Normal Saline Batch Failure and Paediatric Isotonic Solution
A 600-bed hospital pharmacy prepares 1 L batches of 0.90% w/v sodium chloride (M = 58.44 g mol⁻¹) using a Class A volumetric flask and a balance readable to 0.001 g. The specification is 290 mOsm kg⁻¹, but the in-house freezing-point osmometer reads 255 mOsm kg⁻¹ for the latest batch. The NaCl stock bottle was opened three weeks earlier in a humid store at 80% relative humidity, and the volumetric flask was calibrated at 20 °C while the room is at 30 °C. The pharmacist must decide whether to discard the batch, add more NaCl, or dilute it. She must also prepare 250 mL of an isotonic solution containing 1.0% w/v glucose plus NaCl for a paediatric patient. Kf(water) = 1.86 K kg mol⁻¹; the van't Hoff factor for NaCl at this concentration is 1.87. No certified osmometer standard is available that day.
1. Calculate the expected osmolality and freezing point of a correctly prepared 0.90% w/v NaCl solution. From the measured 255 mOsm kg⁻¹, calculate the actual NaCl concentration and decide whether the error is more likely to be a weighing/stock problem or a volumetric-flask temperature effect.
Read the worked answer
0.90% w/v = 9.0 g L⁻¹, so
[C = \frac{9.0}{58.44} = 0.154\ \text{M}.]
With (i = 1.87), osmolality (= 1.87\times154 = 288) mOsm kg⁻¹, close to the 290 specification. The freezing-point depression is
[\Delta T_f = iK_fm \approx 1.87\times1.86\times0.154 = 0.536\ \text{K},]
so the correct solution freezes near −0.536 °C. From the measured 255 mOsm kg⁻¹:
[C = \frac{255}{1.87\times1000} = 0.136\ \text{M} = 7.97\ \text{g L}^{-1} = 0.797%\ \text{w/v}.]
The batch is short by about 1.03 g NaCl per litre, or 11.5%. Thermal expansion of water between 20 and 30 °C is only about 0.26%, and expansion of the glass flask is smaller still (about 0.01%), so temperature cannot explain an 11.5% deficit. The error is in the mass of NaCl delivered: damp or caked stock, a balance or transcription error, or a stock of lower purity. The osmometer itself should also be checked later against a 290 mOsm standard, but the size of the discrepancy points first to the weighing step.
2. The pharmacist proposes adding 1.03 g of extra NaCl per litre to bring the batch up to 290 mOsm kg⁻¹. Evaluate this proposal and state the correct course of action.
Read the worked answer
The arithmetic is internally consistent: 1.03 g L⁻¹ is 17.6 mmol L⁻¹, and (17.6\times1.87 = 33) mOsm kg⁻¹, which would raise 255 to about 288. But that calculation is only valid if the sole problem is a known mass deficit in an accurately measured volume and if (i = 1.87) still applies. Here the cause is unknown. If the stock is damp, its NaCl content is uncertain, so the added mass is not a reliable number of moles; if the flask volume is wrong, adding salt compounds the error. A sterile parenteral batch also cannot simply be topped up after compounding without re-sterilisation, re-labelling and re-testing. Correct action: quarantine and discard the batch; dry the stock at about 110 °C or assay it by argentometric titration; check balance and flask calibration; prepare a fresh batch; verify osmolality and pH; document the investigation. Do not 'correct' an unknown-cause batch by calculation alone.
3. Calculate the mass of NaCl needed to make 250 mL of an isotonic solution that also contains 1.0% w/v glucose (M = 180.16). Target 290 mOsm kg⁻¹; ignore volume non-additivity.
Read the worked answer
Glucose: 1.0% w/v = 10 g L⁻¹, so
[C = \frac{10}{180.16} = 0.0555\ \text{M}.]
Glucose does not dissociate, so it contributes 55.5 mOsm L⁻¹. The NaCl must supply
[290 - 55.5 = 234.5\ \text{mOsm L}^{-1}.]
[C_{\text{NaCl}} = \frac{234.5}{1.87} = 125.4\ \text{mmol L}^{-1}.]
Mass concentration (= 125.4\times58.44 = 7.33) g L⁻¹. For 0.250 L:
[m = 7.33\times0.250 = 1.83\ \text{g NaCl}.]
So weigh 2.50 g glucose and 1.83 g NaCl, dissolve in water and make up to 250 mL. Check: the NaCl contributes (0.125\times1.87\times1000 = 234) mOsm L⁻¹ and the total is about 289–290 mOsm L⁻¹. The result should still be confirmed with the osmometer.