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Practice problems

Worked examples of the math behind InfusionFox's calculators. Each problem shows the setup, the unit cancellation, and the arithmetic. Linked to the relevant calculator and clinical background. Answers collapsed by default.

19 problems

CRI math

CRI math Intro

Fentanyl CRI for a 26.4 lb dog

You need to start a fentanyl CRI at 5 µg/kg/hr in a 26.4 lb dog. Your fentanyl stock is 50 µg/mL. What pump rate (mL/hr) do you set?

mL/hr
Hint

The dose is per kg but the weight is in lb. That conversion is your first step. From there, the answer needs to be in mL/hr. Work backwards from there to figure out which units have to cancel.

Another hint

Three steps. (1) Convert lb to kg using 2.2 lb/kg. (2) Convert the dose to an hourly drug amount (µg/hr). (3) Divide by the stock concentration (µg/mL) so the µg cancel and you're left with mL/hr.

Show worked answer

  1. Convert the patient's weight from lb to kg using the standard clinical factor of 2.2 lb/kg.

    $$\frac{26.4 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 12 \,kg$$

  2. Now dose × weight to get the hourly drug requirement in µg/hr.

    $$5 \,\tfrac{\mu g}{\cancel{kg}\cdot hr} \times 12\,\cancel{kg} = 60 \,\tfrac{\mu g}{hr}$$

  3. Divide by the stock concentration to convert µg/hr to mL/hr. The µg in the numerator cancels the µg in the denominator of the concentration, leaving mL/hr.

    $$\frac{60 \,\cancel{\mu g}/hr}{50 \,\cancel{\mu g}/mL} = 1.2 \,\tfrac{mL}{hr}$$
Answer

Pump rate: 1.2 mL/hr.

Clinical background Fentanyl clinical background Related calculator Fentanyl CRI calculator

Fluid therapy

Fluid therapy Intro

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)?

mL
mL/hr
Most appropriate replacement fluid
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

  1. Convert the patient's weight from lb to kg.

    $$\frac{44 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 20 \,kg$$

  2. Estimated deficit (in liters) = % dehydration × body weight. Each kg of body weight is ~1 L of fluid, so % × kg gives liters directly.

    $$0.08 \times 20\,kg = 1.6 \,L = 1{,}600 \,mL$$

  3. Divide by the replacement window to get an hourly rate.

    $$\frac{1{,}600 \,mL}{12 \,hr} \approx 133 \,\tfrac{mL}{hr}$$

  4. 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.

  5. 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.

Answer

Plan to replace 1,600 mL of deficit. At a 12-hour window, replacement alone is ≈ 133 mL/hr.

Clinical background Fluid therapy clinical background Related calculator Fluid therapy calculator

Electrolytes

Electrolytes Intro

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?

mL
mL/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

  1. 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.

  2. Divide the 30 mEq target by the stock concentration. mEq cancels, leaving mL.

    $$\frac{30 \,\cancel{mEq}}{2 \,\cancel{mEq}/mL} = 15 \,mL$$

  3. Now for the max pump rate. First convert the patient's weight from lb to kg.

    $$\frac{33 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 15 \,kg$$

  4. The maximum infusion rate of KCl is 0.5 mEq/kg/hr. Multiply by the patient's weight to get the per-hour cap.

    $$0.5 \,\tfrac{mEq}{\cancel{kg}\cdot hr} \times 15\,\cancel{kg} = 7.5 \,\tfrac{mEq}{hr}$$

  5. Convert mEq/hr to mL/hr using the bag's final concentration (30 mEq in 1,000 mL).

    $$\frac{7.5 \,\cancel{mEq}/hr}{30 \,\cancel{mEq}/L} = 0.25 \,\tfrac{L}{hr} = 250 \,\tfrac{mL}{hr}$$
Answer

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.

Clinical background Hypokalemia clinical background Related calculator Hypokalemia (KCl) calculator
Electrolytes Advanced

Free-water deficit in a hypernatremic dog

A 66 lb dog presents with severe dehydration and serum Na of 170 mEq/L. You want to correct gradually to a target of 145 mEq/L. Estimate the free-water deficit. Use total body water = 60 % of body weight.

L
Most appropriate fluid for correction
Hint

Convert lb to kg first. The TBW estimate (60 %) is a fraction of the kg weight. You're estimating how much pure water you'd need to dilute existing sodium down to the target.

Another hint

(1) lb → kg via 2.2 lb/kg. (2) TBW (L) ≈ 0.6 × weight (kg). (3) Free-water deficit (L) ≈ TBW × ((current Na ÷ target Na) − 1). The Na ratio is dimensionless so the answer comes out in liters. Don't forget the safety check on correction rate (no faster than 0.5–1 mEq/L/hr drop).

Show worked answer

  1. Convert the patient's weight from lb to kg.

    $$\frac{66 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 30 \,kg$$

  2. Total body water (TBW) is the fluid compartment we're diluting. For a 30 kg dog, TBW ≈ 0.6 × 30 = 18 L.

    $$0.6 \times 30\,kg = 18 \,L$$

  3. Free-water deficit (L) ≈ TBW × ((current Na ÷ target Na) − 1). The unit ratios are dimensionless, so the answer comes out in liters.

    $$18\,L \times \left(\tfrac{170}{145} - 1\right) = 18 \times 0.1724 \approx 3.1\,L$$

  4. That's the free-water deficit alone (volume of pure water needed to bring Na to target). In practice you'd give a hypotonic crystalloid like D5W, 0.45 % NaCl, or maintenance solutions (not pure water), and correct at no faster than 0.5–1 mEq/L/hr to avoid cerebral edema.

