Lab Values for Nurses: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Lab values & electrolytes

Method

  1. Compare the value with the laboratory’s own reference range, then with the patient’s baseline and trend.
  2. Check the sample: hemolysis, a delay, or blood drawn near an IV can give false results.
  3. Link the value to its typical signs and to the drugs and conditions that cause it.
  4. Look for interactions: low potassium or magnesium increases digoxin toxicity; dehydration raises lithium.
  5. Act on the patient: assess, hold a drug if the level is unsafe, read back critical values and inform the prescriber.

Arterial blood gas interpretation

Normal ranges
pH 7.35–7.45, PaCO₂ 35–45 mmHg (4.7–6.0 kPa), HCO₃⁻ 22–26 mmol/L
ROME
Respiratory: PaCO₂ opposite to pH; Metabolic: HCO₃⁻ equal to pH
Winter’s formula
PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2
Metabolic alkalosis
PaCO₂ ≈ 0.7·HCO₃⁻ + 21
Anion gap
AG = Na⁺ − (Cl⁻ + HCO₃⁻) ≈ 8–12
Albumin-corrected gap
AG + 2.5·(4.0 − alb g/dL) = AG + 0.25·(40 − alb g/L)
Delta ratio
(AG − 12)/(24 − HCO₃⁻)
P/F ratio
PaO₂/FiO₂
Alveolar gas equation
PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8
Serum osmolality
2·Na⁺ + glucose + urea (2·Na⁺ + glucose/18 + BUN/2.8)

Method

  1. Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
  2. Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
  3. 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.
  4. In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
  5. 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
Renal adjustment
dose × CrCl/normal; Dettli Q = 1 − fe·(1 − CrCl/normal)

Method

  1. Convert the half-life to a rate constant: k = 0.693/t½.
  2. Turn per-kg values into totals: Vd (L) = L/kg × weight, CL (L/h) = L/kg/h × weight.
  3. Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
  4. For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
  5. 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 124 mEq/L, a headache and new confusion. Which nursing measure is most appropriate?

  1. AKeep the patient flat and limit neurological checks to once a shift
  2. BPut seizure precautions in place and monitor neurological status
  3. CGive the next dose of the prescribed thiazide diuretic early
  4. DEncourage the patient to drink plenty of plain water
2.Lab values & electrolytesHard

The nurse teaches a patient starting lithium. Which statement shows a need for further teaching?

  1. A"I will check with the pharmacist before taking ibuprofen"
  2. B"I will drink plenty of fluids, especially in hot weather"
  3. C"I will tell my doctor if I have vomiting or diarrhea"
  4. D"I will cut out salt from my diet completely to stay healthy"
3.Arterial blood gas interpretationMedium
pH7.357.457.44PaCO₂mmHg354560HCO₃⁻mmol/L222639uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.44, PaCO₂ 60 mmHg (8.0 kPa) and HCO₃⁻ 39 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.

pH 7.44PaCO₂ 60 mmHg (8.0 kPa)HCO₃⁻ 39 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard
pH7.357.457.15PaCO₂mmHg354584HCO₃⁻mmol/L222628acutechronicacute on chronic

A practice gas shows a respiratory acidosis: pH 7.15, PaCO₂ 84 mmHg (11.2 kPa), HCO₃⁻ 28 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.

pH 7.15PaCO₂ 84 mmHg (11.2 kPa)HCO₃⁻ 28 mmol/L
  1. 1acute
  2. 2chronic
  3. 3acute on chronic
5.Clinical pharmacokineticsMedium
Vd = 7 L/kgC = 0.8 mcg/LLD = ? mcg66 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 66 kg patient is to be loaded with digoxin tablets to reach a level of 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.

66 kgVd = 7 L/kgC = 0.8 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​t = ?50%

Practice exercise (made-up values, not clinical guidance). A theophylline infusion is started without a loading dose. The half-life in this patient is 9 h. How long t does it take to reach 50% of the steady-state level?

t½ = 9 h50% Css
t = h
7.Clinical nursingMedium

In UK guidance, neutropenic sepsis is commonly defined by a temperature above 38 °C or signs of sepsis with a neutrophil count at or below:

  1. A150 × 109/L
  2. B5 × 109/L
  3. C0.5 × 109/L
  4. D15 × 109/L
8.Clinical nursingHard

A nurse phones a doctor about a deteriorating patient using SBAR. Which statement belongs in the "R" section?

  1. A"Please review him within the next 15 minutes."
  2. B"He was admitted three days ago with pneumonia."
  3. C"His blood pressure is 85/50 and he is confused."
  4. D"I am calling about Mr Smith, who is breathless."

3Solutions

1.Lab values & electrolytes

Answer: B · Put seizure precautions in place and monitor neurological status

Why this is the answer

Hyponatremia can cause cerebral edema and seizures. The nurse monitors neurological status, takes seizure precautions and reports the result. Extra water and thiazides can lower the sodium further.Checked against: Low blood sodium · MedlinePlus (NIH) (2025)

Answer

B · Put seizure precautions in place and monitor neurological status
2.Lab values & electrolytes

Answer: D · "I will cut out salt from my diet completely to stay healthy"

Why this is the answer

A sudden fall in sodium intake makes the kidneys keep more sodium and lithium, raising the lithium level towards toxicity. Patients keep a normal, steady salt intake, drink enough fluids, report fluid loss and avoid NSAIDs unless approved.

Answer

D · "I will cut out salt from my diet completely to stay healthy"
3.Arterial blood gas interpretation

Answer: fully compensated

Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.

ROME, value by value

pH 7.44: within 7.35–7.45
PaCO₂ 60 mmHg: high (35–45)
HCO₃⁻ 39 mmol/L: high (22–26)

Compensation

Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 46.3–50.3 mmHg, so the measured 60 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: acute

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.

Expected HCO₃⁻

acute: 24 + 1·4.4 = 28.4 mmol/L
chronic: 24 + 3.5·4.4 = 39.4 mmol/L

Compare

measured: 28 mmol/L → acute

Answer

acute
5.Clinical pharmacokinetics

Answer: LD = 528 mcg

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 = 7·66 = 462 L

Amount of active drug needed

Vd·C = 462·0.8 = 369.6 mcg

Loading dose

LD = 369.6/(1·0.7) = 528 mcg

Answer

LD = 528 mcg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: t = 9 h

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.

Rate constant

k = 0.693/9 = 0.07702 h⁻¹

Time to the given fraction of steady state

t = −ln(1 − 0.5)/0.07702 = 0.6931/0.07702 = 9 h

Answer

t = 9 h

That is 1 half-lives.

7.Clinical nursing

Answer: C · 0.5 × 109/L

Why this is the answer

Neutropenic sepsis is defined by a neutrophil count of 0.5 × 109/L or lower with fever or other signs of sepsis. It needs empirical antibiotics within one hour.

Answer

C · 0.5 × 109/L
8.Clinical nursing

Answer: A · "Please review him within the next 15 minutes."

Why this is the answer

In SBAR, R is the recommendation or request, such as a review within a time frame. The other statements belong to Assessment, Background and Situation.

Answer

A · "Please review him within the next 15 minutes."

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