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
Which ECG change is linked to hypocalcemia?
ATall, peaked T waves
BA prominent delta wave
CA prolonged QT interval
DA shortened QT interval
2.Lab values & electrolytesHard
After a crush injury, a patient's urine dipstick is strongly positive for blood, but microscopy shows almost no red cells. CK is very high. What best explains this?
AA kidney stone is scraping the ureter lining
BMyoglobin from damaged muscle is in the urine
CBleeding is coming from the bladder lining
DThe sample was contaminated with menstrual blood
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 152 mmol/L, glucose 10.1 mmol/L and urea 15.4 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 152 mmol/Lglucose 10.1 mmol/Lurea 15.4 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 35% oxygen shows PaO₂ 84 mmHg (11.2 kPa) and PaCO₂ 60 mmHg (8.0 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 25 mg/L at 2 h and 14.4 mg/L at 10 h. Assume first-order elimination. Find the elimination rate constant k.
C1 = 25 mg/L (2 h)C2 = 14.4 mg/L (10 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 53-year-old woman weighs 66 kg and has a serum creatinine of 2.2 mg/dL (194 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 levetiracetam dose is 1000 mg for a creatinine clearance of 120 mL/min, and a fraction fe = 0.66 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.
A patient with dysphagia after a stroke has been cleared for a modified diet. How should they be positioned for meals?
ALying on their side with the knees bent
BSemi-reclined at about 20 degrees, with the head supported on pillows
CSitting fully upright, and staying upright for a time after eating
DLying flat on their back with the head turned to one side
8.Clinical nursingHard
A patient who speaks little English needs to give consent for a procedure. Their 14-year-old child offers to interpret. What is the best action?
ALet the child interpret to save time
BUse gestures and proceed
CArrange a professional interpreter
DAsk the patient to sign the form without explanation
3Solutions
1.Lab values & electrolytes
Answer: C · A prolonged QT interval
Why this is the answer
Low calcium prolongs the QT interval, which can lead to dangerous arrhythmias. High calcium shortens the QT interval, and peaked T waves are typical of hyperkalemia.
Answer
C · A prolonged QT interval
2.Lab values & electrolytes
Answer: B · Myoglobin from damaged muscle is in the urine
Why this is the answer
The dipstick blood test detects heme, which is present in both hemoglobin and myoglobin. Positive "blood" without red cells, together with a very high CK, indicates myoglobinuria from rhabdomyolysis, which can damage the kidneys. IV fluids and close monitoring of urine output and potassium are needed.Checked against:Myoglobin urine test · MedlinePlus (NIH) (2025); Rhabdomyolysis · MedlinePlus (NIH) (2025); Creatine Kinase · MedlinePlus (NIH) (2023)
Answer
B · Myoglobin from damaged muscle is in the urine
3.Arterial blood gas interpretation
Answer: Osm = 330 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 = 2·152 + 10.1 + 15.4 = 329.5 ≈ 330 mOsm/kg
Answer
Osm = 330 mOsm/kg
4.Arterial blood gas interpretation
Answer: A–a gradient = 90.55 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(25/14.4)/8 = 0.5516/8 = 0.06896 h⁻¹
Answer
k = 0.06896 h⁻¹
6.Clinical pharmacokinetics
Answer: D = 509 mg
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 − 53)·66/(72·2.2)·0.85 = 30.81 mL/min
Dose fraction (Dettli)
Q = 1 − 0.66·(1 − 30.81/120) = 0.5095
Adjusted dose
D = 0.5095·1000 = 509.5 ≈ 509 mg
Answer
D = 509 mg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: C · Sitting fully upright, and staying upright for a time after eating
Why this is the answer
Sitting upright during and after meals reduces the risk of aspiration. The patient should also be supervised and given the consistency recommended by the speech and language therapist.
Answer
C · Sitting fully upright, and staying upright for a time after eating
8.Clinical nursing
Answer: C · Arrange a professional interpreter
Why this is the answer
Professional interpreters provide accurate, impartial translation and protect confidentiality. Children and family members should not be used for important clinical discussions such as consent.