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 0.5 mg. Draw up the volume in the 10 mL syringe: what volume V is that?
1:10 0000.5 mg
V = mL
2.Pharmacy calculationsHard
Practice exercise (made-up values, not clinical guidance). You must prepare 500 mL of 8% 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%8%500 mL
V = mL
3.Dosage calculationsMedium
A patient weighs 30 kg. The prescribed dose is 25 mg/kg per day, given in 4 equal doses. How many milligrams D are given per dose?
30 kg25 mg/kg/day4× per day
D = mg
4.Dosage calculationsHard
You must prepare 500 mL of a 5% (w/v) solution from a 10% (w/v) stock solution. What volume of stock V₁ do you need?
C₁ = 10%C₂ = 5%V₂ = 500 mL
V₁ = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 8 mg/L at 2 h and 2.5 mg/L at 7 h. Assume first-order elimination. Find the half-life t½.
C1 = 8 mg/L (2 h)C2 = 2.5 mg/L (7 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). An 81-year-old woman weighs 94 kg and has a serum creatinine of 2.4 mg/dL (212 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual vancomycin regimen is 1000 mg every 12 h for a creatinine clearance of 100 mL/min. Keeping the dose, the interval is lengthened in proportion to the fall in clearance. Find the new dosing interval τ. Round to one decimal place.
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 = 0.5/0.1 = 5 mL
Answer
V = 5 mL
2.Pharmacy calculations
Answer: V = 125 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%: 8 − 3 = 5 parts
3%: 23 − 8 = 15 parts
total: 20 parts
Share the volume
V = 500·5/20 = 125 mL
Answer
V = 125 mL
Check: 125 mL of 23% + 375 mL of 3% = 500 mL.
3.Dosage calculations
Answer: D = 187.5 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
25·30 = 750 mg
Per dose
D = 750/4 = 187.5 mg
Answer
D = 187.5 mg
4.Dosage calculations
Answer: V₁ = 250 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 = 5·500/10 = 250 mL
Answer
V₁ = 250 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: τ = 44 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.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 81)·94/(72·2.4)·0.85 = 27.28 mL/min
New interval
τ = 12·100/27.28 = 43.99 ≈ 44 h
Answer
τ = 44 h
In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.