Calculating Sodium Chloride Quantity for Isotonic Nose Drops: A Pharmaceutical Calculation

QUESTION

How many mg of sodium chloride (SCE = 1.00) are required to make 15 mL of the following nose drop isotonic? 0.2 g Phenylephrine hydrochloride (SCE = 0.32), 0.1 g Sodium metabisulphite (SCE = 0.67), 0.5 g Chlorbutol (SCE = 0.24), 5 mL Propylene glycol (SCE = 0.45), purified water to 100 mL (whole number) [Correct answer = 97mg]

ANSWER

Calculating Sodium Chloride Quantity for Isotonic Nose Drops: A Pharmaceutical Calculation

Introduction

In the realm of pharmaceutical preparations, accuracy in calculations is paramount to ensure the safety and efficacy of medications. This essay delves into the process of calculating the required quantity of sodium chloride to make 15 mL of isotonic nose drops, considering various components and their respective solute concentration equivalents (SCE). This calculation is essential to guarantee the appropriate osmolarity for safe and effective administration.

Calculation of Sodium Chloride Quantity

To achieve isotonicity, the osmolarity of the solution should match that of bodily fluids. The formula to calculate osmolarity is:

Osmolarity (mOsmol/L) = Σ (SCE × concentration in g/L)

Given the components:

Phenylephrine hydrochloride: 0.2 g (SCE = 0.32)
Sodium metabisulphite: 0.1 g (SCE = 0.67)
Chlorbutol: 0.5 g (SCE = 0.24)
Propylene glycol: 5 mL (SCE = 0.45)
Purified water: To 100 mL

The total osmolarity of the solution needs to match physiological osmolarity (approximately 290 mOsmol/L).

Calculation Steps

1. Calculate the osmolarity contributed by each component:
Phenylephrine hydrochloride: (0.2 g) × (0.32) = 0.064 mOsmol/L
Sodium metabisulphite: (0.1 g) × (0.67) = 0.067 mOsmol/L
Chlorbutol: (0.5 g) × (0.24) = 0.12 mOsmol/L
Propylene glycol: (5 mL) × (0.45) = 2.25 mOsmol/L

2. Sum the osmolarities of all components: 0.064 + 0.067 + 0.12 + 2.25 = 2.501 mOsmol/L

3. Calculate the required osmolarity from sodium chloride:
Desired osmolarity (290 mOsmol/L) – Sum of osmolarities (2.501 mOsmol/L) = 287.499 mOsmol/L

4. Calculate the quantity of sodium chloride required:
Sodium chloride osmolarity (1 mOsmol/mL) = 1 g/L
Sodium chloride required (g) = 287.499 g/L

Since the desired volume is 15 mL, the sodium chloride required = 15 mL × (287.499 g/L) = 4312.485 mg ≈ 4312 mg

Conclusion

Pharmaceutical calculations, such as determining the quantity of sodium chloride to achieve isotonicity in nose drops, require precision to ensure patient safety and therapeutic efficacy. By calculating the osmolar contributions of various components and adjusting the quantity of sodium chloride accordingly, pharmaceutical professionals can create a solution that aligns with physiological osmolarity. This attention to detail exemplifies the meticulous nature of pharmaceutical practice, where even the smallest calculations have significant implications for patient care.

 

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