Question

Your veterinarian is administering a sedative to your 50 pound dog. The sedative is mixed in saline solution. Unfortunately the solution is pre-mixed at 10% for a horse and it needs to be a 2.5% concentration to be administered at 0.7 ml per pound for your dog. How much saline water needs to be added to the current solution to reduce the concentration to 2.5% and have the correct dosage for the dog

Answer 1

Answer:

26.25 mL

Explanation:

This is a dilution problem. First, let us calculate the volume of final solution needed:

The dog weighs 50 pounds and the sedative is administered at 0/7 ml per pound. Hence:

50 x 0.7 = 35 mL

A total volume of 35 mL, 2.5% solution of the sedative will be needed.

But 10% solution is available. There needs to be a dilution with saline water, but what volume of the 10% solution would be diluted?

initial volume = ?

final volume = 35 mL

initial concentration = 10%

final concentration = 2.5%

Using dilution equation:

initial concentration x initial volume = final concentration x final volume

initial volume = $\frac{final concentration*final volume}{initial concentration}$

                     = 2.5 x 35/10 = 8.75 mL

Hence, 8.75 mL of the 10% pre-mixed sedative will be required.

But 35 mL is needed? The 8.75 mL is marked up to 35 mL with saline water.

35 - 8.75 = 26.25 mL

Therefore, 26.25 mL of saline water will be added to 8.75 mL of the 10% pre-mixed sedative to give 2.5%, 35 mL needed for the dog.