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 suspected pulmonary TB needs sputum samples for acid-fast bacilli. Which instruction is correct?
AGive one sample after brushing with mouthwash so it is clean
BGive several samples of nasal mucus blown into a tissue
CGive several deep-cough samples, often first thing in the morning
DGive one sample of saliva collected from the mouth at any time of day
2.Lab values & electrolytesHard
Two sets of blood cultures were taken from separate sites. Only one bottle in one set grows coagulase-negative staphylococci, and the patient has no line or implant. What is the most likely interpretation?
AProof that the patient has infective endocarditis
BInfection with a resistant organism such as MRSA
CProbable skin contamination of the sample
DCertain bloodstream infection needing treatment
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 24 mmHg (3.2 kPa). Use the rule that in acute respiratory alkalosis HCO₃⁻ falls by 2 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic acidosis: pH 7.20, PaCO₂ 21 mmHg (2.8 kPa), HCO₃⁻ 8 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.20PaCO₂ 21 mmHg (2.8 kPa)HCO₃⁻ 8 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). Vancomycin is given as a continuous infusion to a patient whose vancomycin clearance is 4 L/h. What infusion rate R in mg/h gives a steady-state level of 25 mg/L?
CL = 4 L/hCss = 25 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 92 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?
92 kgVd = 0.5 L/kgC = 6 → 15 mg/LS = 0.8
LD = mg
7.Clinical nursingMedium
A patient has lost more than 10% of their body weight without trying over the last few months and has a low BMI. What is the most appropriate action?
AEncourage a low-calorie diet
BRestrict fluids to encourage appetite and give a low-protein diet instead
CWeigh the patient again in six months and review at that point
DRefer to the dietitian and start a nutrition care plan with food charts
8.Clinical nursingHard
A patient has a full-thickness burn all the way round the forearm. Which complication must be watched for closely?
ALoss of circulation to the hand as swelling builds under tight burned skin
BDelayed hair growth on the arm
CHypoglycaemia caused by the body using glucose to heal the burn
DExcessive bleeding from the burn surface needing an urgent blood transfusion
3Solutions
1.Lab values & electrolytes
Answer: C · Give several deep-cough samples, often first thing in the morning
Why this is the answer
Usually a series of three sputum samples is collected, at least one in the early morning, when secretions have pooled overnight. The patient rinses with water (not mouthwash), takes deep breaths and coughs deeply to bring up sputum, not saliva. Airborne precautions are used.Checked against:Sputum Culture · MedlinePlus (NIH) (2024); Acid-Fast Bacillus (AFB) Tests · MedlinePlus (NIH) (2025)
Answer
C · Give several deep-cough samples, often first thing in the morning
2.Lab values & electrolytes
Answer: C · Probable skin contamination of the sample
Why this is the answer
Coagulase-negative staphylococci are common skin flora. Growth in only one of several bottles, without a catheter or implant, most often reflects contamination during collection. Good skin antisepsis reduces contamination, which can lead to unnecessary antibiotics.Checked against:Blood culture · MedlinePlus (NIH) (2025)
Answer
C · Probable skin contamination of the sample
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 20.8 mmol/L
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.
Change in PaCO₂
|24 − 40| = 16 mmHg → 1.6 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1.6 = 20.8 mmol/L
Answer
HCO₃⁻ = 20.8 mmol/L
4.Arterial blood gas interpretation
Answer: appropriate compensation
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 PaCO₂
1.5·8 + 8 = 20 → 18–22 mmHg
Compare
measured: 21 mmHg → appropriate compensation
Answer
appropriate compensation
5.Clinical pharmacokinetics
Answer: R = 100 mg/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.
Infusion rate
R = 4·25 = 100 mg/h
Answer
R = 100 mg/h
6.Clinical pharmacokinetics
Answer: LD = 517.5 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 = 0.5·92 = 46 L
Rise needed
15 − 6 = 9 mg/L
Loading dose
LD = 46·9/(0.8·1) = 517.5 mg
Answer
LD = 517.5 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · Refer to the dietitian and start a nutrition care plan with food charts
Why this is the answer
Large unplanned weight loss with a low BMI means high malnutrition risk. Management includes dietitian referral, food and fluid charts, support with eating and regular weighing.
Answer
D · Refer to the dietitian and start a nutrition care plan with food charts
8.Clinical nursing
Answer: A · Loss of circulation to the hand as swelling builds under tight burned skin
Why this is the answer
Circumferential full-thickness burns form a tight eschar that acts like a tourniquet as oedema develops. Neurovascular checks are needed, and escharotomy may be required.
Answer
A · Loss of circulation to the hand as swelling builds under tight burned skin