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 on levothyroxine has palpitations, weight loss and a TSH of 0.05 mIU/L. What is the most likely explanation?
AThe patient has stopped the drug
BThe levothyroxine dose is too high
CThe levothyroxine dose is too low
DThe thyroid is now completely normal
2.Lab values & electrolytesHard
The nurse teaches a patient with severe neutropenia after chemotherapy. Which statement shows a need for further teaching?
A"I'll avoid people who have a cold or the flu"
B"I'll wash my hands before eating and after the toilet"
C"I'll check my temperature rectally to be accurate"
D"I'll call the clinic if my temperature rises"
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 8 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 8 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 21% oxygen shows PaO₂ 63 mmHg (8.4 kPa) and PaCO₂ 38 mmHg (5.1 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 110 mg of gentamicin. The level back-extrapolated to time zero is 6 mg/L and the half-life is 2.5 h. Find the clearance CL in L/h.
D = 110 mgC₀ = 6 mg/Lt½ = 2.5 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 64 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.045 L/kg/h. Each dose is 200 mg of aminophylline. Find the average steady-state level Css,avg.
Which advice is important for a patient on long-term oral corticosteroids?
ATake them only when symptoms flare
BDo not stop them suddenly
CDouble the dose if a dose is missed
DTake them on an empty stomach at night
8.Clinical nursingHard
Why is morphine more likely to cause toxicity in a patient with severe kidney impairment?
AKidney disease makes the opioid receptors disappear
BIts active metabolites build up because the kidneys excrete them
CThe kidneys normally activate morphine, so more of it becomes active
DKidney failure makes the stomach absorb more morphine
3Solutions
1.Lab values & electrolytes
Answer: B · The levothyroxine dose is too high
Why this is the answer
A very low TSH with symptoms of excess thyroid hormone suggests over-replacement. The dose is usually reduced, because long-term over-replacement raises the risk of atrial fibrillation and bone loss, especially in older adults.Checked against:TSH test · MedlinePlus (NIH) (2026)
Answer
B · The levothyroxine dose is too high
2.Lab values & electrolytes
Answer: C · "I'll check my temperature rectally to be accurate"
Why this is the answer
Rectal temperatures, suppositories and enemas are avoided in neutropenia, because injury to the rectal mucosa can let bacteria into the blood. Prompt reporting of fever, hand hygiene and avoiding sick contacts are correct.
Answer
C · "I'll check my temperature rectally to be accurate"
3.Arterial blood gas interpretation
Answer: PaCO₂ = 20 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.
Winter’s formula
PaCO₂ = 1.5·8 + 8 = 20 mmHg
range: 18–22 mmHg
Answer
PaCO₂ = 20 mmHg
4.Arterial blood gas interpretation
Answer: A–a gradient = 39.23 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.
Volume of distribution
Vd = 110/6 = 18.33 L
Rate constant
k = 0.693/2.5 = 0.2773 h⁻¹
Clearance
CL = 0.2773·18.33 = 5.083 L/h
Answer
CL = 5.083 L/h
6.Clinical pharmacokinetics
Answer: Css,avg = 6.944 mg/L
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.
Long-term steroids suppress the adrenal glands, so stopping suddenly can cause adrenal crisis. Patients should carry a steroid alert card or medical ID and follow sick-day guidance.
Answer
B · Do not stop them suddenly
8.Clinical nursing
Answer: B · Its active metabolites build up because the kidneys excrete them
Why this is the answer
Morphine metabolites, including morphine-6-glucuronide, are cleared by the kidneys and accumulate in renal impairment, causing sedation and respiratory depression. Alternative opioids or lower doses are often chosen.
Answer
B · Its active metabolites build up because the kidneys excrete them