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
Before giving prescribed IV potassium chloride, which assessment is most important?
AHearing in both ears
BPupil size and reaction to light
CSkin turgor on the forearm
DUrine output and kidney function
2.Lab values & electrolytesHard
A patient with a round face, central weight gain and purple stretch marks has a glucose of 210 mg/dL (11.7 mmol/L) and a potassium of 3.1 mEq/L. Which condition fits best?
AAdrenal insufficiency
BHypothyroidism
CCushing syndrome
DDiabetes insipidus
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 130 mmol/L, glucose 30.8 mmol/L and urea 20.1 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 130 mmol/Lglucose 30.8 mmol/Lurea 20.1 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 30% oxygen shows PaO₂ 55 mmHg (7.3 kPa) and PaCO₂ 33 mmHg (4.4 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A 68 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.045 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 10 mg/L?
68 kgCL = 0.045 L/kg/hCss = 10 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 46-year-old woman weighs 82 kg and has a serum creatinine of 1.3 mg/dL (115 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 vancomycin regimen is 1000 mg every 12 h for a creatinine clearance of 100 mL/min. Keeping the dose, the interval is lengthened in proportion to the fall in clearance. Find the new dosing interval τ. Round to one decimal place.
A patient is started on spironolactone while already taking an ACE inhibitor. Which blood result should be watched most closely?
AGlucose, because both drugs cause hyperglycaemia
BSodium, because both drugs raise it sharply
CCalcium, because both drugs lower it
DPotassium, because both drugs can raise it
8.Clinical nursingHard
A nurse teaches a patient with asthma about a new inhaled steroid "preventer". Which statement shows a need for further teaching?
A"I'll keep my reliever inhaler with me."
B"I will only use it when I'm wheezy."
C"I will use it every day even when I feel well."
D"I will rinse my mouth after using it."
3Solutions
1.Lab values & electrolytes
Answer: D · Urine output and kidney function
Why this is the answer
The kidneys excrete potassium, so poor urine output or kidney failure can lead to dangerous hyperkalemia during replacement. IV potassium is always diluted and given by infusion pump, never as an IV push.Checked against:High potassium level · MedlinePlus (NIH) (2025)
Answer
D · Urine output and kidney function
2.Lab values & electrolytes
Answer: C · Cushing syndrome
Why this is the answer
Excess cortisol raises glucose, causes central obesity, a round face and purple striae, and has some mineralocorticoid effect that lowers potassium. Adrenal insufficiency causes low glucose and high potassium.Checked against:Hypercortisolism (Cushing Syndrome) · StatPearls (NCBI Bookshelf) (2025); Hypokalemia · StatPearls (NCBI Bookshelf) (2025)
Answer
C · Cushing syndrome
3.Arterial blood gas interpretation
Answer: Osm = 311 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 = 2·130 + 30.8 + 20.1 = 310.9 ≈ 311 mOsm/kg
Answer
Osm = 311 mOsm/kg
4.Arterial blood gas interpretation
Answer: A–a gradient = 117.7 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
CL = 0.045·68 = 3.06 L/h
Infusion rate
R = CL·Css/S = 3.06·10/0.8 = 38.25 mg/h
Answer
R = 38.25 mg/h
6.Clinical pharmacokinetics
Answer: τ = 17.1 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.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 46)·82/(72·1.3)·0.85 = 70 mL/min
New interval
τ = 12·100/70 = 17.14 ≈ 17.1 h
Answer
τ = 17.1 h
In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.
7.Clinical nursing
Answer: D · Potassium, because both drugs can raise it
Why this is the answer
Spironolactone and ACE inhibitors both reduce potassium excretion, so the combination can cause dangerous hyperkalaemia. Potassium and renal function should be checked regularly.
Answer
D · Potassium, because both drugs can raise it
8.Clinical nursing
Answer: B · "I will only use it when I'm wheezy."
Why this is the answer
Inhaled corticosteroids must be used regularly to reduce airway inflammation and prevent attacks. The short-acting reliever is used for symptoms.