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
During a red cell transfusion, when are serious acute reactions most likely, so the nurse stays with the patient?
AThe next morning
BAfter about 2 hours
CIn the first 15 minutes
DAfter the last drop is given
2.Lab values & electrolytesHard
Why can an HbA1c result be misleading in a patient who received a large blood transfusion two weeks ago?
ATransfusions always raise HbA1c to falsely high levels
BHbA1c is accurate in every situation
CHbA1c measures sugar on white cells, not red cells
DMany red cells are donor cells, not exposed to the patient’s glucose
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 152 mmol/L, glucose 9.8 mmol/L and urea 12.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 152 mmol/Lglucose 9.8 mmol/Lurea 12.6 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
A practice gas shows pH 7.32, PaCO₂ 28 mmHg (3.7 kPa) and HCO₃⁻ 14 mmol/L, with Na⁺ 140 and Cl⁻ 107 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.
Practice exercise (made-up values, not clinical guidance). A 78 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.05 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 12 mg/L?
78 kgCL = 0.05 L/kg/hCss = 12 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin IV every 12 h; the vancomycin clearance is 6 L/h. Each dose is 1000 mg of vancomycin. Find the average steady-state level Css,avg.
CL = 6 L/hτ = 12 hS = 1F = 1D = 1000 mg
Css,avg = mg/L
7.Clinical nursingMedium
A postoperative patient has a heart rate of 125/min, blood pressure 88/50 mmHg and a blood-soaked dressing. What is happening?
AHypovolaemic shock from bleeding
BAnxiety only
CNormal pain response
DSeptic shock from a urinary infection
8.Clinical nursingHard
A nurse teaches a patient with hypertension about their medicines. Which statement shows a need for further teaching?
A"I'll keep taking my tablets even if I feel well."
B"I'll have my blood pressure checked regularly."
C"I'll get up slowly from sitting."
D"I'll stop my tablets once my readings are normal."
3Solutions
1.Lab values & electrolytes
Answer: C · In the first 15 minutes
Why this is the answer
Acute hemolytic and anaphylactic reactions usually appear early, so the transfusion is started slowly and the nurse stays with the patient for about the first 15 minutes, checking vital signs before, during and after according to policy.
Answer
C · In the first 15 minutes
2.Lab values & electrolytes
Answer: D · Many red cells are donor cells, not exposed to the patient’s glucose
Why this is the answer
HbA1c forms as glucose attaches to hemoglobin over the life of a red cell. Transfused donor cells and conditions that shorten red cell life give falsely low values, so a glucose log or other markers are used instead.
Answer
D · Many red cells are donor cells, not exposed to the patient’s glucose
3.Arterial blood gas interpretation
Answer: Osm = 326 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 = 2·152 + 9.8 + 12.6 = 326.4 ≈ 326 mOsm/kg
Answer
Osm = 326 mOsm/kg
4.Arterial blood gas interpretation
Answer: added normal-gap acidosis
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 = 140 − (107 + 14) = 19
Delta ratio
(19 − 12)/(24 − 14) = 0.7
added normal-gap acidosis
Answer
added normal-gap acidosis
5.Clinical pharmacokinetics
Answer: R = 58.5 mg/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.
Clearance
CL = 0.05·78 = 3.9 L/h
Infusion rate
R = CL·Css/S = 3.9·12/0.8 = 58.5 mg/h
Answer
R = 58.5 mg/h
6.Clinical pharmacokinetics
Answer: Css,avg = 13.89 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
A · Hypovolaemic shock from bleeding
8.Clinical nursing
Answer: D · "I'll stop my tablets once my readings are normal."
Why this is the answer
Hypertension usually needs long-term treatment; stopping because readings are normal often makes blood pressure rise again. Patients should discuss any changes with their prescriber.
Answer
D · "I'll stop my tablets once my readings are normal."