Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
Check the result: levels fall by half every half-life, and steady state takes about five half-lives.
2Practice problems
1.Pharmacy calculationsMedium
Practice exercise (made-up values, not clinical guidance). A vial contains ceftriaxone 2000 mg as a powder. The practice label says to add 19.2 mL of sterile water; the powder adds 0.8 mL to the volume. The prescribed dose is 1600 mg. What volume V do you draw up? Round to one decimal place.
2000 mg19.2 mL+0.8 mL1600 mg
V = mL
2.Pharmacy calculationsHard
Practice exercise (made-up values, not clinical guidance). You have 30 g of 20% zinc oxide ointment. How many grams m of plain ointment base (0%) must you add to make a 15% ointment? Round to the nearest whole gram.
30 g · 20%0%15%
m = g
3.Dosage calculationsMedium
A patient weighing 20 kg has a body surface area of 0.81 m². The prescription is 175 mg/m². Find the dose D in mg.
20 kgBSA = 0.81 m²175 mg/m²
D = mg
4.Dosage calculationsHard
A patient is 149 cm tall and weighs 48 kg. The prescription is 50 mg/m². Using the Mosteller formula BSA = √(height (cm) × weight (kg)/3600), find the dose D in mg.
149 cm48 kg50 mg/m²
D = mg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 78 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.03 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 10 mg/L?
78 kgCL = 0.03 L/kg/hCss = 10 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 3 h. The target is a peak of 8 mg/L and a trough of 1 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 = 8 mg/LCtrough = 1 mg/L
τ = h
3Solutions
1.Pharmacy calculations
Answer: V = 16 mL
Reconstitution: the powder takes up space, so the final volume is the diluent plus the powder volume. Concentration = drug in the vial/final volume, and the volume to give = dose/concentration. On a multi-strength label, use the concentration that belongs to the amount of diluent you actually added. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Final volume
19.2 + 0.8 = 20 mL
Concentration
2000/20 = 100 mg/mL
Volume to give
V = 1600/100 = 16 mL
Answer
V = 16 mL
2.Pharmacy calculations
Answer: m = 10 g
Alligation: to mix a higher strength H and a lower strength L into a target strength T, take (T − L) parts of H and (H − T) parts of L. The parts are crossed: each stock gets the distance from the target to the other stock. Water or a plain base counts as 0%. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Alligation parts
20%: 15 − 0 = 15 parts
0%: 20 − 15 = 5 parts
Scale to the amount you have
m = 30·5/15 = 10 g
Answer
m = 10 g
Check: 30 g + 10 g = 40 g of 15% ointment.
3.Dosage calculations
Answer: D = 141.8 mg
Body-surface-area dosing: many cytotoxic and some paediatric doses are prescribed per square metre of body surface area: dose (mg) = dose per m² × BSA (m²). The Mosteller formula estimates BSA = √(height (cm) × weight (kg)/3600).
Dose
D = 175·0.81 = 141.8 mg
Answer
D = 141.8 mg
4.Dosage calculations
Answer: D = 70.47 mg
Body-surface-area dosing: many cytotoxic and some paediatric doses are prescribed per square metre of body surface area: dose (mg) = dose per m² × BSA (m²). The Mosteller formula estimates BSA = √(height (cm) × weight (kg)/3600).
Body surface area
BSA = √(149·48/3600) = √1.987 = 1.409 m²
Dose
D = 50·1.409 = 70.47 mg
Answer
D = 70.47 mg
5.Clinical pharmacokinetics
Answer: R = 29.25 mg/h
Clearance and volume: at steady state the dosing rate equals the elimination rate, so the maintenance rate = CL × Css. A loading dose is LD = Vd × C/(S × F), with S the salt factor (fraction of active drug in the salt) and F the bioavailability (1 for IV). From two levels, k = ln(C₁/C₂)/Δt; for an IV bolus Vd = dose/C₀ and CL = k × Vd. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Clearance
CL = 0.03·78 = 2.34 L/h
Infusion rate
R = CL·Css/S = 2.34·10/0.8 = 29.25 mg/h
Answer
R = 29.25 mg/h
6.Clinical pharmacokinetics
Answer: τ = 9 h
Designing a regimen: with rapid IV doses the level falls from peak to trough during the interval, so τ = ln(Cpeak/Ctrough)/k and the dose that restores the peak is Vd × (Cpeak − Ctrough). The average steady-state level is Css,avg = S·F·dose/(CL·τ). An infusion rises as C = (R/CL)(1 − e−kt). For a drug cleared by the kidneys, the dose is scaled with the creatinine clearance. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Rate constant
k = 0.693/3 = 0.231 h⁻¹
Dosing interval
τ = ln(8/1)/0.231 = 2.079/0.231 = 9 h
Answer
τ = 9 h
In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.