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 healthy patient has a mildly raised CK after running a marathon yesterday, without dark urine or symptoms. What is the most likely explanation?
AAn acute myocardial infarction has occurred
BThe patient has developed acute pancreatitis
CThe patient has developed acute hepatitis
DStrenuous exercise raised CK from skeletal muscle
2.Lab values & electrolytesHard
A lithium level of 1.0 mEq/L was drawn 2 hours after the patient’s morning dose, and the prescriber plans to raise the dose. What should the nurse point out?
AThe level is toxic, so lithium should be stopped for good
BThe level is too low, so a large increase is clearly safe
CLithium levels can be drawn at any time of day without affecting the result
DThe sample was taken too soon after the dose, so it is not a valid level
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.31, PaCO₂ 39 mmHg (5.2 kPa) and HCO₃⁻ 19 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.31PaCO₂ 39 mmHg (5.2 kPa)HCO₃⁻ 19 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows pH 7.29, PaCO₂ 28 mmHg (3.7 kPa) and HCO₃⁻ 13 mmol/L, with Na⁺ 146 and Cl⁻ 115 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 160 mg of gentamicin. The level back-extrapolated to time zero is 9 mg/L and the half-life is 3.5 h. Find the clearance CL in L/h.
D = 160 mgC₀ = 9 mg/Lt½ = 3.5 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). An 87 kg patient has a theophylline level of 6 mg/L; the target is 15 mg/L. The drug is given intravenously as aminophylline (S = 0.8), and the volume of distribution is 0.5 L/kg. What extra loading dose LD raises the level to the target?
87 kgVd = 0.5 L/kgC = 6 → 15 mg/LS = 0.8
LD = mg
7.Clinical nursingMedium
Why is an ankle–brachial pressure index (ABPI) measured before applying compression bandages to a leg ulcer?
ATo measure how deep the ulcer is and how much exudate it produces
BTo exclude significant arterial disease, as compression could cut off blood flow
CTo work out the patient's fluid needs
DTo check for a urinary tract infection that could delay ulcer healing
8.Clinical nursingHard
A patient with several days of vomiting and diarrhoea has low blood pressure, low urine output and a rising creatinine. Which description fits best?
AIntrinsic kidney injury from glomerulonephritis, treated with high-dose steroids
BPost-renal acute kidney injury from obstruction, treated with a catheter
CChronic kidney disease, treated with long-term dialysis three times a week
DPre-renal acute kidney injury from poor perfusion, treated by restoring fluid volume
3Solutions
1.Lab values & electrolytes
Answer: D · Strenuous exercise raised CK from skeletal muscle
Why this is the answer
Strenuous exercise, IM injections and muscle trauma can raise CK from skeletal muscle. A very high CK with dark urine suggests rhabdomyolysis, and heart damage is assessed with troponin, not CK.Checked against:Creatine Kinase · MedlinePlus (NIH) (2023); Creatine phosphokinase test · MedlinePlus (NIH) (2025)
Answer
D · Strenuous exercise raised CK from skeletal muscle
2.Lab values & electrolytes
Answer: D · The sample was taken too soon after the dose, so it is not a valid level
Why this is the answer
Lithium levels are standardized at about 12 hours after the last dose, usually just before the morning dose. A sample 2 hours after a dose catches the absorption peak and is not comparable with the target range, so the dose should not be changed on it.
Answer
D · The sample was taken too soon after the dose, so it is not a valid level
3.Arterial blood gas interpretation
Answer: uncompensated
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 7.31: acidaemia (below 7.35)
PaCO₂ 39 mmHg: normal (35–45)
HCO₃⁻ 19 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 34.5–38.5 mmHg, so the measured 39 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: added normal-gap 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.
Anion gap
AG = 146 − (115 + 13) = 18
Delta ratio
(18 − 12)/(24 − 13) = 0.5455
added normal-gap acidosis
Answer
added normal-gap acidosis
5.Clinical pharmacokinetics
Answer: CL = 3.521 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 = 160/9 = 17.78 L
Rate constant
k = 0.693/3.5 = 0.198 h⁻¹
Clearance
CL = 0.198·17.78 = 3.521 L/h
Answer
CL = 3.521 L/h
6.Clinical pharmacokinetics
Answer: LD = 489.4 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 = 0.5·87 = 43.5 L
Rise needed
15 − 6 = 9 mg/L
Loading dose
LD = 43.5·9/(0.8·1) = 489.4 mg
Answer
LD = 489.4 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: B · To exclude significant arterial disease, as compression could cut off blood flow
Why this is the answer
ABPI compares ankle and arm blood pressures. A low result suggests arterial insufficiency, where compression could cause ischaemia; venous ulcers with adequate arterial supply are treated with compression.
Answer
B · To exclude significant arterial disease, as compression could cut off blood flow
8.Clinical nursing
Answer: D · Pre-renal acute kidney injury from poor perfusion, treated by restoring fluid volume
Why this is the answer
Fluid loss reduces kidney perfusion, causing pre-renal AKI. Prompt fluid replacement can restore function before the kidney tissue itself is damaged.
Answer
D · Pre-renal acute kidney injury from poor perfusion, treated by restoring fluid volume