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 has a sodium of 152 mEq/L. Which findings would the nurse expect?
AMuscle twitching with a positive Trousseau’s sign
BBradycardia with peaked T waves
CWeight gain with swollen ankles
DThirst, dry mucous membranes and restlessness
2.Lab values & electrolytesHard
A patient taking apixaban asks why they do not need regular INR tests like their neighbor on warfarin. What is the best explanation?
AApixaban is monitored with an aPTT every day instead
BThe INR is checked secretly together with the other routine blood tests
CIts effect is predictable, so routine clotting tests are not needed
DApixaban does not actually thin the blood
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 132 mmol/L, glucose 22.8 mmol/L and urea 24.7 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 132 mmol/Lglucose 22.8 mmol/Lurea 24.7 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 35% oxygen shows PaO₂ 83 mmHg (11.1 kPa) and PaCO₂ 44 mmHg (5.9 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 220 mg of gentamicin. The level back-extrapolated to time zero is 12 mg/L and the half-life is 3.5 h. Find the clearance CL in L/h.
D = 220 mgC₀ = 12 mg/Lt½ = 3.5 h
CL = L/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 77-year-old man weighs 78 kg and has a serum creatinine of 1.4 mg/dL (124 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual levetiracetam dose is 1000 mg every 12 h for a creatinine clearance of 100 mL/min. Keeping the interval, the dose is reduced in proportion to the creatinine clearance. Find the adjusted dose D. Round to the nearest whole number.
A dying patient can no longer drink and has a dry mouth. What is the most appropriate nursing care?
AStarting a fast intravenous infusion to correct the dryness
BLeaving the mouth alone to avoid disturbing them
CFrequent mouth care that keeps the lips and mouth moist
DInserting a nasogastric tube for fluids
8.Clinical nursingHard
An older patient found on the floor after lying there for many hours has dark brown urine and a very high creatine kinase. What is the main risk?
AA stroke
BAcute kidney injury from rhabdomyolysis
CA urinary tract infection from lying in urine
DHypoglycaemia
3Solutions
1.Lab values & electrolytes
Answer: D · Thirst, dry mucous membranes and restlessness
Why this is the answer
Hypernatremia usually means a water deficit, causing thirst, dry mouth, restlessness, agitation and, if severe, seizures. It is corrected slowly to avoid cerebral edema.
Answer
D · Thirst, dry mucous membranes and restlessness
2.Lab values & electrolytes
Answer: C · Its effect is predictable, so routine clotting tests are not needed
Why this is the answer
Direct oral anticoagulants such as apixaban have predictable effects at fixed doses, so routine INR testing is not needed. Kidney and liver function, adherence and signs of bleeding are monitored, and the INR does not reliably measure their effect.
Answer
C · Its effect is predictable, so routine clotting tests are not needed
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 = 2·132 + 22.8 + 24.7 = 311.5 ≈ 312 mOsm/kg
Answer
Osm = 312 mOsm/kg
4.Arterial blood gas interpretation
Answer: A–a gradient = 111.5 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.
Volume of distribution
Vd = 220/12 = 18.33 L
Rate constant
k = 0.693/3.5 = 0.198 h⁻¹
Clearance
CL = 0.198·18.33 = 3.631 L/h
Answer
CL = 3.631 L/h
6.Clinical pharmacokinetics
Answer: D = 488 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.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 77)·78/(72·1.4) = 48.75 mL/min
Adjusted dose
D = 1000·48.75/100 = 487.5 ≈ 488 mg
Answer
D = 488 mg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: C · Frequent mouth care that keeps the lips and mouth moist
Why this is the answer
Regular mouth care with moist swabs, water sprays and lip balm relieves a dry mouth in the last days of life. Families can be shown how to help.
Answer
C · Frequent mouth care that keeps the lips and mouth moist
8.Clinical nursing
Answer: B · Acute kidney injury from rhabdomyolysis
Why this is the answer
Prolonged pressure on muscles causes rhabdomyolysis, releasing myoglobin and potassium. Myoglobin can block kidney tubules; treatment includes IV fluids and close monitoring of urine output and electrolytes.