Isotonic Solutions

Tonicity and osmosis

Body cells are surrounded by a thin membrane that lets water pass through but not most dissolved substances. Water moves across this membrane from a weaker solution to a stronger one. This movement is called osmosis.

The strength of a solution in terms of osmosis is its tonicity.

A solution of 0.9% w/v sodium chloride (normal saline) is isotonic with blood plasma. About 5% glucose is also roughly isotonic. Blood plasma freezes at about −0.52°C, and any solution that freezes at that temperature is isotonic with it.

Why tonicity matters

Solutions that are put into delicate tissues must be close to isotonic:

• Injections that are far from isotonic can damage blood cells, cause pain and harm tissue.
• Eye drops that are not isotonic sting and can irritate or damage the eye.
• Nasal drops can irritate the lining of the nose and damage the tiny hairs (cilia) that clear it.

Oral medicines and solutions for intact skin do not need to be isotonic.

Sodium chloride equivalents

Different drugs affect osmotic pressure by different amounts. To compare them, each drug is given a sodium chloride equivalent, written E.

The sodium chloride equivalent (E) of a drug is the weight of sodium chloride that has the same osmotic effect as 1 g of the drug.

Some typical values:

Values differ slightly between reference books. In an exam or in practice, use the value from the table you are given.

Making a solution isotonic

1. Work out the sodium chloride needed for the volume to be isotonic: 0.9 g × (volume in mL ÷ 100)
2. Work out the sodium chloride equivalent of the drug: weight of drug × E
3. Subtract: NaCl to add = (step 1) − (step 2)

Then dissolve the drug and the sodium chloride and make up to volume with water.

Example 1. Make 100 mL of 1% w/v atropine sulfate eye drops isotonic. E = 0.13.

1. NaCl for 100 mL = 0.9 g
2. Drug = 1 g, and 1 × 0.13 = 0.13 g NaCl equivalent
3. NaCl to add = 0.9 − 0.13 = 0.77 g

Example 2. Make 30 mL of 2% w/v pilocarpine nitrate eye drops isotonic. E = 0.23.

1. NaCl for 30 mL = 0.9 × 30 ÷ 100 = 0.27 g
2. Drug = 2 × 30 ÷ 100 = 0.6 g, and 0.6 × 0.23 = 0.138 g NaCl equivalent
3. NaCl to add = 0.27 − 0.138 = 0.132 g

Example 3. Make 60 mL of 1% w/v ephedrine hydrochloride nasal drops isotonic. E = 0.30.

1. NaCl for 60 mL = 0.9 × 60 ÷ 100 = 0.54 g
2. Drug = 0.6 g, and 0.6 × 0.30 = 0.18 g NaCl equivalent
3. NaCl to add = 0.54 − 0.18 = 0.36 g

Example 4: two drugs. Make 100 mL containing 0.5% atropine sulfate (E = 0.13) and 1% pilocarpine nitrate (E = 0.23) isotonic.

1. NaCl for 100 mL = 0.9 g
2. Atropine sulfate: 0.5 × 0.13 = 0.065 g. Pilocarpine nitrate: 1 × 0.23 = 0.23 g. Total = 0.295 g
3. NaCl to add = 0.9 − 0.295 = 0.605 g

If the answer is zero or negative

If the drug alone gives 0.9 g or more of sodium chloride equivalent per 100 mL, the solution is already isotonic or hypertonic. No sodium chloride is added. Refer a hypertonic result to the pharmacist.

The strength that is isotonic by itself

A drug solution is isotonic by itself at:

isotonic strength (% w/v) = 0.9 ÷ E

Example. Boric acid, E = 0.50. 0.9 ÷ 0.50 = 1.8% w/v.

Using another substance instead of sodium chloride

Sometimes a different substance is used to adjust tonicity. Divide the sodium chloride weight you need by the E of the new substance.

Example. In Example 1, 0.77 g of sodium chloride was needed. How much boric acid (E = 0.50) gives the same effect?

• 0.77 ÷ 0.50 = 1.54 g of boric acid

Good practice

• Eye drops, nasal drops and injections must also be sterile, with the right preparation method and containers. Follow the pharmacist's instructions.
• Use purified water (or water for injections for injections) to make up to volume.
• Check the E value against the reference table, and check the working with the pharmacist.