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 type 1 diabetes has the flu, is eating little and has a glucose of 290 mg/dL (16.1 mmol/L). Which advice is correct?
AWait 3 days before checking the glucose again
BKeep taking insulin and check blood or urine ketones
CDrink only water and avoid all carbohydrate foods
DStop all insulin until you are eating normally again
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
CHigh hemoglobin and high hematocrit
DNormal hemoglobin with a high platelet count only
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.57, PaCO₂ 43 mmHg (5.7 kPa) and HCO₃⁻ 38 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.57PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 38 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 28% oxygen shows PaO₂ 67 mmHg (8.9 kPa) and PaCO₂ 28 mmHg (3.7 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A 78 kg patient receives a theophylline infusion. The clearance is 0.045 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 12 mg/L?
78 kgCL = 0.045 L/kg/hCss = 12 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 65 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.03 L/kg/h. Each dose is 250 mg of aminophylline. Find the average steady-state level Css,avg.
A pressure ulcer shows full-thickness tissue loss with exposed bone. Which category (stage) is this?
ACategory 3
BCategory 2
CCategory 1
DCategory 4
8.Clinical nursingHard
Which ECG change is classically associated with severe hyperkalaemia?
AProminent U waves
BA shortened PR interval
CST depression only
DTall, peaked T waves
3Solutions
1.Lab values & electrolytes
Answer: B · Keep taking insulin and check blood or urine ketones
Why this is the answer
During illness, stress hormones raise glucose, so insulin is continued even when eating less. Glucose and ketones are checked often, fluids are taken, and the provider is contacted if ketones are moderate or high or vomiting prevents fluids.Checked against:Diabetes - when you are sick · MedlinePlus (NIH) (2026)
Answer
B · Keep taking insulin and check blood or urine ketones
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: 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 7.57: alkalaemia (above 7.45)
PaCO₂ 43 mmHg: normal (35–45)
HCO₃⁻ 38 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 45.6–49.6 mmHg, so the measured 43 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: A–a gradient = 97.64 mmHg
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.
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.
Clearance
CL = 0.045·78 = 3.51 L/h
Infusion rate
R = 3.51·12 = 42.12 mg/h
Answer
R = 42.12 mg/h
6.Clinical pharmacokinetics
Answer: Css,avg = 12.82 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.
Category 4 involves full-thickness tissue loss with exposed bone, tendon or muscle. Category 3 is full-thickness skin loss without exposed bone or tendon. US (NPIAP) terminology calls these stages rather than categories.
Answer
D · Category 4
8.Clinical nursing
Answer: D · Tall, peaked T waves
Why this is the answer
Hyperkalaemia causes peaked T waves, then a widened QRS and possibly a sine-wave pattern and arrest. Hypokalaemia causes flat T waves and U waves.