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 starts lisinopril (an ACE inhibitor). Which laboratory values should the nurse expect to be monitored?
APotassium and creatinine
BCalcium and phosphate
CTSH and free T4
DALT and bilirubin
2.Lab values & electrolytesHard
Which set of results best shows that diabetic ketoacidosis is resolving?
AA closing anion gap with a rising bicarbonate and pH
BA falling glucose with a still widening anion gap
CA falling glucose with a falling serum bicarbonate
DA rising potassium with a falling bicarbonate
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.27, PaCO₂ 18 mmHg (2.4 kPa) and HCO₃⁻ 8 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.27PaCO₂ 18 mmHg (2.4 kPa)HCO₃⁻ 8 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 50% oxygen shows PaO₂ 58 mmHg (7.7 kPa) and PaCO₂ 55 mmHg (7.3 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A patient has a lidocaine clearance of 600 mL/min. What infusion rate R in mg/min keeps the steady-state level at 2.5 mg/L?
CL = 600 mL/minCss = 2.5 mg/L
R = mg/min
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 51 kg patient has a digoxin level of 0.3 mcg/L; the target is 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?
51 kgVd = 7 L/kgC = 0.3 → 0.8 mcg/LF = 0.7
LD = mcg
7.Clinical nursingMedium
An older patient is at high risk of falls on the ward. Which nursing action is most appropriate?
AUse bed rails for every patient automatically
BRemove the patient's glasses to prevent breakage
CRestrict the patient to bed at all times
DKeep the call bell and belongings within reach and review medicines that cause drowsiness
8.Clinical nursingHard
Which nursing assessment best helps detect early deterioration in a patient after major surgery?
ARegular full sets of observations with an early warning score and trend review
BAsking the patient only whether they feel well
CMeasuring temperature once a day
DChecking the wound only at discharge
3Solutions
1.Lab values & electrolytes
Answer: A · Potassium and creatinine
Why this is the answer
ACE inhibitors reduce aldosterone, so potassium can rise, and they lower the filtration pressure in the kidney, so creatinine can rise. Both are checked before and soon after starting or increasing the dose.Checked against:Hyperkalemia · StatPearls (NCBI Bookshelf) (2025)
Answer
A · Potassium and creatinine
2.Lab values & electrolytes
Answer: A · A closing anion gap with a rising bicarbonate and pH
Why this is the answer
DKA has resolved when the ketoacidosis has cleared: ketones fall, the anion gap normalizes, bicarbonate rises and the pH returns towards normal. Glucose alone is a poor guide, because it falls earlier.
Answer
A · A closing anion gap with a rising bicarbonate and pH
3.Arterial blood gas interpretation
Answer: partially compensated
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 7.27: acidaemia (below 7.35)
PaCO₂ 18 mmHg: low (35–45)
HCO₃⁻ 8 mmol/L: low (22–26)
Compensation
partially compensated
Answer
partially compensated
4.Arterial blood gas interpretation
Answer: A–a gradient = 229.8 mmHg
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.
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 in L/min
600 mL/min = 0.6 L/min
Infusion rate
R = 0.6·2.5 = 1.5 mg/min
Answer
R = 1.5 mg/min
6.Clinical pharmacokinetics
Answer: LD = 255 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.
Volume of distribution
Vd = 7·51 = 357 L
Rise needed
0.8 − 0.3 = 0.5 mcg/L
Loading dose
LD = 357·0.5/(1·0.7) = 255 mcg
Answer
LD = 255 mcg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · Keep the call bell and belongings within reach and review medicines that cause drowsiness
Why this is the answer
Falls prevention includes assessing risk factors, reviewing sedating drugs, providing glasses, footwear and walking aids, keeping essentials in reach and supporting toileting.
Answer
D · Keep the call bell and belongings within reach and review medicines that cause drowsiness
8.Clinical nursing
Answer: A · Regular full sets of observations with an early warning score and trend review
Why this is the answer
Track-and-trigger systems such as NEWS2 combine respiratory rate, SpO2, blood pressure, pulse, consciousness and temperature to identify deterioration early.
Answer
A · Regular full sets of observations with an early warning score and trend review