Common Registration Assessment, Part 1

Sodium chloride equivalents and isotonic solutions

Practise using a stated sodium chloride equivalent to work out how much tonicity adjustor to add, with three original worked questions.

Published 10 October 2026. Independent revision material, not GPhC questions, not for patient care.

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A sodium chloride equivalent, often written as E, tells you the weight of sodium chloride that has the same effect on tonicity as 1 g of the drug. The question gives you E; you never have to look it up.

An isotonic solution matches 0.9% w/v sodium chloride. The method is to work out how much sodium chloride the drug represents, subtract that from the 0.9% target, and add the difference.

The examples are fictional and practise the arithmetic only.

The method

  1. Work out the weight of drug in the volume, from its percentage strength.
  2. Multiply by E to find the sodium chloride it represents.
  3. Work out the target: 0.9 g of sodium chloride per 100 mL of the preparation.
  4. Subtract the drug contribution from the target.
  5. Add the difference as sodium chloride or another stated adjustor.

Three practice questions

Work each one on paper first, then open the answer. All drugs and patients are fictional.

Question 1

A fictional 30 mL eye preparation contains 1.5% w/v of a drug with a sodium chloride equivalent of 0.16. How much sodium chloride must be added to make it isotonic?

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Answer: 0.198 g

Working: Drug weight = 1.5 g per 100 mL, so 0.45 g in 30 mL. NaCl represented = 0.45 x 0.16 = 0.072 g. Target = 0.9 g per 100 mL, so 0.27 g in 30 mL. Add 0.27 - 0.072 = 0.198 g.

Question 2

A fictional solution is 0.5% w/v of a drug with E = 0.2. How much sodium chloride is needed per 100 mL to make it isotonic?

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Answer: 0.8 g

Working: NaCl represented = 0.5 x 0.2 = 0.1 g per 100 mL. Target = 0.9 g per 100 mL. Add 0.9 - 0.1 = 0.8 g.

Question 3

A fictional 50 mL preparation contains 1% w/v of drug A (E = 0.18) and 0.5% w/v of drug B (E = 0.22). How much sodium chloride makes it isotonic?

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Answer: 0.305 g

Working: Drug A = 0.5 g in 50 mL, representing 0.5 x 0.18 = 0.09 g NaCl. Drug B = 0.25 g, representing 0.25 x 0.22 = 0.055 g. Together 0.145 g. Target in 50 mL = 0.45 g. Add 0.45 - 0.145 = 0.305 g.

Keep the two amounts separate

Write the drug contribution and the 0.9% target on two lines before subtracting. Most errors here come from mixing them up or forgetting to scale the target to the volume.

Where marks are lost

  • Applying E to the percentage instead of the actual weight in the volume.
  • Using 0.9 g for any volume rather than scaling it to the volume in the question.
  • Adding the drug contribution instead of subtracting it from the target.

Frequently asked questions

Where does the E value come from?

The question supplies it, or it comes from a reference table you are given. It is a measured property of the drug, not something you calculate in the exam.

Why is 0.9% the target?

A 0.9% w/v sodium chloride solution has the same tonicity as body fluids, so preparations adjusted to it are comfortable on sensitive tissues such as the eye.

Sources

The questions above are original and use fictional drugs. PreRegExamPrep is not affiliated with or endorsed by the General Pharmaceutical Council.

Practise until the method is automatic

Try 15 free questions with worked answers. No sign-up required.

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More calculation topics

See also the formula sheet, the eight sample questions and the approved calculator page.