  5. Sanity check on the correction rate: dropping Na from 170 to 145 is a 25 mEq/L change. At 0.5 mEq/L/hr that needs 50 hours; at 1 mEq/L/hr, 25 hours.

  6. Why a hypotonic fluid? You're trying to drop the serum sodium, which means giving fluid with a lower Na concentration than the patient's plasma. Isotonic crystalloids (0.9 % NaCl, LRS) wouldn't move serum Na meaningfully, and hypertonic saline would make it worse. 0.45 % NaCl and D5W are both reasonable choices in dogs; in practice the choice between them depends on whether you also need volume support (0.45 % NaCl) versus pure free water (D5W).

Answer

Free-water deficit ≈ 3.1 L. Correct slowly over 25–50 hr (≤ 0.5–1 mEq/L/hr drop in Na) to avoid cerebral edema.

Clinical background Hypernatremia clinical background Related calculator Hypernatremia calculator
Electrolytes Advanced

Phosphate replacement: K-Phos addition with concurrent KCl

A 55 lb DKA dog has serum phosphorus of 1.2 mg/dL (severe). You want to give K-Phos at 0.03 mmol/kg/hr. K-Phos stock is 3 mmol/mL phosphate, with 4.4 mEq of potassium per mL (the same vial supplies both). The patient is already on a fluid line containing 20 mEq/L KCl at 50 mL/hr. Is the combined potassium delivery safe (cap 0.5 mEq/kg/hr)?

mL/hr
mEq/kg/hr
Is this safe?
Hint

Convert lb to kg first. Both the phosphate dose and the K cap depend on kg. Then three sub-problems: pump rate of K-Phos, K from the K-Phos, K from the maintenance line. Sum the K and compare to the cap.

Another hint

(1) lb → kg via 2.2 lb/kg. (2) K-Phos rate (mL/hr): phosphate dose × weight ÷ K-Phos phosphate concentration. (3) K from K-Phos: that rate × 4.4 mEq/mL. (4) K from maintenance line: 20 mEq/L × 50 mL/hr (watch your L/mL conversion). Sum, divide by weight, compare to 0.5 mEq/kg/hr.

Show worked answer

  1. Convert the patient's weight from lb to kg.

    $$\frac{55 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 25 \,kg$$

  2. Compute the K-Phos rate in mL/hr from the phosphate dose.

    $$0.03 \,\tfrac{mmol}{\cancel{kg}\cdot hr} \times 25\,\cancel{kg} = 0.75 \,\tfrac{mmol}{hr}$$

  3. Convert mmol/hr to mL/hr using the K-Phos phosphate concentration. mmol cancels.

    $$\frac{0.75 \,\cancel{mmol}/hr}{3 \,\cancel{mmol}/mL} = 0.25 \,\tfrac{mL}{hr}$$

  4. Potassium delivered by the K-Phos: each mL has 4.4 mEq K. So 0.25 mL/hr × 4.4 mEq/mL gives the K from K-Phos.

    $$0.25 \,\tfrac{\cancel{mL}}{hr} \times 4.4 \,\tfrac{mEq}{\cancel{mL}} = 1.1 \,\tfrac{mEq}{hr}$$

  5. Potassium from the maintenance line: 20 mEq/L × 50 mL/hr. Convert L to mL (× 1000) so the mL cancel.

    $$\frac{20 \,mEq}{1{,}000 \,\cancel{mL}} \times 50 \,\tfrac{\cancel{mL}}{hr} = 1.0 \,\tfrac{mEq}{hr}$$

  6. Sum the two sources and divide by weight to check against the 0.5 mEq/kg/hr cap.

    $$\frac{1.1 + 1.0 \,mEq/hr}{25\,kg} = 0.084 \,\tfrac{mEq}{kg\cdot hr}$$

  7. 0.084 mEq/kg/hr is well below the 0.5 mEq/kg/hr cap, so combined delivery is safe.

Answer

Run K-Phos at 0.25 mL/hr. Combined K delivery is ≈ 2.1 mEq/hr (0.08 mEq/kg/hr), comfortably below the 0.5 mEq/kg/hr cap.

Clinical background Hypophosphatemia clinical background Related calculator Hypophosphatemia (K-Phos) calculator
Electrolytes Clinical

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.

L
mL/hr
mEq/L
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

  1. Total body water is 60% of body weight for dogs and cats. For a 10 kg dog:

    $$\text{TBW} = 0.6 \times 10 \,kg = 6 \,L$$

  2. Apply the DiBartola free water deficit formula: TBW × [(present Na / reference Na) − 1]. The ratio quantifies how much the patient has concentrated.

    $$\text{deficit} = 6 \,L \times \left(\frac{168}{145} - 1\right) = 6 \times 0.159 \approx 0.95 \,L$$

  3. Convert to mL and spread over 48 hr (the standard DiBartola replacement window).

    $$\frac{950 \,mL}{48 \,hr} \approx 20 \,\tfrac{mL}{hr}$$

  4. 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.

    $$\text{predicted } \Delta\text{Na}_{24} = (168 - 145) \times \frac{24}{48} \approx 11.5 \,\tfrac{mEq/L}{24\,hr}$$

  5. 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.

Answer

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.

