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
How does normal pregnancy usually affect serum creatinine?
AIt rises, because GFR falls
BIt falls, because GFR rises
CIt stays exactly the same
DIt rises with every week
2.Lab values & electrolytesHard
A patient without symptoms has a TSH of 7.5 mIU/L on two occasions with a normal free T4. Which term fits best?
ASubclinical hypothyroidism
BSecondary hypothyroidism
CSubclinical hyperthyroidism
DOvert primary hypothyroidism
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 146 mmol/L, glucose 16.7 mmol/L and urea 20.5 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 146 mmol/Lglucose 16.7 mmol/Lurea 20.5 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
Practice values: Na⁺ 144, Cl⁻ 112, HCO₃⁻ 22 mmol/L and albumin 22 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Practice exercise (made-up values, not clinical guidance). A 53 kg patient is to be loaded with digoxin tablets to reach a level of 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.
53 kgVd = 7 L/kgC = 0.8 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin IV every 12 h; the vancomycin clearance is 3 L/h. Each dose is 750 mg of vancomycin. Find the average steady-state level Css,avg.
CL = 3 L/hτ = 12 hS = 1F = 1D = 750 mg
Css,avg = mg/L
7.Clinical nursingMedium
A rising serum creatinine most often indicates:
AImproved liver function
BReduced kidney function
CDehydration of the skin only
DIncreased muscle strength
8.Clinical nursingHard
A patient rescued from a house fire has soot around the mouth, singed nasal hairs and a hoarse voice. What is the main concern?
AHypothermia from the fire service water
BA minor throat infection
CDehydration from sweating
DAirway swelling from inhalation injury
3Solutions
1.Lab values & electrolytes
Answer: B · It falls, because GFR rises
Why this is the answer
In pregnancy, kidney blood flow and GFR rise by about 50%, so creatinine falls below non-pregnant values. A creatinine that looks "normal" for a non-pregnant adult may indicate reduced kidney function in pregnancy.
Answer
B · It falls, because GFR rises
2.Lab values & electrolytes
Answer: A · Subclinical hypothyroidism
Why this is the answer
A raised TSH with a normal free T4 is subclinical hypothyroidism; the pituitary is working harder to keep T4 normal. Overt hypothyroidism has a low free T4, and subclinical hyperthyroidism has a low TSH.Checked against:TSH (Thyroid-stimulating hormone) Test · MedlinePlus (NIH) (2024)
Answer
A · Subclinical hypothyroidism
3.Arterial blood gas interpretation
Answer: Osm = 329 mOsm/kg
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.
Calculated osmolality
Osm = 2·146 + 16.7 + 20.5 = 329.2 ≈ 329 mOsm/kg
Answer
Osm = 329 mOsm/kg
4.Arterial blood gas interpretation
Answer: AGc = 14.5 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 = 144 − (112 + 22) = 10
Correction for albumin
10 + 0.25·(40 − 22) = 14.5 mmol/L
Answer
AGc = 14.5 mmol/L
Above about 12 after correction: a high anion gap.
5.Clinical pharmacokinetics
Answer: LD = 424 mcg
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 = 7·53 = 371 L
Amount of active drug needed
Vd·C = 371·0.8 = 296.8 mcg
Loading dose
LD = 296.8/(1·0.7) = 424 mcg
Answer
LD = 424 mcg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: Css,avg = 20.83 mg/L
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.
Creatinine is a muscle waste product filtered by the kidneys. A rise shows the glomerular filtration rate is falling, as in acute kidney injury.
Answer
B · Reduced kidney function
8.Clinical nursing
Answer: D · Airway swelling from inhalation injury
Why this is the answer
These signs suggest inhalation injury, which can cause rapid airway swelling and obstruction. Early senior anaesthetic review is needed, as early intubation may be required. Carbon monoxide poisoning must also be considered.