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
Practice exercise (made-up values, not clinical guidance). Vancomycin is given as a continuous infusion to a patient whose vancomycin clearance is 4.5 L/h. What infusion rate R in mg/h gives a steady-state level of 20 mg/L?
CL = 4.5 L/hCss = 20 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A vancomycin infusion runs at 100 mg/h with no loading dose. The patient's clearance is 5 L/h and the half-life is 6 h. What is the level C after 10 h?
R = 100 mg/hCL = 5 L/ht½ = 6 ht = 10 h
C = mg/L
7.Clinical nursingMedium
A patient with type 2 diabetes is acutely unwell, very dehydrated and has a blood glucose of 40 mmol/L without significant ketones. What is the most likely diagnosis?
ADiabetes insipidus
BDiabetic ketoacidosis with dehydration
CHyperosmolar hyperglycaemic state
DHypoglycaemia
8.Clinical nursingHard
A patient with a central venous catheter develops fever and rigors shortly after the line is flushed, with no other obvious source. What should be suspected?
AA catheter-related bloodstream infection
BNormal response to line use
CAn allergic reaction to the saline flush
DA pulmonary embolism
3Solutions
1.Lab values & electrolytes
Answer: D · IV regular insulin given with glucose
Why this is the answer
Insulin activates the sodium–potassium pump and moves potassium into the cells; glucose is given with it to prevent hypoglycemia. Calcium gluconate protects the heart but does not lower the potassium level.Checked against:High potassium level · MedlinePlus (NIH) (2025)
Answer
D · IV regular insulin given with glucose
2.Lab values & electrolytes
Answer: C · "I can have a light breakfast before the test"
Why this is the answer
The OGTT is done after an overnight fast of at least 8 hours; water is allowed. Eating before the test makes the results invalid. Patients rest during the test and avoid smoking, which can affect glucose.Checked against:Blood Glucose Test · MedlinePlus (NIH) (2024); Glucose tolerance test - non-pregnant · MedlinePlus (NIH) (2025)
Answer
C · "I can have a light breakfast before the test"
3.Arterial blood gas interpretation
Answer: PaCO₂ = 26 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 = 1.5·12 + 8 = 26 mmHg
range: 24–28 mmHg
Answer
PaCO₂ = 26 mmHg
4.Arterial blood gas interpretation
Answer: AGc = 18.25 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.
Anion gap
AG = 144 − (108 + 20) = 16
Correction for albumin
16 + 0.25·(40 − 31) = 18.25 mmol/L
Answer
AGc = 18.25 mmol/L
Above about 12 after correction: a high anion gap.
5.Clinical pharmacokinetics
Answer: R = 90 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.
Infusion rate
R = 4.5·20 = 90 mg/h
Answer
R = 90 mg/h
6.Clinical pharmacokinetics
Answer: C = 13.7 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.
Steady-state level
Css = 100/5 = 20 mg/L
Rate constant
k = 0.693/6 = 0.1155 h⁻¹
Level after the given time
C = 20·(1 − e−0.1155·10) = 13.7 mg/L
Answer
C = 13.7 mg/L
7.Clinical nursing
Answer: C · Hyperosmolar hyperglycaemic state
Why this is the answer
HHS usually affects people with type 2 diabetes. It develops over days with extreme hyperglycaemia, dehydration and high osmolality, but little ketosis.
Answer
C · Hyperosmolar hyperglycaemic state
8.Clinical nursing
Answer: A · A catheter-related bloodstream infection
Why this is the answer
Rigors after flushing suggest bacteria colonising the catheter. Blood cultures are taken (often from the line and a peripheral vein) and the team decides whether to remove the line.