How many milliliters of a 5% solution of boric acid should be used in compounding a given prescription? (Details below)

Question

Oxymetazoline hydrochloride 1/2%

Boric acid solution qs

Purified water ad 15 mL

Make isotonic solution

Sig. for the nose, as decongestant

Please provide step-by-step rationale, and emphasize how one can tell which formula would work for this problem.

Answer 1

To determine how many milliliters of a 5% solution of boric acid should be used in compounding the given prescription, we need to make an isotonic solution. An isotonic solution has the same osmotic pressure as bodily fluids, typically equivalent to a 0.9% NaCl (sodium chloride) solution.

Step-by-Step Rationale

1. Determine the isotonicity requirement:
2. We need to make the final solution isotonic, which is equivalent to 0.9% NaCl.
3. Identify the components and their concentrations:
4. Oxymetazoline hydrochloride: 0.5% (1/2%)
5. Boric acid solution: quantity sufficient (qs) to make the solution isotonic
6. Purified water ad: up to 15 mL total volume
7. Calculate the contribution of oxymetazoline hydrochloride to isotonicity:
8. Oxymetazoline hydrochloride (O) does not directly contribute to the tonicity like NaCl or boric acid. Instead, we will assume that it adds minimal tonicity and focus on the boric acid to achieve the isotonic solution.
9. Determine the required amount of NaCl for isotonicity:
10. Isotonic solution = 0.9% NaCl
11. For 15 mL of solution:
12. 0.9% of 15 mL = 0.135 grams of NaCl
13. Convert the NaCl requirement to boric acid equivalent:
14. Boric acid has an E-value (equivalent value) which indicates how much boric acid is needed to match the tonicity of NaCl.
15. The E-value of boric acid is approximately 0.52 (meaning 1 gram of boric acid is equivalent to 0.52 grams of NaCl).
16. Calculate the amount of boric acid needed:
17. Required NaCl equivalent = 0.135 grams
18. Boric acid needed = 0.135 grams / 0.52
19. Boric acid needed ≈ 0.260 grams
20. Calculate the volume of 5% boric acid solution needed:
21. 5% boric acid solution means 5 grams of boric acid in 100 mL of solution.
22. To find out how many milliliters of the 5% solution contains 0.260 grams of boric acid:
23. (5 grams / 100 mL) = 0.05 grams per mL
24. Volume required = 0.260 grams / 0.05 grams per mL
25. Volume required = 5.2 mL

Final Answer

The pharmacist should use 5.2 milliliters of the 5% solution of boric acid to compound the given prescription and make it isotonic. This ensures the final 15 mL solution is isotonic and suitable for nasal use as a decongestant.