Module 7 of 14 · 15 min
Electrolytes & isotonicity
Potassium, magnesium, calcium, phosphate and bicarbonate are ordered in mEq or mmol but supplied as salts in mg or % — a mix-up can be fatal. Exams also ask for osmolarity and the sodium chloride needed to make eye drops isotonic.
🧠 Weight ÷ MW = moles; moles × charge = equivalents; moles × pieces = osmoles.
📌 Memorize these
- $\text{mmol}$
- $\text{mg}\div\text{MW}$use the MW of the whole salt, incl. water of hydration
- Na⁺, K⁺, Cl⁻, HCO₃⁻
- $1\,\text{mmol} = 1\,\text{mEq}$monovalent
- Ca²⁺, Mg²⁺
- $1\,\text{mmol} = 2\,\text{mEq}$divalent
- Particles
- NaCl → 2 · CaCl₂ → 3 · dextrose → 1
- $0.9\%$ NaCl
- $154\,\text{mmol/L}$, $\approx 308\,\text{mOsm/L}$ideal; measured ≈ 286
- Isotonic
- $0.9\,\text{g NaCl}$ per $100\,\text{mL}$
A millimole (mmol) counts particles: mg ÷ molecular weight (MW, in mg/mmol). A milliequivalent (mEq) counts charge: mmol × valence. For monovalent ions (Na⁺, K⁺, Cl⁻, HCO₃⁻) mEq = mmol; for divalent ions (Ca²⁺, Mg²⁺) 1 mmol = 2 mEq. For a salt, use the valence of the ion you are asked about, and the MW of the WHOLE salt as supplied — including any water of hydration (CaCl₂·2H₂O, MgSO₄·7H₂O).
Osmolarity counts every dissolved particle: mOsm = mmol × number of particles the compound dissociates into (NaCl → 2, CaCl₂ → 3, dextrose → 1). These ideal values overestimate slightly — measured 0.9% NaCl is about 286 mOsm/L, not 308 — but exams use the ideal calculation unless a dissociation factor is given. Body fluids are about 280–300 mOsm/L.
Isotonic means the same tonicity as 0.9% NaCl. The sodium chloride equivalent (E value) of a drug is the grams of NaCl that 1 g of drug is osmotically equal to. To make a solution isotonic: NaCl needed $= 0.9\%\times V - (\text{g drug}\times E)$.
If a different tonicity agent is used (e.g. boric acid), divide the NaCl amount by that agent's own E value to get grams of the substitute.
🧭 The method
- 1Get the mass of salt in mg (from % w/v × volume, or as given).
- 2mmol = mg ÷ MW of the salt as supplied (check for hydrates).
- 3mEq = mmol × valence of the ion asked about; mOsm = mmol × particles.
- 4Isotonicity: NaCl to add = 0.009 × volume (mL) − (g drug × E).
- 5Convert back to mL of product (mEq ÷ mEq/mL) when a volume is needed.
Before each reveal, predict the next step — then check.
How many mEq of potassium are in 10 mL of a 14.9% w/v potassium chloride solution? (KCl MW 74.5)
- 1
Mass of KCl in the volume.
$$10\,\text{mL}\times\frac{14.9\,\text{g}}{100\,\text{mL}} = 1.49\,\text{g} = 1490\,\text{mg}$$ Convert mg to mmol with the molecular weight.
$$1490\,\cancel{\text{mg}}\times\frac{1\,\text{mmol}}{74.5\,\cancel{\text{mg}}} = 20\,\text{mmol KCl}$$Each KCl gives one K⁺, valence 1.
$$20\,\text{mmol}\times 1 = 20\,\text{mEq K}^+$$Sense check: this is why '2 mEq/mL KCl concentrate' and 14.9% are the same product.
$$\frac{20\,\text{mEq}}{10\,\text{mL}} = 2\,\text{mEq/mL}$$
Same method — the last steps are yours. Work on paper, then reveal.
Magnesium sulfate injection is 50% w/v of MgSO₄·7H₂O (MW 246.5). How many mEq of magnesium per mL?
- 1
mg of salt per mL.
$$\frac{50\,\text{g}}{100\,\text{mL}} = 500\,\text{mg/mL}$$ - 2
Your turn
Now you: mmol per mL — use the heptahydrate MW.
Work it out on paper, then reveal.
$$\frac{500\,\text{mg}}{246.5\,\text{mg/mmol}} = 2.03\,\text{mmol/mL}$$ - 3
Your turn
Now you: Mg²⁺ is divalent.
Work it out on paper, then reveal.
$$2.03\,\text{mmol}\times 2 = 4.06\,\text{mEq/mL}$$
Type your number. Your first try decides when this module comes back for review.
