Vd = 0.3 L/kg1 → 6 mg/LD = ? mg82 kg

Practice exercise (made-up values, not clinical guidance). An 82 kg patient on regular tobramycin doses is at steady state. The target is a peak of 6 mg/L and a trough of 1 mg/L, and the volume of distribution is 0.3 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?

82 kgVd = 0.3 L/kgCpeak = 6 mg/LCtrough = 1 mg/L
D = mg

Answer key

  1. 1. tss = 400 h
  2. 2. k = 0.3466 h⁻¹
  3. 3. CL = 9.098 L/h
  4. 4. R = 1.7 mg/min
  5. 5. D = 267 mg
  6. 6. D = 123 mg

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