Clinical background Hypernatremia clinical background Related calculator Hypernatremia water deficit calculator
Electrolytes Clinical

pRBC volume and infusion rates for an anemic dog

A 25 kg dog has a PCV of 15% from chronic IMHA. You decide to transfuse pRBC to a target PCV of 25%. Donor pRBC PCV is the default 80%. Calculate the total pRBC volume needed, the slow trial rate for the first 30 minutes, and the main rate to finish the transfusion over a 4-hour total window.

mL
mL/hr
mL/hr
Hint

Volume needed depends on three things: the patient's total blood volume, how much PCV needs to rise, and how concentrated the donor product is. The math is one equation; the trick is knowing the species blood-volume constant.

Another hint

Blood volume = 90 mL/kg in dogs (60 mL/kg in cats). The slow trial is 0.5 mL/kg/hr for 30 minutes, designed to deliver 0.25 mL/kg total during the trial. The main rate finishes the remainder over 3.5 hours (4 hours total minus the 30-min trial).

Show worked answer

  1. Compute total blood volume. Dogs have ~90 mL/kg blood volume.

    $$\text{blood volume} = 90 \,\tfrac{mL}{\cancel{kg}} \times 25\,\cancel{kg} = 2{,}250 \,mL$$

  2. Compute the PCV rise needed.

    $$\Delta\text{PCV} = 25\% - 15\% = 10 \text{ points}$$

  3. Apply the transfusion volume formula: total pRBC volume = (PCV rise / donor PCV) × patient blood volume. The donor PCV in the denominator captures how concentrated each mL of product is.

    $$\text{volume} = \frac{10}{80} \times 2{,}250 \,mL \approx 281 \,mL$$

  4. Slow trial rate: 0.5 mL/kg/hr for the first 30 minutes. This delivers 0.25 mL/kg total during the trial, slow enough that acute reactions (anaphylaxis, hemolysis, febrile non-hemolytic) can be detected before a meaningful volume has gone in.

    $$\text{slow trial} = 0.5 \,\tfrac{mL}{kg \cdot hr} \times 25 \,kg = 12.5 \,\tfrac{mL}{hr}$$

  5. Volume given during the slow trial = 0.25 mL/kg × 25 kg = 6.25 mL. Remaining volume = 281 − 6.25 ≈ 275 mL.

  6. Main rate to finish in 3.5 more hours (4 hr total window):

    $$\text{main rate} = \frac{275 \,mL}{3.5 \,hr} \approx 78.5 \,\tfrac{mL}{hr}$$

  7. Monitor TPR every 15 min during the slow trial and every 30 min thereafter. Aim to complete the transfusion within 4 hours to limit bacterial contamination risk in an open product.

Answer

Total pRBC volume ≈ 281 mL. Slow trial at 12.5 mL/hr for 30 min, then main rate at ~78.5 mL/hr for 3.5 hr. Total infusion time 4 hr.

Clinical background Transfusion clinical background Related calculator Transfusion calculator

Multi-drug CRIs

Multi-drug CRIs Clinical

Build an MLK bag to last 24 hours

You're starting an MLK CRI on a 22 lb dog recovering from a splenectomy. You'd like to run the infusion at 1 mL/kg/hr and want enough drug in the bag to last 24 hours. Target doses: morphine 0.2 mg/kg/hr, lidocaine 1.6 mg/kg/hr, ketamine 0.4 mg/kg/hr (all within published ranges). Stock concentrations: morphine 5 mg/mL, lidocaine 20 mg/mL (2 %), ketamine 100 mg/mL. What pump rate (mL/hr) do you set, what bag size should you choose, and how much of each drug do you add?

mL/hr
Standard bag size to use
mL
mL
mL
Could you run this same protocol in a cat?
Hint

Convert weight to kg first. Then the planned duration (24 hr) is what you multiply each per-hour quantity by: pump rate × duration tells you how much fluid you need (which sets the bag size), and dose × weight × duration tells you the mg of each drug.

Another hint

(1) lb → kg via 2.2 lb/kg. (2) Pump rate (mL/hr) = weight × per-kg rate. (3) Bag volume needed = pump rate × 24 hr; round up to the next standard bag size. (4) Each drug: total mg = dose × weight × 24 hr; stock volume = total mg ÷ stock concentration. (5) Remove the combined drug volume from the bag.

Show worked answer

  1. Convert the patient's weight from lb to kg.

    $$\frac{22 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 10 \,kg$$

  2. Pump rate (mL/hr) = weight × per-kg rate. kg cancels.

    $$10\,\cancel{kg} \times 1 \,\tfrac{mL}{\cancel{kg}\cdot hr} = 10 \,\tfrac{mL}{hr}$$

  3. Bag volume needed for a 24-hour infusion = pump rate × duration. hr cancels, leaves mL.

    $$10 \,\tfrac{mL}{\cancel{hr}} \times 24 \,\cancel{hr} = 240 \,mL$$

  4. 240 mL isn't a standard bag size. Round up to the next standard size (a 250 mL bag) so you have enough fluid for the full 24 hours. The drug amounts below are still calculated for 24 hr of infusion; the extra ~10 mL of carrier doesn't change the per-hour dose because that's set by the pump rate, not the bag size.

