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 has taken a proton pump inhibitor (such as omeprazole) for many years and now has muscle cramps and tremor. Which electrolyte is most likely to be low?
AChloride
BSodium
CPhosphate
DMagnesium
2.Lab values & electrolytesHard
A stable patient's hemoglobin has been 13–14 g/dL for days. Today's result is 8.2 g/dL, but the patient feels well, the vital signs are normal and there is no bleeding. What should the nurse suspect first?
AA sample taken from the wrong patient
BSudden massive internal bleeding
CAcute bone marrow failure overnight
DA normal day-to-day variation in Hb
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 25 mmHg (3.3 kPa). Use the rule that in chronic respiratory alkalosis HCO₃⁻ falls by 5 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
Practice exercise (made-up values, not clinical guidance). A patient has a lidocaine clearance of 700 mL/min. What infusion rate R in mg/min keeps the steady-state level at 4 mg/L?
CL = 700 mL/minCss = 4 mg/L
R = mg/min
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 91 kg patient on regular gentamicin doses is at steady state. The target is a peak of 10 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?
91 kgVd = 0.3 L/kgCpeak = 10 mg/LCtrough = 1 mg/L
D = mg
7.Clinical nursingMedium
Which is the correct technique for using a metered-dose inhaler without a spacer?
APress the canister and wait one minute before breathing in
BBreathe in quickly through the nose while pressing the canister
CBreathe out, then breathe in slowly and deeply while pressing the canister, then hold
DPress the canister twice quickly, then breathe in fast and breathe straight out again
8.Clinical nursingHard
A patient with poorly controlled hyperthyroidism develops a high fever, very fast heart rate, agitation and vomiting during an infection. What is likely?
A sudden large change with no clinical change suggests a specimen error, such as wrong blood in the tube or dilution with IV fluid. The nurse assesses the patient, informs the provider and arranges a repeat sample with careful identification, while staying alert for true bleeding.
Answer
A · A sample taken from the wrong patient
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 16.5 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
|25 − 40| = 15 mmHg → 1.5 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 5·1.5 = 16.5 mmol/L
Answer
HCO₃⁻ = 16.5 mmol/L
4.Arterial blood gas interpretation
Answer: AGc = 20.5 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.
Anion gap
AG = 134 − (98 + 18) = 18
Correction for albumin
18 + 0.25·(40 − 30) = 20.5 mmol/L
Answer
AGc = 20.5 mmol/L
Above about 12 after correction: a high anion gap.
5.Clinical pharmacokinetics
Answer: R = 2.8 mg/min
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.
Clearance in L/min
700 mL/min = 0.7 L/min
Infusion rate
R = 0.7·4 = 2.8 mg/min
Answer
R = 2.8 mg/min
6.Clinical pharmacokinetics
Answer: D = 245.7 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 = 0.3·91 = 27.3 L
Dose
D = 27.3·(10 − 1) = 245.7 mg
Answer
D = 245.7 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: C · Breathe out, then breathe in slowly and deeply while pressing the canister, then hold
Why this is the answer
The patient breathes out, seals the lips around the mouthpiece, starts breathing in slowly, presses the canister and continues a deep breath, then holds it for about 10 seconds. Spacers make coordination easier.
Answer
C · Breathe out, then breathe in slowly and deeply while pressing the canister, then hold
8.Clinical nursing
Answer: B · Thyroid storm
Why this is the answer
Thyroid storm is a life-threatening worsening of hyperthyroidism, often triggered by infection or surgery. It needs urgent treatment with cooling, beta-blockade and antithyroid drugs.