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.

New worksheetPractise online with hints

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
110100t (h)CC₀t½ = ?C₀/2

Practice exercise (made-up values, not clinical guidance). The elimination rate constant of tobramycin in a patient is 0.231 h⁻¹. Find the half-life t½.

k = 0.231 h⁻¹
t½ = 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
VCss​ = 25 mg/LR = ? mg/hCL = 50 mL/min

Practice exercise (made-up values, not clinical guidance). Vancomycin is given as a continuous infusion to a patient whose vancomycin clearance is 50 mL/min. What infusion rate R in mg/h gives a steady-state level of 25 mg/L?

CL = 50 mL/minCss = 25 mg/L
R = mg/h
4.Medium
110t (h)C (mg/L)C₀ = 7 mg/Lt½ = 2.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 200 mg of tobramycin. The level back-extrapolated to time zero is 7 mg/L and the half-life is 2.5 h. Find the volume of distribution Vd.

D = 200 mgC₀ = 7 mg/Lt½ = 2.5 h
Vd = L
5.Hard
VCss,avg​ = ?500 mg · q12hCL = 0.035 L/kg/h95 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 95 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.035 L/kg/h. Each dose is 500 mg of theophylline. Find the average steady-state level Css,avg.

CL = 0.035 L/kg/hτ = 12 hS = 1F = 195 kgD = 500 mg
Css,avg = mg/L
6.Hard
Vd = 7 L/kg0.5 → 0.8 mcg/LLD = ? mcg95 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 95 kg patient has a digoxin level of 0.5 mcg/L; the target is 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?

95 kgVd = 7 L/kgC = 0.5 → 0.8 mcg/LF = 0.7
LD = mcg

Answer key

  1. 1. t½ = 3.001 h
  2. 2. k = 0.3466 h⁻¹
  3. 3. R = 75 mg/h
  4. 4. Vd = 28.57 L
  5. 5. Css,avg = 12.53 mg/L
  6. 6. LD = 285 mcg

Full step-by-step solutions: practise this topic on examproblems.com.