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 0.15 mg. Draw up the volume in the 1 mL syringe: what volume V is that?
1:10000.15 mg
V = mL
2.Pharmacy calculationsHard
Practice exercise (made-up values, not clinical guidance). You have 150 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.
150 g · 20%0%15%
m = g
3.Dosage calculationsMedium
A patient weighs 70 kg. The prescribed dose is 2 mg/kg per day, given in 4 equal doses. How many milligrams D are given per dose?
70 kg2 mg/kg/day4× per day
D = mg
4.Dosage calculationsHard
A child weighing 20 kg is prescribed 15 mg/kg per dose of an oral suspension that contains 100 mg in 5 mL. What volume V do you give?
20 kg15 mg/kg100 mg/5 mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 96 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 15 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.
96 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 73 kg patient takes oral aminophylline (S = 0.8, F = 1) every 12 h. The theophylline clearance is 0.04 L/kg/h. What dose D in mg gives an average steady-state level of 12 mg/L?
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 = 0.15/1 = 0.15 mL
Answer
V = 0.15 mL
2.Pharmacy calculations
Answer: m = 50 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 = 150·5/15 = 50 g
Answer
m = 50 g
Check: 150 g + 50 g = 200 g of 15% ointment.
3.Dosage calculations
Answer: D = 35 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.
Total per day
2·70 = 140 mg
Per dose
D = 140/4 = 35 mg
Answer
D = 35 mg
4.Dosage calculations
Answer: V = 15 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.
Dose from weight
15·20 = 300 mg
Volume
V = (300/100)·5 = 15 mL
Answer
V = 15 mL
5.Clinical pharmacokinetics
Answer: LD = 1096 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 = 0.7·96 = 67.2 L
Amount of active drug needed
Vd·C = 67.2·15 = 1008 mg
Loading dose
LD = 1008/(0.92·1) = 1096 mg
Answer
LD = 1096 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: D = 525.6 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.
Clearance
CL = 0.04·73 = 2.92 L/h
Dose
D = Css,avg·CL·τ/(S·F) = 12·2.92·12/(0.8·1) = 525.6 mg