  5. Total morphine over 24 hr = dose × weight × duration. Two cancellations: kg cancels kg, hr cancels hr, leaving mg.

    $$0.2 \,\tfrac{mg}{\cancel{kg}\cdot\cancel{hr}} \times 10\,\cancel{kg} \times 24\,\cancel{hr} = 48 \,mg$$

  6. Volume of morphine stock to draw = mg ÷ concentration.

    $$\frac{50 \,\cancel{mg}}{5 \,\cancel{mg}/mL} = 10 \,mL$$

  7. Lidocaine and ketamine follow the same pattern.

    $$\text{Lidocaine: } 1.6 \times 10 \times 25 = 400 \,mg \;\to\; \tfrac{400}{20} = 20 \,mL$$
  8. $$\text{Ketamine: } 0.4 \times 10 \times 25 = 100 \,mg \;\to\; \tfrac{100}{100} = 1 \,mL$$

  9. Total drug volume = 10 + 20 + 1 = 31 mL. Remove 31 mL of saline from the 250 mL bag, then add the three drugs. The bag now holds 50 mg morphine, 400 mg lidocaine, and 100 mg ketamine in 250 mL (enough to last 25 hr at 10 mL/hr, comfortably more than the 24 hr of planned infusion). Pre-loading the bag for its full duration keeps the drug concentration exactly on target throughout the infusion.

  10. Note: this protocol is dog-only. Cats are uniquely sensitive to lidocaine cardiotoxicity (myocardial depression, arrhythmias, and CNS effects can occur at the MLK lidocaine dose). For feline multimodal analgesia, consider a single-agent fentanyl, buprenorphine, or hydromorphone CRI instead.

Answer

Pump rate 10 mL/hr in a 250 mL bag (240 mL needed for 24 hr, rounded up; the bag lasts 25 hr at this rate). Add 50 mg morphine (10 mL), 400 mg lidocaine (20 mL), and 100 mg ketamine (1 mL) after removing 31 mL of saline. Loading the bag for its full duration keeps the drug concentration exactly on target.

Clinical background MLK CRI clinical background Related calculator MLK CRI calculator
Multi-drug CRIs Advanced

Controlled-drug waste from a partially used MLK bag

You built the MLK bag from problem 4 (50 mg morphine, 400 mg lidocaine, 100 mg ketamine in 250 mL). The patient was extubated and recovered well after ~15 hours and the CRI was discontinued with 100 mL remaining in the bag. How much of each controlled drug needs to be logged as waste?

mg
mg
DEA schedule of morphine
Hint

The bag is a uniform mixture. If you know the fraction of the bag that wasn't given, that same fraction of every drug in the bag was wasted.

Another hint

Wasted fraction = volume remaining ÷ bag volume. Then for each drug: mg wasted = total mg in bag × wasted fraction. Lidocaine isn't federally controlled, but the math is the same.

Show worked answer

  1. First, identify the wasted-volume fraction. 100 mL of a 250 mL bag = 100 / 250 = 0.4, or 40 % of the bag was wasted.

  2. Each drug in the bag is wasted proportionally. Morphine wasted = total mg × wasted fraction. mg stays, the dimensionless fraction multiplies through.

    $$50 \,mg \times 0.4 = 20 \,mg \text{ morphine (C-II)}$$

  3. Same approach for ketamine.

    $$100 \,mg \times 0.4 = 40 \,mg \text{ ketamine (C-III)}$$

  4. Lidocaine is not federally controlled, but you can compute it the same way for completeness.

    $$400 \,mg \times 0.4 = 160 \,mg \text{ lidocaine}$$

  5. Optional: stock-equivalent volumes for logs that record mL. Divide each wasted mg by its stock concentration.

    $$\tfrac{20\,\cancel{mg}}{5\,\cancel{mg}/mL} = 4\,mL \text{ morphine stock-equivalent}$$

  6. The physical waste is 100 mL of diluted bag mixture, not undiluted stock. Most controlled-drug logs want the mg figure since that's what reconciles against your inventory.

  7. On the schedules: morphine is DEA Schedule II (C-II), a full µ-agonist opioid with high abuse potential and accepted medical use. Ketamine is Schedule III (C-III). Both require controlled-drug logging; lidocaine is not federally scheduled. State scheduling can differ, and individual practices may have stricter internal policies.

Answer

Log 20 mg morphine (C-II) and 40 mg ketamine (C-III) as waste. Lidocaine is not federally controlled but 160 mg was discarded with the bag.

Clinical background MLK CRI clinical background Related calculator MLK CRI calculator (waste section)

Vasopressor CRIs

Vasopressor CRIs Clinical

Dopamine CRI from a 400 mg / 250 mL bag

A 55 lb dog has an intraoperative MAP of 55 mmHg despite fluid loading. You've prepared a dopamine bag: 400 mg of dopamine in 250 mL of NaCl. What pump rate (mL/hr) delivers 5 µg/kg/min?

mL/hr
MAP target for adequate organ perfusion
Hint

Three unit mismatches to fix: lb vs kg, mg vs µg in the bag, and min vs hr for the dose. Convert weight first, then untangle the drug-amount units before doing any arithmetic.

Another hint

(1) lb → kg via 2.2 lb/kg. (2) Bag concentration: 400 mg ÷ 250 mL = 1,600 µg/mL once you put it in matching units. (3) Pump rate (mL/hr) = (dose × weight × 60) ÷ bag concentration. The 60 converts µg/min to µg/hr.

