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 recurrent gout starts allopurinol. Which laboratory result shows that the treatment is achieving its goal?
AA falling serum potassium
BA rising white cell count
CA rising serum calcium
DA falling serum uric acid
2.Lab values & electrolytesHard
A patient has taken metformin for 8 years and now has numbness in the feet, a hemoglobin of 10.2 g/dL (102 g/L) and an MCV of 108 fL. Which deficiency is most likely?
AVitamin D (calciferol) deficiency
BVitamin B12 deficiency
CIron deficiency from gut loss
DVitamin K deficiency
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.59, PaCO₂ 41 mmHg (5.5 kPa) and HCO₃⁻ 38 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.59PaCO₂ 41 mmHg (5.5 kPa)HCO₃⁻ 38 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.23, PaCO₂ 82 mmHg (10.9 kPa), HCO₃⁻ 33 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 7.23PaCO₂ 82 mmHg (10.9 kPa)HCO₃⁻ 33 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 78 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.03 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 10 mg/L?
78 kgCL = 0.03 L/kg/hCss = 10 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 51 kg patient has a phenytoin level of 7.5 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?
51 kgVd = 0.7 L/kgC = 7.5 → 15 mg/LS = 0.92
LD = mg
7.Clinical nursingMedium
A patient with type 2 diabetes is acutely unwell, very dehydrated and has a blood glucose of 40 mmol/L without significant ketones. What is the most likely diagnosis?
AHyperosmolar hyperglycaemic state
BDiabetes insipidus
CHypoglycaemia
DDiabetic ketoacidosis with dehydration
8.Clinical nursingHard
A postoperative patient has a heart rate of 115/min, temperature 38.6 °C, respiratory rate 26/min and blood pressure 85/50 mmHg despite a fluid bolus. What does this most likely indicate?
AAnxiety about discharge
BMild dehydration only
CA normal inflammatory response in the first days after surgery
DSeptic shock requiring urgent critical care review
3Solutions
1.Lab values & electrolytes
Answer: D · A falling serum uric acid
Why this is the answer
Allopurinol reduces uric acid production by blocking xanthine oxidase. Over time, a lower serum uric acid reduces gout attacks and dissolves urate crystals. Kidney function is checked, and a rash must be reported because of rare severe skin reactions.
Answer
D · A falling serum uric acid
2.Lab values & electrolytes
Answer: B · Vitamin B12 deficiency
Why this is the answer
Long-term metformin reduces vitamin B12 absorption in the gut. B12 deficiency causes macrocytic anemia and peripheral neuropathy, which can be mistaken for diabetic neuropathy. Periodic B12 testing is recommended for long-term users.
Answer
B · Vitamin B12 deficiency
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 7.59: alkalaemia (above 7.45)
PaCO₂ 41 mmHg: normal (35–45)
HCO₃⁻ 38 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 45.6–49.6 mmHg, so the measured 41 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 + 1·4.2 = 28.2 mmol/L
chronic: 24 + 3.5·4.2 = 38.7 mmol/L
Compare
measured: 33 mmol/L → acute on chronic
Answer
acute on chronic
5.Clinical pharmacokinetics
Answer: R = 29.25 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 = 0.03·78 = 2.34 L/h
Infusion rate
R = CL·Css/S = 2.34·10/0.8 = 29.25 mg/h
Answer
R = 29.25 mg/h
6.Clinical pharmacokinetics
Answer: LD = 291 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 = 0.7·51 = 35.7 L
Rise needed
15 − 7.5 = 7.5 mg/L
Loading dose
LD = 35.7·7.5/(0.92·1) = 291 mg
Answer
LD = 291 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: A · Hyperosmolar hyperglycaemic state
Why this is the answer
HHS usually affects people with type 2 diabetes. It develops over days with extreme hyperglycaemia, dehydration and high osmolality, but little ketosis.
Answer
A · Hyperosmolar hyperglycaemic state
8.Clinical nursing
Answer: D · Septic shock requiring urgent critical care review
Why this is the answer
Infection signs with hypotension not responding to fluid suggest septic shock. Vasopressors and critical care input are likely needed, along with antibiotics and source control.
Answer
D · Septic shock requiring urgent critical care review