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 had a total thyroidectomy yesterday. Which laboratory value is most important to monitor in the first days after surgery?
ASerum sodium
BSerum glucose
CSerum calcium
DSerum potassium
2.Lab values & electrolytesHard
A patient on an IV insulin infusion has a blood glucose of 65 mg/dL (3.6 mmol/L). What is the best action?
AStop the infusion and leave the glucose recheck until the morning
BGive oral potassium to protect the heart
CFollow the protocol for low glucose and inform the provider
DContinue at the same rate and recheck in four hours
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 28% oxygen shows PaO₂ 59 mmHg (7.9 kPa) and PaCO₂ 42 mmHg (5.6 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A 95 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 12 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.
95 kgVd = 0.5 L/kgC = 12 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 98 kg patient on regular gentamicin doses is at steady state. The target is a peak of 6 mg/L and a trough of 0.5 mg/L, and the volume of distribution is 0.25 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?
Which oxygen device delivers a precise, fixed concentration of oxygen?
ASimple face mask
BNasal cannula
CNon-rebreathe mask
DVenturi mask
8.Clinical nursingHard
A patient on IV vancomycin develops flushing and an itchy red rash on the face and neck during a fast infusion. What is the most likely cause and response?
AInfusion-related reaction; stop or slow the infusion and inform the prescriber
BNormal response; increase the infusion rate
CLiver failure; stop all medicines permanently
DSunburn; apply cream and continue
3Solutions
1.Lab values & electrolytes
Answer: C · Serum calcium
Why this is the answer
The parathyroid glands can be injured or removed during thyroidectomy, so parathyroid hormone falls and hypocalcemia can follow. The nurse monitors calcium and checks for tingling around the mouth, muscle cramps and Chvostek or Trousseau signs.Checked against:Hypoparathyroidism · MedlinePlus (NIH) (2026)
Answer
C · Serum calcium
2.Lab values & electrolytes
Answer: C · Follow the protocol for low glucose and inform the provider
Why this is the answer
A glucose below 70 mg/dL on an insulin infusion is hypoglycemia. The nurse follows the hospital’s protocol, which typically stops or reduces the insulin and gives glucose, rechecks often and informs the provider.Checked against:Low blood sugar · MedlinePlus (NIH) (2026)
Answer
C · Follow the protocol for low glucose and inform the provider
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 = 88.14 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 = 0.5·95 = 47.5 L
Amount of active drug needed
Vd·C = 47.5·12 = 570 mg
Loading dose
LD = 570/(0.8·1) = 712.5 mg
Answer
LD = 712.5 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: D = 134.8 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 = 0.25·98 = 24.5 L
Dose
D = 24.5·(6 − 0.5) = 134.8 mg
Answer
D = 134.8 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · Venturi mask
Why this is the answer
A Venturi mask entrains a set amount of room air, giving a fixed oxygen percentage regardless of the patient's breathing pattern. This makes it useful for patients at risk of hypercapnia.
Answer
D · Venturi mask
8.Clinical nursing
Answer: A · Infusion-related reaction; stop or slow the infusion and inform the prescriber
Why this is the answer
Rapid vancomycin infusion can release histamine and cause flushing and rash ("vancomycin infusion reaction"). Slowing the rate usually prevents it, but anaphylaxis must be excluded.
Answer
A · Infusion-related reaction; stop or slow the infusion and inform the prescriber