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 has burning on urination and frequency. The dipstick is positive for nitrites and leukocyte esterase, and a urine culture is ordered. When should the sample be taken?
AAt any time, as antibiotics do not matter
BBefore the first dose of antibiotic
CAfter the symptoms have fully settled
DAfter 3 days of antibiotic treatment
2.Lab values & electrolytesHard
Which finding most strongly suggests that a patient’s raised creatinine is due to acute kidney injury rather than long-standing chronic kidney disease?
AThe patient has had a high creatinine for many years
BThe creatinine was normal three days ago and has doubled since
CThe patient has small kidneys on an ultrasound scan
DThe patient has long-standing anemia
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 141 mmol/L, glucose 29.2 mmol/L and urea 18.2 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 141 mmol/Lglucose 29.2 mmol/Lurea 18.2 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.32, PaCO₂ 66 mmHg (8.8 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 7.32PaCO₂ 66 mmHg (8.8 kPa)HCO₃⁻ 33 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 90 mg of tobramycin. The level back-extrapolated to time zero is 7 mg/L and the half-life is 3 h. Find the clearance CL in L/h.
D = 90 mgC₀ = 7 mg/Lt½ = 3 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 98 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.05 L/kg/h. What dose D in mg gives an average steady-state level of 12 mg/L?
Which side effect is associated with SGLT2 inhibitors such as dapagliflozin?
ABradycardia
BSevere weight gain and fluid retention
CGenital fungal infections
DHair loss
8.Clinical nursingHard
A patient with small cell lung cancer has hyponatraemia, concentrated urine and no signs of dehydration. Which condition and first measure fit best?
ACushing's syndrome, managed with potassium
BDiabetes insipidus, managed with extra fluids
CSIADH, often managed first with fluid restriction
DDehydration, managed with an IV fluid bolus
3Solutions
1.Lab values & electrolytes
Answer: B · Before the first dose of antibiotic
Why this is the answer
The culture sample is taken before antibiotics, so the organism can be grown and its sensitivities tested. A sample taken after antibiotics may give a false negative culture.
Answer
B · Before the first dose of antibiotic
2.Lab values & electrolytes
Answer: B · The creatinine was normal three days ago and has doubled since
Why this is the answer
A rapid rise from a normal baseline over days is the hallmark of acute kidney injury. Long-standing raised values, small kidneys and chronic anemia point to chronic kidney disease.Checked against:Acute Kidney Injury · StatPearls (NCBI Bookshelf) (2026)
Answer
B · The creatinine was normal three days ago and has doubled since
3.Arterial blood gas interpretation
Answer: Osm = 329 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 = 2·141 + 29.2 + 18.2 = 329.4 ≈ 329 mOsm/kg
Answer
Osm = 329 mOsm/kg
4.Arterial blood gas interpretation
Answer: 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 + 1·2.6 = 26.6 mmol/L
chronic: 24 + 3.5·2.6 = 33.1 mmol/L
Compare
measured: 33 mmol/L → chronic
Answer
chronic
5.Clinical pharmacokinetics
Answer: CL = 2.971 L/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.
Volume of distribution
Vd = 90/7 = 12.86 L
Rate constant
k = 0.693/3 = 0.231 h⁻¹
Clearance
CL = 0.231·12.86 = 2.971 L/h
Answer
CL = 2.971 L/h
6.Clinical pharmacokinetics
Answer: D = 705.6 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.
Clearance
CL = 0.05·98 = 4.9 L/h
Dose
D = Css,avg·CL·τ/(S·F) = 12·4.9·12/(1·1) = 705.6 mg
Answer
D = 705.6 mg
7.Clinical nursing
Answer: C · Genital fungal infections
Why this is the answer
SGLT2 inhibitors cause glycosuria, which increases the risk of genital thrush and urinary infections. They can also rarely cause DKA with only modestly raised glucose.
Answer
C · Genital fungal infections
8.Clinical nursing
Answer: C · SIADH, often managed first with fluid restriction
Why this is the answer
In SIADH, excess ADH makes the kidneys retain water, so blood sodium is diluted while urine is concentrated. Small cell lung cancer is a classic cause; fluid restriction is a common first step.
Answer
C · SIADH, often managed first with fluid restriction