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 is on an IV insulin infusion protocol in the ICU. How often is the blood glucose usually checked?
AOnce a day before breakfast
BOnce every 12 hours
COnly if symptoms occur
DAbout every hour
2.Lab values & electrolytesHard
A patient with acute kidney injury has a potassium of 6.8 mEq/L despite medical treatment, a pH of 7.12 and pulmonary edema not responding to diuretics. Which treatment is most likely needed?
AMore oral fluid
BA potassium supplement
CAn IV magnesium dose
DUrgent dialysis
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 143 mmol/L, glucose 29.2 mmol/L and urea 15.3 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 143 mmol/Lglucose 29.2 mmol/Lurea 15.3 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.31, PaCO₂ 29 mmHg (3.9 kPa), HCO₃⁻ 14 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.31PaCO₂ 29 mmHg (3.9 kPa)HCO₃⁻ 14 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 76 kg patient receives a theophylline infusion. The clearance is 0.04 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 10 mg/L?
76 kgCL = 0.04 L/kg/hCss = 10 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). An 82 kg patient has a digoxin level of 0.5 mcg/L; the target is 1 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?
82 kgVd = 7 L/kgC = 0.5 → 1 mcg/LF = 0.7
LD = mcg
7.Clinical nursingMedium
A patient has a fentanyl patch for chronic pain. Which advice is correct?
AApply the patch to broken skin for faster effect
BKeep using the old patch with the new one
CAvoid heat sources such as hot water bottles over the patch
DCut the patch in half if the pain improves, to lower the dose
8.Clinical nursingHard
A patient has a swollen, warm, tender calf on one side. Which action should the nurse avoid?
AChecking whether VTE prophylaxis was given
BMassaging the calf
CMeasuring and comparing calf circumference
DReporting the findings to the doctor
3Solutions
1.Lab values & electrolytes
Answer: D · About every hour
Why this is the answer
IV insulin acts quickly and has a short half-life, so glucose is usually checked hourly (or as the protocol specifies) and the rate adjusted. This reduces the risk of hypoglycemia and keeps glucose in the target range.
Answer
D · About every hour
2.Lab values & electrolytes
Answer: D · Urgent dialysis
Why this is the answer
Indications for urgent dialysis in AKI include refractory hyperkalemia, severe metabolic acidosis, fluid overload not responding to diuretics, uremic complications and some poisonings. This patient has three of them.Checked against:Acute tubular necrosis · MedlinePlus (NIH) (2025)
Answer
D · Urgent dialysis
3.Arterial blood gas interpretation
Answer: Osm = 331 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 = 2·143 + 29.2 + 15.3 = 330.5 ≈ 331 mOsm/kg
Answer
Osm = 331 mOsm/kg
4.Arterial blood gas interpretation
Answer: appropriate compensation
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
1.5·14 + 8 = 29 → 27–31 mmHg
Compare
measured: 29 mmHg → appropriate compensation
Answer
appropriate compensation
5.Clinical pharmacokinetics
Answer: R = 30.4 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 = 0.04·76 = 3.04 L/h
Infusion rate
R = 3.04·10 = 30.4 mg/h
Answer
R = 30.4 mg/h
6.Clinical pharmacokinetics
Answer: LD = 410 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 = 7·82 = 574 L
Rise needed
1 − 0.5 = 0.5 mcg/L
Loading dose
LD = 574·0.5/(1·0.7) = 410 mcg
Answer
LD = 410 mcg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: C · Avoid heat sources such as hot water bottles over the patch
Why this is the answer
Heat increases absorption of fentanyl through the skin and can cause overdose. Old patches are removed and folded before disposal, and patches should not be cut unless the product allows it.
Answer
C · Avoid heat sources such as hot water bottles over the patch
8.Clinical nursing
Answer: B · Massaging the calf
Why this is the answer
Unilateral calf swelling and tenderness suggests DVT. Massaging the leg is avoided in case it dislodges the clot; the patient needs medical review and investigation.