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 on a heparin infusion has blood in the urine and an aPTT far above the target. The nurse stops the infusion and informs the provider. Which antidote should the nurse anticipate?
AVitamin K (phytonadione)
BProtamine sulfate
CCalcium gluconate
DNaloxone
2.Lab values & electrolytesHard
A patient with diabetes and chronic kidney disease had a CT scan with IV contrast. Two days later, the creatinine has risen from 1.4 to 2.0 mg/dL (124 to 177 µmol/L). Which action is most appropriate?
ARestrict all fluids to protect the kidneys from more contrast
BIgnore it, because contrast never affects kidney function
CReport it, monitor urine output and hold metformin until reviewed
DGive extra metformin, because the glucose rises after contrast
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 14 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 14 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 40% oxygen shows PaO₂ 55 mmHg (7.3 kPa) and PaCO₂ 42 mmHg (5.6 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). An 84 kg patient is to be loaded with digoxin tablets to reach a level of 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.
84 kgVd = 7 L/kgC = 0.8 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 66 kg patient has a phenytoin level of 6 mg/L; the target is 15 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?
66 kgVd = 0.7 L/kgC = 6 → 15 mg/LS = 0.92
LD = mg
7.Clinical nursingMedium
A nurse finds a patient lying on the floor beside the bed. What should be done first?
ALift the patient straight back into bed so they are comfortable
BPhone the patient's family
CComplete the incident report form
DCheck the patient for injuries before moving them
8.Clinical nursingHard
A patient has a single hot, swollen, very painful knee and a fever. Why is urgent joint aspiration important?
ATo prevent osteoporosis
BTo measure the patient's cholesterol
CTo check for rheumatoid arthritis antibodies in the joint fluid
DTo rule out septic arthritis, which can rapidly destroy the joint
Answer: C · Report it, monitor urine output and hold metformin until reviewed
Why this is the answer
A creatinine rise of 0.3 mg/dL or more within 48 hours meets the criteria for AKI, and contrast is a recognized risk factor, especially in CKD and diabetes. The nurse reports it, monitors fluid balance and holds metformin until kidney function is reviewed, as metformin can accumulate.Checked against:Acute Kidney Injury · StatPearls (NCBI Bookshelf) (2026); Abdominal CT scan · MedlinePlus (NIH) (2024); Acute kidney failure · MedlinePlus (NIH) (2026)
Answer
C · Report it, monitor urine output and hold metformin until reviewed
3.Arterial blood gas interpretation
Answer: PaCO₂ = 29 mmHg
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Winter’s formula
PaCO₂ = 1.5·14 + 8 = 29 mmHg
range: 27–31 mmHg
Answer
PaCO₂ = 29 mmHg
4.Arterial blood gas interpretation
Answer: A–a gradient = 177.7 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.
Volume of distribution
Vd = 7·84 = 588 L
Amount of active drug needed
Vd·C = 588·0.8 = 470.4 mcg
Loading dose
LD = 470.4/(1·0.7) = 672 mcg
Answer
LD = 672 mcg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: LD = 452 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.
Volume of distribution
Vd = 0.7·66 = 46.2 L
Rise needed
15 − 6 = 9 mg/L
Loading dose
LD = 46.2·9/(0.92·1) = 452 mg
Answer
LD = 452 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · Check the patient for injuries before moving them
Why this is the answer
After a fall, the patient is assessed first, including airway, breathing, circulation, head injury and possible fractures, before deciding how to move them safely. The fall is then reported and the care plan reviewed.
Answer
D · Check the patient for injuries before moving them
8.Clinical nursing
Answer: D · To rule out septic arthritis, which can rapidly destroy the joint
Why this is the answer
Septic arthritis can destroy cartilage within days. Aspiration allows microscopy and culture, and helps distinguish infection from crystal arthritis such as gout.
Answer
D · To rule out septic arthritis, which can rapidly destroy the joint