Show worked answer

  1. Convert the patient's weight from lb to kg.

    $$\frac{55 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 25 \,kg$$

  2. Next, find the bag's final concentration in µg/mL. 400 mg = 400,000 µg.

    $$\frac{400{,}000 \,\mu g}{250 \,mL} = 1{,}600 \,\tfrac{\mu g}{mL}$$

  3. Compute the µg per minute the patient needs: dose × weight. kg cancels.

    $$5 \,\tfrac{\mu g}{\cancel{kg}\cdot min} \times 25\,\cancel{kg} = 125 \,\tfrac{\mu g}{min}$$

  4. Convert µg/min to µg/hr by multiplying by 60 min/hr.

    $$125 \,\tfrac{\mu g}{\cancel{min}} \times 60\,\tfrac{\cancel{min}}{hr} = 7{,}500 \,\tfrac{\mu g}{hr}$$

  5. Divide by the bag concentration to convert µg/hr to mL/hr. µg cancels.

    $$\frac{7{,}500 \,\cancel{\mu g}/hr}{1{,}600 \,\cancel{\mu g}/mL} \approx 4.7 \,\tfrac{mL}{hr}$$

  6. Titrate to a MAP of at least 65 mmHg; that's the conventional perfusion threshold below which organ blood flow becomes pressure-dependent. Below 50–60 mmHg renal autoregulation is lost and AKI risk rises sharply.

Answer

Run the dopamine bag at ≈ 4.7 mL/hr to deliver 5 µg/kg/min.

Clinical background Dopamine clinical background Related calculator Dopamine CRI calculator
Vasopressor CRIs Clinical

Norepinephrine titration: changing the dose

A 39.6 lb dog is on a norepinephrine CRI from a 4 mg in 250 mL NaCl bag (final concentration 16 µg/mL) at 0.05 µg/kg/min. MAP is only 58 mmHg and you want to titrate up to 0.2 µg/kg/min. What's the new pump rate (mL/hr)?

mL/hr
Which adrenergic receptor predominantly drives norepi's pressor effect?
Hint

Convert weight first. The starting dose is a distractor. When you titrate, you recalculate from the new dose. Same math, new number.

Another hint

(1) lb → kg via 2.2 lb/kg. (2) Pump rate (mL/hr) = (dose × weight × 60) ÷ bag concentration. The 60 is min → hr. Bag concentration is already given in µg/mL so no further conversion.

Show worked answer

  1. Convert the patient's weight from lb to kg.

    $$\frac{39.6 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 18 \,kg$$

  2. Compute µg/min at the new dose.

    $$0.2 \,\tfrac{\mu g}{\cancel{kg}\cdot min} \times 18\,\cancel{kg} = 3.6 \,\tfrac{\mu g}{min}$$

  3. Convert µg/min to µg/hr.

    $$3.6 \,\tfrac{\mu g}{\cancel{min}} \times 60\,\tfrac{\cancel{min}}{hr} = 216 \,\tfrac{\mu g}{hr}$$

  4. Divide by bag concentration to get mL/hr.

    $$\frac{216 \,\cancel{\mu g}/hr}{16 \,\cancel{\mu g}/mL} = 13.5 \,\tfrac{mL}{hr}$$

  5. Shortcut for next time: at this bag concentration and weight, mL/hr ≈ (dose in µg/kg/min) × 67.5. Useful for fast bedside titration once the first dose is set up.

  6. Why norepinephrine raises blood pressure: it's a potent α-1 agonist with modest β-1 activity. The α-1 effect is what drives the pressor response (peripheral vasoconstriction → increased systemic vascular resistance). The β-1 effect adds modest inotropy. Unlike dopamine, norepi has minimal effect at dopaminergic receptors. This is why it's preferred for vasodilatory shock where the primary problem is loss of vascular tone.

Answer

Set the pump to 13.5 mL/hr to deliver 0.2 µg/kg/min.

Clinical background Norepinephrine clinical background Related calculator Norepinephrine CRI calculator

DKA management

DKA management Clinical

Regular insulin CRI for a 33 lb DKA dog

A 33 lb dog presents in DKA with BG 480 mg/dL. You prepare a regular insulin CRI bag using the standard 2.2 U/kg method: add 2.2 U/kg to 250 mL of 0.9 % NaCl, then run the first 50 mL through the line to saturate the tubing (insulin binds to plastic). At what mL/hr should you start the infusion to deliver 0.05 U/kg/hr?

mL/hr
Which insulin do you use for a DKA CRI?
Hint

Convert weight first. Both the loading dose (2.2 U/kg) and the hourly dose (0.05 U/kg/hr) need kg. Then the 50 mL line prime is the trick: you've thrown away 50 mL but kept all the insulin.

Another hint

(1) lb → kg via 2.2 lb/kg. (2) Total U in bag = 2.2 U/kg × kg. (3) Effective concentration = total U ÷ remaining mL (200, not 250). (4) Target hourly dose ÷ effective concentration → mL/hr.

Show worked answer

  1. Convert the patient's weight from lb to kg.

    $$\frac{33 \,\cancel{lb}}{2.2 \,\cancel{lb}/kg} = 15 \,kg$$

  2. Total units added to the bag = 2.2 U/kg × weight.

    $$2.2 \,\tfrac{U}{\cancel{kg}} \times 15 \,\cancel{kg} = 33 \,U$$

  3. After priming 50 mL through the line, 200 mL of bag remains. The 33 U are now in 200 mL of effective volume.

    $$\frac{33 \,U}{200 \,mL} = 0.165 \,\tfrac{U}{mL}$$

  4. Target hourly dose: 0.05 U/kg/hr × 15 kg.

    $$0.05 \,\tfrac{U}{\cancel{kg}\cdot hr} \times 15\,\cancel{kg} = 0.75 \,\tfrac{U}{hr}$$

  5. Pump rate = hourly dose ÷ concentration. U cancels, leaves mL/hr.

    $$\frac{0.75 \,\cancel{U}/hr}{0.165 \,\cancel{U}/mL} \approx 4.5 \,\tfrac{mL}{hr}$$

  6. Why regular insulin? Only short-acting (regular) insulin is suitable for IV CRI in DKA. Its rapid onset and short half-life let you titrate against bedside BG checks. The longer-acting preparations (NPH, glargine, detemir) and the meal-time analogs (lispro, aspart) are designed for subcutaneous use; their kinetics make precise IV titration impossible.

