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
An older adult with chronic kidney disease takes glyburide (glibenclamide, a sulfonylurea) and has repeated low glucose readings. Why is this a concern?
AThe drug makes the glucose meter read falsely low
BThe drug stops working entirely when the kidneys fail
CThe drug causes high potassium that lowers the glucose
DThe drug and its metabolites build up and cause long lows
2.Lab values & electrolytesHard
A patient with severe hypertriglyceridemia has a measured sodium of 128 mEq/L, but a measured serum osmolality is normal, and the patient has no symptoms. What is the most likely explanation?
APseudohyponatremia, an artifact caused by very high lipids
BHyponatremia caused by drinking far too much water
CTrue hyponatremia from SIADH that needs fluid restriction
DHyponatremia caused by severe hyperglycemia in diabetes
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 77 mmHg (10.3 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
At sea level, a practice gas taken on 24% oxygen shows PaO₂ 50 mmHg (6.7 kPa) and PaCO₂ 42 mmHg (5.6 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 vancomycin, the level is 35 mg/L at 2 h and 13.9 mg/L at 10 h. Assume first-order elimination. Find the half-life t½.
C1 = 35 mg/L (2 h)C2 = 13.9 mg/L (10 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 94 kg patient on regular tobramycin doses is at steady state. The target is a peak of 7 mg/L and a trough of 1 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?
94 kgVd = 0.3 L/kgCpeak = 7 mg/LCtrough = 1 mg/L
D = mg
7.Clinical nursingMedium
What should the nurse regularly check for a patient with a syringe driver?
AOnly that the battery is charged and the device is switched on
BOnly the patient's weight, measured once a week on the same scales
CThe infusion site, the rate and volume remaining, and the patient's symptoms
DNothing, once the syringe driver is running
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
BLying flat or slightly head-down, holding the breath or breathing out
CSitting fully upright and taking a deep breath in as the line comes out
DStanding up and walking
3Solutions
1.Lab values & electrolytes
Answer: D · The drug and its metabolites build up and cause long lows
Why this is the answer
Glyburide and its active metabolites are cleared by the kidneys, so they accumulate in CKD and can cause severe, prolonged hypoglycemia. Older adults are at greater risk. The prescriber is informed, and a different drug is usually chosen.
Answer
D · The drug and its metabolites build up and cause long lows
2.Lab values & electrolytes
Answer: A · Pseudohyponatremia, an artifact caused by very high lipids
Why this is the answer
Very high lipids or proteins take up part of the plasma volume, so some laboratory methods report a falsely low sodium. The osmolality and the sodium measured by direct methods are normal. Treating this "hyponatremia" would be harmful.
Answer
A · Pseudohyponatremia, an artifact caused by very high lipids
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 27.7 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.
Change in PaCO₂
|77 − 40| = 37 mmHg → 3.7 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·3.7 = 27.7 mmol/L
Answer
HCO₃⁻ = 27.7 mmol/L
4.Arterial blood gas interpretation
Answer: A–a gradient = 68.62 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(35/13.9)/8 = 0.9235/8 = 0.1154 h⁻¹
Half-life
t½ = 0.693/0.1154 = 6.005 h
Answer
t½ = 6.005 h
6.Clinical pharmacokinetics
Answer: D = 169.2 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 = 0.3·94 = 28.2 L
Dose
D = 28.2·(7 − 1) = 169.2 mg
Answer
D = 169.2 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: C · The infusion site, the rate and volume remaining, and the patient's symptoms
Why this is the answer
Nurses check that the infusion is running as planned, the site has no redness or swelling, and the symptoms are controlled. Problems are acted on quickly to avoid distress.
Answer
C · The infusion site, the rate and volume remaining, and the patient's symptoms
8.Clinical nursing
Answer: B · 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
B · Lying flat or slightly head-down, holding the breath or breathing out