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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 calcium of 11.8 mg/dL (2.95 mmol/L) from bone metastases. Which nursing measure is appropriate?

  1. AEncourage fluids and mobility as tolerated
  2. BOffer extra milk and cheese at each meal
  3. CGive calcium carbonate antacids for nausea
  4. DEncourage bed rest and limit oral fluids
2.Lab values & electrolytesHard

Ten days after chemotherapy, a patient has a platelet count of 9 × 10³/µL and reports a sudden severe headache and blurred vision. What is the priority?

  1. ADim the lights and encourage rest, as headaches are common after chemotherapy
  2. BAssess neurological status and report it at once as possible brain bleeding
  3. CEncourage the patient to drink more fluids and check the blood pressure later
  4. DGive acetaminophen (paracetamol) for the headache and reassess in 1 hour
3.Arterial blood gas interpretationMedium
pH7.357.457.30PaCO₂mmHg354551HCO₃⁻mmol/L222624uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.30, PaCO₂ 51 mmHg (6.8 kPa) and HCO₃⁻ 24 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.30PaCO₂ 51 mmHg (6.8 kPa)HCO₃⁻ 24 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

A practice gas shows a metabolic alkalosis with HCO₃⁻ 36 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.

HCO₃⁻ 36 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Vd = 0.5 L/kgC = 10 mg/LLD = ? mg58 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 58 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 10 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.

58 kgVd = 0.5 L/kgC = 10 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
VCss,avg​ = ?400 mg · q12hCL = 0.04 L/kg/h65 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 65 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.04 L/kg/h. Each dose is 400 mg of theophylline. Find the average steady-state level Css,avg.

CL = 0.04 L/kg/hτ = 12 hS = 1F = 165 kgD = 400 mg
Css,avg = mg/L
7.Clinical nursingMedium

Which advice is important for a patient on long-term oral corticosteroids?

  1. ATake them only when symptoms flare
  2. BDouble the dose if a dose is missed
  3. CDo not stop them suddenly
  4. DTake them on an empty stomach at night
8.Clinical nursingHard

Which set of findings best distinguishes cardiogenic shock from hypovolaemic shock?

  1. ACool, pale skin
  2. BReduced urine output
  3. CTachycardia, hypotension and a weak, thready peripheral pulse
  4. DRaised jugular venous pressure and lung crackles

3Solutions

1.Lab values & electrolytes

Answer: A · Encourage fluids and mobility as tolerated

Why this is the answer

Hydration helps the kidneys excrete calcium and prevents stones, and weight-bearing mobility slows calcium loss from bone. Calcium-containing foods and antacids are not encouraged. The nurse also monitors for confusion, constipation and arrhythmias.Checked against: Hypercalcemia · MedlinePlus (NIH) (2026)

Answer

A · Encourage fluids and mobility as tolerated
2.Lab values & electrolytes

Answer: B · Assess neurological status and report it at once as possible brain bleeding

Why this is the answer

Platelet counts usually reach their lowest point (nadir) about 7–14 days after chemotherapy. With a count below 10 × 10³/µL, a sudden severe headache or visual change may signal intracranial bleeding, which is an emergency. The nurse assesses neurological status and reports it immediately.Checked against: Bleeding and Bruising (Thrombocytopenia) and Cancer Treatment · National Cancer Institute (NIH) (2022); Pancytopenia · StatPearls (NCBI Bookshelf) (2023)

Answer

B · Assess neurological status and report it at once as possible brain bleeding
3.Arterial blood gas interpretation

Answer: uncompensated

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 7.30: acidaemia (below 7.35)
PaCO₂ 51 mmHg: high (35–45)
HCO₃⁻ 24 mmol/L: normal (22–26)

Compensation

uncompensated

Answer

uncompensated
4.Arterial blood gas interpretation

Answer: PaCO₂ = 46.2 mmHg

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 PaCO₂

0.7·36 + 21 = 46.2 mmHg

Answer

PaCO₂ = 46.2 mmHg

The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.

5.Clinical pharmacokinetics

Answer: LD = 362.5 mg

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 = 0.5·58 = 29 L

Amount of active drug needed

Vd·C = 29·10 = 290 mg

Loading dose

LD = 290/(0.8·1) = 362.5 mg

Answer

LD = 362.5 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: Css,avg = 12.82 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.

Clearance

CL = 0.04·65 = 2.6 L/h

Average steady-state level

Css,avg = S·F·D/(CL·τ) = 1·1·400/(2.6·12) = 12.82 mg/L

Answer

Css,avg = 12.82 mg/L
7.Clinical nursing

Answer: C · Do not stop them suddenly

Why this is the answer

Long-term steroids suppress the adrenal glands, so stopping suddenly can cause adrenal crisis. Patients should carry a steroid alert card or medical ID and follow sick-day guidance.

Answer

C · Do not stop them suddenly
8.Clinical nursing

Answer: D · Raised jugular venous pressure and lung crackles

Why this is the answer

Both types cause tachycardia, hypotension, cool skin and low urine output. In cardiogenic shock the failing heart causes venous congestion and pulmonary oedema, so large fluid boluses may worsen it.

Answer

D · Raised jugular venous pressure and lung crackles

ExamProblems · examproblems.com/courses/lab-values-for-nurses