11.3 Compounding Calculations, Isotonicity & Milliequivalents

Compounding Calculations, Isotonicity & Milliequivalents

This is the most technical corner of the calculations syllabus. The questions are formulaic, so learn the four formulas below and the values that go with them, and these marks are reliable.

Milliequivalents and millimoles

mEq = (mg × valence) ÷ atomic or molecular weight. Equivalently, 1 mEq of a substance weighs (molecular weight ÷ valence) milligrams.

Worked example 1 (KCl): How many milligrams of potassium chloride provide 20 mEq of potassium? MW of KCl = 74.5, valence = 1.

1. Weight of 1 mEq = 74.5 ÷ 1 = 74.5 mg
2. 20 mEq × 74.5 mg = 1,490 mg (1.49 g) of KCl

Worked example 2 (NaCl): How many mEq of sodium are in 1 g of sodium chloride? MW of NaCl = 58.5, valence = 1.

1. mEq = 1,000 mg × 1 ÷ 58.5 = 17.1 mEq of Na⁺ (and 17.1 mEq of Cl⁻)

mmol vs mEq: millimoles count particles; milliequivalents count charge. For monovalent ions (Na⁺, K⁺, Cl⁻) 1 mmol = 1 mEq. For divalent ions (Ca²⁺, Mg²⁺, SO₄²⁻) 1 mmol = 2 mEq. Confusing the two for a divalent ion halves or doubles the answer — a favourite distractor.

Osmolarity basics: osmolarity ≈ molarity × number of particles per formula unit. Sodium chloride 0.9% w/v contains 9 g/L ÷ 58.5 = 0.154 mol/L = 154 mmol/L, and since NaCl splits into 2 ions, its osmolarity is 154 × 2 = 308 mOsm/L — roughly isotonic with plasma (≈ 285–310 mOsm/L).

Isotonicity: freezing-point depression method

Blood and tears freeze at −0.52°C, so an isotonic solution must depress the freezing point by ΔTf = 0.52°C. The drug already contributes some depression; the tonicity agent only needs to close the gap. Sodium chloride 1% w/v depresses the freezing point by 0.58°C.

Worked example 3: Render 30 mL of atropine sulfate 1% eye drops isotonic using NaCl. ΔTf of atropine sulfate 1% = 0.08°C.

1. Gap to close: 0.52 − 0.08 = 0.44°C
2. NaCl concentration needed: 0.44 ÷ 0.58 = 0.759% w/v
3. For 30 mL: 0.759 × 30 ÷ 100 = 0.228 → 0.23 g of NaCl

Isotonicity: sodium chloride equivalent (E-value) method

The E value is the weight of NaCl with the same osmotic effect as 1 g of the drug. For atropine sulfate, E = 0.13.

Worked example 4 (same prescription, second method):

1. Drug present: 1% of 30 mL = 0.3 g atropine sulfate
2. Its NaCl equivalent: 0.3 × 0.13 = 0.039 g
3. NaCl needed for isotonicity if no drug present: 0.9% × 30 mL = 0.27 g
4. NaCl to add: 0.27 − 0.039 = 0.231 → 0.23 g

Same answer as the freezing-point method — as it must be. Pick whichever method the question's data supports.

Buffers and pH in extemporaneous formulation

Eye drops and liquid orals are buffered so the pH sits near the drug's stability optimum and physiological comfort (tears ≈ pH 7.4). The governing relationship is the Henderson–Hasselbalch equation: for a weak acid buffer, pH = pKa + log([salt] ÷ [acid]). An acetate buffer (pKa 4.76) with equal salt and acid gives pH 4.76; a 10:1 salt-to-acid ratio raises it to pH 5.76. The exam point is conceptual: equal parts → pH = pKa; tenfold ratio change → one pH unit.

Displacement volumes for suppositories

A suppository mould holds a fixed volume. Adding drug displaces base, so you must subtract what the drug displaces using the displacement value (DV) — the grams of drug that displace 1 g of base (the Pharmaceutical Codex convention). Because the DV sits in the denominator, base displaced = drug weight ÷ DV; multiplying instead of dividing is the single most common error on these items. Calibrate first: pour blank (base-only) suppositories to find the mould capacity.

Worked example 5: Prepare 10 suppositories each containing 300 mg of a drug with DV = 1.5, in a mould calibrated at 2 g of base per blank suppository.

1. Total base capacity: 10 × 2 g = 20 g
2. Total drug: 10 × 300 mg = 3 g
3. Base displaced: 3 g ÷ 1.5 = 2 g
4. Base to weigh: 20 − 2 = 18 g

Sanity-check the direction: a DV above 1 means the drug is denser than the base, so it must displace less base than its own weight. An answer where the displaced base exceeds the drug weight (here, 4.5 g displaced by 3 g of drug) is arithmetically impossible for DV > 1.

In real practice add an overage (prepare for 11–12 moulds) to cover pouring losses; if a question asks for overage, apply the same arithmetic to the larger number.

Potency correction for raw materials

Raw materials are rarely 100% pure. Corrected weight = required weight ÷ potency (as a decimal).

Worked example 6: A formula needs 500 mg of an active ingredient supplied at 95% potency. Weigh 500 ÷ 0.95 = 526.3 mg. If the material is instead labelled on a dried basis or as a salt (e.g. 83% of the salt is active base), divide by that fraction the same way.

Error traps

• Valence errors: using valence 1 for Ca²⁺ or Mg²⁺ doubles the mEq answer.
• Ignoring the drug's contribution: in isotonicity questions, subtracting nothing for the drug gives 0.27 g instead of 0.23 g in the atropine example.
• Mixing methods' constants: 0.52°C belongs to freezing point; 0.9% belongs to NaCl equivalence. Do not cross them.
• Percentage strength slips: 1% w/v means 1 g per 100 mL — in 30 mL that is 0.3 g, not 3 g.