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 with cirrhosis has an albumin of 2.2 g/dL (22 g/L). Which finding is most likely related?
AIncreased urine output
BHigh blood pressure
CEdema and ascites
DHyperkalemia
2.Lab values & electrolytesHard
Which cerebrospinal fluid (CSF) pattern most strongly suggests bacterial meningitis?
ACloudy fluid, low glucose, high protein, many neutrophils
BClear fluid, normal glucose, normal protein, no white cells
CClear fluid, normal glucose, mildly raised protein, lymphocytes
DClear fluid, high glucose, low protein, many red cells
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.65, PaCO₂ 37 mmHg (4.9 kPa) and HCO₃⁻ 39 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.65PaCO₂ 37 mmHg (4.9 kPa)HCO₃⁻ 39 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.31, PaCO₂ 27 mmHg (3.6 kPa), HCO₃⁻ 13 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.31PaCO₂ 27 mmHg (3.6 kPa)HCO₃⁻ 13 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 180 mg of gentamicin. The level back-extrapolated to time zero is 6 mg/L and the half-life is 3 h. Find the clearance CL in L/h.
D = 180 mgC₀ = 6 mg/Lt½ = 3 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 2 h. The target is a peak of 10 mg/L and a trough of 0.5 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.
t½ = 2 hCpeak = 10 mg/LCtrough = 0.5 mg/L
τ = h
7.Clinical nursingMedium
Why are people with diabetes advised to have regular eye screening?
ATo measure their blood glucose through the eye
BTo prevent conjunctivitis
CTo detect diabetic retinopathy early, before sight is lost
DTo check whether they need reading glasses for close work
8.Clinical nursingHard
A postoperative patient on IV fluids has a sodium of 128 mmol/L and is becoming confused. Which fluid could make this worse if given in large amounts?
AHartmann's solution (similar to lactated Ringer's)
B5% glucose (D5W)
CHuman albumin solution
D0.9% sodium chloride
3Solutions
1.Lab values & electrolytes
Answer: C · Edema and ascites
Why this is the answer
Albumin provides most of the oncotic pressure that keeps fluid in the vessels. When it is low, fluid leaks into the tissues and abdomen, causing edema and ascites, together with portal hypertension in cirrhosis.Checked against:Albumin Blood Test · MedlinePlus (NIH) (2024)
Answer
C · Edema and ascites
2.Lab values & electrolytes
Answer: A · Cloudy fluid, low glucose, high protein, many neutrophils
Why this is the answer
Bacterial meningitis typically gives cloudy CSF with many neutrophils, high protein and low glucose, because bacteria and white cells use glucose. Viral meningitis usually shows lymphocytes with a normal glucose; option C is normal CSF.Checked against:CSF glucose test · MedlinePlus (NIH) (2025); CSF cell count · MedlinePlus (NIH) (2025); CSF total protein · MedlinePlus (NIH) (2025)
Answer
A · Cloudy fluid, low glucose, high protein, many neutrophils
3.Arterial blood gas interpretation
Answer: uncompensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.65: alkalaemia (above 7.45)
PaCO₂ 37 mmHg: normal (35–45)
HCO₃⁻ 39 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 46.3–50.3 mmHg, so the measured 37 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: appropriate compensation
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
1.5·13 + 8 = 27.5 → 25.5–29.5 mmHg
Compare
measured: 27 mmHg → appropriate compensation
Answer
appropriate compensation
5.Clinical pharmacokinetics
Answer: CL = 6.931 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 = 180/6 = 30 L
Rate constant
k = 0.693/3 = 0.231 h⁻¹
Clearance
CL = 0.231·30 = 6.931 L/h
Answer
CL = 6.931 L/h
6.Clinical pharmacokinetics
Answer: τ = 8.6 h
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.
Rate constant
k = 0.693/2 = 0.3466 h⁻¹
Dosing interval
τ = ln(10/0.5)/0.3466 = 2.996/0.3466 = 8.644 ≈ 8.6 h
Answer
τ = 8.6 h
In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.
7.Clinical nursing
Answer: C · To detect diabetic retinopathy early, before sight is lost
Why this is the answer
Diabetic retinopathy often has no symptoms until it is advanced. Screening allows early treatment that can prevent blindness.
Answer
C · To detect diabetic retinopathy early, before sight is lost
8.Clinical nursing
Answer: B · 5% glucose (D5W)
Why this is the answer
The glucose in 5% glucose is metabolised, leaving free water that dilutes the plasma sodium further. Postoperative patients often have raised ADH, so large volumes of hypotonic fluid can cause dangerous hyponatraemia.