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
A solution has a ratio strength of 1:1000. You need 0.44 mg. Draw up the volume in the 1 mL syringe: what volume V is that?
1:10000.44 mg
V = mL
2.Pharmacy calculationsHard
Practice exercise (made-up values, not clinical guidance). You must prepare 600 mL of 4% chlorhexidine from a 20% stock and purified water. How many mL V of water do you need? Round to the nearest whole mL.
20%0%4%600 mL
V = mL
3.Dosage calculationsMedium
A child weighing 18 kg is prescribed 2 mg/kg of a drug. The stock solution contains 10 mg/mL. Draw up the dose: what volume V do you give?
18 kg2 mg/kg10 mg/mL
V = mL
4.Dosage calculationsHard
A patient is 161 cm tall and weighs 51 kg. The prescription is 25 mg/m². Using the Mosteller formula BSA = √(height (cm) × weight (kg)/3600), find the dose D in mg.
161 cm51 kg25 mg/m²
D = mg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 9 mg/L at 2 h and 1.4 mg/L at 10 h. Assume first-order elimination. Find the half-life t½.
C1 = 9 mg/L (2 h)C2 = 1.4 mg/L (10 h)
t½ = h
6.Clinical pharmacokineticsHard
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 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 = 2.5 hCpeak = 6 mg/LCtrough = 1 mg/L
τ = h
3Solutions
1.Pharmacy calculations
Answer: V = 0.44 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.
Strength in mg/mL
1000 mg/1000 mL = 1 mg/mL
Volume
V = 0.44/1 = 0.44 mL
Answer
V = 0.44 mL
2.Pharmacy calculations
Answer: V = 480 mL
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%: 4 − 0 = 4 parts
0%: 20 − 4 = 16 parts
total: 20 parts
Water
V = 600·16/20 = 480 mL
Answer
V = 480 mL
Check: 120 mL of 20% contains 24 g of drug, which is 4% of 600 mL.
3.Dosage calculations
Answer: V = 3.6 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 = 2·18 = 36 mg
Volume
V = 36/10 = 3.6 mL
On the syringe
On this 10 mL syringe every small line is 0.2 mL.
Answer
V = 3.6 mL
4.Dosage calculations
Answer: D = 37.76 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 = √(161·51/3600) = √2.281 = 1.51 m²
Dose
D = 25·1.51 = 37.76 mg
Answer
D = 37.76 mg
5.Clinical pharmacokinetics
Answer: t½ = 2.98 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.
Elimination rate constant
k = ln(9/1.4)/8 = 1.861/8 = 0.2326 h⁻¹
Half-life
t½ = 0.693/0.2326 = 2.98 h
Answer
t½ = 2.98 h
6.Clinical pharmacokinetics
Answer: τ = 6.5 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 = 0.693/2.5 = 0.2773 h⁻¹
Dosing interval
τ = ln(6/1)/0.2773 = 1.792/0.2773 = 6.462 ≈ 6.5 h
Answer
τ = 6.5 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.