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

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

A patient with kidney stones and fatigue has a calcium of 11.6 mg/dL (2.9 mmol/L), a high parathyroid hormone (PTH) and a low phosphate. Which condition is most likely?

  1. AVitamin D deficiency
  2. BCancer-related hypercalcemia
  3. CPrimary hyperparathyroidism
  4. DHypoparathyroidism
3.Arterial blood gas interpretationMedium
pH7.357.457.37PaCO₂mmHg354572HCO₃⁻mmol/L222640uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.37, PaCO₂ 72 mmHg (9.6 kPa) and HCO₃⁻ 40 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.37PaCO₂ 72 mmHg (9.6 kPa)HCO₃⁻ 40 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard
pH7.357.457.32PaCO₂mmHg354571HCO₃⁻mmol/L222635acutechronicacute on chronic

A practice gas shows a respiratory acidosis: pH 7.32, PaCO₂ 71 mmHg (9.5 kPa), HCO₃⁻ 35 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 7.32PaCO₂ 71 mmHg (9.5 kPa)HCO₃⁻ 35 mmol/L
  1. 1acute
  2. 2chronic
  3. 3acute on chronic
5.Clinical pharmacokineticsMedium
VCss​ = 12 mg/LR = ? mg/hCL = 0.045 L/kg/h58 kg

Practice exercise (made-up values, not clinical guidance). A 58 kg patient receives a theophylline infusion. The clearance is 0.045 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 12 mg/L?

58 kgCL = 0.045 L/kg/hCss = 12 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Vd = 7 L/kg0.5 → 0.8 mcg/LLD = ? mcg58 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 58 kg patient has a digoxin level of 0.5 mcg/L; the target is 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?

58 kgVd = 7 L/kgC = 0.5 → 0.8 mcg/LF = 0.7
LD = mcg
7.Clinical nursingMedium

Which electrolyte disturbance is most dangerous for the heart rhythm?

  1. ALow phosphate
  2. BLow sodium
  3. CHigh chloride
  4. DHigh or low potassium
8.Clinical nursingHard

A patient with a confirmed pulmonary embolism is started on anticoagulation. Which statement best explains its purpose?

  1. AIt thins the blood so that the lungs absorb more oxygen
  2. BIt is mainly given to lower the blood pressure
  3. CIt dissolves the clot completely within minutes of the first dose being given
  4. DIt stops the clot growing and prevents new clots while the body breaks it down

3Solutions

1.Lab values & electrolytes

Answer: D · 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

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

Answer: C · Primary hyperparathyroidism

Why this is the answer

In primary hyperparathyroidism, usually from an adenoma, PTH is high despite high calcium, and phosphate is low because PTH increases its excretion. Cancer-related hypercalcemia typically has a suppressed PTH.Checked against: Parathyroid Hormone (PTH) Test · MedlinePlus (NIH) (2023); Hyperparathyroidism · MedlinePlus (NIH) (2026)

Answer

C · Primary hyperparathyroidism
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 7.37: within 7.35–7.45
PaCO₂ 72 mmHg: high (35–45)
HCO₃⁻ 40 mmol/L: high (22–26)

Compensation

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: chronic

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 + 1·3.1 = 27.1 mmol/L
chronic: 24 + 3.5·3.1 = 34.85 mmol/L

Compare

measured: 35 mmol/L → chronic

Answer

chronic
5.Clinical pharmacokinetics

Answer: R = 31.32 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 = 0.045·58 = 2.61 L/h

Infusion rate

R = 2.61·12 = 31.32 mg/h

Answer

R = 31.32 mg/h
6.Clinical pharmacokinetics

Answer: LD = 174 mcg

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 = 7·58 = 406 L

Rise needed

0.8 − 0.5 = 0.3 mcg/L

Loading dose

LD = 406·0.3/(1·0.7) = 174 mcg

Answer

LD = 174 mcg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: D · High or low potassium

Why this is the answer

Both hyperkalaemia and hypokalaemia can cause life-threatening arrhythmias. Potassium must never be given as an undiluted rapid IV bolus.

Answer

D · High or low potassium
8.Clinical nursing

Answer: D · It stops the clot growing and prevents new clots while the body breaks it down

Why this is the answer

Anticoagulants prevent clot extension and new clots; the body's own fibrinolysis gradually breaks down the existing clot. Thrombolysis, which actively dissolves clots, is reserved for massive PE with shock.

Answer

D · It stops the clot growing and prevents new clots while the body breaks it down

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