Answer

Start the insulin CRI at ≈ 4.5 mL/hr.

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Blood gas · Basic

Blood gas · Basic Intro

Identify the primary disturbance: vomiting/diarrhea dog

A 20 kg dog with 3 days of vomiting and diarrhea has the following arterial blood gas: pH 7.25, PCO₂ 28 mm Hg, HCO₃⁻ 12 mEq/L. What is the primary acid-base disturbance?

Primary disturbance
Hint

Start with pH. Is the patient acidemic (pH < 7.35) or alkalemic (pH > 7.46)? That tells you the direction. Then look at HCO₃⁻ and PCO₂. Which one moved in the direction that explains the pH?

Another hint

Acidemia + low HCO₃⁻ = metabolic acidosis. Acidemia + high PCO₂ = respiratory acidosis. The low PCO₂ here is the body's attempt to compensate (hyperventilation blowing off CO₂), not the primary process.

Show worked answer

  1. pH is 7.25, which is below the dog reference range (7.35–7.46). The patient is acidemic.

  2. HCO₃⁻ is 12 mEq/L (reference 19–26), which is markedly low. A low HCO₃⁻ would push pH down. This explains the acidemia.

  3. PCO₂ is 28 mm Hg (reference 31–43), which is also low. A low PCO₂ would push pH up, so it cannot be causing the acidemia. It's the compensatory response (hyperventilation).

  4. Primary disturbance: metabolic acidosis with respiratory compensation. Loss of HCO₃⁻ in diarrheal fluid is the classic mechanism here.

Answer

Metabolic acidosis (with appropriate respiratory compensation).

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Blood gas · Basic Intro

Anesthetized dog hypoventilating

A dog under isoflurane anesthesia has not been mechanically ventilated. An arterial blood gas drawn 30 minutes into the procedure shows: pH 7.25, PCO₂ 65 mm Hg, HCO₃⁻ 24 mEq/L. What is the primary disturbance, and is the bicarbonate response consistent with a simple disorder?

Primary disturbance
mEq/L
Hint

Acidemia with high PCO₂ is respiratory acidosis. The next question is acute vs chronic. A dog 30 minutes into anesthesia has had no time for renal compensation, which takes 2–5 days.

Another hint

Apply the acute respiratory acidosis rule: HCO₃⁻ rises 0.15 mEq/L per 1 mm Hg rise in PCO₂. Baseline HCO₃⁻ for dogs is about 22, baseline PCO₂ about 37.

Show worked answer

  1. pH 7.25 is acidemic. PCO₂ 65 is markedly elevated; HCO₃⁻ 24 is within reference range but at the upper end. Primary disturbance is respiratory acidosis driven by inhalant-induced hypoventilation.

  2. Time course is acute (30 minutes). Renal compensation takes 2–5 days to fully develop, so HCO₃⁻ should rise only modestly via intracellular non-bicarbonate buffer titration.

  3. Apply the dog acute respiratory acidosis rule: PCO₂ went up by 65 − 37 = 28 mm Hg. Expected HCO₃⁻ rise = 0.15 × 28 = 4.2 mEq/L. Expected HCO₃⁻ ≈ 22 + 4.2 = 26 mEq/L.

    $$\Delta \text{HCO}_3^- = 0.15 \,\tfrac{mEq/L}{\cancel{mm\,Hg}} \times 28 \,\cancel{mm\,Hg} = 4.2 \,mEq/L$$

  4. Observed HCO₃⁻ of 24 is within ±2 of the expected 26, consistent with simple acute respiratory acidosis. The fix is mechanical ventilation, not bicarbonate.

Answer

Acute respiratory acidosis from inhalant-induced hypoventilation. Expected HCO₃⁻ ≈ 26 mEq/L; observed 24 is within range. Simple disorder. Treat by ventilating, not by giving bicarbonate.

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Blood gas · Basic Clinical

DKA dog with low PCO₂: simple or mixed?

A 15 kg dog presents in diabetic ketoacidosis. Arterial blood gas: pH 7.10, PCO₂ 15 mm Hg, HCO₃⁻ 8 mEq/L. Apply the dog metabolic acidosis compensation rule and decide whether this is a simple disorder or a mixed disorder.

mm Hg
Verdict
Hint

First confirm metabolic acidosis is present (it is: low pH, low HCO₃⁻). Then apply the dog rule: PCO₂ should fall 0.7 mm Hg for every 1 mEq/L fall in HCO₃⁻ below the reference midpoint.

Another hint

Baseline HCO₃⁻ ≈ 22, observed is 8, so HCO₃⁻ has fallen by 14. Expected PCO₂ drop = 0.7 × 14 = 9.8. Expected PCO₂ = 37 − 9.8 ≈ 27 mm Hg. Compare to observed PCO₂ of 15.

Show worked answer

  1. Confirm primary disturbance. pH 7.10 = severe acidemia. HCO₃⁻ 8 (very low) explains the acidosis; PCO₂ 15 is low, consistent with compensation. Primary: metabolic acidosis.

