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%3f = ?

Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin as a constant IV infusion without a loading dose. The half-life is 8 h. What percentage f of the steady-state level has been reached after 24 h?

t½ = 8 ht = 24 h
f = %
2.Easy
10100t (h)C (mg/L)25 mg/L2 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the vancomycin level is 25 mg/L. The half-life is 8 h, and the drug is eliminated by first-order kinetics. What is the level C 2 h later?

C₀ = 25 mg/Lt½ = 8 ht = 2 h
C = mg/L
3.Medium
Vd = 0.7 L/kgC = 20 mg/LLD = ? mg78 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 78 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.

78 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
4.Medium
110100t (h)C (mg/L)2 h25 mg/L14 h8.8 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 25 mg/L at 2 h and 8.8 mg/L at 14 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 25 mg/L (2 h)C2 = 8.8 mg/L (14 h)
k = h⁻¹
5.Hard
Vd = 0.25 L/kg1 → 7 mg/LD = ? mg64 kg

Practice exercise (made-up values, not clinical guidance). A 64 kg patient on regular gentamicin doses is at steady state. The target is a peak of 7 mg/L and a trough of 1 mg/L, and the volume of distribution is 0.25 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?

64 kgVd = 0.25 L/kgCpeak = 7 mg/LCtrough = 1 mg/L
D = mg
6.Hard
t (h)C (mg/L)Css​t = ?95%

Practice exercise (made-up values, not clinical guidance). A theophylline infusion is started without a loading dose. The half-life in this patient is 9 h. How long t does it take to reach 95% of the steady-state level?

t½ = 9 h95% Css
t = h

Answer key

  1. 1. f = 87.5 %
  2. 2. C = 21.02 mg/L
  3. 3. LD = 1187 mg
  4. 4. k = 0.08701 h⁻¹
  5. 5. D = 96 mg
  6. 6. t = 38.9 h

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