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
110100t (h)CC₀t½ = ?C₀/2

Practice exercise (made-up values, not clinical guidance). The elimination rate constant of digoxin in a patient is 0.0173 h⁻¹. Find the half-life t½.

k = 0.0173 h⁻¹
t½ = h
2.Easy
110100t (h)CC₀t½ = 1.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, lidocaine has an elimination half-life of 1.5 h. Find the elimination rate constant k in h⁻¹.

t½ = 1.5 h
k = h⁻¹
3.Medium
Vd = 0.7 L/kgC = 15 mg/LLD = ? mg67 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 67 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 15 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.

67 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
4.Medium
110100t (h)C (mg/L)1 h25 mg/L13 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 1 h and 8.8 mg/L at 13 h. Assume first-order elimination. Find the half-life t½.

C1 = 25 mg/L (1 h)C2 = 8.8 mg/L (13 h)
t½ = h
5.Hard
0.1110t (h)C (mg/L)Cpeak​ = 6τ = ?Ctrough​ = 0.5

Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 3 h. The target is a peak of 6 mg/L and a trough of 0.5 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.

t½ = 3 hCpeak = 6 mg/LCtrough = 0.5 mg/L
τ = h
6.Hard
0.1110t (h)C (mg/L)Cpeak​ = 6τ = ?Ctrough​ = 0.5

Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 2.5 h. The target is a peak of 6 mg/L and a trough of 0.5 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.

t½ = 2.5 hCpeak = 6 mg/LCtrough = 0.5 mg/L
τ = h

Answer key

  1. 1. t½ = 40.07 h
  2. 2. k = 0.4621 h⁻¹
  3. 3. LD = 764.7 mg
  4. 4. t½ = 7.966 h
  5. 5. τ = 10.8 h
  6. 6. τ = 9 h

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