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's levothyroxine dose was changed today. When is the TSH usually rechecked?
AAfter about 6–8 weeks
BAfter about 24 hours
CAfter about 3 days
DAfter about 1 year
2.Lab values & electrolytesHard
A patient in DKA is receiving IV fluids and insulin. The glucose has fallen to about 220 mg/dL (12.2 mmol/L), but the anion gap is still high. What change would the nurse expect?
AAdd dextrose to the IV fluids and continue the insulin
BStop the insulin infusion because the glucose is near normal
CStop the IV fluids and start oral fluids only
DGive an IV bolus of sodium bicarbonate at once
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 20 mmHg (2.7 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⁻.
At sea level, a practice gas taken on 35% oxygen shows PaO₂ 79 mmHg (10.5 kPa) and PaCO₂ 28 mmHg (3.7 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 10 mg/L at 1 h and 3.3 mg/L at 5 h. Assume first-order elimination. Find the half-life t½.
C1 = 10 mg/L (1 h)C2 = 3.3 mg/L (5 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 59 kg patient has a digoxin level of 0.4 mcg/L; the target is 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?
59 kgVd = 7 L/kgC = 0.4 → 0.8 mcg/LF = 0.7
LD = mcg
7.Clinical nursingMedium
Which technique is correct when suctioning through a tracheostomy?
ASuction every 15 minutes whether or not secretions are present
BApply suction only while withdrawing the catheter, for no more than about 10–15 seconds
CKeep suctioning for at least a minute each time to clear all the secretions
DApply suction continuously while inserting the catheter as far as it will go
8.Clinical nursingHard
A patient with a tracheostomy suddenly cannot breathe and the tube appears blocked. What is the first action?
ARemove the inner tube (cannula) and check whether it is blocked
BWait for the doctor before doing anything
CGive oxygen through the mouth only
DDeflate the cuff and leave the patient
3Solutions
1.Lab values & electrolytes
Answer: A · After about 6–8 weeks
Why this is the answer
Levothyroxine has a half-life of about a week, so a steady state takes about 6 weeks. The TSH is therefore rechecked about 6–8 weeks after a dose change; checking earlier gives a misleading result.Checked against:Hypothyroidism · MedlinePlus (NIH) (2024)
Answer
A · After about 6–8 weeks
2.Lab values & electrolytes
Answer: A · Add dextrose to the IV fluids and continue the insulin
Why this is the answer
The glucose usually falls before the ketoacidosis resolves. Dextrose is added so insulin can continue, clearing ketones without causing hypoglycemia, until the anion gap closes. Stopping insulin too early lets ketoacidosis return.
Answer
A · Add dextrose to the IV fluids and continue the insulin
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 20 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
|20 − 40| = 20 mmHg → 2 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·2 = 20 mmol/L
Answer
HCO₃⁻ = 20 mmol/L
4.Arterial blood gas interpretation
Answer: A–a gradient = 135.5 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.
Elimination rate constant
k = ln(10/3.3)/4 = 1.109/4 = 0.2772 h⁻¹
Half-life
t½ = 0.693/0.2772 = 2.501 h
Answer
t½ = 2.501 h
6.Clinical pharmacokinetics
Answer: LD = 236 mcg
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 = 7·59 = 413 L
Rise needed
0.8 − 0.4 = 0.4 mcg/L
Loading dose
LD = 413·0.4/(1·0.7) = 236 mcg
Answer
LD = 236 mcg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: B · Apply suction only while withdrawing the catheter, for no more than about 10–15 seconds
Why this is the answer
Suctioning only on withdrawal and keeping it brief reduces hypoxia and mucosal damage. Suction is performed when needed, not routinely, and the patient is pre-oxygenated if required.
Answer
B · Apply suction only while withdrawing the catheter, for no more than about 10–15 seconds
8.Clinical nursing
Answer: A · Remove the inner tube (cannula) and check whether it is blocked
Why this is the answer
Following emergency algorithms, remove the inner cannula first, attempt suction and call for help. If that fails, further steps include cuff deflation and changing the tube.
Answer
A · Remove the inner tube (cannula) and check whether it is blocked