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 hyperkalemia is prescribed an oral potassium binder such as patiromer or sodium zirconium cyclosilicate. How does this drug lower potassium?
AIt protects the heart muscle against high potassium levels
BIt moves potassium into cells in the next few minutes
CIt stops the kidneys from absorbing water in the tubules
DIt binds potassium in the gut so it leaves in the stool
2.Lab values & electrolytesHard
During surgery under general anesthesia, a patient develops a rising end-tidal CO2, muscle rigidity and tachycardia, then a high potassium and a rising temperature. Which condition is most likely?
AMalignant hyperthermia
BAcute hemolytic transfusion reaction
CHypoglycemia from fasting
DHypovolemic shock
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 149 mmol/L, glucose 20.6 mmol/L and urea 19.5 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 149 mmol/Lglucose 20.6 mmol/Lurea 19.5 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
Practice values: Na⁺ 137, Cl⁻ 99, HCO₃⁻ 18 mmol/L and albumin 21 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 10 mg/L at 1 h and 1.8 mg/L at 6 h. Assume first-order elimination. Find the half-life t½.
C1 = 10 mg/L (1 h)C2 = 1.8 mg/L (6 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 64 kg patient takes oral theophylline (F = 1) every 8 h. The clearance is 0.04 L/kg/h. What dose D in mg gives an average steady-state level of 15 mg/L?
A pressure ulcer shows full-thickness tissue loss with exposed bone. Which category (stage) is this?
ACategory 3
BCategory 1
CCategory 4
DCategory 2
8.Clinical nursingHard
A patient with long-standing COPD has an arterial blood gas with pH 7.37, a raised PaCO2 and a raised bicarbonate. How is this best described?
AAcute respiratory acidosis
BUncompensated metabolic alkalosis
CCompensated respiratory acidosis
DMixed metabolic and respiratory alkalosis
3Solutions
1.Lab values & electrolytes
Answer: D · It binds potassium in the gut so it leaves in the stool
Why this is the answer
Potassium binders exchange potassium in the gut for another ion, so potassium is removed from the body in the stool. They act over hours, so they are not used alone for emergencies; calcium protects the heart, and insulin with glucose shifts potassium into cells quickly.Checked against:High potassium level · MedlinePlus (NIH) (2025)
Answer
D · It binds potassium in the gut so it leaves in the stool
2.Lab values & electrolytes
Answer: A · Malignant hyperthermia
Why this is the answer
Malignant hyperthermia is a genetic hypermetabolic reaction to volatile anesthetics or succinylcholine. Muscle cells release calcium, producing CO2, heat, rigidity, hyperkalemia and very high CK. It is treated with dantrolene and stopping the trigger.
Answer
A · Malignant hyperthermia
3.Arterial blood gas interpretation
Answer: Osm = 338 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 = 2·149 + 20.6 + 19.5 = 338.1 ≈ 338 mOsm/kg
Answer
Osm = 338 mOsm/kg
4.Arterial blood gas interpretation
Answer: AGc = 24.75 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.
Anion gap
AG = 137 − (99 + 18) = 20
Correction for albumin
20 + 0.25·(40 − 21) = 24.75 mmol/L
Answer
AGc = 24.75 mmol/L
Above about 12 after correction: a high anion gap.
5.Clinical pharmacokinetics
Answer: t½ = 2.021 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.
Elimination rate constant
k = ln(10/1.8)/5 = 1.715/5 = 0.343 h⁻¹
Half-life
t½ = 0.693/0.343 = 2.021 h
Answer
t½ = 2.021 h
6.Clinical pharmacokinetics
Answer: D = 307.2 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.
Clearance
CL = 0.04·64 = 2.56 L/h
Dose
D = Css,avg·CL·τ/(S·F) = 15·2.56·8/(1·1) = 307.2 mg
Answer
D = 307.2 mg
7.Clinical nursing
Answer: C · Category 4
Why this is the answer
Category 4 involves full-thickness tissue loss with exposed bone, tendon or muscle. Category 3 is full-thickness skin loss without exposed bone or tendon. US (NPIAP) terminology calls these stages rather than categories.
Answer
C · Category 4
8.Clinical nursing
Answer: C · Compensated respiratory acidosis
Why this is the answer
Chronic CO2 retention is offset by the kidneys retaining bicarbonate, bringing the pH back to normal. A normal pH with high PaCO2 and high bicarbonate is a compensated respiratory acidosis.