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 oxacillin 2000 mg as a powder. The practice label says to add 11.5 mL of sterile water; the powder adds 1 mL to the volume. What is the concentration C in mg/mL after reconstitution?
2000 mg11.5 mL+1 mL
C = mg/mL
2.Pharmacy calculationsHard
Practice exercise (made-up values, not clinical guidance). You must prepare 500 mL of 5% sodium chloride by mixing 23% and 3% sodium chloride. How many mL V of the 23% solution do you need? Round to the nearest whole mL.
23%3%5%500 mL
V = mL
3.Dosage calculationsMedium
A patient weighs 80 kg and is prescribed 10 mg/kg as a single dose. Find the dose D in mg.
80 kg10 mg/kg
D = mg
4.Dosage calculationsHard
The prescription is 2000 units of a drug that comes as 5000 units/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?
2000 units5000 units/mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 63 kg patient receives a theophylline infusion. The clearance is 0.045 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 15 mg/L?
63 kgCL = 0.045 L/kg/hCss = 15 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 93 kg patient has a theophylline level of 4.5 mg/L; the target is 15 mg/L. The drug is given intravenously as aminophylline (S = 0.8), and the volume of distribution is 0.5 L/kg. What extra loading dose LD raises the level to the target?
93 kgVd = 0.5 L/kgC = 4.5 → 15 mg/LS = 0.8
LD = mg
3Solutions
1.Pharmacy calculations
Answer: C = 160 mg/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
11.5 + 1 = 12.5 mL
Concentration
C = 2000/12.5 = 160 mg/mL
Answer
C = 160 mg/mL
2.Pharmacy calculations
Answer: V = 50 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
23%: 5 − 3 = 2 parts
3%: 23 − 5 = 18 parts
total: 20 parts
Share the volume
V = 500·2/20 = 50 mL
Answer
V = 50 mL
Check: 50 mL of 23% + 450 mL of 3% = 500 mL.
3.Dosage calculations
Answer: D = 800 mg
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.
Dose
10·80 = 800 mg
Answer
D = 800 mg
4.Dosage calculations
Answer: V = 0.4 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.
Volume
V = 2000/5000 = 0.4 mL
On the syringe
On this 1 mL syringe every small line is 0.01 mL.
Answer
V = 0.4 mL
5.Clinical pharmacokinetics
Answer: R = 42.52 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 = 0.045·63 = 2.835 L/h
Infusion rate
R = 2.835·15 = 42.52 mg/h
Answer
R = 42.52 mg/h
6.Clinical pharmacokinetics
Answer: LD = 610.3 mg
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.
Volume of distribution
Vd = 0.5·93 = 46.5 L
Rise needed
15 − 4.5 = 10.5 mg/L
Loading dose
LD = 46.5·10.5/(0.8·1) = 610.3 mg
Answer
LD = 610.3 mg
In practice the dose is rounded to a convenient amount.