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 woman with hyperemesis gravidarum has vomited for 3 days. Which urine finding would the nurse expect?
ABilirubin
BRed cell casts
CKetones
DNitrites
2.Lab values & electrolytesHard
The nurse teaches a patient about the HbA1c test. Which statement shows a need for further teaching?
A"My daily finger-stick readings still matter between the tests"
B"The test shows my average glucose over the past few months"
C"A lower result means less risk of damage to my eyes and kidneys"
D"I must not eat anything for 12 hours before the HbA1c test"
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.17, PaCO₂ 43 mmHg (5.7 kPa) and HCO₃⁻ 15 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.17PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 15 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 36 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 36 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 50 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.045 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 12 mg/L?
50 kgCL = 0.045 L/kg/hCss = 12 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 53-year-old man weighs 77 kg and has a serum creatinine of 2.2 mg/dL (194 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 levetiracetam dose is 1000 mg for a creatinine clearance of 120 mL/min, and a fraction fe = 0.66 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.
A patient who injects drugs has a fever, a new heart murmur and small dark lines under the nails. What is the likely diagnosis?
AInfective endocarditis
BMyocardial infarction
CRheumatic fever
DPericarditis
8.Clinical nursingHard
A patient faints during cannulation, with a slow pulse and pallor, and recovers quickly when laid flat. There is no rash, wheeze or swelling. What is most likely?
AA stroke
BAnaphylaxis
CVasovagal syncope
DA hypoglycaemic coma
3Solutions
1.Lab values & electrolytes
Answer: C · Ketones
Why this is the answer
Prolonged vomiting and poor intake lead to fat breakdown and ketonuria, along with dehydration and electrolyte loss such as hypokalemia. Urine ketones help assess severity and response to IV fluids and antiemetics.Checked against:Ketones urine test · MedlinePlus (NIH) (2025)
Answer
C · Ketones
2.Lab values & electrolytes
Answer: D · "I must not eat anything for 12 hours before the HbA1c test"
Why this is the answer
HbA1c reflects glucose attached to hemoglobin over about 2–3 months, so fasting is not needed. Lower HbA1c values are linked to fewer microvascular complications, and home glucose monitoring is still useful for day-to-day decisions.Checked against:The A1C Test & Diabetes · NIDDK (NIH) (2018); Hemoglobin A1C (HbA1c) Test · MedlinePlus (NIH) (2025)
Answer
D · "I must not eat anything for 12 hours before the HbA1c test"
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 7.17: acidaemia (below 7.35)
PaCO₂ 43 mmHg: normal (35–45)
HCO₃⁻ 15 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 28.5–32.5 mmHg, so the measured 43 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 = 46.2 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
0.7·36 + 21 = 46.2 mmHg
Answer
PaCO₂ = 46.2 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: R = 33.75 mg/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.
Clearance
CL = 0.045·50 = 2.25 L/h
Infusion rate
R = CL·Css/S = 2.25·12/0.8 = 33.75 mg/h
Answer
R = 33.75 mg/h
6.Clinical pharmacokinetics
Answer: D = 573 mg
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 − 53)·77/(72·2.2) = 42.29 mL/min
Dose fraction (Dettli)
Q = 1 − 0.66·(1 − 42.29/120) = 0.5726
Adjusted dose
D = 0.5726·1000 = 572.6 ≈ 573 mg
Answer
D = 573 mg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: A · Infective endocarditis
Why this is the answer
Infective endocarditis causes fever and a new or changing murmur, sometimes with splinter haemorrhages. Several sets of blood cultures are taken before long-course IV antibiotics.
Answer
A · Infective endocarditis
8.Clinical nursing
Answer: C · Vasovagal syncope
Why this is the answer
Vasovagal syncope is a reflex drop in heart rate and blood pressure triggered by pain, fear or needles. Anaphylaxis would usually include skin, airway or breathing features and a fast pulse.