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

How does the nurse start a 24-hour urine collection at 08:00?

  1. ACollect only the first urine passed on waking, and repeat this the next morning
  2. BSave the 08:00 urine as the first sample, then collect until 08:00 the next day
  3. CHave the patient void and discard that urine, then save all urine until 08:00 the next day
  4. DSave three samples spread across the day and discard all of the urine passed at night
2.Lab values & electrolytesHard

A 5-year-old has new bruising and petechiae two weeks after a viral illness. Platelets are 12 × 10³/µL; hemoglobin, white cell count and clotting times are normal. Which condition is most likely?

  1. AImmune thrombocytopenia (ITP)
  2. BDisseminated intravascular coagulation
  3. CHemophilia A
  4. DAcute leukemia
3.Arterial blood gas interpretationMedium

A practice gas shows a metabolic acidosis with HCO₃⁻ 14 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.

HCO₃⁻ 14 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
pH7.357.457.21PaCO₂mmHg354518HCO₃⁻mmol/L22267appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.21, PaCO₂ 18 mmHg (2.4 kPa), HCO₃⁻ 7 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.21PaCO₂ 18 mmHg (2.4 kPa)HCO₃⁻ 7 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
0.1110t (h)C (mg/L)1 h7 mg/L7 h0.9 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 7 mg/L at 1 h and 0.9 mg/L at 7 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 7 mg/L (1 h)C2 = 0.9 mg/L (7 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
VCss,avg​ = ?400 mg · q12hCL = 0.035 L/kg/h76 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 76 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.035 L/kg/h. Each dose is 400 mg of theophylline. Find the average steady-state level Css,avg.

CL = 0.035 L/kg/hτ = 12 hS = 1F = 176 kgD = 400 mg
Css,avg = mg/L
7.Clinical nursingMedium

Why does a patient with a suspected TIA need urgent specialist assessment?

  1. AThe symptoms will come back permanently within an hour
  2. BThe risk of a full stroke is highest in the days after a TIA
  3. CA TIA always turns into a brain tumour unless it is treated
  4. DThe patient needs urgent surgery on the brain
8.Clinical nursingHard

An acutely ill patient without COPD is on 15 L/min oxygen, and their SpO2 is now 100% and stable. What should be done?

  1. AStop the oxygen and the saturation monitoring because the patient is now well
  2. BReduce the oxygen to keep the SpO2 within the prescribed target range
  3. CIncrease the oxygen further to build up a reserve
  4. DKeep the oxygen at 15 L/min for at least a week to be on the safe side

3Solutions

1.Lab values & electrolytes

Answer: C · Have the patient void and discard that urine, then save all urine until 08:00 the next day

Why this is the answer

The collection starts with an empty bladder, so the first void is discarded and the time recorded. All urine for the next 24 hours is saved, including a final void at the end time. If urine is discarded by mistake, the collection usually has to restart.Checked against: Urine 24-hour volume · MedlinePlus (NIH) (2025)

Answer

C · Have the patient void and discard that urine, then save all urine until 08:00 the next day
2.Lab values & electrolytes

Answer: A · Immune thrombocytopenia (ITP)

Why this is the answer

ITP is immune destruction of platelets, often after a viral infection in children. It causes isolated thrombocytopenia with normal other counts and clotting tests. Leukemia usually affects other cell lines, DIC prolongs clotting times and hemophilia prolongs the aPTT.Checked against: Immune thrombocytopenic purpura (ITP) · MedlinePlus (NIH) (2026)

Answer

A · Immune thrombocytopenia (ITP)
3.Arterial blood gas interpretation

Answer: PaCO₂ = 29 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.

Winter’s formula

PaCO₂ = 1.5·14 + 8 = 29 mmHg
range: 27–31 mmHg

Answer

PaCO₂ = 29 mmHg
4.Arterial blood gas interpretation

Answer: appropriate compensation

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

1.5·7 + 8 = 18.5 → 16.5–20.5 mmHg

Compare

measured: 18 mmHg → appropriate compensation

Answer

appropriate compensation
5.Clinical pharmacokinetics

Answer: k = 0.3419 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(7/0.9)/6 = 2.051/6 = 0.3419 h⁻¹

Answer

k = 0.3419 h⁻¹
6.Clinical pharmacokinetics

Answer: Css,avg = 12.53 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 = 0.035·76 = 2.66 L/h

Average steady-state level

Css,avg = S·F·D/(CL·τ) = 1·1·400/(2.66·12) = 12.53 mg/L

Answer

Css,avg = 12.53 mg/L
7.Clinical nursing

Answer: B · The risk of a full stroke is highest in the days after a TIA

Why this is the answer

The risk of stroke is highest soon after a TIA. Rapid assessment and treatment, such as antiplatelets, statins and carotid imaging, reduce this risk.

Answer

B · The risk of a full stroke is highest in the days after a TIA
8.Clinical nursing

Answer: B · Reduce the oxygen to keep the SpO2 within the prescribed target range

Why this is the answer

Oxygen is prescribed to a target range, usually 94–98% for most patients. Once the patient is stable it should be weaned down, as unnecessary hyperoxia has no benefit and can be harmful.

Answer

B · Reduce the oxygen to keep the SpO2 within the prescribed target range

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