  2. Apply the dog metabolic acidosis rule. HCO₃⁻ fell from baseline ~22 to observed 8, a drop of 14 mEq/L.

    $$\Delta \text{HCO}_3^- = 22 - 8 = 14 \,mEq/L$$

  3. Expected PCO₂ drop = 0.7 × 14 = 9.8 mm Hg. Expected PCO₂ from baseline 37 = 37 − 9.8 ≈ 27 mm Hg.

    $$\text{expected PCO}_2 = 37 - (0.7 \times 14) \approx 27 \,mm\,Hg$$

  4. Observed PCO₂ is 15, well below the expected 27 (±2 → 25–29). PCO₂ is lower than compensation alone should produce.

  5. Conclusion: mixed disorder. Primary metabolic acidosis from DKA, plus a concurrent respiratory alkalosis (the patient is hyperventilating beyond what compensation requires, which could reflect sepsis, pain, anxiety, or central drive from the ketoacidosis itself).

Answer

Mixed disorder: metabolic acidosis (DKA) plus a concurrent respiratory alkalosis. PCO₂ of 15 is below the expected 27 ±2 for simple compensation.

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Blood gas · Basic Clinical

High-AG vs normal-AG metabolic acidosis

Two dogs both present with metabolic acidosis (HCO₃⁻ 12 mEq/L in both). Dog A: Na 145, Cl 110. Dog B: Na 145, Cl 120. Compute the anion gap for each, classify each as high-AG or normal-AG, and pick the differential that fits each.

mEq/L
mEq/L
Dog A fits
Dog B fits
Hint

AG = Na − (Cl + HCO₃⁻). Dog reference range is 13–25 mEq/L. The anion gap reflects unmeasured anions, mostly proteins plus any organic acids that have accumulated.

Another hint

High AG = organic acid added (lactate, ketones, ethylene glycol metabolites, uremic acids). Normal AG with low HCO₃⁻ = HCO₃⁻ lost and replaced by Cl⁻ (GI loss, RTA, dilution).

Show worked answer

  1. Dog A: AG = 145 − (110 + 12) = 23 mEq/L. Upper end of normal (reference 13–25), bordering on high.

    $$\text{AG}_A = 145 - (110 + 12) = 23 \,mEq/L$$

  2. Dog B: AG = 145 − (120 + 12) = 13 mEq/L. Low-normal.

    $$\text{AG}_B = 145 - (120 + 12) = 13 \,mEq/L$$

  3. Dog A has a borderline-high AG with low HCO₃⁻ and a normal chloride. Pattern fits an organic acid: lactate (sepsis, hypoperfusion), ketones (DKA), uremic acids, ethylene glycol.

  4. Dog B has a normal AG with hyperchloremia. Pattern fits HCO₃⁻ loss replaced by chloride: small bowel diarrhea, renal tubular acidosis, dilutional acidosis.

  5. Both dogs have the same HCO₃⁻ but very different mechanisms. The anion gap is the key distinguishing test.

Answer

Dog A: AG ≈ 23 (high-normal), high-AG pattern fits an organic acidosis (lactic acidosis from sepsis). Dog B: AG = 13 (normal), hyperchloremic, fits HCO₃⁻ loss from diarrhea.

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Blood gas · Basic Clinical

Cat with metabolic acidosis and normal PCO₂

A 4 kg cat with chronic kidney disease has an arterial blood gas: pH 7.22, PCO₂ 33 mm Hg, HCO₃⁻ 14 mEq/L. PCO₂ is right at the cat reference midpoint. If you apply the dog rule for expected respiratory compensation, you'd predict PCO₂ around 25 mm Hg. Does the observed PCO₂ of 33 mean this cat has a mixed disorder?

Conclusion
Hint

This is a species-specific physiology question, not a math question. What does DiBartola say about the feline kidney's adaptive response to metabolic acidosis compared with the dog and the human?

Another hint

DiBartola Ch. 12 p. 304 explicitly warns that dog and human compensation formulas should not be extrapolated to cats with metabolic acidosis. The feline kidney lacks the adaptive ammoniagenesis response, and the respiratory compensation response also appears blunted.

Show worked answer

  1. First the math check: cat reference HCO₃⁻ midpoint ≈ 18. Observed HCO₃⁻ 14, a drop of 4 mEq/L. If we (wrongly) applied the dog rule: expected PCO₂ drop = 0.7 × 4 = 2.8 mm Hg. Expected PCO₂ ≈ 31 − 2.8 ≈ 28 mm Hg. Observed 33 is above this. The dog rule predicts a mismatch.

  2. But the dog rule should not be applied. DiBartola Ch. 12 p. 304: "the feline kidney apparently is unable to adapt to metabolic acidosis ... cats may not compensate for metabolic acidosis to the same extent (if at all) as do dogs and humans. Thus formulas for dogs or humans should not be extrapolated for use in cats."

  3. The cat may simply not be hyperventilating in response to the metabolic acidosis. A normal PCO₂ in a cat with metabolic acidosis is NOT by itself evidence of a mixed disorder.

  4. Clinical interpretation in this cat depends on history, physical exam, and ancillary labs, not on a borrowed compensation formula. CKD is sufficient to explain the metabolic acidosis (impaired renal acid excretion).

Answer

No mixed disorder is established by this blood gas. The cat may not be compensating because cats often don't compensate for metabolic acidosis. Dog compensation formulas should not be extrapolated to cats in this setting (DiBartola Ch. 12).

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Blood gas · Basic Advanced

Normal pH but abnormal PCO₂ and HCO₃⁻

A 30 kg dog with chronic bronchitis presents for acute illness. Arterial blood gas: pH 7.40, PCO₂ 55 mm Hg, HCO₃⁻ 33 mEq/L. The pH is normal. Does this dog have an acid-base disorder?

