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 type 2 diabetes has a target HbA1c below 7% (53 mmol/mol). Today's result is 7.8% (62 mmol/mol). What is the best interpretation?
AAverage glucose over the past months has been above the target
BThe result shows today's fasting glucose is above the target
CThe patient has had frequent episodes of severe hypoglycemia
DGlucose control has been excellent for the past several months
2.Lab values & electrolytesHard
A patient on hemodialysis receives epoetin, but the hemoglobin has not risen in 2 months. Which results should be checked first?
ATSH and the free T3 level
BAmylase and lipase levels
CFerritin and transferrin saturation
DTroponin and creatine kinase
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.60, PaCO₂ 42 mmHg (5.6 kPa) and HCO₃⁻ 40 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.60PaCO₂ 42 mmHg (5.6 kPa)HCO₃⁻ 40 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.28, PaCO₂ 69 mmHg (9.2 kPa), HCO₃⁻ 31 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.28PaCO₂ 69 mmHg (9.2 kPa)HCO₃⁻ 31 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 95 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.045 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 15 mg/L?
95 kgCL = 0.045 L/kg/hCss = 15 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin IV every 12 h; the vancomycin clearance is 5 L/h. What dose D in mg gives an average steady-state level of 20 mg/L?
CL = 5 L/hτ = 12 hS = 1F = 1Css,avg = 20 mg/L
D = mg
7.Clinical nursingMedium
A patient started on an ACE inhibitor develops a persistent dry cough. What is the likely cause?
AAsthma caused by a beta-blocker
BA recognised side effect of the ACE inhibitor
CPulmonary embolism
DAn allergic reaction to a recently given penicillin
8.Clinical nursingHard
A patient after a total hip replacement suddenly has severe hip pain, a shortened leg and the leg is rotated. What is the most likely complication?
AUrinary retention
BWound infection
CDislocation of the prosthesis
DDeep vein thrombosis in the calf
3Solutions
1.Lab values & electrolytes
Answer: A · Average glucose over the past months has been above the target
Why this is the answer
An HbA1c of 7.8% shows that average glucose has been above the individual target over about 2–3 months. The nurse explores diet, activity, medication use and barriers, and the treatment plan may be adjusted.Checked against:Hemoglobin A1C (HbA1c) Test · MedlinePlus (NIH) (2025)
Answer
A · Average glucose over the past months has been above the target
2.Lab values & electrolytes
Answer: C · Ferritin and transferrin saturation
Why this is the answer
Iron deficiency is the most common reason for a poor response to erythropoiesis-stimulating agents, because the marrow needs iron to make hemoglobin. Ferritin and transferrin saturation are checked, and IV iron is often given in dialysis patients. Infection and inflammation can also blunt the response.Checked against:Anemia of Chronic Kidney Disease · StatPearls (NCBI Bookshelf) (2024); Erythropoietin Stimulating Agents · StatPearls (NCBI Bookshelf) (2024)
Answer
C · Ferritin and transferrin saturation
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 7.60: alkalaemia (above 7.45)
PaCO₂ 42 mmHg: normal (35–45)
HCO₃⁻ 40 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 47–51 mmHg, so the measured 42 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: acute on chronic
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.
Expected HCO₃⁻
acute: 24 + 1·2.9 = 26.9 mmol/L
chronic: 24 + 3.5·2.9 = 34.15 mmol/L
Compare
measured: 31 mmol/L → acute on chronic
Answer
acute on chronic
5.Clinical pharmacokinetics
Answer: R = 80.16 mg/h
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 = 0.045·95 = 4.275 L/h
Infusion rate
R = CL·Css/S = 4.275·15/0.8 = 80.16 mg/h
Answer
R = 80.16 mg/h
6.Clinical pharmacokinetics
Answer: D = 1200 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.
Dose
D = Css,avg·CL·τ/(S·F) = 20·5·12/(1·1) = 1200 mg
Answer
D = 1200 mg
7.Clinical nursing
Answer: B · A recognised side effect of the ACE inhibitor
Why this is the answer
Dry cough is a common side effect of ACE inhibitors such as ramipril. An angiotensin II receptor blocker is often substituted.
Answer
B · A recognised side effect of the ACE inhibitor
8.Clinical nursing
Answer: C · Dislocation of the prosthesis
Why this is the answer
Hip dislocation after arthroplasty causes sudden pain, shortening and rotation. It needs urgent review and reduction; patients are taught movement precautions.