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 woman who is 30 weeks pregnant has a new blood pressure of 152/98 mm Hg. Which urine finding would most support preeclampsia?
AGlucose in the urine
BNitrites in the urine
CKetones in the urine
DProtein in the urine
2.Lab values & electrolytesHard
A patient has eaten very little for 2 weeks and has received broad-spectrum antibiotics. The INR is now 2.3 without warfarin, and the platelets are normal. What is the most likely cause?
AHemophilia A
BIron-deficiency anemia
CVitamin K deficiency
DImmune thrombocytopenia
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 14 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 14 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 37 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 37 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 80 mg of tobramycin. The level back-extrapolated to time zero is 6 mg/L and the half-life is 2.5 h. Find the clearance CL in L/h.
D = 80 mgC₀ = 6 mg/Lt½ = 2.5 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 62 kg patient has a phenytoin level of 4.5 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?
62 kgVd = 0.7 L/kgC = 4.5 → 15 mg/LS = 0.92
LD = mg
7.Clinical nursingMedium
What happens in third-degree (complete) heart block?
AEvery impulse reaches the ventricles, but each one is slightly delayed
BThe ventricles beat very fast and regularly
CThe atria beat chaotically with no P waves, and the ventricular rate is irregular
DNo impulses from the atria reach the ventricles, so they beat independently
8.Clinical nursingHard
Which action is correct when caring for a chest drain with an underwater seal?
ALift the bottle above the patient to empty it
BKeep the drainage bottle below the level of the chest
CClamp the tube routinely during every transfer or bed move
DRemove the water from the bottle to prevent spills
3Solutions
1.Lab values & electrolytes
Answer: D · Protein in the urine
Why this is the answer
Preeclampsia is new hypertension after 20 weeks of pregnancy with proteinuria or other signs of organ involvement, such as low platelets, raised liver enzymes or rising creatinine. Proteinuria reflects damage to the glomerular filter.
Answer
D · Protein in the urine
2.Lab values & electrolytes
Answer: C · Vitamin K deficiency
Why this is the answer
Vitamin K comes from the diet and from gut bacteria. Poor intake plus antibiotics can cause vitamin K deficiency, which lowers factors II, VII, IX and X and prolongs the PT/INR. Vitamin K replacement corrects it.
Answer
C · Vitamin K deficiency
3.Arterial blood gas interpretation
Answer: PaCO₂ = 29 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 = 1.5·14 + 8 = 29 mmHg
range: 27–31 mmHg
Answer
PaCO₂ = 29 mmHg
4.Arterial blood gas interpretation
Answer: PaCO₂ = 46.9 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.
Expected PaCO₂
0.7·37 + 21 = 46.9 mmHg
Answer
PaCO₂ = 46.9 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: CL = 3.697 L/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.
Volume of distribution
Vd = 80/6 = 13.33 L
Rate constant
k = 0.693/2.5 = 0.2773 h⁻¹
Clearance
CL = 0.2773·13.33 = 3.697 L/h
Answer
CL = 3.697 L/h
6.Clinical pharmacokinetics
Answer: LD = 495.3 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 = 0.7·62 = 43.4 L
Rise needed
15 − 4.5 = 10.5 mg/L
Loading dose
LD = 43.4·10.5/(0.92·1) = 495.3 mg
Answer
LD = 495.3 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · No impulses from the atria reach the ventricles, so they beat independently
Why this is the answer
In complete heart block the atria and ventricles are dissociated, and a slow escape rhythm drives the ventricles. Patients may be dizzy or collapse and often need pacing.
Answer
D · No impulses from the atria reach the ventricles, so they beat independently
8.Clinical nursing
Answer: B · Keep the drainage bottle below the level of the chest
Why this is the answer
The bottle stays below chest level so fluid and air cannot return to the pleural space. Routine clamping can cause a tension pneumothorax.
Answer
B · Keep the drainage bottle below the level of the chest