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
After pituitary surgery, a patient passes about 400 mL of pale urine per hour with a specific gravity of 1.002, and the sodium has risen to 149 mEq/L. Which condition is most likely?
ADiabetes insipidus
BHyperglycemia with ketoacidosis
CSIADH
DAcute kidney injury with oliguria
2.Lab values & electrolytesHard
A patient with no kidney problems and a normal ECG has a potassium of 6.2 mEq/L. The laboratory comments that the sample was hemolyzed. What is the best interpretation?
ATreat immediately with insulin and glucose without further checks
BThe patient has kidney failure
CThe potassium is certainly normal and needs no follow-up
DThe result may be falsely high; assess the patient and arrange a repeat sample
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 143 mmol/L, glucose 7.2 mmol/L and urea 18.9 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 143 mmol/Lglucose 7.2 mmol/Lurea 18.9 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
Practice values: Na⁺ 139, Cl⁻ 88 and HCO₃⁻ 15 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 139 mmol/LCl⁻ 88 mmol/LHCO₃⁻ 15 mmol/L
delta ratio =
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 61 kg patient is to be loaded with digoxin tablets to reach a level of 1 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.
61 kgVd = 7 L/kgC = 1 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 76 kg patient on regular tobramycin doses is at steady state. The target is a peak of 6 mg/L and a trough of 0.5 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?
Why are alternate sutures sometimes removed first when taking out stitches?
ABecause the remaining sutures dissolve on their own within a day
BTo reduce the cost of dressings
CTo make the procedure faster
DTo check the wound stays closed before removing the rest
8.Clinical nursingHard
A patient taking regular ibuprofen suddenly develops severe upper abdominal pain, a rigid abdomen and becomes tachycardic. What is most likely?
AIrritable bowel syndrome
BConstipation
CA perforated peptic ulcer
DIndigestion that will settle with antacids
3Solutions
1.Lab values & electrolytes
Answer: A · Diabetes insipidus
Why this is the answer
Surgery near the pituitary can stop the release of antidiuretic hormone. Without it, the kidneys pass large volumes of dilute urine with a low specific gravity, and the sodium rises. SIADH is the opposite: concentrated urine and a low sodium. The nurse measures intake and output closely and reports early.Checked against:Diabetes insipidus · MedlinePlus (NIH) (2025); Urine specific gravity test · MedlinePlus (NIH) (2025); Sodium blood test · MedlinePlus (NIH) (2025)
Answer
A · Diabetes insipidus
2.Lab values & electrolytes
Answer: D · The result may be falsely high; assess the patient and arrange a repeat sample
Why this is the answer
Red cells are rich in potassium. When they break in the tube (from a tight tourniquet, fist clenching or a small needle), potassium leaks out and the result is falsely high. The nurse still assesses the patient and repeats the test, because true hyperkalemia must not be missed.Checked against:Potassium Blood Test · MedlinePlus (NIH) (2024)
Answer
D · The result may be falsely high; assess the patient and arrange a repeat sample
3.Arterial blood gas interpretation
Answer: Osm = 312 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 = 2·143 + 7.2 + 18.9 = 312.1 ≈ 312 mOsm/kg
Answer
Osm = 312 mOsm/kg
4.Arterial blood gas interpretation
Answer: delta ratio = 2.667
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 = 139 − (88 + 15) = 36
Delta ratio
(36 − 12)/(24 − 15) = 24/9 = 2.667
Answer
delta ratio = 2.667
Above 2: an added metabolic alkalosis.
5.Clinical pharmacokinetics
Answer: LD = 610 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 = 7·61 = 427 L
Amount of active drug needed
Vd·C = 427·1 = 427 mcg
Loading dose
LD = 427/(1·0.7) = 610 mcg
Answer
LD = 610 mcg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: D = 125.4 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 = 0.3·76 = 22.8 L
Dose
D = 22.8·(6 − 0.5) = 125.4 mg
Answer
D = 125.4 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · To check the wound stays closed before removing the rest
Why this is the answer
Removing alternate sutures allows the nurse to see whether the wound edges stay together. If they begin to separate, the remaining sutures can be left and the team informed.
Answer
D · To check the wound stays closed before removing the rest
8.Clinical nursing
Answer: C · A perforated peptic ulcer
Why this is the answer
Sudden severe pain with a board-like abdomen suggests perforation and peritonitis. The patient needs nil by mouth, IV access, analgesia and urgent surgical review.