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 taking furosemide has a potassium of 2.9 mEq/L. Which finding would the nurse expect?
ATall, peaked T waves on the ECG
BMuscle weakness and leg cramps
CDiarrhea and abdominal cramping from overactive bowels
DTingling around the mouth with a positive Chvostek’s sign
2.Lab values & electrolytesHard
Which set of results is most typical of iron-deficiency anemia?
ALow hemoglobin, large red cells (high MCV) and high ferritin
BLow hemoglobin, small red cells (low MCV) and low ferritin
CNormal hemoglobin with a high platelet count only
DHigh hemoglobin and high hematocrit
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 137 mmol/L, glucose 13.3 mmol/L and urea 8.4 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 137 mmol/Lglucose 13.3 mmol/Lurea 8.4 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 45 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 45 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 150 mg of tobramycin. The level back-extrapolated to time zero is 7 mg/L and the half-life is 2 h. Find the clearance CL in L/h.
D = 150 mgC₀ = 7 mg/Lt½ = 2 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 90 kg patient on regular tobramycin doses is at steady state. The target is a peak of 8 mg/L and a trough of 0.5 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?
A patient who injects drugs has a fever, a new heart murmur and small dark lines under the nails. What is the likely diagnosis?
APericarditis
BMyocardial infarction
CRheumatic fever
DInfective endocarditis
8.Clinical nursingHard
An older patient's blood pressure is 142/80 mmHg lying and 116/72 mmHg after standing for three minutes, and they feel dizzy. What does this show?
ANormal blood pressure variation
BHypertensive emergency
CWhite coat hypertension
DPostural hypotension
3Solutions
1.Lab values & electrolytes
Answer: B · Muscle weakness and leg cramps
Why this is the answer
Low potassium causes muscle weakness, cramps, reduced bowel movement (constipation, even ileus) and arrhythmias. Peaked T waves are seen in hyperkalemia, and perioral tingling suggests low calcium.Checked against:Heart failure - fluids and diuretics · MedlinePlus (NIH); Low blood potassium · MedlinePlus (NIH) (2025)
Answer
B · Muscle weakness and leg cramps
2.Lab values & electrolytes
Answer: B · Low hemoglobin, small red cells (low MCV) and low ferritin
Why this is the answer
Iron deficiency causes a microcytic anemia (low MCV) with low iron stores (low ferritin). Large red cells suggest vitamin B12 or folate deficiency.Checked against:Ferritin Blood Test · MedlinePlus (NIH) (2025); RBC indices · MedlinePlus (NIH) (2026)
Answer
B · Low hemoglobin, small red cells (low MCV) and low ferritin
3.Arterial blood gas interpretation
Answer: Osm = 296 mOsm/kg
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.
Calculated osmolality
Osm = 2·137 + 13.3 + 8.4 = 295.7 ≈ 296 mOsm/kg
Answer
Osm = 296 mOsm/kg
4.Arterial blood gas interpretation
Answer: PaCO₂ = 52.5 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
0.7·45 + 21 = 52.5 mmHg
Answer
PaCO₂ = 52.5 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: CL = 7.427 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 = 150/7 = 21.43 L
Rate constant
k = 0.693/2 = 0.3466 h⁻¹
Clearance
CL = 0.3466·21.43 = 7.427 L/h
Answer
CL = 7.427 L/h
6.Clinical pharmacokinetics
Answer: D = 168.8 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 = 0.25·90 = 22.5 L
Dose
D = 22.5·(8 − 0.5) = 168.8 mg
Answer
D = 168.8 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · 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
D · Infective endocarditis
8.Clinical nursing
Answer: D · Postural hypotension
Why this is the answer
A fall in systolic pressure of 20 mmHg or more (here 26 mmHg) on standing, especially with symptoms, indicates postural hypotension. Causes include medicines, dehydration and autonomic problems.