KCl supplementation: serum K 3.2 mEq/L
A 33 lb dog has a serum potassium of 3.2 mEq/L. Using the standard sliding scale (add 30 mEq KCl per liter of fluid for K 3.0–3.5), what volume of 2 mEq/mL KCl injection do you draw up to add to a 1 L bag of LRS? And what's the maximum safe pump rate so you don't exceed 0.5 mEq/kg/hr?
Hint
Two pieces. The volume of KCl is a one-step unit conversion from the target mEq using the stock concentration. The rate cap needs the patient's weight in kg, then a mEq/hr → mL/hr step using the bag's final mEq/L.
Another hint
Volume: divide the 30 mEq target by 2 mEq/mL so mEq cancels and you're left with mL. Rate chain: (1) lb → kg via 2.2 lb/kg. (2) 0.5 mEq/kg/hr × kg = mEq/hr cap. (3) Divide mEq/hr by 30 mEq/L to get L/hr, then convert L to mL.
Show worked answer
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From the sliding scale, K 3.0–3.5 mEq/L needs 30 mEq KCl added per liter of fluid. For a 1 L bag, that's 30 mEq. The stock is 2 mEq/mL, so we need to convert mEq to mL.
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Divide the 30 mEq target by the stock concentration. mEq cancels, leaving mL.
$$\frac{30 \,\cancel{mEq}}{2 \,\cancel{mEq}/mL} = 15 \,mL$$ -
Now for the max pump rate. First convert the patient's weight from lb to kg.
$$\frac{33 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 15 \,kg$$ -
The maximum infusion rate of KCl is 0.5 mEq/kg/hr. Multiply by the patient's weight to get the per-hour cap.
$$0.5 \,\tfrac{mEq}{\cancel{kg}\cdot hr} \times 15\,\cancel{kg} = 7.5 \,\tfrac{mEq}{hr}$$ -
Convert mEq/hr to mL/hr using the bag's final concentration (30 mEq in 1,000 mL).
$$\frac{7.5 \,\cancel{mEq}/hr}{30 \,\cancel{mEq}/L} = 0.25 \,\tfrac{L}{hr} = 250 \,\tfrac{mL}{hr}$$
Draw up 15 mL of 2 mEq/mL KCl and add it to the 1 L bag (delivers 30 mEq). Max pump rate ≈ 250 mL/hr. Do not exceed 0.5 mEq/kg/hr.