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
Before giving prescribed IV potassium chloride, which assessment is most important?
AHearing in both ears
BPupil size and reaction to light
CSkin turgor on the forearm
DUrine output and kidney function
2.Lab values & electrolytesHard
A patient with long-term heavy alcohol use is admitted with alcohol withdrawal. Which group of electrolyte problems is most common?
ALow magnesium, low phosphate and low potassium
BHigh magnesium, high phosphate and high potassium
CHigh sodium, high calcium and high glucose
DLow chloride, high bicarbonate and high calcium
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 138 mmol/L, glucose 14.1 mmol/L and urea 11.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 138 mmol/Lglucose 14.1 mmol/Lurea 11.6 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.25, PaCO₂ 76 mmHg (10.1 kPa), HCO₃⁻ 32 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.25PaCO₂ 76 mmHg (10.1 kPa)HCO₃⁻ 32 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 68 kg patient is to be loaded with digoxin tablets to reach a level of 1 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.
68 kgVd = 7 L/kgC = 1 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). An 81 kg patient on regular gentamicin doses is at steady state. The target is a peak of 6 mg/L and a trough of 1 mg/L, and the volume of distribution is 0.3 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?
81 kgVd = 0.3 L/kgCpeak = 6 mg/LCtrough = 1 mg/L
D = mg
7.Clinical nursingMedium
Which blood results are typical of Addison's disease (primary adrenal insufficiency)?
ALow sodium and high potassium
BHigh sodium and low potassium
CLow haemoglobin and high platelets
DHigh glucose and high calcium
8.Clinical nursingHard
A patient with diabetes is found confused with slurred speech and weakness on one side. What should the nurse check first?
AUrine output for the last day
BBody mass index
CCapillary blood glucose
DSerum cholesterol
3Solutions
1.Lab values & electrolytes
Answer: D · Urine output and kidney function
Why this is the answer
The kidneys excrete potassium, so poor urine output or kidney failure can lead to dangerous hyperkalemia during replacement. IV potassium is always diluted and given by infusion pump, never as an IV push.Checked against:High potassium level · MedlinePlus (NIH) (2025)
Answer
D · Urine output and kidney function
2.Lab values & electrolytes
Answer: A · Low magnesium, low phosphate and low potassium
Why this is the answer
Poor diet, vomiting, diarrhea and alcohol's effect on the kidneys commonly cause low magnesium, phosphate and potassium. These worsen arrhythmia and seizure risk, so they are checked and replaced, together with thiamine.
Answer
A · Low magnesium, low phosphate and low potassium
3.Arterial blood gas interpretation
Answer: Osm = 302 mOsm/kg
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.
Calculated osmolality
Osm = 2·138 + 14.1 + 11.6 = 301.7 ≈ 302 mOsm/kg
Answer
Osm = 302 mOsm/kg
4.Arterial blood gas interpretation
Answer: acute on chronic
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.
Expected HCO₃⁻
acute: 24 + 1·3.6 = 27.6 mmol/L
chronic: 24 + 3.5·3.6 = 36.6 mmol/L
Compare
measured: 32 mmol/L → acute on chronic
Answer
acute on chronic
5.Clinical pharmacokinetics
Answer: LD = 680 mcg
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 = 7·68 = 476 L
Amount of active drug needed
Vd·C = 476·1 = 476 mcg
Loading dose
LD = 476/(1·0.7) = 680 mcg
Answer
LD = 680 mcg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: D = 121.5 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 = 0.3·81 = 24.3 L
Dose
D = 24.3·(6 − 1) = 121.5 mg
Answer
D = 121.5 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: A · Low sodium and high potassium
Why this is the answer
Without aldosterone, the kidneys lose sodium and retain potassium. Patients may also have low blood pressure, low glucose and increased skin pigmentation.
Answer
A · Low sodium and high potassium
8.Clinical nursing
Answer: C · Capillary blood glucose
Why this is the answer
Hypoglycaemia can cause focal neurological signs that mimic stroke. A bedside glucose check is quick and should be done in any patient with sudden neurological symptoms.