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 starts lisinopril (an ACE inhibitor). Which laboratory values should the nurse expect to be monitored?
APotassium and creatinine
BCalcium and phosphate
CTSH and free T4
DALT and bilirubin
2.Lab values & electrolytesHard
Two weeks after a sore throat, a child develops dark (cola-colored) urine, puffy eyes and high blood pressure. Urinalysis shows red cell casts and protein. Which condition is most likely?
AKidney stone in the ureter
BPost-streptococcal glomerulonephritis
CPrerenal acute kidney injury
DLower urinary tract infection (cystitis)
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.50, PaCO₂ 21 mmHg (2.8 kPa) and HCO₃⁻ 16 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.
pH 7.50PaCO₂ 21 mmHg (2.8 kPa)HCO₃⁻ 16 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.32, PaCO₂ 73 mmHg (9.7 kPa), HCO₃⁻ 36 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.32PaCO₂ 73 mmHg (9.7 kPa)HCO₃⁻ 36 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 140 mg of gentamicin. The level back-extrapolated to time zero is 6 mg/L and the half-life is 3 h. Find the volume of distribution Vd.
D = 140 mgC₀ = 6 mg/Lt½ = 3 h
Vd = L
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin IV every 12 h; the vancomycin clearance is 5 L/h. Each dose is 1250 mg of vancomycin. Find the average steady-state level Css,avg.
CL = 5 L/hτ = 12 hS = 1F = 1D = 1250 mg
Css,avg = mg/L
7.Clinical nursingMedium
Which oxygen device delivers a precise, fixed concentration of oxygen?
ASimple face mask
BVenturi mask
CNasal cannula
DNon-rebreathe mask
8.Clinical nursingHard
A patient with long-standing COPD has an arterial blood gas with pH 7.37, a raised PaCO2 and a raised bicarbonate. How is this best described?
AAcute respiratory acidosis
BMixed metabolic and respiratory alkalosis
CCompensated respiratory acidosis
DUncompensated metabolic alkalosis
3Solutions
1.Lab values & electrolytes
Answer: A · Potassium and creatinine
Why this is the answer
ACE inhibitors reduce aldosterone, so potassium can rise, and they lower the filtration pressure in the kidney, so creatinine can rise. Both are checked before and soon after starting or increasing the dose.Checked against:Hyperkalemia · StatPearls (NCBI Bookshelf) (2025)
Answer
A · Potassium and creatinine
2.Lab values & electrolytes
Answer: B · Post-streptococcal glomerulonephritis
Why this is the answer
Red cell casts show bleeding from the glomeruli, which is typical of glomerulonephritis. After a streptococcal infection, immune complexes inflame the glomeruli, causing hematuria, proteinuria, edema and hypertension. Cystitis and stones do not cause red cell casts.Checked against:Strep A Test · MedlinePlus (NIH) (2024)
Answer
B · Post-streptococcal glomerulonephritis
3.Arterial blood gas interpretation
Answer: partially compensated
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 7.50: alkalaemia (above 7.45)
PaCO₂ 21 mmHg: low (35–45)
HCO₃⁻ 16 mmol/L: low (22–26)
Compensation
partially compensated
Answer
partially compensated
4.Arterial blood gas interpretation
Answer: chronic
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 HCO₃⁻
acute: 24 + 1·3.3 = 27.3 mmol/L
chronic: 24 + 3.5·3.3 = 35.55 mmol/L
Compare
measured: 36 mmol/L → chronic
Answer
chronic
5.Clinical pharmacokinetics
Answer: Vd = 23.33 L
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 = 140/6 = 23.33 L
Answer
Vd = 23.33 L
6.Clinical pharmacokinetics
Answer: Css,avg = 20.83 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.
A Venturi mask entrains a set amount of room air, giving a fixed oxygen percentage regardless of the patient's breathing pattern. This makes it useful for patients at risk of hypercapnia.
Answer
B · Venturi mask
8.Clinical nursing
Answer: C · Compensated respiratory acidosis
Why this is the answer
Chronic CO2 retention is offset by the kidneys retaining bicarbonate, bringing the pH back to normal. A normal pH with high PaCO2 and high bicarbonate is a compensated respiratory acidosis.