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Isotonicity & E-Value Practice Problems

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Solutions placed in the eye, nose, or bloodstream should be isotonic with body fluids (equivalent to 0.9% NaCl). The sodium chloride equivalent (E-value) tells you how much NaCl produces the same osmotic effect as 1 g of a given drug, which is what you need before you can work out how much NaCl to add.

Formula(1) drug in formulation = %×vol ÷ 100   (2) NaCl equivalent = drug g × E   (3) NaCl for volume = 0.009 × vol   (4) NaCl to add = step 3 − step 2
Worked example

Make 30 mL of 1% pilocarpine HCl (E = 0.24) isotonic. (1) drug = 0.3 g; (2) 0.3 × 0.24 = 0.072 g; (3) 0.009 × 30 = 0.27 g; (4) 0.27 − 0.072 = 0.198 g.

The exam gives you the E-value. The skill is the 4-step setup, not memorizing E-values.

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Isotonicity (E-values)
How many grams of NaCl must be added to make 30 mL of a 1% pilocarpine HCl solution isotonic? (E-value of pilocarpine HCl = 0.24) Round your answer to three decimal places.
g

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How to set up an isotonicity problem

A solution going into the eye, the nose, or a vein should exert the same osmotic pressure as body fluids, which for pharmacy purposes means the same as 0.9% sodium chloride. A drug dissolved in water already contributes some osmotic pressure. The sodium chloride equivalent, or E-value, says how much: it is the grams of NaCl that have the same osmotic effect as 1 gram of the drug. From it you can work out how much NaCl the drug has already "used up" and how much you still need to add.

1. drug in the formula = % × volume  →  2. NaCl equivalent = drug g × E  →  3. NaCl to make the volume isotonic alone = 0.009 × volume  →  4. NaCl to add = step 3 − step 2

The exam gives the E-value. The skill is the four-step setup, and the discipline is keeping every intermediate to full precision, because the final answer is a small number of grams and rounding at step 2 is enough to move its last digit.

Worked example 1

How many grams of sodium chloride are needed to make 30 mL of a 2% pilocarpine HCl solution isotonic? (E = 0.24)

drug = 2% × 30 = 0.6 g

NaCl equivalent of the drug = 0.6 × 0.24 = 0.144 g

NaCl for the volume alone = 0.009 × 30 = 0.27 g

NaCl to add = 0.27 − 0.144 = 0.126 g

Worked example 2

How many grams of sodium chloride are needed to make 15 mL of 2% lidocaine HCl isotonic? (E = 0.22)

drug = 2% × 15 = 0.3 g

NaCl equivalent of the drug = 0.3 × 0.22 = 0.066 g

NaCl for the volume alone = 0.009 × 15 = 0.135 g

NaCl to add = 0.135 − 0.066 = 0.069 g

When the answer comes out negative

If step 2 is bigger than step 3, the drug alone already makes the solution hypertonic, and no amount of NaCl fixes that. The E-value method only tells you how to raise tonicity. A 10% phenylephrine solution, for example, is hypertonic on its own. A question with that setup is testing whether you recognize it, and the answer is "none; the solution is already hypertonic," not a negative number of grams.

Using a different tonicity agent

Sometimes the agent to be added isn't sodium chloride, for example dextrose or boric acid, because NaCl is incompatible with the drug. Work steps 1 to 4 exactly as above to get the grams of NaCl that would be needed, then divide by the E-value of the substitute agent to convert. If the substitute's E-value is 0.18, then 0.126 g of NaCl is equivalent to 0.126 ÷ 0.18 = 0.7 g of the substitute. The question supplies that second E-value too.

Where students lose the points

10 Isotonicity & E-Value practice problems with worked solutions

Every problem below is worked start to finish. Try it yourself first, then check the steps. For a fresh problem every time, use the practice box above. It generates unlimited variations of these same question types.

Problem 1: Isotonicity (E-values)

How many grams of NaCl must be added to make 15 mL of a 0.5% ephedrine sulfate solution isotonic? (E-value of ephedrine sulfate = 0.23) Round your answer to three decimal places.

Answer 0.118 g

Solution
  1. Step 1: drug in the formulation: 15 mL × 0.5% = 0.075 g.
  2. Step 2: NaCl equivalent of the drug: 0.075 g × E(0.23) = 0.01725 g.
  3. Step 3: NaCl to make 15 mL isotonic on its own: 15 mL × 0.009 g/mL = 0.135 g.
  4. Step 4: NaCl to add = 0.135 − 0.01725 = 0.11775 g.
  5. Rounded as the question asks: 0.118 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 2: Isotonicity (E-values)

How many grams of NaCl must be added to make 30 mL of a 2% lidocaine HCl solution isotonic? (E-value of lidocaine HCl = 0.22) Round your answer to three decimal places.

Answer 0.138 g

Solution
  1. Step 1: drug in the formulation: 30 mL × 2% = 0.6 g.
  2. Step 2: NaCl equivalent of the drug: 0.6 g × E(0.22) = 0.132 g.
  3. Step 3: NaCl to make 30 mL isotonic on its own: 30 mL × 0.009 g/mL = 0.27 g.
  4. Step 4: NaCl to add = 0.27 − 0.132 = 0.138 g.
  5. Rounded as the question asks: 0.138 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 3: Isotonicity (E-values)

How many grams of NaCl must be added to make 30 mL of a 0.25% phenylephrine HCl solution isotonic? (E-value of phenylephrine HCl = 0.32) Round your answer to three decimal places.

