Notice that the main skill required for these calculations is the appropriate use of units. Make sure that at all stages, you use the correct units for all values and that calculations do not use mixed units (e.g. mg and μg or mL and L). It is useful to remember that …

- 1000 ng = 1 μg
- 1000 μg = 1mg
- 1μg/L = 1 ng/mL
- 1 mg/L = 1µg/mL

**Question 1)** A drug has a desirable concentration range of 150-250 µg/L. Its mean volume of distribution is 0.72 L/Kg. What bolus i.v. dose should be administered to a patient weighing 65 Kg? (Answer in units of mg)

**Question 2)** A drug has a mean volume of distribution of 0.91 L/Kg and a desirable concentration range of 20-40pg/L. A 5mg bolus i.v. dose of this drug is to be administered to a patient weighing 80Kg. Is this dose likely to be satisfactory?

**Question 3)** An initial concentration of 12.5 ng/mL arises following the bolus i.v. administration of 200 lig of a drug. Calculate the volume of distribution of this drug in this patient (in units of L).

**Question 4)** A dose (5mg) of a salt that contains 70% by weight of the parent drug is administered by bolus i.v. injection. The volume of distribution of the drug is 70 L. Calculate the initial concentration of the drug (in units of ng/ mL).

**Question 5)** 0.5 mg of a drug is administered by bolus i.v. injection. The initial concentration is 2Ong,/mL. Calculate the volume of distribution of the drug (in units of L).

**Solution**

** **

- Target concentration = midrange value = 200 μg/L

Predicted V = 0.72 L/Kg x 65 Kg = 46.8 L

D = C x V = 200 μg/L x 46.8L

= 9,360 μg = 9.4 mg

- Predicted V = 0.91L/Kg x 80Kg = 72.8L

C = D/V = 5mg / 72.8L = 0.069 mg/L = 69 μg/L (Remember ling = 1000 μg)

Unsatisfactory – the dose is excessive.

- V = D / C = 200 μg / 12.5 ng/mL = 200 μg / 12.5 μg/L = 16L (Mixed mass units need to be adjusted)

- C = (D x S) / V = (5mg x 0.7) / 70L = 3.5mg / 70L = 0.05mg/L = 0.05 μg/mL = 50 ng/mL

(Note Salt factor is included to reduce the calculated concentration)

- V = D/C = 0.5mg / 20 ng/mL = 500 μg / 20 μg/L = 25 L

(Mixed mass units to be adjusted)