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 type 2 diabetes has a target HbA1c below 7% (53 mmol/mol). Today's result is 7.8% (62 mmol/mol). What is the best interpretation?
AAverage glucose over the past months has been above the target
BThe patient has had frequent episodes of severe hypoglycemia
CThe result shows today's fasting glucose is above the target
DGlucose control has been excellent for the past several months
2.Lab values & electrolytesHard
A patient has marked generalized edema. Results show heavy proteinuria, serum albumin 1.8 g/dL (18 g/L) and total cholesterol 340 mg/dL (8.8 mmol/L). Which condition fits best?
ALiver cirrhosis only
BPrerenal kidney injury
CNephrotic syndrome
DAcute pyelonephritis
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.60, PaCO₂ 41 mmHg (5.5 kPa) and HCO₃⁻ 39 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.60PaCO₂ 41 mmHg (5.5 kPa)HCO₃⁻ 39 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 34 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 34 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 200 mg of tobramycin. The level back-extrapolated to time zero is 9 mg/L and the half-life is 2.5 h. Find the clearance CL in L/h.
D = 200 mgC₀ = 9 mg/Lt½ = 2.5 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 3 h. The target is a peak of 7 mg/L and a trough of 1 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.
t½ = 3 hCpeak = 7 mg/LCtrough = 1 mg/L
τ = h
7.Clinical nursingMedium
Which pattern is typical of relapsing–remitting multiple sclerosis?
ASudden severe headache with neck stiffness
BConstant weakness that starts in the feet and spreads upwards after a viral illness
CAttacks of symptoms, such as visual loss or numbness, that improve between relapses
DA single stroke that leaves permanent weakness
8.Clinical nursingHard
A patient with myasthenia gravis develops increasing difficulty swallowing and weak, shallow breathing. What is the main danger?
ASevere hypertension from anxiety
BRespiratory failure from a myasthenic crisis
CA urinary tract infection
DA sudden rise in blood glucose from the stress
3Solutions
1.Lab values & electrolytes
Answer: A · Average glucose over the past months has been above the target
Why this is the answer
An HbA1c of 7.8% shows that average glucose has been above the individual target over about 2–3 months. The nurse explores diet, activity, medication use and barriers, and the treatment plan may be adjusted.Checked against:Hemoglobin A1C (HbA1c) Test · MedlinePlus (NIH) (2025)
Answer
A · Average glucose over the past months has been above the target
2.Lab values & electrolytes
Answer: C · Nephrotic syndrome
Why this is the answer
Nephrotic syndrome is heavy proteinuria with low albumin, edema and high lipids. Low albumin reduces oncotic pressure, causing edema, and the liver increases lipoprotein production. There is also a higher risk of clots and infection.Checked against:Nephrotic syndrome · MedlinePlus (NIH) (2025)
Answer
C · Nephrotic syndrome
3.Arterial blood gas interpretation
Answer: uncompensated
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 7.60: alkalaemia (above 7.45)
PaCO₂ 41 mmHg: normal (35–45)
HCO₃⁻ 39 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 46.3–50.3 mmHg, so the measured 41 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: PaCO₂ = 44.8 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
0.7·34 + 21 = 44.8 mmHg
Answer
PaCO₂ = 44.8 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: CL = 6.161 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 = 200/9 = 22.22 L
Rate constant
k = 0.693/2.5 = 0.2773 h⁻¹
Clearance
CL = 0.2773·22.22 = 6.161 L/h
Answer
CL = 6.161 L/h
6.Clinical pharmacokinetics
Answer: τ = 8.4 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.
Rate constant
k = 0.693/3 = 0.231 h⁻¹
Dosing interval
τ = ln(7/1)/0.231 = 1.946/0.231 = 8.422 ≈ 8.4 h
Answer
τ = 8.4 h
In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.
7.Clinical nursing
Answer: C · Attacks of symptoms, such as visual loss or numbness, that improve between relapses
Why this is the answer
MS causes demyelination in the brain and spinal cord. Many patients first have attacks such as optic neuritis or sensory symptoms, with partial or full recovery between relapses.
Answer
C · Attacks of symptoms, such as visual loss or numbness, that improve between relapses
8.Clinical nursing
Answer: B · Respiratory failure from a myasthenic crisis
Why this is the answer
In a myasthenic crisis, weakness of respiratory and bulbar muscles can cause aspiration and respiratory failure. Breathing function is monitored closely and ventilation may be needed.