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 healthy woman at 28 weeks of pregnancy has a mildly lower hemoglobin than before pregnancy, with a normal MCV and ferritin. What is the most likely explanation?
AThe fetus is destroying the mother's red cells in the circulation
BIron is being lost through the kidneys during the pregnancy
CPregnancy hormones stop the bone marrow from making red cells
DPlasma volume rises more than red cell mass, diluting the blood
2.Lab values & electrolytesHard
A patient with suspected Cushing syndrome takes dexamethasone at night. The next morning, the cortisol is still high. What does this suggest?
ACortisol is being produced without normal feedback control
BThe patient has primary hypothyroidism instead
CThe adrenal glands are not able to produce enough cortisol
DThe result is normal, as dexamethasone raises cortisol
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 25 mmHg (3.3 kPa). Use the rule that in chronic respiratory alkalosis HCO₃⁻ falls by 5 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 100 mg of gentamicin. The level back-extrapolated to time zero is 6 mg/L and the half-life is 3.5 h. Find the clearance CL in L/h.
D = 100 mgC₀ = 6 mg/Lt½ = 3.5 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin IV every 12 h; the vancomycin clearance is 3 L/h. What dose D in mg gives an average steady-state level of 20 mg/L?
CL = 3 L/hτ = 12 hS = 1F = 1Css,avg = 20 mg/L
D = mg
7.Clinical nursingMedium
Which oxygen device delivers a precise, fixed concentration of oxygen?
ASimple face mask
BNasal cannula
CVenturi mask
DNon-rebreathe mask
8.Clinical nursingHard
Why should "units" never be abbreviated to "U" or "IU" on an insulin prescription?
APharmacists cannot dispense insulin written that way
BThe letters can be misread as a zero, causing a tenfold overdose
CIt changes the type of insulin given
DThe abbreviation is only allowed on prescriptions for children
3Solutions
1.Lab values & electrolytes
Answer: D · Plasma volume rises more than red cell mass, diluting the blood
Why this is the answer
In pregnancy, plasma volume increases by more than red cell mass, so the hemoglobin falls slightly (physiologic or dilutional anemia). Iron deficiency is still common in pregnancy and is checked with the MCV and ferritin.Checked against:Anemia · StatPearls (NCBI Bookshelf) (2026)
Answer
D · Plasma volume rises more than red cell mass, diluting the blood
2.Lab values & electrolytes
Answer: A · Cortisol is being produced without normal feedback control
Why this is the answer
Dexamethasone normally suppresses ACTH and lowers morning cortisol. Failure to suppress suggests autonomous cortisol excess (Cushing syndrome). Further tests then find the source, such as a pituitary or adrenal tumor.
Answer
A · Cortisol is being produced without normal feedback control
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 16.5 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
|25 − 40| = 15 mmHg → 1.5 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 5·1.5 = 16.5 mmol/L
Answer
HCO₃⁻ = 16.5 mmol/L
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 = 142 − (112 + 10) = 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: CL = 3.301 L/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.
Volume of distribution
Vd = 100/6 = 16.67 L
Rate constant
k = 0.693/3.5 = 0.198 h⁻¹
Clearance
CL = 0.198·16.67 = 3.301 L/h
Answer
CL = 3.301 L/h
6.Clinical pharmacokinetics
Answer: D = 720 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.
Dose
D = Css,avg·CL·τ/(S·F) = 20·3·12/(1·1) = 720 mg
Answer
D = 720 mg
7.Clinical nursing
Answer: C · Venturi mask
Why this is the answer
A Venturi mask entrains a set amount of room air, giving a fixed oxygen percentage regardless of the patient's breathing pattern. This makes it useful for patients at risk of hypercapnia.
Answer
C · Venturi mask
8.Clinical nursing
Answer: B · The letters can be misread as a zero, causing a tenfold overdose
Why this is the answer
"U" can be misread as 0, so "10U" might be read as 100, and "IU" can be mistaken for "IV" or "10". Writing "units" in full is a standard safety rule (ISMP), because insulin errors can cause severe hypoglycaemia.
Answer
B · The letters can be misread as a zero, causing a tenfold overdose