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 starts lisinopril (an ACE inhibitor). Which laboratory values should the nurse expect to be monitored?

  1. ACalcium and phosphate
  2. BALT and bilirubin
  3. CTSH and free T4
  4. DPotassium and creatinine
2.Lab values & electrolytesHard

After a crush injury, a patient's urine dipstick is strongly positive for blood, but microscopy shows almost no red cells. CK is very high. What best explains this?

  1. AMyoglobin from damaged muscle is in the urine
  2. BA kidney stone is scraping the ureter lining
  3. CBleeding is coming from the bladder lining
  4. DThe sample was contaminated with menstrual blood
3.Arterial blood gas interpretationMedium

Practice values: Na⁺ 143 mmol/L, glucose 7.5 mmol/L and urea 24.7 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 143 mmol/Lglucose 7.5 mmol/Lurea 24.7 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
pH7.357.457.28PaCO₂mmHg354566HCO₃⁻mmol/L222630acutechronicacute on chronic

A practice gas shows a respiratory acidosis: pH 7.28, PaCO₂ 66 mmHg (8.8 kPa), HCO₃⁻ 30 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 7.28PaCO₂ 66 mmHg (8.8 kPa)HCO₃⁻ 30 mmol/L
  1. 1acute
  2. 2chronic
  3. 3acute on chronic
5.Clinical pharmacokineticsMedium
Vd = 0.5 L/kgC = 10 mg/LLD = ? mg56 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 56 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.

56 kgVd = 0.5 L/kgC = 10 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​24 hC = ?

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion runs at 125 mg/h with no loading dose. The patient's clearance is 5 L/h and the half-life is 10 h. What is the level C after 24 h?

R = 125 mg/hCL = 5 L/ht½ = 10 ht = 24 h
C = mg/L
7.Clinical nursingMedium

When should discharge planning usually begin?

  1. AAfter the patient has been home for a week
  2. BOn or soon after admission
  3. COnly after the patient asks to go home
  4. DOn the morning of discharge
8.Clinical nursingHard

A patient has a full-thickness burn all the way round the forearm. Which complication must be watched for closely?

  1. AExcessive bleeding from the burn surface needing an urgent blood transfusion
  2. BDelayed hair growth on the arm
  3. CLoss of circulation to the hand as swelling builds under tight burned skin
  4. DHypoglycaemia caused by the body using glucose to heal the burn

3Solutions

1.Lab values & electrolytes

Answer: D · Potassium and creatinine

Why this is the answer

ACE inhibitors reduce aldosterone, so potassium can rise, and they lower the filtration pressure in the kidney, so creatinine can rise. Both are checked before and soon after starting or increasing the dose.Checked against: Hyperkalemia · StatPearls (NCBI Bookshelf) (2025)

Answer

D · Potassium and creatinine
2.Lab values & electrolytes

Answer: A · Myoglobin from damaged muscle is in the urine

Why this is the answer

The dipstick blood test detects heme, which is present in both hemoglobin and myoglobin. Positive "blood" without red cells, together with a very high CK, indicates myoglobinuria from rhabdomyolysis, which can damage the kidneys. IV fluids and close monitoring of urine output and potassium are needed.Checked against: Myoglobin urine test · MedlinePlus (NIH) (2025); Rhabdomyolysis · MedlinePlus (NIH) (2025); Creatine Kinase · MedlinePlus (NIH) (2023)

Answer

A · Myoglobin from damaged muscle is in the urine
3.Arterial blood gas interpretation

Answer: Osm = 318 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 = 2·143 + 7.5 + 24.7 = 318.2 ≈ 318 mOsm/kg

Answer

Osm = 318 mOsm/kg
4.Arterial blood gas interpretation

Answer: acute on 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 + 1·2.6 = 26.6 mmol/L
chronic: 24 + 3.5·2.6 = 33.1 mmol/L

Compare

measured: 30 mmol/L → acute on chronic

Answer

acute on chronic
5.Clinical pharmacokinetics

Answer: LD = 350 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 = 0.5·56 = 28 L

Amount of active drug needed

Vd·C = 28·10 = 280 mg

Loading dose

LD = 280/(0.8·1) = 350 mg

Answer

LD = 350 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: C = 20.26 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.

Steady-state level

Css = 125/5 = 25 mg/L

Rate constant

k = 0.693/10 = 0.06931 h⁻¹

Level after the given time

C = 25·(1 − e−0.06931·24) = 20.26 mg/L

Answer

C = 20.26 mg/L
7.Clinical nursing

Answer: B · On or soon after admission

Why this is the answer

Discharge planning starts early so that needs such as equipment, care packages, medicines and follow-up are ready when the patient is fit to leave.

Answer

B · On or soon after admission
8.Clinical nursing

Answer: C · Loss of circulation to the hand as swelling builds under tight burned skin

Why this is the answer

Circumferential full-thickness burns form a tight eschar that acts like a tourniquet as oedema develops. Neurovascular checks are needed, and escharotomy may be required.

Answer

C · Loss of circulation to the hand as swelling builds under tight burned skin

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