Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
Clinical pharmacokinetics
Rate constant and half-life
k = ln 2/t½ = 0.693/t½
First-order decay
C = C₀·e−kt = C₀·(1/2)t/t½
Two levels
k = ln(C₁/C₂)/Δt
Volume and clearance
Vd = dose/C₀, CL = k·Vd
Loading dose
LD = Vd·C/(S·F)
Maintenance rate
R = CL·Css/(S·F)
Average steady state
Css,avg = S·F·D/(CL·τ)
Steady state
1 − (1/2)ⁿ after n half-lives; ≈ 97% after 5
Infusion
C = (R/CL)(1 − e−kt)
Interval and dose from peak and trough
τ = ln(Cpeak/Ctrough)/k, D = Vd·(Cpeak − Ctrough)
Cockcroft–Gault
CrCl = (140 − age) × kg/(72 × SCr mg/dL), × 0.85 for women
Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
Check the result: levels fall by half every half-life, and steady state takes about five half-lives.
2Practice problems
1.Lab values & electrolytesMedium
A patient with hyponatremia from SIADH starts tolvaptan. Why is the serum sodium checked frequently in the first day?
AThe sodium always falls further
BThe sodium can rise too quickly
CThe drug causes severe hyperkalemia
DThe drug lowers blood glucose sharply
2.Lab values & electrolytesHard
A patient with chronic kidney disease stage 4 has a high phosphate, a low-normal calcium and a very high parathyroid hormone (PTH). What does this show?
APrimary hyperparathyroidism from a tumor
BSecondary hyperparathyroidism from CKD
CHypoparathyroidism after neck surgery
DVitamin D toxicity from supplements
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 145 mmol/L, glucose 11.9 mmol/L and urea 3.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 145 mmol/Lglucose 11.9 mmol/Lurea 3.6 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 21% oxygen shows PaO₂ 70 mmHg (9.3 kPa) and PaCO₂ 48 mmHg (6.4 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A 65 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.04 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 15 mg/L?
65 kgCL = 0.04 L/kg/hCss = 15 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 41-year-old woman weighs 92 kg and has a serum creatinine of 3.1 mg/dL (274 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual digoxin dose is 250 mcg for a creatinine clearance of 100 mL/min, and a fraction fe = 0.7 of the drug is excreted unchanged by the kidneys. Using the Dettli dose fraction Q = 1 − fe·(1 − CrCl/normal CrCl), find the adjusted dose D. Round to the nearest whole number.
Which electrolyte disturbance is most dangerous for the heart rhythm?
AHigh chloride
BHigh or low potassium
CLow sodium
DLow phosphate
8.Clinical nursingHard
A patient's urine output has been very low for 12 hours and creatinine has risen sharply. Which drug on the chart is most likely to need review?
AAn emollient cream
BA vitamin supplement
CParacetamol at usual doses
DA non-steroidal anti-inflammatory drug such as ibuprofen
3Solutions
1.Lab values & electrolytes
Answer: B · The sodium can rise too quickly
Why this is the answer
Tolvaptan blocks ADH receptors in the kidney, causing free water loss. Sodium can rise too fast, risking osmotic demyelination, so treatment is started in hospital with frequent sodium checks. Fluid restriction is usually avoided at the start.
Answer
B · The sodium can rise too quickly
2.Lab values & electrolytes
Answer: B · Secondary hyperparathyroidism from CKD
Why this is the answer
In CKD, phosphate retention and reduced activation of vitamin D lower calcium, which stimulates the parathyroid glands. The result is high PTH with high phosphate, causing bone disease. Phosphate binders, diet and active vitamin D are used.Checked against:Parathyroid Hormone (PTH) Test · MedlinePlus (NIH) (2023)
Answer
B · Secondary hyperparathyroidism from CKD
3.Arterial blood gas interpretation
Answer: Osm = 306 mOsm/kg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
Calculated osmolality
Osm = 2·145 + 11.9 + 3.6 = 305.5 ≈ 306 mOsm/kg
Answer
Osm = 306 mOsm/kg
4.Arterial blood gas interpretation
Answer: A–a gradient = 19.73 mmHg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
Clearance and volume: at steady state the dosing rate equals the elimination rate, so the maintenance rate = CL × Css. A loading dose is LD = Vd × C/(S × F), with S the salt factor (fraction of active drug in the salt) and F the bioavailability (1 for IV). From two levels, k = ln(C₁/C₂)/Δt; for an IV bolus Vd = dose/C₀ and CL = k × Vd. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Clearance
CL = 0.04·65 = 2.6 L/h
Infusion rate
R = CL·Css/S = 2.6·15/0.8 = 48.75 mg/h
Answer
R = 48.75 mg/h
6.Clinical pharmacokinetics
Answer: D = 136 mcg
Designing a regimen: with rapid IV doses the level falls from peak to trough during the interval, so τ = ln(Cpeak/Ctrough)/k and the dose that restores the peak is Vd × (Cpeak − Ctrough). The average steady-state level is Css,avg = S·F·dose/(CL·τ). An infusion rises as C = (R/CL)(1 − e−kt). For a drug cleared by the kidneys, the dose is scaled with the creatinine clearance. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 41)·92/(72·3.1)·0.85 = 34.69 mL/min
Dose fraction (Dettli)
Q = 1 − 0.7·(1 − 34.69/100) = 0.5428
Adjusted dose
D = 0.5428·250 = 135.7 ≈ 136 mcg
Answer
D = 136 mcg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: B · High or low potassium
Why this is the answer
Both hyperkalaemia and hypokalaemia can cause life-threatening arrhythmias. Potassium must never be given as an undiluted rapid IV bolus.
Answer
B · High or low potassium
8.Clinical nursing
Answer: D · A non-steroidal anti-inflammatory drug such as ibuprofen
Why this is the answer
In acute kidney injury, nephrotoxic drugs (such as NSAIDs and aminoglycosides) and drugs that reduce renal perfusion (such as ACE inhibitors and diuretics) are reviewed and often stopped.
Answer
D · A non-steroidal anti-inflammatory drug such as ibuprofen