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
Which instruction is correct for a patient starting oral iron tablets?
ATake each tablet with milk to protect the stomach
BTake the tablet with an antacid to improve absorption
CYour urine will turn bright orange while you take it
DYour stools may turn dark green or black
2.Lab values & electrolytesHard
A young man has a swollen knee after minor trauma. His aPTT is prolonged, but his PT/INR and platelet count are normal. Which condition fits best?
AVitamin K deficiency from poor diet
BDisseminated intravascular coagulation
CImmune thrombocytopenia (ITP)
DHemophilia A (factor VIII deficiency)
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 143 mmol/L, glucose 12.9 mmol/L and urea 17.7 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 143 mmol/Lglucose 12.9 mmol/Lurea 17.7 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
Practice values: Na⁺ 139, Cl⁻ 109, HCO₃⁻ 14 mmol/L and albumin 38 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 6 mg/L at 2 h and 1.9 mg/L at 7 h. Assume first-order elimination. Find the half-life t½.
C1 = 6 mg/L (2 h)C2 = 1.9 mg/L (7 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 61 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.04 L/kg/h. Each dose is 200 mg of aminophylline. Find the average steady-state level Css,avg.
A postoperative patient has a heart rate of 125/min, blood pressure 88/50 mmHg and a blood-soaked dressing. What is happening?
AAnxiety only
BSeptic shock from a urinary infection
CHypovolaemic shock from bleeding
DNormal pain response
8.Clinical nursingHard
A patient with a central venous catheter needs the line removed. Which position reduces the risk of air embolism during removal of a neck or chest line?
ALying on the right side with the legs raised
BStanding up and walking
CLying flat or slightly head-down, holding the breath or breathing out
DSitting fully upright and taking a deep breath in as the line comes out
3Solutions
1.Lab values & electrolytes
Answer: D · Your stools may turn dark green or black
Why this is the answer
Oral iron commonly darkens the stools, which is harmless, but black tarry stools with abdominal pain should be reported. Milk, antacids and tea reduce absorption; vitamin C or orange juice can improve it.
Answer
D · Your stools may turn dark green or black
2.Lab values & electrolytes
Answer: D · Hemophilia A (factor VIII deficiency)
Why this is the answer
Hemophilia A is an X-linked lack of factor VIII, which belongs to the intrinsic pathway, so only the aPTT is prolonged. Bleeding into joints and muscles is typical. Vitamin K deficiency and DIC prolong the PT as well, and ITP lowers the platelets.
Answer
D · Hemophilia A (factor VIII deficiency)
3.Arterial blood gas interpretation
Answer: Osm = 317 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 + 12.9 + 17.7 = 316.6 ≈ 317 mOsm/kg
Answer
Osm = 317 mOsm/kg
4.Arterial blood gas interpretation
Answer: AGc = 16.5 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 = 139 − (109 + 14) = 16
Correction for albumin
16 + 0.25·(40 − 38) = 16.5 mmol/L
Answer
AGc = 16.5 mmol/L
Above about 12 after correction: a high anion gap.
5.Clinical pharmacokinetics
Answer: t½ = 3.014 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.
Elimination rate constant
k = ln(6/1.9)/5 = 1.15/5 = 0.23 h⁻¹
Half-life
t½ = 0.693/0.23 = 3.014 h
Answer
t½ = 3.014 h
6.Clinical pharmacokinetics
Answer: Css,avg = 8.197 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.
Tachycardia, hypotension and visible blood loss suggest haemorrhagic (hypovolaemic) shock. Apply pressure, call for urgent help and prepare for fluids or blood.
Answer
C · Hypovolaemic shock from bleeding
8.Clinical nursing
Answer: C · Lying flat or slightly head-down, holding the breath or breathing out
Why this is the answer
A flat or head-down position and a Valsalva manoeuvre or breath-hold raise central venous pressure, reducing the risk of air being drawn in. An occlusive dressing is then applied.
Answer
C · Lying flat or slightly head-down, holding the breath or breathing out