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 ventricular arrhythmia shows a twisting pattern (torsades de pointes) on the monitor. Which electrolyte is commonly given IV as treatment?
ASodium bicarbonate
BPotassium phosphate
CCalcium carbonate
DMagnesium sulfate
2.Lab values & electrolytesHard
A 5-year-old has new bruising and petechiae two weeks after a viral illness. Platelets are 12 × 10³/µL; hemoglobin, white cell count and clotting times are normal. Which condition is most likely?
AImmune thrombocytopenia (ITP)
BHemophilia A
CDisseminated intravascular coagulation
DAcute leukemia
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 6 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 6 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 35% oxygen shows PaO₂ 87 mmHg (11.6 kPa) and PaCO₂ 29 mmHg (3.9 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 220 mg of tobramycin. The level back-extrapolated to time zero is 8 mg/L and the half-life is 2 h. Find the clearance CL in L/h.
D = 220 mgC₀ = 8 mg/Lt½ = 2 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 57 kg patient on regular tobramycin doses is at steady state. The target is a peak of 10 mg/L and a trough of 1 mg/L, and the volume of distribution is 0.25 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?
Why are patients with atrial fibrillation often given anticoagulants?
ATo lower the blood pressure
BTo reduce the risk of stroke from clots in the atria
CTo convert the rhythm back to normal
DTo slow the heart rate directly and prevent palpitations
8.Clinical nursingHard
In a patient with a head injury, which is usually the earliest sign of rising intracranial pressure?
AA change in level of consciousness
BA slow pulse with high blood pressure
CA fixed, dilated pupil on both sides
DIrregular breathing pauses
3Solutions
1.Lab values & electrolytes
Answer: D · Magnesium sulfate
Why this is the answer
IV magnesium sulfate is the standard drug treatment for torsades de pointes, even when serum magnesium is normal. Low magnesium and low potassium both prolong the QT interval and raise the risk.
Answer
D · Magnesium sulfate
2.Lab values & electrolytes
Answer: A · Immune thrombocytopenia (ITP)
Why this is the answer
ITP is immune destruction of platelets, often after a viral infection in children. It causes isolated thrombocytopenia with normal other counts and clotting tests. Leukemia usually affects other cell lines, DIC prolongs clotting times and hemophilia prolongs the aPTT.Checked against:Immune thrombocytopenic purpura (ITP) · MedlinePlus (NIH) (2026)
Answer
A · Immune thrombocytopenia (ITP)
3.Arterial blood gas interpretation
Answer: PaCO₂ = 17 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 = 1.5·6 + 8 = 17 mmHg
range: 15–19 mmHg
Answer
PaCO₂ = 17 mmHg
4.Arterial blood gas interpretation
Answer: A–a gradient = 126.3 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 = 220/8 = 27.5 L
Rate constant
k = 0.693/2 = 0.3466 h⁻¹
Clearance
CL = 0.3466·27.5 = 9.531 L/h
Answer
CL = 9.531 L/h
6.Clinical pharmacokinetics
Answer: D = 128.3 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.
Volume of distribution
Vd = 0.25·57 = 14.25 L
Dose
D = 14.25·(10 − 1) = 128.3 mg
Answer
D = 128.3 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: B · To reduce the risk of stroke from clots in the atria
Why this is the answer
In AF the atria do not contract effectively, so blood can pool and clot, especially in the left atrial appendage. A clot can travel to the brain, so anticoagulation reduces stroke risk.
Answer
B · To reduce the risk of stroke from clots in the atria
8.Clinical nursing
Answer: A · A change in level of consciousness
Why this is the answer
Subtle changes in consciousness, such as restlessness, confusion or a falling GCS, are often the first signs. Pupil changes and Cushing's triad (hypertension, bradycardia, irregular breathing) are late signs.