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 hyperkalemia received IV regular insulin with dextrose. Which result must the nurse monitor closely over the next hours?
ABlood glucose, because hypoglycemia can occur
BSerum calcium, because it falls sharply
CHemoglobin, because insulin causes red cell loss
DSerum sodium, because it rises quickly
2.Lab values & electrolytesHard
A patient's low calcium does not improve despite several days of calcium replacement. Which other electrolyte should be checked and corrected?
ASodium
BChloride
CBicarbonate
DMagnesium
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 55 mmHg (7.3 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic acidosis: pH 7.09, PaCO₂ 34 mmHg (4.5 kPa), HCO₃⁻ 10 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.09PaCO₂ 34 mmHg (4.5 kPa)HCO₃⁻ 10 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 200 mg of gentamicin. The level back-extrapolated to time zero is 7 mg/L and the half-life is 3.5 h. Find the volume of distribution Vd.
D = 200 mgC₀ = 7 mg/Lt½ = 3.5 h
Vd = L
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 3.5 h. The target is a peak of 6 mg/L and a trough of 1 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.
t½ = 3.5 hCpeak = 6 mg/LCtrough = 1 mg/L
τ = h
7.Clinical nursingMedium
A patient's abdomen is rigid and board-like, and they lie very still because movement worsens the pain. What does this suggest?
AConstipation
BIndigestion
CPeritonitis
DUrinary tract infection
8.Clinical nursingHard
A patient after a total hip replacement suddenly has severe hip pain, a shortened leg and the leg is rotated. What is the most likely complication?
ADislocation of the prosthesis
BWound infection
CDeep vein thrombosis in the calf
DUrinary retention
3Solutions
1.Lab values & electrolytes
Answer: A · Blood glucose, because hypoglycemia can occur
Why this is the answer
Insulin shifts potassium into cells but also lowers glucose, and hypoglycemia can occur hours later, especially in kidney failure. Glucose is checked frequently, and potassium is rechecked, as the effect lasts only a few hours.Checked against:High potassium level · MedlinePlus (NIH) (2025)
Answer
A · Blood glucose, because hypoglycemia can occur
2.Lab values & electrolytes
Answer: D · Magnesium
Why this is the answer
Low magnesium reduces parathyroid hormone secretion and the body's response to it, so hypocalcemia does not correct until magnesium is replaced. The same applies to refractory hypokalemia.Checked against:Magnesium Blood Test · MedlinePlus (NIH) (2023)
Answer
D · Magnesium
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 25.5 mmol/L
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.
Change in PaCO₂
|55 − 40| = 15 mmHg → 1.5 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·1.5 = 25.5 mmol/L
Answer
HCO₃⁻ = 25.5 mmol/L
4.Arterial blood gas interpretation
Answer: added respiratory acidosis
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
1.5·10 + 8 = 23 → 21–25 mmHg
Compare
measured: 34 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
5.Clinical pharmacokinetics
Answer: Vd = 28.57 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 = 200/7 = 28.57 L
Answer
Vd = 28.57 L
6.Clinical pharmacokinetics
Answer: τ = 9 h
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.
Rate constant
k = 0.693/3.5 = 0.198 h⁻¹
Dosing interval
τ = ln(6/1)/0.198 = 1.792/0.198 = 9.047 ≈ 9 h
Answer
τ = 9 h
In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.
7.Clinical nursing
Answer: C · Peritonitis
Why this is the answer
Peritonitis, often from perforation, causes guarding, rigidity and pain worsened by movement. It is a surgical emergency.
Answer
C · Peritonitis
8.Clinical nursing
Answer: A · Dislocation of the prosthesis
Why this is the answer
Hip dislocation after arthroplasty causes sudden pain, shortening and rotation. It needs urgent review and reduction; patients are taught movement precautions.