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's potassium is unexpectedly 5.8 mEq/L. The sample was taken after the tourniquet was on for several minutes while the patient pumped their fist. What is the best interpretation?
AThe result may be falsely high and should be repeated with better technique
BTourniquet time affects only the glucose result, not the potassium
CFist pumping lowers potassium, so the true level is even higher
DThe result confirms true hyperkalemia and needs no further checking
2.Lab values & electrolytesHard
In the first 24 hours after a major burn, which laboratory pattern is most expected?
ALow potassium and low hematocrit
BLow sodium and low hematocrit
CHigh potassium and high hematocrit
DHigh calcium and high albumin
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 18 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 18 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 21% oxygen shows PaO₂ 72 mmHg (9.6 kPa) and PaCO₂ 46 mmHg (6.1 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A 93 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.
93 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin IV every 12 h; the vancomycin clearance is 6 L/h. What dose D in mg gives an average steady-state level of 15 mg/L?
CL = 6 L/hτ = 12 hS = 1F = 1Css,avg = 15 mg/L
D = mg
7.Clinical nursingMedium
Which group of symptoms suggests hypercalcaemia?
AThirst, constipation, confusion and abdominal pain
BTingling lips, muscle cramps and spasms of the hands and feet
CWeight gain, feeling cold and a slow pulse
DSweating, tremor and sudden hunger
8.Clinical nursingHard
Several hours after thyroid surgery, a patient has increasing neck swelling, stridor and difficulty breathing. What is most likely?
ANormal postoperative swelling
BHypocalcaemia from parathyroid damage
CA thyroid storm
DA haematoma compressing the airway
3Solutions
1.Lab values & electrolytes
Answer: A · The result may be falsely high and should be repeated with better technique
Why this is the answer
Repeated fist clenching and a long tourniquet time release potassium from muscle and can cause hemolysis, both of which give a falsely high potassium. The patient is assessed and the test is repeated with correct technique; true hyperkalemia must not be missed.
Answer
A · The result may be falsely high and should be repeated with better technique
2.Lab values & electrolytes
Answer: C · High potassium and high hematocrit
Why this is the answer
Early after a burn, damaged cells release potassium, and plasma shifts into the tissues, which concentrates the blood and raises the hematocrit. Later, as fluid returns and the kidneys excrete potassium, hypokalemia can develop.
Answer
C · High potassium and high hematocrit
3.Arterial blood gas interpretation
Answer: PaCO₂ = 35 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 = 1.5·18 + 8 = 35 mmHg
range: 33–37 mmHg
Answer
PaCO₂ = 35 mmHg
4.Arterial blood gas interpretation
Answer: A–a gradient = 20.23 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 = 0.7·93 = 65.1 L
Amount of active drug needed
Vd·C = 65.1·20 = 1302 mg
Loading dose
LD = 1302/(0.92·1) = 1415 mg
Answer
LD = 1415 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: D = 1080 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.
Dose
D = Css,avg·CL·τ/(S·F) = 15·6·12/(1·1) = 1080 mg
Answer
D = 1080 mg
7.Clinical nursing
Answer: A · Thirst, constipation, confusion and abdominal pain
Why this is the answer
High calcium causes polyuria and thirst, constipation, abdominal pain, kidney stones and confusion. Tingling and cramps suggest low calcium instead.
Answer
A · Thirst, constipation, confusion and abdominal pain
8.Clinical nursing
Answer: D · A haematoma compressing the airway
Why this is the answer
A rapidly expanding neck haematoma after thyroid surgery is an airway emergency. Call for urgent help; local protocols often require equipment to release the wound clips or sutures to be kept at the bedside.