Section 2.14: Pharmaceutical Calculations: Osmolarity, Milliequivalents, and Alligations

Pharmaceutical Calculations: Osmolarity, Milliequivalents, and Alligations

Advanced pharmaceutical calculations are essential for compounding isotonic formulations, determining electrolyte balances in Total Parenteral Nutrition (TPN), and mixing bulk solutions of varying strengths. This section details the principles of osmolarity, milliequivalent chemical activity, sodium chloride equivalent ($E$-value) tonicity adjustments, and alligation grid calculations.

Osmolarity versus Osmolality

Osmotic pressure dictates the movement of water across biological membranes. Injectable and ophthalmic formulations must be formulated to minimize tissue damage, pain, or hemolysis of red blood cells:

• Osmolarity: The concentration of osmotically active particles per Liter of solution (expressed in $\text{mOsmol/L}$).
• Osmolality: The concentration of osmotically active particles per Kilogram of solvent (expressed in $\text{mOsmol/kg}$).

In clinical practice, body fluids have an osmolality of approximately $285\text{ to }300\text{ mOsmol/kg}$. Formulations that deviate significantly from this range are:

• Hypotonic ($<280\text{ mOsmol/L}$): Cause water to enter cells, potentially leading to cell swelling and hemolysis.
• Hypertonic ($>310\text{ mOsmol/L}$): Cause water to leave cells, leading to crenation. Hypertonic infusions must be administered through a central venous catheter rather than a peripheral line to prevent thrombophlebitis.

Osmolarity Calculation Equation

$$\text{mOsmol/L} = \frac{\text{Weight of solute (g/L)}}{\text{Molecular Weight (g/mol)}} \cdot \text{Dissociation Particles } (i) \cdot 1000$$

where $i$ is the number of species formed upon dissociation (e.g., Dextrose $i = 1$; $\text{NaCl } i = 2$; $\text{CaCl}_2\text{ } i = 3$).

Milliequivalents (mEq) Dosing

Milliequivalents express the chemical activity of electrolytes. It accounts for both the molar concentration and the electrical valence of the ions:

$$\text{mEq} = \frac{\text{Weight (mg)} \cdot \text{Valence}}{\text{Molecular Weight}}$$ $$\text{mEq} = \text{mmol} \cdot \text{Valence}$$

Physiologically, electrolytes are measured in milliequivalents because it represents charge matching. For instance, in TPN formulation, balancing the positive charges of cations ($Na^+$, $K^+$, $Mg^{2+}$, $Ca^{2+}$) with the negative charges of anions ($Cl^-$, acetate, phosphate) is essential to maintain electrical neutrality and prevent precipitation of salts.

Calcium Dosing: Gluconate versus Chloride

Clinical pharmacists must distinguish between Calcium Gluconate and Calcium Chloride:

• Calcium Gluconate: Preferred for peripheral administration because it is less irritating to veins and carries a lower risk of tissue necrosis if extravasation occurs. A $1\text{ g}$ vial contains $93\text{ mg}$ ($4.65\text{ mEq}$) of elemental calcium.
• Calcium Chloride: More irritating and typically reserved for central line administration in emergency resuscitation. A $1\text{ g}$ vial contains $273\text{ mg}$ ($13.6\text{ mEq}$) of elemental calcium (almost three times the amount of elemental calcium as gluconate).

Tonicity Adjustment and Sodium Chloride Equivalent ($E$-value)

An isotonic solution contains osmotic activity equivalent to a $0.9\%\text{ w/v}$ Sodium Chloride solution ($0.9\text{ g/100 mL}$). The Sodium Chloride Equivalent ($E$-value) of a drug is the mass of sodium chloride (in grams) that has the same osmotic effect as $1\text{ gram}$ of the drug.

$$\text{E-value} \approx 58.5 \cdot \frac{i_{\text{drug}}}{\text{MW}_{\text{drug}} \cdot 1.8}$$

where $1.8$ accounts for the dissociation of sodium chloride (which is $90\%$ dissociated in solution, so $i = 1.8$).

Five-Step Method to Calculate Tonicity Adjustments:

1. Calculate the mass (in grams) of the drug in the formulation.
2. Multiply the mass of the drug by its $E$-value to determine the sodium chloride equivalent mass of the drug.
3. Calculate the total mass of sodium chloride needed to make the volume of the formulation isotonic in the absence of any drug ($\text{Volume (mL)} \cdot 0.009\text{ g/mL}$).
4. Subtract the drug's sodium chloride equivalent mass (Step 2) from the total sodium chloride required (Step 3). This represents the mass of sodium chloride to add.
5. If a tonicity agent other than sodium chloride (e.g., dextrose) is used, divide the mass of sodium chloride to add (Step 4) by the $E$-value of that agent.

The Alligation Alternate Method

Alligation alternate is a rapid grid method used to find the ratio of two starting strengths (high $H$ and low $L$) required to make an intermediate target strength ($T$):

• Parts of High = $|T - L|$ (placed adjacent to High)
• Parts of Low = $|H - T|$ (placed adjacent to Low) Convert the parts into total volume or weight based on requirements.

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Step-by-Step Worked Clinical Calculations

Example 1: Calculating Theoretical Osmolarity

Calculate the theoretical osmolarity of a $0.9\%\text{ w/v}$ Sodium Chloride solution ($\text{MW} = 58.5\text{ g/mol}$). Assume sodium chloride dissociates completely into two particles ($i = 2$).

1. Express the concentration in grams per Liter: $$0.9\%\text{ w/v} = 9.0\text{ g/L}$$
2. Apply the osmolarity equation: $$\text{mOsmol/L} = \frac{9.0\text{ g/L}}{58.5\text{ g/mol}} \cdot 2 \cdot 1000 \approx 308\text{ mOsmol/L}$$

Example 2: Electrolyte Milliequivalent Calculation

How many milliequivalents of calcium are present in a $10\text{ mL}$ ampule of a $10\%\text{ w/v}$ Calcium Gluconate injection ($\text{MW} = 430\text{ g/mol}$, valence of calcium = $2$)?

1. Calculate the mass of calcium gluconate in the ampule: $$\text{Mass} = 10\%\text{ w/v} \cdot 10\text{ mL} = 1\text{ g} = 1000\text{ mg}$$
2. Calculate milliequivalents: $$\text{mEq} = \frac{1000\text{ mg} \cdot 2}{430\text{ g/mol}} \approx 4.65\text{ mEq}$$

Example 3: Alligation Alternate Calculation

A pharmacist needs to prepare $500\text{ mL}$ of a $20\%$ dextrose solution by mixing a $50\%$ dextrose solution and sterile water ($0\%$ dextrose). Calculate the volume of each component required.

1. Set up alligation parts:
   • Parts of $50\%$ dextrose = $|20 - 0| = 20\text{ parts}$
   • Parts of water = $|50 - 20| = 30\text{ parts}$
   • Total parts = $20 + 30 = 50\text{ parts}$
2. Calculate the required volumes: $$\text{Volume of 50\% dextrose} = \frac{20\text{ parts}}{50\text{ total parts}} \cdot 500\text{ mL} = 200\text{ mL}$$ $$\text{Volume of water} = \frac{30\text{ parts}}{50\text{ total parts}} \cdot 500\text{ mL} = 300\text{ mL}$$