Compounding 500 mL of 0.90% Saline from 23.4% Stock
A nurse on a pediatric floor needs 500 mL of 0.90% w/v sodium chloride irrigation, but the unit stocks only 23.4% w/v sodium chloride concentrate and sterile water. The pharmacist approves the dilution if the calculation is documented. The nurse must also confirm that the final solution is about 0.154 M NaCl, using a molar mass of 58.44 g/mol. Two syringes are available: a 10 mL syringe graduated in 0.2 mL increments and a 50 mL syringe graduated in 1 mL increments. A colleague suggests simply adding 19 mL of the concentrate to 500 mL of sterile water and shaking it, arguing that the difference is too small to matter in irrigation fluid. The nurse has to write the calculation, verify the molarity, and justify both the measuring device and the mixing procedure before signing off.
1. Calculate the volume of 23.4% stock needed, showing the dilution relationship and a sanity check.
Read the worked answer
Use the dilution relationship, which states that the amount of solute is unchanged: C₁V₁ = C₂V₂, with both concentrations in the same units (% w/v here).
(23.4)(V₁) = (0.90)(500 mL) V₁ = 450 ÷ 23.4 = 19.23 mL, so measure 19.2 mL of concentrate.
Ratio sanity check: 23.4% is 26 times more concentrated than 0.90% (23.4 ÷ 0.90 = 26). The volume needed is therefore 500 mL ÷ 26 = 19.2 mL. A dilution from a much stronger stock always requires far less stock than final volume; an answer larger than 500 mL would immediately signal an inverted ratio.
Critical definition: V₂ is the final total volume, not the volume of solvent. The 19.2 mL of concentrate is transferred to a container, and sterile water is added up to the 500 mL mark. That distinction is the entire difference between a correct product and the colleague's incorrect method.
2. Verify that 0.90% w/v NaCl corresponds to about 0.154 M, and state the assumptions behind the conversion.
Read the worked answer
0.90% w/v means 0.90 g of NaCl per 100 mL, which is 9.0 g per liter.
Moles per liter: 9.0 g ÷ 58.44 g/mol = 0.154 mol, so the concentration is 0.154 M.
Check on the actual batch: 500 mL contains 0.90 g/100 mL × 500 mL = 4.50 g; 4.50 ÷ 58.44 = 0.0770 mol; 0.0770 mol ÷ 0.500 L = 0.154 M. The same value results because molarity is intensive and does not depend on batch size.
Assumptions to state: the label is % w/v (grams per 100 mL of solution), not % w/w (grams per 100 g of solution); the dilution is volumetric, so final volume is brought to the mark rather than assumed additive; and the molar mass used is for anhydrous NaCl. For the dilute final product, w/v and w/w are close, but the 23.4% concentrate is dense enough that the distinction is real and belongs in the documentation.
3. Evaluate the colleague's suggested procedure, then choose the syringe and describe the correct sequence.
Read the worked answer
The colleague's method produces the wrong concentration. Adding 19 mL of concentrate to 500 mL of water gives a total volume of roughly 519 mL. The solute mass is 23.4 g/100 mL × 19 mL = 4.45 g, so the final concentration is 4.45 ÷ 519 × 100 = 0.86% - about 5% below target. The error is systematic, always low, and it comes from adding solvent to a fixed volume instead of diluting to a fixed volume.
Correct sequence: measure 19.2 mL of concentrate, transfer it to a graduated container, add sterile water to the 500 mL calibration mark, and mix by inversion.
Syringe choice: 19.2 mL cannot be measured well with the 50 mL syringe. A 1 mL graduation gives roughly ±0.5 mL uncertainty, about 2.6% of the drawn volume. The 10 mL syringe graduated in 0.2 mL allows the volume to be assembled as 10.0 mL plus 9.2 mL, keeping uncertainty near ±0.2 mL, about 1%, consistent with the three-significant-figure calculation. Choose the 10 mL syringe with two draws and verify final volume in a graduated cylinder.