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 1:10 000. You need 0.78 mg. Draw up the volume in the 10 mL syringe: what volume V is that?
1:10 0000.78 mg
V = mL
2.Pharmacy calculationsHard
Practice exercise (made-up values, not clinical guidance). You must prepare 300 mL of 3% hydrogen peroxide from a 30% stock and purified water. How many mL V of water do you need? Round to the nearest whole mL.
30%0%3%300 mL
V = mL
3.Dosage calculationsMedium
A child weighing 14 kg is prescribed 20 mg/kg of a drug. The stock solution contains 50 mg/mL. Draw up the dose: what volume V do you give?
14 kg20 mg/kg50 mg/mL
V = mL
4.Dosage calculationsHard
The prescription is 410 units of a drug that comes as 1000 units/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?
410 units1000 units/mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 8 mg/L at 1 h and 2.5 mg/L at 6 h. Assume first-order elimination. Find the half-life t½.
C1 = 8 mg/L (1 h)C2 = 2.5 mg/L (6 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 60 kg patient has a digoxin level of 0.6 mcg/L; the target is 1 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?
60 kgVd = 7 L/kgC = 0.6 → 1 mcg/LF = 0.7
LD = mcg
3Solutions
1.Pharmacy calculations
Answer: V = 7.8 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/10 000 mL = 0.1 mg/mL
Volume
V = 0.78/0.1 = 7.8 mL
Answer
V = 7.8 mL
2.Pharmacy calculations
Answer: V = 270 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
30%: 3 − 0 = 3 parts
0%: 30 − 3 = 27 parts
total: 30 parts
Water
V = 300·27/30 = 270 mL
Answer
V = 270 mL
Check: 30 mL of 30% contains 9 g of drug, which is 3% of 300 mL.
3.Dosage calculations
Answer: V = 5.6 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 = 20·14 = 280 mg
Volume
V = 280/50 = 5.6 mL
On the syringe
On this 10 mL syringe every small line is 0.2 mL.
Answer
V = 5.6 mL
4.Dosage calculations
Answer: V = 0.41 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 = 410/1000 = 0.41 mL
On the syringe
On this 1 mL syringe every small line is 0.01 mL.
Answer
V = 0.41 mL
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(8/2.5)/5 = 1.163/5 = 0.2326 h⁻¹
Half-life
t½ = 0.693/0.2326 = 2.98 h
Answer
t½ = 2.98 h
6.Clinical pharmacokinetics
Answer: LD = 240 mcg
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 = 7·60 = 420 L
Rise needed
1 − 0.6 = 0.4 mcg/L
Loading dose
LD = 420·0.4/(1·0.7) = 240 mcg
Answer
LD = 240 mcg
In practice the dose is rounded to a convenient amount.