Free water deficit and safe correction rate
A 10 kg dog presents after being locked in a hot car (free water loss without proportional sodium loss). Serum Na is 168 mEq/L. Using a reference normal of 145 mEq/L, calculate the free water deficit, the hourly D5W replacement rate over 48 hr, and confirm the predicted rate of Na correction stays under the 12 mEq/L per 24 hr ceiling.
Hint
The deficit formula uses total body water, not body weight directly. TBW is roughly 0.6 × body weight (kg) for dogs and cats. Using the raw body weight in place of TBW overstates the deficit by ~67%.
Another hint
Three steps. (1) TBW = 0.6 × 10 = 6 L. (2) Deficit (L) = TBW × [(present Na / reference Na) − 1]. (3) Hourly rate = deficit in mL ÷ 48 hr. For the correction-rate check, predicted Na drop over 24 hr ≈ (168 − 145) × (24 / 48), since the deficit is being replaced over 48 hr so half the correction happens in the first 24.
Show worked answer
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Total body water is 60% of body weight for dogs and cats. For a 10 kg dog:
$$\text{TBW} = 0.6 \times 10 \,kg = 6 \,L$$ -
Apply the DiBartola free water deficit formula: TBW × [(present Na / reference Na) − 1]. The ratio quantifies how much the patient has concentrated.
$$\text{deficit} = 6 \,L \times \left(\frac{168}{145} - 1\right) = 6 \times 0.159 \approx 0.95 \,L$$ -
Convert to mL and spread over 48 hr (the standard DiBartola replacement window).
$$\frac{950 \,mL}{48 \,hr} \approx 20 \,\tfrac{mL}{hr}$$ -
Check the correction rate against the 10–12 mEq/L per 24 hr ceiling. Over the full 48 hr the Na drops by 23 mEq/L (168 → 145), so half of that, ≈ 11.5 mEq/L, happens in the first 24 hr.
$$\text{predicted } \Delta\text{Na}_{24} = (168 - 145) \times \frac{24}{48} \approx 11.5 \,\tfrac{mEq/L}{24\,hr}$$ -
11.5 mEq/L per 24 hr is just under the 12 mEq/L ceiling. The 48-hour replacement schedule is safe. If the predicted drop had exceeded 12, the response would be to lengthen the replacement window (e.g., to 72 hours) rather than slow the infusion within the same window.
Water deficit ≈ 0.95 L. D5W at 20 mL/hr × 48 hr. Predicted Na drop ≈ 11.5 mEq/L per 24 hr, just under the 12 mEq/L ceiling, so the 48-hour window is safe. Use 5% dextrose in water as the replacement fluid since this is pure free-water loss.