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 who has been vomiting for two days has a BUN of 32 mg/dL (urea 11.4 mmol/L) and a normal creatinine. What is the most likely cause?
AOverhydration
BLow-protein diet
CDehydration
DLiver failure
2.Lab values & electrolytesHard
A patient on a thiazide diuretic has a sodium of 129 mEq/L, a potassium of 3.2 mEq/L and feels weak. Which explanation fits best?
AThe results are typical of kidney failure
BThe patient is overhydrated from IV fluids only
CThe diuretic is causing sodium and potassium loss
DThe patient has primary hyperaldosteronism with high sodium
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.27, PaCO₂ 43 mmHg (5.7 kPa) and HCO₃⁻ 19 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.27PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 19 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 35% oxygen shows PaO₂ 81 mmHg (10.8 kPa) and PaCO₂ 32 mmHg (4.3 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 30 mg/L at 2 h and 17.2 mg/L at 10 h. Assume first-order elimination. Find the half-life t½.
C1 = 30 mg/L (2 h)C2 = 17.2 mg/L (10 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 52-year-old man weighs 93 kg and has a serum creatinine of 2.4 mg/dL (212 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 gentamicin regimen is 80 mg every 8 h for a creatinine clearance of 100 mL/min. Keeping the dose, the interval is lengthened in proportion to the fall in clearance. Find the new dosing interval τ. Round to one decimal place.
Which mask is required for staff during an aerosol-generating procedure on a patient with a respiratory infection spread by aerosols?
AA standard fluid-repellent surgical mask with a visor
BA cloth face covering
CNo mask, as long as gloves are worn
DA fit-tested respirator mask (such as FFP3 or N95)
8.Clinical nursingHard
A patient after abdominal surgery has not passed urine for eight hours, has lower abdominal discomfort and a palpable bladder. What is the most likely problem?
AAcute kidney injury with no urine production
BDehydration only
CAcute urinary retention
DUrinary tract infection without retention
3Solutions
1.Lab values & electrolytes
Answer: C · Dehydration
Why this is the answer
When the body is short of fluid, the kidneys reabsorb more urea, so the BUN rises out of proportion to creatinine (a BUN-to-creatinine ratio above 20:1). Liver failure and a low-protein diet lower the BUN.
Answer
C · Dehydration
2.Lab values & electrolytes
Answer: C · The diuretic is causing sodium and potassium loss
Why this is the answer
Thiazide diuretics commonly cause hyponatremia and hypokalemia, especially in older adults. The nurse reports the results, and the prescriber reviews the drug and electrolyte replacement.
Answer
C · The diuretic is causing sodium and potassium loss
3.Arterial blood gas interpretation
Answer: uncompensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.27: acidaemia (below 7.35)
PaCO₂ 43 mmHg: normal (35–45)
HCO₃⁻ 19 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 34.5–38.5 mmHg, so the measured 43 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: A–a gradient = 128.5 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.
Elimination rate constant
k = ln(30/17.2)/8 = 0.5563/8 = 0.06954 h⁻¹
Half-life
t½ = 0.693/0.06954 = 9.968 h
Answer
t½ = 9.968 h
6.Clinical pharmacokinetics
Answer: τ = 16.9 h
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 − 52)·93/(72·2.4) = 47.36 mL/min
New interval
τ = 8·100/47.36 = 16.89 ≈ 16.9 h
Answer
τ = 16.9 h
In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.
7.Clinical nursing
Answer: D · A fit-tested respirator mask (such as FFP3 or N95)
Why this is the answer
Aerosol-generating procedures, such as intubation, release small infectious particles. Respirators that are fit-tested protect the wearer, while surgical masks mainly protect against droplets.
Answer
D · A fit-tested respirator mask (such as FFP3 or N95)
8.Clinical nursing
Answer: C · Acute urinary retention
Why this is the answer
Anaesthesia, opioids and pain can cause retention. A bladder scan confirms residual volume; catheterisation is often needed.