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 chest pain that started one hour ago has a normal first troponin. What is the correct interpretation?
AInfarction is not ruled out; troponin is repeated per protocol
BTroponin is not useful in patients with chest pain
CInfarction is ruled out, so no further troponin tests are needed
DThe patient can be discharged home straight away
2.Lab values & electrolytesHard
A patient on warfarin with a stable INR of 2.4 starts a course of an antibiotic such as ciprofloxacin. A week later the INR is 4.6. What is the most likely explanation?
AA laboratory error is the only possible explanation
BThe antibiotic has cured the patient’s need for anticoagulation
CThe patient has eaten more vitamin K–rich food
DA drug interaction that increases the effect of warfarin
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 147 mmol/L, glucose 23.5 mmol/L and urea 11.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 147 mmol/Lglucose 23.5 mmol/Lurea 11.6 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.22, PaCO₂ 45 mmHg (6.0 kPa), HCO₃⁻ 18 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.22PaCO₂ 45 mmHg (6.0 kPa)HCO₃⁻ 18 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 51 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 15 mg/L?
51 kgCL = 0.03 L/kg/hCss = 15 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A theophylline infusion is started without a loading dose. The half-life in this patient is 6 h. How long t does it take to reach 90% of the steady-state level?
t½ = 6 h90% Css
t = h
7.Clinical nursingMedium
Which examination findings suggest a large pleural effusion on one side?
ADull percussion and reduced breath sounds over the lung base
BA hyper-resonant chest with the trachea pulled towards that side
CWheeze heard over both lungs
DNormal breath sounds and normal percussion
8.Clinical nursingHard
Why must blood samples for a cross-match be labelled at the bedside immediately after they are taken?
ATo prevent "wrong blood in tube" errors that can cause fatal reactions
BTo let the patient sign the label
CTo keep the sample warm so the cells do not break down in transit
DTo speed up the laboratory's computer system and reduce the waiting time
3Solutions
1.Lab values & electrolytes
Answer: A · Infarction is not ruled out; troponin is repeated per protocol
Why this is the answer
Troponin may still be normal early after the onset of symptoms. Serial measurements, together with the ECG and the symptoms, are needed to rule infarction in or out. Many protocols repeat high-sensitivity troponin after 1–3 hours.Checked against:Troponin Test · MedlinePlus (NIH) (2023)
Answer
A · Infarction is not ruled out; troponin is repeated per protocol
2.Lab values & electrolytes
Answer: D · A drug interaction that increases the effect of warfarin
Why this is the answer
Several antibiotics increase warfarin’s effect by reducing its breakdown or by changing gut bacteria that make vitamin K. The INR is monitored more often when such drugs are started. More vitamin K in the diet would lower the INR.Checked against:Taking warfarin · MedlinePlus (NIH) (2025)
Answer
D · A drug interaction that increases the effect of warfarin
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 = 2·147 + 23.5 + 11.6 = 329.1 ≈ 329 mOsm/kg
Answer
Osm = 329 mOsm/kg
4.Arterial blood gas interpretation
Answer: added respiratory acidosis
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 PaCO₂
1.5·18 + 8 = 35 → 33–37 mmHg
Compare
measured: 45 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
5.Clinical pharmacokinetics
Answer: R = 28.69 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 = 0.03·51 = 1.53 L/h
Infusion rate
R = CL·Css/S = 1.53·15/0.8 = 28.69 mg/h
Answer
R = 28.69 mg/h
6.Clinical pharmacokinetics
Answer: t = 19.93 h
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.
Rate constant
k = 0.693/6 = 0.1155 h⁻¹
Time to the given fraction of steady state
t = −ln(1 − 0.9)/0.1155 = 2.303/0.1155 = 19.93 h
Answer
t = 19.93 h
That is 3.322 half-lives.
7.Clinical nursing
Answer: A · Dull percussion and reduced breath sounds over the lung base
Why this is the answer
Fluid in the pleural space muffles breath sounds and makes percussion stony dull. A large effusion may push the trachea away from the affected side.
Answer
A · Dull percussion and reduced breath sounds over the lung base
8.Clinical nursing
Answer: A · To prevent "wrong blood in tube" errors that can cause fatal reactions
Why this is the answer
A sample labelled with the wrong patient's details can lead to incompatible blood being issued. Labelling at the bedside after positive identification is a key transfusion safety step.
Answer
A · To prevent "wrong blood in tube" errors that can cause fatal reactions