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 chronic kidney disease has a potassium of 5.6 mEq/L. Which food should the nurse advise the patient to limit?
AApples
BCabbage
CBananas
DWhite rice
2.Lab values & electrolytesHard
After a crush injury, a patient's urine dipstick is strongly positive for blood, but microscopy shows almost no red cells. CK is very high. What best explains this?
AThe sample was contaminated with menstrual blood
BBleeding is coming from the bladder lining
CA kidney stone is scraping the ureter lining
DMyoglobin from damaged muscle is in the urine
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.29, PaCO₂ 62 mmHg (8.3 kPa) and HCO₃⁻ 29 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.
pH 7.29PaCO₂ 62 mmHg (8.3 kPa)HCO₃⁻ 29 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 24% oxygen shows PaO₂ 68 mmHg (9.1 kPa) and PaCO₂ 51 mmHg (6.8 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 9 mg/L at 2 h and 1.4 mg/L at 10 h. Assume first-order elimination. Find the half-life t½.
C1 = 9 mg/L (2 h)C2 = 1.4 mg/L (10 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A theophylline infusion is started without a loading dose. The half-life in this patient is 9 h. How long t does it take to reach 95% of the steady-state level?
t½ = 9 h95% Css
t = h
7.Clinical nursingMedium
A patient with type 2 diabetes is acutely unwell, very dehydrated and has a blood glucose of 40 mmol/L without significant ketones. What is the most likely diagnosis?
AHypoglycaemia
BHyperosmolar hyperglycaemic state
CDiabetic ketoacidosis with dehydration
DDiabetes insipidus
8.Clinical nursingHard
Two days after a bowel resection, a patient has rising pain, fever, tachycardia and fluid that looks like faeces in the wound drain. What is most likely?
AConstipation
BUrinary retention
CAnastomotic leak
DNormal postoperative recovery
3Solutions
1.Lab values & electrolytes
Answer: C · Bananas
Why this is the answer
Bananas, oranges, potatoes, tomatoes, dried fruit and beans are high in potassium. Apples, white rice and cabbage are lower-potassium choices that are commonly allowed on a potassium-restricted diet.Checked against:Potassium in diet · MedlinePlus (NIH) (2026); Chronic kidney disease · MedlinePlus (NIH) (2025)
Answer
C · Bananas
2.Lab values & electrolytes
Answer: D · Myoglobin from damaged muscle is in the urine
Why this is the answer
The dipstick blood test detects heme, which is present in both hemoglobin and myoglobin. Positive "blood" without red cells, together with a very high CK, indicates myoglobinuria from rhabdomyolysis, which can damage the kidneys. IV fluids and close monitoring of urine output and potassium are needed.Checked against:Myoglobin urine test · MedlinePlus (NIH) (2025); Rhabdomyolysis · MedlinePlus (NIH) (2025); Creatine Kinase · MedlinePlus (NIH) (2023)
Answer
D · Myoglobin from damaged muscle is in the urine
3.Arterial blood gas interpretation
Answer: partially compensated
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 7.29: acidaemia (below 7.35)
PaCO₂ 62 mmHg: high (35–45)
HCO₃⁻ 29 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
4.Arterial blood gas interpretation
Answer: A–a gradient = 39.37 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(9/1.4)/8 = 1.861/8 = 0.2326 h⁻¹
Half-life
t½ = 0.693/0.2326 = 2.98 h
Answer
t½ = 2.98 h
6.Clinical pharmacokinetics
Answer: t = 38.9 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 = 0.693/9 = 0.07702 h⁻¹
Time to the given fraction of steady state
t = −ln(1 − 0.95)/0.07702 = 2.996/0.07702 = 38.9 h
Answer
t = 38.9 h
That is 4.322 half-lives.
7.Clinical nursing
Answer: B · Hyperosmolar hyperglycaemic state
Why this is the answer
HHS usually affects people with type 2 diabetes. It develops over days with extreme hyperglycaemia, dehydration and high osmolality, but little ketosis.
Answer
B · Hyperosmolar hyperglycaemic state
8.Clinical nursing
Answer: C · Anastomotic leak
Why this is the answer
An anastomotic leak lets bowel contents into the abdomen, causing peritonitis and sepsis. It is a surgical emergency that needs urgent review.