A pharmacist receives a special request from an ophthalmologist to prepare a fortified tobramycin ophthalmic solution.
The available solution contains tobramycin, 3 mg/mL.

How many milliliters of a tobramycin injection containing 40 mg/mL must be aseptically added to a 5-mL container of the ophthalmic solution to prepare one 0.5% in concentration?

Respuesta :

Answer:

2/7 ml

Step-by-step explanation:

When we have a solution expressed in Q mg/ml, the concentration is (Q/10)%.

For example, a solution of 3 mg/ml is a 3/10=0.3% concentration solution, a solution of 40 mg/ml is a 40/10=4% concentration solution.

Why?

Q mg/ml means that 1 ml of solution has Q mg of the drug or active component. So, 100 ml have 100*Q mg.

But 100*Q mg = Q/10 grams.

If a solution has Q/10 grams in 100 ml, then it is a (Q/10)% concentration.

Now, we have 5 ml of a solution that contains tobramycin, 3 mg/ml (each ml has 3 mg of tobramycin).

This means that the 5 ml contain 15 mg of  tobramycin.

On the other hand, we have another solution S containing 40 mg/ml of tobramycin. This means that each ml of S contains 40 mg of tobramycin.

Let x be the amount of ml of S we have to add to our first 5 ml.

If 1 ml of S contains 40 mg of  tobramycin, x ml contain 40*x mg.

If we add this x ml to our first 5 ml, we have 5+x ml containing 15+40*x mg of  tobramycin.  

Now, cross-multiply to get how much tobramycin contains 1 ml of this new solution.

5+x ml ____contain____15+40*x mg of  tobramycin

   1 ml ____contains____? mg

And we have

[tex]\large \frac{5+x}{1}=\frac{15+40x}{?}\Rightarrow ?=\frac{15+40x}{5+x}[/tex]

as a result, we have a solution of  

[tex]\large \frac{15+40x}{5+x}\;mg/ml[/tex]

We want to prepare a solution 0.5% in concentration.

Using what we state at the beginning, we need to find a value x such that

[tex]\large \frac{(15+40x)/(5+x)}{10}=0.5[/tex]

solving for x

[tex]\large \frac{(15+40x)/(5+x)}{10}=0.5\Rightarrow \frac{15+40x}{5+x}=5\Rightarrow 15+40x=25+5x\Rightarrow\\\\\Rightarrow 40x-5x=25-15\Rightarrow 35x=10\Rightarrow x=10/35=2/7 \;ml[/tex]

So, we have to add just 2/7 ml of the stronger solution.