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
An older adult takes lisinopril and furosemide and recently started daily ibuprofen for knee pain. The creatinine has risen from 1.0 to 1.8 mg/dL (88 to 159 µmol/L). What is the most likely cause?
AThe NSAID reduces kidney blood flow on top of the other two drugs
BNormal aging of the kidneys over a period of a few weeks
CA laboratory error, since ibuprofen has no effect on the kidney
DMore activity has increased the patient's muscle mass quickly
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?
AAcute aspirin overdose with rapid breathing
BCOPD with chronic CO₂ retention and metabolic compensation
CAcute severe diarrhea with large bicarbonate losses
DAcute diabetic ketoacidosis with severe dehydration and vomiting
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.55, PaCO₂ 50 mmHg (6.7 kPa) and HCO₃⁻ 42 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.55PaCO₂ 50 mmHg (6.7 kPa)HCO₃⁻ 42 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 43 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 43 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient has a lidocaine clearance of 650 mL/min. What infusion rate R in mg/min keeps the steady-state level at 2.5 mg/L?
CL = 650 mL/minCss = 2.5 mg/L
R = mg/min
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 97 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?
97 kgVd = 7 L/kgC = 0.5 → 1 mcg/LF = 0.7
LD = mcg
7.Clinical nursingMedium
A patient on the ward has shingles with open blisters. Which people are most at risk from contact?
AOnly people over 80 who have already had shingles themselves
BPregnant women and immunocompromised people who have not had chickenpox
CPeople who have already had chickenpox in childhood and are now healthy
DNo one, because shingles cannot spread
8.Clinical nursingHard
A patient with COPD on high-flow oxygen becomes drowsy with a bounding pulse and flushed skin. Blood gas shows a raised PaCO2 and low pH. What is the most likely cause?
AMetabolic alkalosis from vomiting
BHypoglycaemia
CHypercapnic respiratory failure, partly due to excessive oxygen
DOpioid allergy
3Solutions
1.Lab values & electrolytes
Answer: A · The NSAID reduces kidney blood flow on top of the other two drugs
Why this is the answer
NSAIDs reduce prostaglandin-driven blood flow to the kidney. Combined with an ACE inhibitor and a diuretic, this can cause acute kidney injury, especially in older adults. The nurse reports the result and the NSAID is usually stopped.Checked against:Creatinine blood test · MedlinePlus (NIH) (2025)
Answer
A · The NSAID reduces kidney blood flow on top of the other two drugs
2.Lab values & electrolytes
Answer: B · 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
B · COPD with chronic CO₂ retention and metabolic compensation
3.Arterial blood gas interpretation
Answer: partially compensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.55: alkalaemia (above 7.45)
PaCO₂ 50 mmHg: high (35–45)
HCO₃⁻ 42 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
4.Arterial blood gas interpretation
Answer: PaCO₂ = 51.1 mmHg
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
0.7·43 + 21 = 51.1 mmHg
Answer
PaCO₂ = 51.1 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: R = 1.625 mg/min
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 in L/min
650 mL/min = 0.65 L/min
Infusion rate
R = 0.65·2.5 = 1.625 mg/min
Answer
R = 1.625 mg/min
6.Clinical pharmacokinetics
Answer: LD = 485 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 = 7·97 = 679 L
Rise needed
1 − 0.5 = 0.5 mcg/L
Loading dose
LD = 679·0.5/(1·0.7) = 485 mcg
Answer
LD = 485 mcg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: B · Pregnant women and immunocompromised people who have not had chickenpox
Why this is the answer
Someone who has never had chickenpox can catch chickenpox from contact with shingles blisters. Non-immune pregnant and immunocompromised people are at particular risk, so lesions are covered and precautions followed.
Answer
B · Pregnant women and immunocompromised people who have not had chickenpox
8.Clinical nursing
Answer: C · Hypercapnic respiratory failure, partly due to excessive oxygen
Why this is the answer
Uncontrolled oxygen in CO2-retaining patients can worsen hypercapnia and respiratory acidosis. Oxygen should be titrated to 88–92% and medical review sought, possibly for non-invasive ventilation.
Answer
C · Hypercapnic respiratory failure, partly due to excessive oxygen