Replacement volume for an 8 % dehydrated dog
A 44 lb dog is estimated to be 8 % dehydrated on physical exam. What replacement volume should you plan for, and at what hourly rate if you want to correct the deficit over 12 hours (replacement only, not including maintenance)?
Hint
Convert weight to kg first. The deficit formula uses kg. Then there are two pieces: a total volume and a per-hour rate. Find the volume first, then split it over the time window.
Another hint
(1) lb → kg via 2.2 lb/kg. (2) Deficit (in L) = % dehydration as a decimal × body weight (kg). This uses the simplification that 1 kg of body weight ≈ 1 L of fluid. (3) Divide by hours for the rate.
Show worked answer
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Convert the patient's weight from lb to kg.
$$\frac{44 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 20 \,kg$$ -
Estimated deficit (in liters) = % dehydration × body weight. Each kg of body weight is ~1 L of fluid, so % × kg gives liters directly.
$$0.08 \times 20\,kg = 1.6 \,L = 1{,}600 \,mL$$ -
Divide by the replacement window to get an hourly rate.
$$\frac{1{,}600 \,mL}{12 \,hr} \approx 133 \,\tfrac{mL}{hr}$$ -
This is the deficit-replacement rate alone. Maintenance fluids (typically 2–3 mL/kg/hr in dogs, so ~40–60 mL/hr for this patient) and ongoing losses would be added on top in a real treatment plan.
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For fluid choice: a balanced isotonic crystalloid like LRS is the standard replacement fluid for dehydration of unknown or mixed cause. 0.9 % NaCl is also acceptable but is mildly acidifying and lacks the potassium and buffer of LRS. Hypotonic fluids (0.45 % NaCl, D5W) are not appropriate for volume replacement; synthetic colloids like hetastarch are for refractory hypovolemia, not routine dehydration.
Plan to replace 1,600 mL of deficit. At a 12-hour window, replacement alone is ≈ 133 mL/hr.