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 levothyroxine has palpitations, weight loss and a TSH of 0.05 mIU/L. What is the most likely explanation?
AThe patient has stopped the drug
BThe levothyroxine dose is too low
CThe levothyroxine dose is too high
DThe thyroid is now completely normal
2.Lab values & electrolytesHard
In the first 24 hours after a major burn, which laboratory pattern is most expected?
ALow potassium and low hematocrit
BHigh potassium and high hematocrit
CLow sodium and low hematocrit
DHigh calcium and high albumin
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 16 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 16 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
Practice values: Na⁺ 133, Cl⁻ 100 and HCO₃⁻ 17 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 133 mmol/LCl⁻ 100 mmol/LHCO₃⁻ 17 mmol/L
delta ratio =
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 220 mg of gentamicin. The level back-extrapolated to time zero is 9 mg/L and the half-life is 2 h. Find the volume of distribution Vd.
D = 220 mgC₀ = 9 mg/Lt½ = 2 h
Vd = L
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 56-year-old man weighs 73 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 atenolol dose is 100 mg for a creatinine clearance of 100 mL/min, and a fraction fe = 0.9 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 on a ward develops wheeze and breathlessness after the first dose of a beta-blocker. Which history makes this most likely?
AOsteoarthritis
BMigraine
CAsthma
DHypothyroidism
8.Clinical nursingHard
A patient has had nausea and vomiting for three days and is taking furosemide. Which electrolyte problem is most likely?
AHypokalaemia
BHypercalcaemia
CHyperkalaemia
DHypernatraemia only
3Solutions
1.Lab values & electrolytes
Answer: C · The levothyroxine dose is too high
Why this is the answer
A very low TSH with symptoms of excess thyroid hormone suggests over-replacement. The dose is usually reduced, because long-term over-replacement raises the risk of atrial fibrillation and bone loss, especially in older adults.Checked against:TSH test · MedlinePlus (NIH) (2026)
Answer
C · The levothyroxine dose is too high
2.Lab values & electrolytes
Answer: B · High potassium and high hematocrit
Why this is the answer
Early after a burn, damaged cells release potassium, and plasma shifts into the tissues, which concentrates the blood and raises the hematocrit. Later, as fluid returns and the kidneys excrete potassium, hypokalemia can develop.Checked against:High potassium level · MedlinePlus (NIH) (2025)
Answer
B · High potassium and high hematocrit
3.Arterial blood gas interpretation
Answer: PaCO₂ = 32 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 = 1.5·16 + 8 = 32 mmHg
range: 30–34 mmHg
Answer
PaCO₂ = 32 mmHg
4.Arterial blood gas interpretation
Answer: delta ratio = 0.5714
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.
Anion gap
AG = 133 − (100 + 17) = 16
Delta ratio
(16 − 12)/(24 − 17) = 4/7 = 0.5714
Answer
delta ratio = 0.5714
Below 1: an added normal anion gap acidosis.
5.Clinical pharmacokinetics
Answer: Vd = 24.44 L
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 = 220/9 = 24.44 L
Answer
Vd = 24.44 L
6.Clinical pharmacokinetics
Answer: D = 45 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 − 56)·73/(72·2.2) = 38.71 mL/min
Dose fraction (Dettli)
Q = 1 − 0.9·(1 − 38.71/100) = 0.4484
Adjusted dose
D = 0.4484·100 = 44.84 ≈ 45 mg
Answer
D = 45 mg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: C · Asthma
Why this is the answer
Non-selective and even cardioselective beta-blockers can cause bronchospasm, so they are usually avoided in asthma.
Answer
C · Asthma
8.Clinical nursing
Answer: A · Hypokalaemia
Why this is the answer
Vomiting and loop diuretics both cause potassium loss. Hypokalaemia causes weakness, cramps and arrhythmias and needs monitoring and replacement.