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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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10 Isotonicity & E-Value practice problems with worked solutions

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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.

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