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 chronic kidney disease has a magnesium of 3.8 mEq/L (1.9 mmol/L). Which product is the most likely cause?
AMagnesium-containing laxatives or antacids
BProton pump inhibitors taken long term
CLoop diuretics such as furosemide
DExcess alcohol use over many years
2.Lab values & electrolytesHard
A patient with diabetes had a glucose of 58 mg/dL (3.2 mmol/L), took 15 g of fast-acting carbohydrate and, 15 minutes later, the glucose is 64 mg/dL (3.6 mmol/L). The patient is alert. What should the nurse do next?
AGive another 15 g of fast carbohydrate and recheck
BGive a protein snack only and recheck tomorrow
CWait until the next meal before giving anything
DGive a long-acting insulin dose to stabilize it
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.52, PaCO₂ 51 mmHg (6.8 kPa) and HCO₃⁻ 40 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.52PaCO₂ 51 mmHg (6.8 kPa)HCO₃⁻ 40 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.28, PaCO₂ 69 mmHg (9.2 kPa), HCO₃⁻ 31 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 7.28PaCO₂ 69 mmHg (9.2 kPa)HCO₃⁻ 31 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 69 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.
69 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 72 kg patient has a phenytoin level of 7.5 mg/L; the target is 15 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?
72 kgVd = 0.7 L/kgC = 7.5 → 15 mg/LS = 0.92
LD = mg
7.Clinical nursingMedium
A patient with MRSA colonisation is admitted. Which precautions are appropriate?
AContact precautions with gloves, apron and good hand hygiene
BNo precautions are needed for colonisation
CAirborne precautions with an FFP3 respirator mask for all staff
DIsolation in a negative-pressure room for six months
8.Clinical nursingHard
A potassium result of 6.9 mmol/L comes back on a well patient with a normal ECG, and the laboratory reports that the sample was haemolysed. What is the best action?
AGive IV insulin and glucose straight away
BIgnore the result because the patient looks well and the ECG is normal
CRepeat the blood test promptly, as haemolysis can falsely raise potassium
DStart a potassium supplement
3Solutions
1.Lab values & electrolytes
Answer: A · Magnesium-containing laxatives or antacids
Why this is the answer
In kidney failure, magnesium from milk of magnesia, magnesium citrate or magnesium-containing antacids can build up to toxic levels. Patients with CKD are taught to avoid these products. Diuretics, PPIs and alcohol cause low magnesium.Checked against:Magnesium Blood Test · MedlinePlus (NIH) (2023)
Answer
A · Magnesium-containing laxatives or antacids
2.Lab values & electrolytes
Answer: A · Give another 15 g of fast carbohydrate and recheck
Why this is the answer
The "15-15 rule" is repeated until the glucose is at least 70 mg/dL (3.9 mmol/L); UK guidance uses 4.0 mmol/L and 15–20 g of carbohydrate. Then a snack or meal with complex carbohydrate and protein is given if the next meal is not soon, to prevent recurrence.
Answer
A · Give another 15 g of fast carbohydrate and recheck
3.Arterial blood gas interpretation
Answer: partially compensated
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 7.52: alkalaemia (above 7.45)
PaCO₂ 51 mmHg: high (35–45)
HCO₃⁻ 40 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
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 + 1·2.9 = 26.9 mmol/L
chronic: 24 + 3.5·2.9 = 34.15 mmol/L
Compare
measured: 31 mmol/L → acute on chronic
Answer
acute on chronic
5.Clinical pharmacokinetics
Answer: LD = 1050 mg
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 = 0.7·69 = 48.3 L
Amount of active drug needed
Vd·C = 48.3·20 = 966 mg
Loading dose
LD = 966/(0.92·1) = 1050 mg
Answer
LD = 1050 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: LD = 410.9 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 = 0.7·72 = 50.4 L
Rise needed
15 − 7.5 = 7.5 mg/L
Loading dose
LD = 50.4·7.5/(0.92·1) = 410.9 mg
Answer
LD = 410.9 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: A · Contact precautions with gloves, apron and good hand hygiene
Why this is the answer
MRSA is spread mainly by direct and indirect contact, so contact precautions, isolation where possible and strict hand hygiene limit transmission.
Answer
A · Contact precautions with gloves, apron and good hand hygiene
8.Clinical nursing
Answer: C · Repeat the blood test promptly, as haemolysis can falsely raise potassium
Why this is the answer
Haemolysis in the tube releases intracellular potassium, giving a falsely high reading. The result still needs prompt repeating and review, because true hyperkalaemia cannot be excluded.
Answer
C · Repeat the blood test promptly, as haemolysis can falsely raise potassium