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 cefuroxime 750 mg as a powder. The practice label says to add 7.2 mL of sterile water; the powder adds 0.3 mL to the volume. What is the concentration C in mg/mL after reconstitution?
750 mg7.2 mL+0.3 mL
C = mg/mL
2.Pharmacy calculationsHard
A vial holds 250 mg of powder. You add 2 mL of diluent, and the powder adds 0.5 mL of volume. The prescription is 160 mg. Draw up the dose: what volume V do you give?
250 mg2 mL + 0.5 mL160 mg
V = mL
3.Dosage calculationsMedium
A child weighing 24 kg is prescribed 7.5 mg/kg of a drug. The stock solution contains 25 mg/mL. Draw up the dose: what volume V do you give?
24 kg7.5 mg/kg25 mg/mL
V = mL
4.Dosage calculationsHard
A patient weighs 26.4 lb. The prescription is 2 mg/kg as a single dose, and the stock contains 200 mg in 5 mL. Using 1 kg = 2.2 lb, what volume V do you give?
26.4 lb2 mg/kg200 mg/5 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 56 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.
56 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 65-year-old man weighs 75 kg and has a serum creatinine of 3.1 mg/dL (274 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual gentamicin regimen is 80 mg every 8 h for a creatinine clearance of 120 mL/min. Keeping the dose, the interval is lengthened in proportion to the fall in clearance. Find the new dosing interval τ. Round to one decimal place.
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
7.2 + 0.3 = 7.5 mL
Concentration
C = 750/7.5 = 100 mg/mL
Answer
C = 100 mg/mL
2.Pharmacy calculations
Answer: V = 1.6 mL
Strength to volume: turn the strength into mg per mL first. A w/v percentage is grams per 100 mL, so x% = 10x mg/mL. A ratio strength 1:n is 1 g in n mL. After reconstitution the concentration is the powder's dose over the final volume (diluent + powder volume). Then volume = dose/concentration.
Final volume and concentration
2 + 0.5 = 2.5 mL
C = 250/2.5 = 100 mg/mL
Volume
V = 160/100 = 1.6 mL
Answer
V = 1.6 mL
3.Dosage calculations
Answer: V = 7.2 mL
Drawing up a dose: volume = (prescribed dose/stock strength) × stock volume. Convert to the same units first, then check that the volume fits the syringe and lands on one of its graduations.
Dose from body weight
D = 7.5·24 = 180 mg
Volume
V = 180/25 = 7.2 mL
On the syringe
On this 10 mL syringe every small line is 0.2 mL.
Answer
V = 7.2 mL
4.Dosage calculations
Answer: V = 0.6 mL
Dose calculation: the amount to give equals what is prescribed divided by what you have, times the quantity it comes in: amount = (desired dose/stock strength) × stock volume. Always keep the units the same (convert g to mg, mg to mcg first) and check that the answer is sensible.
Weight in kg
26.4/2.2 = 12 kg
Dose
2·12 = 24 mg
Volume
V = (24/200)·5 = 0.6 mL
Answer
V = 0.6 mL
5.Clinical pharmacokinetics
Answer: LD = 852.2 mg
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.
Volume of distribution
Vd = 0.7·56 = 39.2 L
Amount of active drug needed
Vd·C = 39.2·20 = 784 mg
Loading dose
LD = 784/(0.92·1) = 852.2 mg
Answer
LD = 852.2 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: τ = 38.1 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.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 65)·75/(72·3.1) = 25.2 mL/min
New interval
τ = 8·120/25.2 = 38.09 ≈ 38.1 h
Answer
τ = 38.1 h
In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.