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 IV fluids running into the left forearm. Where should the nurse take a blood sample?
AFrom any vein, since the fluid does not affect results
BFrom the opposite arm
CFrom just above the IV site in the same arm
DFrom the IV cannula while the fluid is running
2.Lab values & electrolytesHard
A patient's creatinine was 0.9 mg/dL (80 µmol/L) on admission and is 1.3 mg/dL (115 µmol/L) 36 hours later. Using the common KDIGO criteria, what does this show?
AA laboratory error, because creatinine cannot change within 2 days
BAcute kidney injury, because it rose by at least 0.3 mg/dL within 48 hours
CChronic kidney disease, because creatinine has risen above baseline
DNo kidney injury, because the creatinine is still within the normal range
3.Arterial blood gas interpretationMedium
Practice values: Na⁺ 138 mmol/L, glucose 26.1 mmol/L and urea 24 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 138 mmol/Lglucose 26.1 mmol/Lurea 24 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
Practice values: Na⁺ 143, Cl⁻ 107, HCO₃⁻ 22 mmol/L and albumin 27 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 vancomycin, the level is 35 mg/L at 1 h and 17.5 mg/L at 9 h. Assume first-order elimination. Find the half-life t½.
C1 = 35 mg/L (1 h)C2 = 17.5 mg/L (9 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 75 kg patient on regular tobramycin doses is at steady state. The target is a peak of 10 mg/L and a trough of 1 mg/L, and the volume of distribution is 0.25 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?
A patient with pneumonia is breathless and coughing up thick sputum. Which nursing measure helps most with breathing?
ADiscouraging coughing to save energy
BRestricting all oral fluids
CLying the patient flat so that they can rest and conserve energy
DSitting the patient upright and supporting them to clear secretions
8.Clinical nursingHard
Why should chronic hypernatraemia be corrected gradually?
ASlow correction is only needed in children
BLowering the sodium too fast can cause cerebral oedema
CFast correction causes osmotic demyelination from swelling of the kidneys
DFast correction causes severe hyperkalaemia
3Solutions
1.Lab values & electrolytes
Answer: B · From the opposite arm
Why this is the answer
Blood taken near a running infusion is mixed with the IV fluid, causing false results, such as a high glucose or a low hemoglobin. The sample is taken from the other arm, or according to policy if that is not possible.
Answer
B · From the opposite arm
2.Lab values & electrolytes
Answer: B · Acute kidney injury, because it rose by at least 0.3 mg/dL within 48 hours
Why this is the answer
KDIGO defines AKI as a creatinine rise of at least 0.3 mg/dL (26.5 µmol/L) within 48 hours, a rise to 1.5 times baseline within 7 days, or urine output below 0.5 mL/kg/h for 6 hours. A result inside the reference range can still be a significant rise.
Answer
B · Acute kidney injury, because it rose by at least 0.3 mg/dL within 48 hours
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 = 2·138 + 26.1 + 24 = 326.1 ≈ 326 mOsm/kg
Answer
Osm = 326 mOsm/kg
4.Arterial blood gas interpretation
Answer: AGc = 17.25 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 = 143 − (107 + 22) = 14
Correction for albumin
14 + 0.25·(40 − 27) = 17.25 mmol/L
Answer
AGc = 17.25 mmol/L
Above about 12 after correction: a high anion gap.
5.Clinical pharmacokinetics
Answer: t½ = 8 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(35/17.5)/8 = 0.6931/8 = 0.08664 h⁻¹
Half-life
t½ = 0.693/0.08664 = 8 h
Answer
t½ = 8 h
6.Clinical pharmacokinetics
Answer: D = 168.8 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 = 0.25·75 = 18.75 L
Dose
D = 18.75·(10 − 1) = 168.8 mg
Answer
D = 168.8 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: D · Sitting the patient upright and supporting them to clear secretions
Why this is the answer
An upright position improves lung expansion, and adequate fluids, deep breathing and coughing help clear sputum. Oxygen is given to target saturations if needed.
Answer
D · Sitting the patient upright and supporting them to clear secretions
8.Clinical nursing
Answer: B · Lowering the sodium too fast can cause cerebral oedema
Why this is the answer
In chronic hypernatraemia, brain cells adapt by holding extra solutes. If sodium falls too quickly, water moves into brain cells and causes cerebral oedema, so correction is controlled.
Answer
B · Lowering the sodium too fast can cause cerebral oedema