Answer 0.246 g

Solution
  1. Step 1: drug in the formulation: 30 mL × 0.25% = 0.075 g.
  2. Step 2: NaCl equivalent of the drug: 0.075 g × E(0.32) = 0.024 g.
  3. Step 3: NaCl to make 30 mL isotonic on its own: 30 mL × 0.009 g/mL = 0.27 g.
  4. Step 4: NaCl to add = 0.27 − 0.024 = 0.246 g.
  5. Rounded as the question asks: 0.246 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 4: Isotonicity (E-values)

How many grams of NaCl must be added to make 10 mL of a 0.5% lidocaine HCl solution isotonic? (E-value of lidocaine HCl = 0.22) Round your answer to three decimal places.

Answer 0.079 g

Solution
  1. Step 1: drug in the formulation: 10 mL × 0.5% = 0.05 g.
  2. Step 2: NaCl equivalent of the drug: 0.05 g × E(0.22) = 0.011 g.
  3. Step 3: NaCl to make 10 mL isotonic on its own: 10 mL × 0.009 g/mL = 0.09 g.
  4. Step 4: NaCl to add = 0.09 − 0.011 = 0.079 g.
  5. Rounded as the question asks: 0.079 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 5: Isotonicity (E-values)

How many grams of NaCl must be added to make 30 mL of a 3% ephedrine sulfate solution isotonic? (E-value of ephedrine sulfate = 0.23) Round your answer to three decimal places.

Answer 0.063 g

Solution
  1. Step 1: drug in the formulation: 30 mL × 3% = 0.9 g.
  2. Step 2: NaCl equivalent of the drug: 0.9 g × E(0.23) = 0.207 g.
  3. Step 3: NaCl to make 30 mL isotonic on its own: 30 mL × 0.009 g/mL = 0.27 g.
  4. Step 4: NaCl to add = 0.27 − 0.207 = 0.063 g.
  5. Rounded as the question asks: 0.063 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 6: Isotonicity (E-values)

How many grams of NaCl must be added to make 10 mL of a 1% lidocaine HCl solution isotonic? (E-value of lidocaine HCl = 0.22) Round your answer to three decimal places.

Answer 0.068 g

Solution
  1. Step 1: drug in the formulation: 10 mL × 1% = 0.1 g.
  2. Step 2: NaCl equivalent of the drug: 0.1 g × E(0.22) = 0.022 g.
  3. Step 3: NaCl to make 10 mL isotonic on its own: 10 mL × 0.009 g/mL = 0.09 g.
  4. Step 4: NaCl to add = 0.09 − 0.022 = 0.068 g.
  5. Rounded as the question asks: 0.068 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 7: Isotonicity (E-values)

How many grams of NaCl must be added to make 50 mL of a 0.5% lidocaine HCl solution isotonic? (E-value of lidocaine HCl = 0.22) Round your answer to three decimal places.

Answer 0.395 g

Solution
  1. Step 1: drug in the formulation: 50 mL × 0.5% = 0.25 g.
  2. Step 2: NaCl equivalent of the drug: 0.25 g × E(0.22) = 0.055 g.
  3. Step 3: NaCl to make 50 mL isotonic on its own: 50 mL × 0.009 g/mL = 0.45 g.
  4. Step 4: NaCl to add = 0.45 − 0.055 = 0.395 g.
  5. Rounded as the question asks: 0.395 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 8: Isotonicity (E-values)

How many grams of NaCl must be added to make 30 mL of a 0.5% pilocarpine HCl solution isotonic? (E-value of pilocarpine HCl = 0.24) Round your answer to three decimal places.

Answer 0.234 g

Solution
  1. Step 1: drug in the formulation: 30 mL × 0.5% = 0.15 g.
  2. Step 2: NaCl equivalent of the drug: 0.15 g × E(0.24) = 0.036 g.
  3. Step 3: NaCl to make 30 mL isotonic on its own: 30 mL × 0.009 g/mL = 0.27 g.
  4. Step 4: NaCl to add = 0.27 − 0.036 = 0.234 g.
  5. Rounded as the question asks: 0.234 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 9: Isotonicity (E-values)

How many grams of NaCl must be added to make 5 mL of a 2% atropine sulfate solution isotonic? (E-value of atropine sulfate = 0.13) Round your answer to three decimal places.

Answer 0.032 g

Solution
  1. Step 1: drug in the formulation: 5 mL × 2% = 0.1 g.
  2. Step 2: NaCl equivalent of the drug: 0.1 g × E(0.13) = 0.013 g.
  3. Step 3: NaCl to make 5 mL isotonic on its own: 5 mL × 0.009 g/mL = 0.045 g.
  4. Step 4: NaCl to add = 0.045 − 0.013 = 0.032 g.
  5. Rounded as the question asks: 0.032 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Problem 10: Isotonicity (E-values)

How many grams of NaCl must be added to make 2 mL of a 1% tetracaine HCl solution isotonic? (E-value of tetracaine HCl = 0.18) Round your answer to three decimal places.

Answer 0.014 g

Solution
  1. Step 1: drug in the formulation: 2 mL × 1% = 0.02 g.
  2. Step 2: NaCl equivalent of the drug: 0.02 g × E(0.18) = 0.0036 g.
  3. Step 3: NaCl to make 2 mL isotonic on its own: 2 mL × 0.009 g/mL = 0.018 g.
  4. Step 4: NaCl to add = 0.018 − 0.0036 = 0.0144 g.
  5. Rounded as the question asks: 0.014 g.

E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value. The skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.

Sources and how far they go

Where a convention on this page comes from a published standard, the standard is named. Where it comes from a textbook or from ordinary practice instead, the entry says so. That difference matters when a preceptor or an exam disagrees with you.

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