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 with pneumonia had a CRP of 180 mg/L on admission and 40 mg/L after 4 days of antibiotics. What does this most likely show?

  1. AThe patient has developed a new infection
  2. BThe inflammation is settling with treatment
  3. CThe infection has spread to another organ
  4. DThe antibiotic is causing liver damage
2.Lab values & electrolytesHard

Which patient’s result should the nurse report first?

  1. AHbA1c 7.4% (57 mmol/mol) in a patient with type 2 diabetes
  2. BCreatinine 1.6 mg/dL (141 µmol/L) in a patient with known stable CKD
  3. CTotal cholesterol slightly above target
  4. DGlucose 42 mg/dL (2.3 mmol/L) in a patient who is sweating and confused
3.Arterial blood gas interpretationMedium

Practice values: Na⁺ 128 mmol/L, glucose 29.2 mmol/L and urea 3.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 128 mmol/Lglucose 29.2 mmol/Lurea 3.6 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 141, Cl⁻ 99, HCO₃⁻ 22 mmol/L and albumin 34 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 141 mmol/LCl⁻ 99 mmol/LHCO₃⁻ 22 mmol/Lalbumin 34 g/L
AGc = mmol/L
5.Clinical pharmacokineticsMedium
VCss​ = 10 mg/LR = ? mg/h · S = 0.8CL = 0.045 L/kg/h97 kg

Practice exercise (made-up values, not clinical guidance). A 97 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.045 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 10 mg/L?

97 kgCL = 0.045 L/kg/hCss = 10 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
VCss,avg​ = 10 mg/LD = ? mg · q8hCL = 0.04 L/kg/h86 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). An 86 kg patient takes oral theophylline (F = 1) every 8 h. The clearance is 0.04 L/kg/h. What dose D in mg gives an average steady-state level of 10 mg/L?

CL = 0.04 L/kg/hτ = 8 hS = 1F = 186 kgCss,avg = 10 mg/L
D = mg
7.Clinical nursingMedium

Which blood result often rises in an upper GI bleed because blood is digested in the gut?

  1. AUrea (blood urea nitrogen)
  2. BCalcium
  3. CGlucose
  4. DSodium
8.Clinical nursingHard

A postoperative patient has a heart rate of 115/min, temperature 38.6 °C, respiratory rate 26/min and blood pressure 85/50 mmHg despite a fluid bolus. What does this most likely indicate?

  1. ASeptic shock requiring urgent critical care review
  2. BMild dehydration only
  3. CA normal inflammatory response in the first days after surgery
  4. DAnxiety about discharge

3Solutions

1.Lab values & electrolytes

Answer: B · The inflammation is settling with treatment

Why this is the answer

CRP has a short half-life of about 19 hours, so it falls within days when inflammation is controlled. A falling CRP, together with clinical improvement, supports a response to treatment; a CRP that stays high or rises again needs review.Checked against: C-reactive protein · MedlinePlus (NIH) (2025)

Answer

B · The inflammation is settling with treatment
2.Lab values & electrolytes

Answer: D · Glucose 42 mg/dL (2.3 mmol/L) in a patient who is sweating and confused

Why this is the answer

Symptomatic hypoglycemia can quickly lead to seizures and brain injury, so it is treated straight away according to protocol. The other results are chronic or stable and can be discussed later.Checked against: Low blood sugar · MedlinePlus (NIH) (2026)

Answer

D · Glucose 42 mg/dL (2.3 mmol/L) in a patient who is sweating and confused
3.Arterial blood gas interpretation

Answer: Osm = 289 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 = 2·128 + 29.2 + 3.6 = 288.8 ≈ 289 mOsm/kg

Answer

Osm = 289 mOsm/kg
4.Arterial blood gas interpretation

Answer: AGc = 21.5 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 = 141 − (99 + 22) = 20

Correction for albumin

20 + 0.25·(40 − 34) = 21.5 mmol/L

Answer

AGc = 21.5 mmol/L

Above about 12 after correction: a high anion gap.

5.Clinical pharmacokinetics

Answer: R = 54.56 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 = 0.045·97 = 4.365 L/h

Infusion rate

R = CL·Css/S = 4.365·10/0.8 = 54.56 mg/h

Answer

R = 54.56 mg/h
6.Clinical pharmacokinetics

Answer: D = 275.2 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.

Clearance

CL = 0.04·86 = 3.44 L/h

Dose

D = Css,avg·CL·τ/(S·F) = 10·3.44·8/(1·1) = 275.2 mg

Answer

D = 275.2 mg
7.Clinical nursing

Answer: A · Urea (blood urea nitrogen)

Why this is the answer

Blood proteins digested in the upper GI tract are absorbed and converted to urea, so urea rises out of proportion to creatinine. Urea is part of the Glasgow-Blatchford bleeding score.

Answer

A · Urea (blood urea nitrogen)
8.Clinical nursing

Answer: A · Septic shock requiring urgent critical care review

Why this is the answer

Infection signs with hypotension not responding to fluid suggest septic shock. Vasopressors and critical care input are likely needed, along with antibiotics and source control.

Answer

A · Septic shock requiring urgent critical care review

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