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 0.67 mg. Draw up the volume in the 1 mL syringe: what volume V is that?
1:10000.67 mg
V = mL
2.Pharmacy calculationsHard
Practice exercise (made-up values, not clinical guidance). You have 50 g of 5% coal tar ointment. How many grams m of 1% coal tar ointment must you add to make a 2% ointment? Round to the nearest whole gram.
50 g · 5%1%2%
m = g
3.Dosage calculationsMedium
A child weighing 30 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?
30 kg7.5 mg/kg25 mg/mL
V = mL
4.Dosage calculationsHard
You must prepare 100 mL of a 5% (w/v) solution from a 50% (w/v) stock solution. How much water W must be added to the stock?
C₁ = 50%C₂ = 5%V₂ = 100 mL
W = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). An 83 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.
83 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A theophylline infusion runs at 24 mg/h with no loading dose. The patient's clearance is 2 L/h and the half-life is 8 h. What is the level C after 6 h?
R = 24 mg/hCL = 2 L/ht½ = 8 ht = 6 h
C = mg/L
3Solutions
1.Pharmacy calculations
Answer: V = 0.67 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 = 0.67/1 = 0.67 mL
Answer
V = 0.67 mL
2.Pharmacy calculations
Answer: m = 150 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
5%: 2 − 1 = 1 parts
1%: 5 − 2 = 3 parts
Scale to the amount you have
m = 50·3/1 = 150 g
Answer
m = 150 g
Check: 50 g + 150 g = 200 g of 2% ointment.
3.Dosage calculations
Answer: V = 9 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 = 7.5·30 = 225 mg
Volume
V = 225/25 = 9 mL
On the syringe
On this 10 mL syringe every small line is 0.2 mL.
Answer
V = 9 mL
4.Dosage calculations
Answer: W = 90 mL
Solution strengths: a percentage strength is grams per 100 mL, so 1% = 1 g/100 mL = 10 mg/mL. A ratio strength 1:1000 means 1 g in 1000 mL = 1 mg/mL; 1:10 000 is 0.1 mg/mL. For a dilution the amount of drug stays the same: C₁·V₁ = C₂·V₂.
Volume of stock
V₁ = 5·100/50 = 10 mL
Water to add
W = 100 − 10 = 90 mL
Answer
W = 90 mL
5.Clinical pharmacokinetics
Answer: LD = 947.3 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 = 0.7·83 = 58.1 L
Amount of active drug needed
Vd·C = 58.1·15 = 871.5 mg
Loading dose
LD = 871.5/(0.92·1) = 947.3 mg
Answer
LD = 947.3 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: C = 4.865 mg/L
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.