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 conscious patient with a blood glucose of 54 mg/dL (3.0 mmol/L) has taken 15 g of fast-acting carbohydrate. When should the glucose be rechecked?
AAt the next meal
BAfter about 2 hours
CThe next morning
DAfter about 15 minutes
2.Lab values & electrolytesHard
Which set of results best shows that diabetic ketoacidosis is resolving?
AA falling glucose with a falling serum bicarbonate
BA rising potassium with a falling bicarbonate
CA closing anion gap with a rising bicarbonate and pH
DA falling glucose with a still widening anion gap
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.29, PaCO₂ 41 mmHg (5.5 kPa) and HCO₃⁻ 19 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.29PaCO₂ 41 mmHg (5.5 kPa)HCO₃⁻ 19 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 32 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 32 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 93 kg patient is to be loaded with digoxin tablets to reach a level of 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.
93 kgVd = 7 L/kgC = 0.8 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 69-year-old woman weighs 61 kg and has a serum creatinine of 1.4 mg/dL (124 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 digoxin dose is 250 mcg for a creatinine clearance of 100 mL/min, and a fraction fe = 0.7 of the drug is excreted unchanged by the kidneys. Using the Dettli dose fraction Q = 1 − fe·(1 − CrCl/normal CrCl), find the adjusted dose D. Round to the nearest whole number.
Which signs suggest a blood transfusion reaction in the first minutes?
AImproved colour and energy
BFever, chills, back pain and low blood pressure
CSlow, steady rise in haemoglobin over the next day
DMild hunger
8.Clinical nursingHard
A patient on an opioid PCA pump has a respiratory rate of 7/min and is difficult to rouse. What should be done first?
ARecord the observations and recheck them at the next four-hourly round
BStop the PCA, rouse the patient, give oxygen and get naloxone given
CEncourage the patient to press the PCA button more often for comfort
DIncrease the PCA bolus dose because the patient may be in pain
3Solutions
1.Lab values & electrolytes
Answer: D · After about 15 minutes
Why this is the answer
The rule of 15 is 15 g of fast-acting carbohydrate, a recheck after 15 minutes, and a repeat if the glucose is still below 70 mg/dL (3.9 mmol/L). A snack or meal then helps prevent a recurrence.Checked against:Low blood sugar - self-care · MedlinePlus (NIH) (2026)
Answer
D · After about 15 minutes
2.Lab values & electrolytes
Answer: C · A closing anion gap with a rising bicarbonate and pH
Why this is the answer
DKA has resolved when the ketoacidosis has cleared: ketones fall, the anion gap normalizes, bicarbonate rises and the pH returns towards normal. Glucose alone is a poor guide, because it falls earlier.
Answer
C · A closing anion gap with a rising bicarbonate and pH
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 7.29: acidaemia (below 7.35)
PaCO₂ 41 mmHg: normal (35–45)
HCO₃⁻ 19 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 34.5–38.5 mmHg, so the measured 41 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: PaCO₂ = 43.4 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
0.7·32 + 21 = 43.4 mmHg
Answer
PaCO₂ = 43.4 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: LD = 744 mcg
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 = 7·93 = 651 L
Amount of active drug needed
Vd·C = 651·0.8 = 520.8 mcg
Loading dose
LD = 520.8/(1·0.7) = 744 mcg
Answer
LD = 744 mcg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: D = 139 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.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 69)·61/(72·1.4)·0.85 = 36.52 mL/min
Dose fraction (Dettli)
Q = 1 − 0.7·(1 − 36.52/100) = 0.5556
Adjusted dose
D = 0.5556·250 = 138.9 ≈ 139 mcg
Answer
D = 139 mcg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: B · Fever, chills, back pain and low blood pressure
Why this is the answer
An acute haemolytic or febrile reaction can cause fever, rigors, loin or back pain, breathlessness and hypotension. Stop the transfusion and call for help.
Answer
B · Fever, chills, back pain and low blood pressure
8.Clinical nursing
Answer: B · Stop the PCA, rouse the patient, give oxygen and get naloxone given
Why this is the answer
Over-sedation and slow breathing indicate opioid toxicity. Stop the opioid, support the airway and breathing, give oxygen and naloxone as prescribed, and monitor closely.
Answer
B · Stop the PCA, rouse the patient, give oxygen and get naloxone given