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 weight loss, tremor and heat intolerance. Which thyroid test pattern fits Graves disease?

  1. ANormal TSH with a low free T4
  2. BHigh TSH with a low free T4
  3. CLow TSH with a high free T4
  4. DHigh TSH with a normal free T4
2.Lab values & electrolytesHard

Four laboratory results arrive. Which one should the nurse act on first?

  1. APotassium 6.6 mEq/L in a patient with CKD who has new muscle weakness
  2. BGlucose 140 mg/dL (7.8 mmol/L) after a meal in a patient with diabetes
  3. CSodium 137 mEq/L in a patient admitted with pneumonia yesterday
  4. DHemoglobin 11.8 g/dL (118 g/L) in a patient with chronic kidney disease
3.Arterial blood gas interpretationMedium

Practice values: Na⁺ 136 mmol/L, glucose 31.9 mmol/L and urea 10.1 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 136 mmol/Lglucose 31.9 mmol/Lurea 10.1 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 138, Cl⁻ 112, HCO₃⁻ 12 mmol/L and albumin 20 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 138 mmol/LCl⁻ 112 mmol/LHCO₃⁻ 12 mmol/Lalbumin 20 g/L
AGc = mmol/L
5.Clinical pharmacokineticsMedium
0.1110t (h)C (mg/L)1 h8 mg/L11 h1.1 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 8 mg/L at 1 h and 1.1 mg/L at 11 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 8 mg/L (1 h)C2 = 1.1 mg/L (11 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​t = ?80%

Practice exercise (made-up values, not clinical guidance). A theophylline infusion is started without a loading dose. The half-life in this patient is 8 h. How long t does it take to reach 80% of the steady-state level?

t½ = 8 h80% Css
t = h
7.Clinical nursingMedium

Why is an ankle–brachial pressure index (ABPI) measured before applying compression bandages to a leg ulcer?

  1. ATo measure how deep the ulcer is and how much exudate it produces
  2. BTo check for a urinary tract infection that could delay ulcer healing
  3. CTo work out the patient's fluid needs
  4. DTo exclude significant arterial disease, as compression could cut off blood flow
8.Clinical nursingHard

A nurse teaches a patient with asthma about a new inhaled steroid "preventer". Which statement shows a need for further teaching?

  1. A"I will rinse my mouth after using it."
  2. B"I will only use it when I'm wheezy."
  3. C"I will use it every day even when I feel well."
  4. D"I'll keep my reliever inhaler with me."

3Solutions

1.Lab values & electrolytes

Answer: C · Low TSH with a high free T4

Why this is the answer

In Graves disease, antibodies stimulate the thyroid to make too much hormone. The high free T4 suppresses TSH from the pituitary through negative feedback, so TSH is low. A high TSH with a low T4 fits primary hypothyroidism.Checked against: TSH test · MedlinePlus (NIH) (2026); Thyroxine (T4) Test · MedlinePlus (NIH) (2024); Graves disease · MedlinePlus (NIH) (2026)

Answer

C · Low TSH with a high free T4
2.Lab values & electrolytes

Answer: A · Potassium 6.6 mEq/L in a patient with CKD who has new muscle weakness

Why this is the answer

A potassium of 6.6 mEq/L with weakness can cause life-threatening arrhythmias. The nurse arranges cardiac monitoring and informs the provider at once. Mild anemia in CKD and a mildly raised glucose after a meal are expected; the sodium is normal.Checked against: High potassium level · MedlinePlus (NIH) (2025)

Answer

A · Potassium 6.6 mEq/L in a patient with CKD who has new muscle weakness
3.Arterial blood gas interpretation

Answer: Osm = 314 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 = 2·136 + 31.9 + 10.1 = 314 ≈ 314 mOsm/kg

Answer

Osm = 314 mOsm/kg
4.Arterial blood gas interpretation

Answer: AGc = 19 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 = 138 − (112 + 12) = 14

Correction for albumin

14 + 0.25·(40 − 20) = 19 mmol/L

Answer

AGc = 19 mmol/L

Above about 12 after correction: a high anion gap.

5.Clinical pharmacokinetics

Answer: k = 0.1984 h⁻¹

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(8/1.1)/10 = 1.984/10 = 0.1984 h⁻¹

Answer

k = 0.1984 h⁻¹
6.Clinical pharmacokinetics

Answer: t = 18.58 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 = 0.693/8 = 0.08664 h⁻¹

Time to the given fraction of steady state

t = −ln(1 − 0.8)/0.08664 = 1.609/0.08664 = 18.58 h

Answer

t = 18.58 h

That is 2.322 half-lives.

7.Clinical nursing

Answer: D · To exclude significant arterial disease, as compression could cut off blood flow

Why this is the answer

ABPI compares ankle and arm blood pressures. A low result suggests arterial insufficiency, where compression could cause ischaemia; venous ulcers with adequate arterial supply are treated with compression.

Answer

D · To exclude significant arterial disease, as compression could cut off blood flow
8.Clinical nursing

Answer: B · "I will only use it when I'm wheezy."

Why this is the answer

Inhaled corticosteroids must be used regularly to reduce airway inflammation and prevent attacks. The short-acting reliever is used for symptoms.

Answer

B · "I will only use it when I'm wheezy."

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