Pharmacokinetics Practice Problems
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Core PK calculations connect dose, concentration, and the body's handling of a drug. Half-life, volume of distribution, clearance, loading dose, and the elimination rate constant all derive from a small set of interlocking formulas.
Formulast½ = 0.693 ÷ k • Vd = Dose ÷ C₀ • CL = k × VdLoading dose = C_target × Vd • k = ln(C₁ ÷ C₂) ÷ (t₂ − t₁)
Worked exampleTwo levels: 40 mg/L at hour 0, 10 mg/L at hour 8. k = ln(40 ÷ 10) ÷ 8 = 1.386 ÷ 8 = 0.173 hr⁻¹, so t½ = 0.693 ÷ 0.173 ≈ 4 hr.
Loading dose depends on Vd; maintenance depends on clearance. Exams like to swap the two.
10 Pharmacokinetics practice problems with worked solutions
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Problem 1: PK, Clearance
A drug has an elimination rate constant of 0.25 hr⁻¹ and a Vd of 30 L. What is its total body clearance? Round your answer to one decimal place.
Answer 7.5 L/hr
Solution- CL = k × Vd.
- CL = 0.25 hr⁻¹ × 30 L = 7.5 L/hr.
- Rounded as the question asks: 7.5 L/hr.
Also rearranges to k = CL/Vd and t½ = 0.693·Vd/CL, the same relationship written three ways.
Problem 2: PK, Loading Dose
Target plasma concentration is 20 mg/L and the drug's Vd is 50 L. What IV loading dose is required? (Assume S = 1, F = 1.) Round your answer to the nearest whole number.
Answer 1,000 mg
Solution- LD = (Cp_target × Vd) ÷ (S × F).
- LD = 20 mg/L × 50 L ÷ 1 = 1000 mg.
- Rounded as the question asks: 1,000 mg.
Loading dose depends on Vd; maintenance dose depends on clearance. Exam questions love to make you mix these up.
Problem 3: PK, Half-life
A drug has an elimination rate constant of 0.1 hr⁻¹. What is its half-life? Round your answer to one decimal place.
Answer 6.9 hr
Solution- t½ = 0.693 ÷ k.
- t½ = 0.693 ÷ 0.1 = 6.93 hr.
- Rounded as the question asks: 6.9 hr.
~94% of a drug is eliminated after 4 half-lives; steady state is reached after about 4 to 5 half-lives of dosing.
Problem 4: PK, Volume of Distribution
A 1000 mg IV bolus produces an initial plasma concentration (C₀) of 20 mg/L. What is the volume of distribution? Round your answer to one decimal place.
Answer 50.0 L
Solution- Vd = Dose ÷ C₀.
- Vd = 1000 mg ÷ 20 mg/L = 50 L.
- Rounded as the question asks: 50.0 L.
Vd is theoretical. A Vd far above total body water (about 42 L) means extensive tissue binding (e.g., digoxin ~500 L).
Problem 5: PK, Elimination Rate Constant
Two plasma levels are measured for a drug: 50 mg/L at hour 0 and 10 mg/L at hour 24. What is the elimination rate constant (k)? Round your answer to four decimal places.
Answer 0.0671 hr⁻¹
Solution- k = ln(C₁ ÷ C₂) ÷ (t₂ − t₁).
- C₁ ÷ C₂ = 50 mg/L ÷ 10 mg/L = 5; ln(5) ≈ 1.60944.
- k = 1.60944 ÷ 24 ≈ 0.06706 hr⁻¹.
- Rounded as the question asks: 0.0671 hr⁻¹.
First-order elimination: plot ln(concentration) vs time and k is the negative slope. Half-life then follows from t½ = 0.693 ÷ k. Carry k to at least 4 decimals. It is small, so a truncated k throws every downstream figure off.
Problem 6: PK, Clearance
A drug has an elimination rate constant of 0.25 hr⁻¹ and a Vd of 60 L. What is its total body clearance? Round your answer to one decimal place.
Answer 15.0 L/hr
Solution- CL = k × Vd.
- CL = 0.25 hr⁻¹ × 60 L = 15 L/hr.
- Rounded as the question asks: 15.0 L/hr.
Also rearranges to k = CL/Vd and t½ = 0.693·Vd/CL, the same relationship written three ways.
Problem 7: PK, Loading Dose
Target plasma concentration is 25 mg/L and the drug's Vd is 30 L. What IV loading dose is required? (Assume S = 1, F = 1.) Round your answer to the nearest whole number.
Answer 750 mg
Solution- LD = (Cp_target × Vd) ÷ (S × F).
- LD = 25 mg/L × 30 L ÷ 1 = 750 mg.
- Rounded as the question asks: 750 mg.
Loading dose depends on Vd; maintenance dose depends on clearance. Exam questions love to make you mix these up.
Problem 8: PK, Half-life
A drug has an elimination rate constant of 0.231 hr⁻¹. What is its half-life? Round your answer to one decimal place.
Answer 3.0 hr
Solution- t½ = 0.693 ÷ k.
- t½ = 0.693 ÷ 0.231 = 3 hr.
- Rounded as the question asks: 3.0 hr.
~94% of a drug is eliminated after 4 half-lives; steady state is reached after about 4 to 5 half-lives of dosing.
Problem 9: PK, Volume of Distribution
A 250 mg IV bolus produces an initial plasma concentration (C₀) of 40 mg/L. What is the volume of distribution? Round your answer to one decimal place.
Answer 6.3 L
Solution- Vd = Dose ÷ C₀.
- Vd = 250 mg ÷ 40 mg/L = 6.25 L.
- Rounded as the question asks: 6.3 L.
Vd is theoretical. A Vd far above total body water (about 42 L) means extensive tissue binding (e.g., digoxin ~500 L).
Problem 10: PK, Elimination Rate Constant
Two plasma levels are measured for a drug: 25 mg/L at hour 0 and 10 mg/L at hour 6. What is the elimination rate constant (k)? Round your answer to four decimal places.
Answer 0.1527 hr⁻¹
Solution- k = ln(C₁ ÷ C₂) ÷ (t₂ − t₁).
- C₁ ÷ C₂ = 25 mg/L ÷ 10 mg/L = 2.5; ln(2.5) ≈ 0.91629.
- k = 0.91629 ÷ 6 ≈ 0.15272 hr⁻¹.
- Rounded as the question asks: 0.1527 hr⁻¹.
First-order elimination: plot ln(concentration) vs time and k is the negative slope. Half-life then follows from t½ = 0.693 ÷ k. Carry k to at least 4 decimals. It is small, so a truncated k throws every downstream figure off.