How many mEq of sodium are in 1 L of 0.9% w/v sodium chloride? (NaCl MW 58.5)
0.9% = 9 g per litre.
= 153.8 mEq
How many mEq of bicarbonate are in a 50 mL syringe of 8.4% w/v sodium bicarbonate? (NaHCO₃ MW 84)
% → g → mg → mmol; HCO₃⁻ is monovalent.
= 50 mEq
What is the ideal osmolarity of 0.9% w/v sodium chloride, in mOsm/L? (NaCl MW 58.5, assume complete dissociation)
NaCl dissociates into 2 particles: Na⁺ and Cl⁻.
= 307.7 mOsm/L
How many mg of calcium chloride dihydrate (CaCl₂·2H₂O, MW 147) provide 10 mmol of calcium?
The question is in mmol, not mEq — don't use the valence here.
= 1470 mg
A patient needs 15 mmol of phosphate. Potassium phosphates injection contains 3 mmol phosphate and 4.4 mEq potassium per mL. How many mEq of potassium will the patient also receive?
Find the volume from the phosphate content first; the potassium comes along with it.
= 22 mEq
How many mg of sodium chloride are needed to make 30 mL of a 1% w/v ephedrine sulfate solution isotonic? (E value of ephedrine sulfate = 0.23)
NaCl for 30 mL of plain isotonic solution, minus the NaCl-equivalent already supplied by the drug.
= 201 mg
⚠️ Tricky
How many mEq of magnesium are in 2 mL of magnesium sulfate 50% w/v injection? (MgSO₄·7H₂O, MW 246.5)
50% w/v = 500 mg/mL. Magnesium is Mg²⁺.
= 8.11 mEq
⚠️ The trap: Forgetting that Mg²⁺ is divalent gives 4.06 mEq — half the true amount.
⚠️ Tricky
An order calls for 16 mEq of magnesium as magnesium sulfate. The injection is made from the heptahydrate. How many grams of MgSO₄·7H₂O is that? (MW anhydrous MgSO₄ 120.4; MW MgSO₄·7H₂O 246.5)
mEq → mmol (Mg²⁺), then multiply by the MW of the form actually in the vial.
= 1.97 g
⚠️ The trap: Using the anhydrous MW (120.4) gives 0.96 g.
⚠️ Tricky
How many mmol of chloride are in 10 mL of 10% w/v calcium chloride injection? (CaCl₂·2H₂O, MW 147)
Find mmol of the salt first — then count how many Cl⁻ each formula unit releases.
= 13.61 mmol
⚠️ The trap: Stopping at the mmol of the salt gives 6.80 mmol — half the chloride.
⚠️ Tricky
What is the ideal osmolarity of dextrose 5% / sodium chloride 0.45% with 20 mEq/L potassium chloride? (Dextrose MW 180, NaCl MW 58.5, KCl MW 74.5; assume complete dissociation)
Per litre: grams → mmol for dextrose and NaCl; KCl is already given in mEq. Then count particles for each.
= 471.6 mOsm/L
⚠️ The trap: Counting each salt as one particle (no dissociation) gives 374.7 mOsm/L.
⚠️ Tricky
Rx: 60 mL of a 2% w/v ophthalmic solution of a drug (MW 300, E value 0.16), made isotonic with sodium chloride. You add the sodium chloride as 23.4% w/v concentrate. How many mL of concentrate are needed?
First find the grams of drug actually in 60 mL; the MW is not needed.
= 1.49 mL
⚠️ The trap: Taking '2%' as 2 g of drug in the 60 mL gives 0.22 g NaCl and only 0.94 mL of concentrate.
🕵️ Spot the mistake
How many mEq of calcium are in 10 mL of 10% w/v calcium chloride dihydrate (CaCl₂·2H₂O, MW 147)?
A student wrote
⚠️ Traps examiners set
- Forgetting valence for Ca²⁺ and Mg²⁺ (1 mmol = 2 mEq).
- Using the anhydrous MW for a hydrate (MgSO₄·7H₂O is 246.5, not 120.4).
- Counting particles wrong in osmolarity (NaCl = 2, CaCl₂ = 3, dextrose = 1).
- Isotonicity: forgetting to subtract the drug's own NaCl-equivalent.
- Ignoring the second ion delivered by a salt (K⁺ with potassium phosphate, Na⁺ with sodium bicarbonate).
💬 Explain it in your own words
Why does 1 mmol of calcium carry 2 mEq while 1 mmol of sodium carries 1 — and why does osmolarity care about the number of particles but not their charge?