Interpretation
Hint

Normal pH does not rule out an acid-base disorder. Two disorders pulling pH in opposite directions can normalize it. Look at PCO₂ and HCO₃⁻: both are abnormal, and they're moved in the same direction. Is that consistent with a single disorder?

Another hint

Try the chronic respiratory acidosis rule for dogs: HCO₃⁻ rises 0.35 mEq/L per 1 mm Hg rise in PCO₂. Compare the observed HCO₃⁻ rise to what that rule would predict. If observed > predicted, there's a metabolic alkalosis on top.

Show worked answer

  1. pH is 7.40, exactly normal. That does NOT mean no disorder. Compensation never fully normalizes pH; overcompensation does not occur. A normal pH with both PCO₂ and HCO₃⁻ abnormal raises strong suspicion of a counterbalancing mixed disorder.

  2. PCO₂ 55 is high (reference 31–43); HCO₃⁻ 33 is high (reference 19–26). Both moved in the same direction. A single primary disorder + compensation can't produce both abnormalities pushing pH the SAME way. The two would be in opposite directions if compensation alone were at work.

  3. Apply the chronic respiratory acidosis rule (chronic bronchitis suggests longstanding hypercapnia): HCO₃⁻ should rise 0.35 mEq/L per 1 mm Hg rise in PCO₂.

    $$\Delta \text{PCO}_2 = 55 - 37 = 18 \,mm\,Hg$$

  4. Expected HCO₃⁻ rise from the chronic respiratory rule: 0.35 × 18 = 6.3 mEq/L. Expected HCO₃⁻ ≈ 22 + 6.3 ≈ 28 mEq/L.

    $$\text{expected HCO}_3^- = 22 + (0.35 \times 18) \approx 28 \,mEq/L$$

  5. Observed HCO₃⁻ is 33, well above the predicted 28 (±2 → 26–30). The bicarbonate is higher than chronic respiratory compensation alone explains. There's a concurrent metabolic alkalosis.

  6. Final interpretation: mixed disorder. Chronic respiratory acidosis (the bronchitis baseline) plus a superimposed metabolic alkalosis. Common causes in this setting: loop diuretic use (furosemide for the bronchitis), vomiting, corticosteroid effects.

Answer

Mixed disorder: chronic respiratory acidosis (from the underlying bronchitis) plus a concurrent metabolic alkalosis. Observed HCO₃⁻ of 33 is well above the ~28 predicted by chronic respiratory compensation alone. Normal pH is a clue here, not reassurance.

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Nutrition

Nutrition Clinical

NG-tube feeding ramp for an inappetent cat

An 8 kg cat is hospitalized for hepatic lipidosis. You place an NG tube and plan a 3-day ramp to 100% RER using a liquid diet at 1.0 kcal/mL, delivered as four boluses per day. Calculate RER, the Day 3 daily volume, and the per-bolus volume on Day 3. Compare the per-bolus volume to the 10 mL/kg per-feeding cap.

kcal/day
mL/day
mL
Does Day 3 bolus exceed the 10 mL/kg cap?
Hint

RER is not linear in body weight. It uses an exponent: the metabolic rate scales with body mass to the 3/4 power. You'll need to compute 8 raised to the 0.75 power (or use a calculator).

Another hint

Three steps. (1) RER = 70 × BW^0.75 (BW in kg). For 8 kg, BW^0.75 ≈ 4.76. (2) Day 3 is 100% RER in a 3-day ramp; daily volume = RER kcal/day ÷ diet kcal/mL. (3) Per-bolus = daily ÷ 4 feedings. The 10 mL/kg cap = 10 × 8 = 80 mL per feeding.

Show worked answer

  1. Compute RER from the standard formula. The 0.75 exponent reflects allometric scaling: metabolism rises with body mass but more slowly than linearly.

    $$\text{RER} = 70 \times (8)^{0.75} \approx 70 \times 4.76 \approx 333 \,\tfrac{kcal}{day}$$

  2. The 3-day ramp delivers 33% / 66% / 100% of RER on Days 1, 2, 3 respectively. The ramp is conservative: it gives the GI tract time to tolerate enteral nutrition and reduces refeeding-syndrome risk in chronically inappetent patients.

  3. Day 3 daily kcal target = 100% × 333 = 333 kcal. Convert to volume of liquid diet at 1.0 kcal/mL:

    $$\frac{333 \,\cancel{kcal}}{1.0 \,\cancel{kcal}/mL} = 333 \,\tfrac{mL}{day}$$

  4. Divide by 4 feedings per day for the per-bolus volume.

    $$\frac{333 \,mL}{4 \,\text{feedings}} \approx 83 \,\tfrac{mL}{\text{feeding}}$$

  5. Compare to the 10 mL/kg per-feeding cap. For an 8 kg cat, the cap is 80 mL. 83 mL slightly exceeds the cap.

    $$\text{cap} = 10 \,\tfrac{mL}{\cancel{kg}} \times 8 \,\cancel{kg} = 80 \,\tfrac{mL}{\text{feeding}}$$

  6. The fix is to add a fifth feeding rather than push the individual bolus higher. 333 mL ÷ 5 = ~67 mL per feeding, comfortably below the 80 mL cap. Boluses larger than the cap raise the risk of regurgitation and aspiration.

Answer

RER ≈ 333 kcal/day. Day 3 (100% RER) = 333 mL/day at 1.0 kcal/mL. Split across 4 feedings, that's ~83 mL per bolus, which slightly exceeds the 10 mL/kg per-feeding cap (80 mL). Either add a fifth feeding (drops bolus to ~67 mL) or accept the marginal overage with close monitoring for regurgitation.

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