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 an upper gastrointestinal bleed has a BUN of 48 mg/dL (urea 17 mmol/L) and a normal creatinine. What is the most likely explanation?
ADigested blood adds protein, which the liver turns into urea
BThe liver has stopped making urea because of the bleeding
CThe kidneys have suddenly lost most of their function
DSkeletal muscle is breaking down because of the blood loss
2.Lab values & electrolytesHard
A patient with diabetes and chronic kidney disease had a CT scan with IV contrast. Two days later, the creatinine has risen from 1.4 to 2.0 mg/dL (124 to 177 µmol/L). Which action is most appropriate?
AGive extra metformin, because the glucose rises after contrast
BRestrict all fluids to protect the kidneys from more contrast
CIgnore it, because contrast never affects kidney function
DReport it, monitor urine output and hold metformin until reviewed
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 56 mmHg (7.5 kPa). Use the rule that in chronic respiratory acidosis HCO₃⁻ rises by 3.5 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic alkalosis with HCO₃⁻ 44 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 44 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 210 mg of tobramycin. The level back-extrapolated to time zero is 8 mg/L and the half-life is 3 h. Find the volume of distribution Vd.
D = 210 mgC₀ = 8 mg/Lt½ = 3 h
Vd = L
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 72-year-old man weighs 90 kg and has a serum creatinine of 2 mg/dL (177 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 digoxin dose is 250 mcg for a creatinine clearance of 120 mL/min, and a fraction fe = 0.7 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's NEWS2 score has risen to 7. What is the appropriate response?
AEmergency assessment by a clinical team with critical care skills
BRecord the score, increase observations and wait for the morning ward round
CContinue observations every 12 hours
DGive a sedative and recheck later
8.Clinical nursingHard
A patient on an unfractionated heparin infusion is also started on warfarin. Which tests are used to monitor each drug?
AINR for heparin and APTT for warfarin
BPlatelet count for both drugs
CAPTT for heparin and INR for warfarin
DD-dimer for both drugs
3Solutions
1.Lab values & electrolytes
Answer: A · Digested blood adds protein, which the liver turns into urea
Why this is the answer
Blood in the upper gut is digested, and the absorbed protein is converted to urea, so the BUN rises out of proportion to the creatinine. Reduced kidney blood flow from volume loss can add to the rise.Checked against:BUN - blood test · MedlinePlus (NIH) (2025)
Answer
A · Digested blood adds protein, which the liver turns into urea
2.Lab values & electrolytes
Answer: D · Report it, monitor urine output and hold metformin until reviewed
Why this is the answer
A creatinine rise of 0.3 mg/dL or more within 48 hours meets the criteria for AKI, and contrast is a recognized risk factor, especially in CKD and diabetes. The nurse reports it, monitors fluid balance and holds metformin until kidney function is reviewed, as metformin can accumulate.Checked against:Acute Kidney Injury · StatPearls (NCBI Bookshelf) (2026); Abdominal CT scan · MedlinePlus (NIH) (2024); Acute kidney failure · MedlinePlus (NIH) (2026)
Answer
D · Report it, monitor urine output and hold metformin until reviewed
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 29.6 mmol/L
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.
Change in PaCO₂
|56 − 40| = 16 mmHg → 1.6 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 3.5·1.6 = 29.6 mmol/L
Answer
HCO₃⁻ = 29.6 mmol/L
4.Arterial blood gas interpretation
Answer: PaCO₂ = 51.8 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.
Expected PaCO₂
0.7·44 + 21 = 51.8 mmHg
Answer
PaCO₂ = 51.8 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: Vd = 26.25 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 = 210/8 = 26.25 L
Answer
Vd = 26.25 L
6.Clinical pharmacokinetics
Answer: D = 137 mcg
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 − 72)·90/(72·2) = 42.5 mL/min
Dose fraction (Dettli)
Q = 1 − 0.7·(1 − 42.5/120) = 0.5479
Adjusted dose
D = 0.5479·250 = 137 mcg
Answer
D = 137 mcg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: A · Emergency assessment by a clinical team with critical care skills
Why this is the answer
In NEWS2, a score of 7 or more triggers an emergency response, usually continuous monitoring and urgent review by a team competent in managing critically ill patients.
Answer
A · Emergency assessment by a clinical team with critical care skills
8.Clinical nursing
Answer: C · APTT for heparin and INR for warfarin
Why this is the answer
The APTT reflects the intrinsic pathway and is used to adjust heparin infusions. Warfarin is monitored by the INR, a standardised prothrombin time. Platelets are also checked on heparin to detect HIT.