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 stable adult who is not bleeding receives one unit of packed red cells. By about how much is the hemoglobin expected to rise?
AAbout 1 g/dL (10 g/L)
BAbout 0.1 g/dL (1 g/L)
CAbout 5 g/dL (50 g/L)
DAbout 3 g/dL (30 g/L)
2.Lab values & electrolytesHard
A patient has a sodium of 150 mEq/L, a glucose of 180 mg/dL and a BUN of 28 mg/dL. Using the formula 2 × Na + glucose/18 + BUN/2.8, what is the calculated serum osmolality?
AAbout 300 mOsm/kg, which is normal
BAbout 285 mOsm/kg, which is normal
CAbout 320 mOsm/kg, which is high
DAbout 358 mOsm/kg, which is high
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 16 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 16 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 35% oxygen shows PaO₂ 86 mmHg (11.5 kPa) and PaCO₂ 35 mmHg (4.7 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). An 82 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 15 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.
82 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 97 kg patient has a theophylline level of 4.8 mg/L; the target is 12 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?
97 kgVd = 0.5 L/kgC = 4.8 → 12 mg/LS = 0.8
LD = mg
7.Clinical nursingMedium
After thyroid surgery, a patient has tingling around the mouth and muscle cramps. Which electrolyte problem should be suspected?
AHyperkalaemia
BHypocalcaemia
CHypermagnesaemia
DHypernatraemia
8.Clinical nursingHard
A patient with severe asthma becomes quiet, with a "silent chest", exhaustion and a falling respiratory rate. What does this indicate?
ANormal recovery after a nebuliser
BAnxiety only
CLife-threatening asthma with impending respiratory arrest
DImprovement of the attack as the bronchospasm settles and the airways relax
3Solutions
1.Lab values & electrolytes
Answer: A · About 1 g/dL (10 g/L)
Why this is the answer
In an average-sized adult without ongoing bleeding or hemolysis, one unit of red cells raises the hemoglobin by about 1 g/dL (and the hematocrit by about 3%). A smaller rise suggests ongoing blood loss or red cell destruction.
Answer
A · About 1 g/dL (10 g/L)
2.Lab values & electrolytes
Answer: C · About 320 mOsm/kg, which is high
Why this is the answer
Osmolality = 300 + 10 + 10 = 320 mOsm/kg, above the typical range of about 275–295 mOsm/kg. The high sodium is the main cause, which points to water loss or too little water intake.
Answer
C · About 320 mOsm/kg, which is high
3.Arterial blood gas interpretation
Answer: PaCO₂ = 32 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 = 1.5·16 + 8 = 32 mmHg
range: 30–34 mmHg
Answer
PaCO₂ = 32 mmHg
4.Arterial blood gas interpretation
Answer: A–a gradient = 119.8 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.
Volume of distribution
Vd = 0.5·82 = 41 L
Amount of active drug needed
Vd·C = 41·15 = 615 mg
Loading dose
LD = 615/(0.8·1) = 768.8 mg
Answer
LD = 768.8 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: LD = 436.5 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 = 0.5·97 = 48.5 L
Rise needed
12 − 4.8 = 7.2 mg/L
Loading dose
LD = 48.5·7.2/(0.8·1) = 436.5 mg
Answer
LD = 436.5 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: B · Hypocalcaemia
Why this is the answer
Damage to the parathyroid glands lowers calcium, causing perioral tingling, tetany and positive Chvostek or Trousseau signs.
Answer
B · Hypocalcaemia
8.Clinical nursing
Answer: C · Life-threatening asthma with impending respiratory arrest
Why this is the answer
A silent chest, exhaustion, cyanosis and confusion are life-threatening signs. The patient needs immediate senior and critical care input.
Answer
C · Life-threatening asthma with impending respiratory arrest