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
When is a peak level for an IV aminoglycoside such as gentamicin usually drawn?
AAt any time convenient for the patient
BAbout 30 minutes after the infusion ends
CJust before the next dose is started
DExactly 24 hours after the first dose
2.Lab values & electrolytesHard
A patient with a long-standing lung condition has a serum bicarbonate of 32 mEq/L, an arterial PaCO₂ that is always high and a pH of 7.36. Which condition fits best?
ACOPD with chronic CO₂ retention and metabolic compensation
BAcute aspirin overdose with rapid breathing
CAcute severe diarrhea with large bicarbonate losses
DAcute diabetic ketoacidosis with severe dehydration and vomiting
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 63 mmHg (8.4 kPa). Use the rule that in chronic respiratory acidosis HCO₃⁻ rises by 3.5 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic acidosis: pH 7.29, PaCO₂ 28 mmHg (3.7 kPa), HCO₃⁻ 13 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.29PaCO₂ 28 mmHg (3.7 kPa)HCO₃⁻ 13 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 62 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.
62 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 73-year-old man weighs 69 kg and has a serum creatinine of 2.7 mg/dL (239 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 tobramycin regimen is 120 mg every 8 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's fluid balance chart shows intake of 1,200 mL and output of 2,300 mL over 24 hours. What does this indicate?
AA negative fluid balance of 1,100 mL
BA balanced fluid status
CA positive fluid balance of 1,100 mL
DA positive fluid balance of 3,500 mL
8.Clinical nursingHard
A drowsy patient in recovery is making snoring sounds and the SpO2 is falling. What is the first action?
ALie the patient face down
BWait for the patient to wake up fully
CGive sips of oral fluid to clear the throat and wake them
DOpen the airway with a head tilt–chin lift or jaw thrust
3Solutions
1.Lab values & electrolytes
Answer: B · About 30 minutes after the infusion ends
Why this is the answer
Peak aminoglycoside levels are usually drawn about 30 minutes after a 30-minute infusion ends, to confirm the level is high enough to kill bacteria. Troughs, drawn just before the next dose, check for accumulation. Local protocols may differ, especially with once-daily dosing.
Answer
B · About 30 minutes after the infusion ends
2.Lab values & electrolytes
Answer: A · COPD with chronic CO₂ retention and metabolic compensation
Why this is the answer
In chronic CO₂ retention, as in severe COPD, the kidneys retain bicarbonate over days and bring the pH back towards normal, often to the low end of the range. DKA and diarrhea lower bicarbonate, and aspirin overdose causes rapid breathing that lowers the PaCO₂.Checked against:Respiratory Acidosis · StatPearls (NCBI Bookshelf)
Answer
A · COPD with chronic CO₂ retention and metabolic compensation
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 32.05 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
|63 − 40| = 23 mmHg → 2.3 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 3.5·2.3 = 32.05 mmol/L
Answer
HCO₃⁻ = 32.05 mmol/L
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
1.5·13 + 8 = 27.5 → 25.5–29.5 mmHg
Compare
measured: 28 mmHg → appropriate compensation
Answer
appropriate compensation
5.Clinical pharmacokinetics
Answer: LD = 943.5 mg
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 = 0.7·62 = 43.4 L
Amount of active drug needed
Vd·C = 43.4·20 = 868 mg
Loading dose
LD = 868/(0.92·1) = 943.5 mg
Answer
LD = 943.5 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: τ = 33.6 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 − 73)·69/(72·2.7) = 23.78 mL/min
New interval
τ = 8·100/23.78 = 33.64 ≈ 33.6 h
Answer
τ = 33.6 h
In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.
7.Clinical nursing
Answer: A · A negative fluid balance of 1,100 mL
Why this is the answer
1,200 − 2,300 = −1,100 mL. A negative balance means more fluid is lost than gained, which may cause dehydration.
Answer
A · A negative fluid balance of 1,100 mL
8.Clinical nursing
Answer: D · Open the airway with a head tilt–chin lift or jaw thrust
Why this is the answer
In a drowsy patient the tongue can fall back and obstruct the airway, causing snoring. Simple airway manoeuvres, oxygen and, if needed, an airway adjunct are used while help is called.
Answer
D · Open the airway with a head tilt–chin lift or jaw thrust