Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
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.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
Nursing math & unit conversions
Metric steps
1 kg = 1000 g, 1 g = 1000 mg, 1 mg = 1000 mcg, 1 L = 1000 mL
Weight and length
1 kg = 2.2 lb, 1 lb = 16 oz, 1 in = 2.54 cm, 1 ft = 12 in
Household measures
1 tsp = 5 mL, 1 tbsp = 15 mL, 1 fl oz = 30 mL, 1 cup = 8 fl oz = 240 mL
Write the given quantity with its unit and the unit you want.
Pick the equivalent: 1000 between kg, g, mg and mcg or L and mL; 1 kg = 2.2 lb; 1 tsp = 5 mL, 1 tbsp = 15 mL, 1 fl oz = 30 mL, 1 cup = 240 mL.
Write it as a fraction so that the unwanted unit cancels; for a rate, use one fraction per unit that changes.
Check the direction: a smaller unit gives a bigger number.
Round only the final answer, to the precision the question names.
2Practice problems
1.Clinical assessment scoresMedium
Practice exercise (made-up values, not clinical guidance). A practice patient is assessed with the Morse fall scale: history of falling in the last 3 months: no; secondary diagnosis: no; ambulatory aid: holds on to furniture; IV therapy or saline lock: no; gait: weak; mental status: knows own limits. Find the Morse total MFS.
0–125
MFS =
2.Clinical assessment scoresHard
Practice exercise (made-up values, not clinical guidance). The team aims for a mean arterial pressure of at least 60 mmHg. A patient’s systolic pressure is 100 mmHg. Using MAP = (SBP + 2·DBP)/3, what is the lowest diastolic pressure DBP that still gives a MAP of 60 mmHg?
MAP 60 mmHgSBP 100 mmHg
DBP = mmHg
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.20, PaCO₂ 40 mmHg (5.3 kPa) and HCO₃⁻ 15 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.20PaCO₂ 40 mmHg (5.3 kPa)HCO₃⁻ 15 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 21% oxygen shows PaO₂ 59 mmHg (7.9 kPa) and PaCO₂ 55 mmHg (7.3 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
ATo prevent lumpy fatty swelling that affects absorption
BBecause each site can only be used once
CTo make the insulin work more quickly with each injection
DTo stop the patient getting used to the needle
6.Nursing fundamentalsHard
How should a urine sample be taken from a patient with a urinary catheter?
AFrom the sampling port, using aseptic technique
BBy removing the catheter and collecting the next void
CBy disconnecting the catheter from the bag
DFrom the drainage bag tap
7.Nursing math & unit conversionsMedium
Practice exercise (made-up values, not clinical guidance). A patient’s temperature is 36.9 °C. Convert it to degrees Fahrenheit T with °F = °C × 9/5 + 32. Round to one decimal place.
36.9 °C
T = °F
8.Nursing math & unit conversionsHard
Practice exercise (made-up values, not clinical guidance). During a shift a patient takes 2 cups of tea, ½ tbsp of liquid supplement, 2 fl oz of juice, 2 tsp of cough syrup. Using the equivalents given, find the total oral intake V in mL.
1 cup = 8 fl oz = 240 mL1 tbsp = 15 mL1 fl oz = 30 mL1 tsp = 5 mL
V = mL
3Solutions
1.Clinical assessment scores
Answer: MFS = 40
Morse fall scale: six items. History of falling: no 0, yes 25. Secondary diagnosis: no 0, yes 15. Ambulatory aid: none, bed rest or nurse assistance 0; crutches, cane or walker 15; holding on to furniture 30. IV therapy or saline lock: no 0, yes 20. Gait: normal, bed rest or wheelchair 0; weak 10; impaired 20. Mental status: knows own limits 0; overestimates or forgets limits 15. The total runs from 0 to 125. Risk cut-offs differ between hospitals; a common set is 0–24 low, 25–44 moderate and 45 or more high.
Score each item
history of falling in the last 3 months: no → 0
secondary diagnosis: no → 0
ambulatory aid: holds on to furniture → 30
IV therapy or saline lock: no → 0
gait: weak → 10
mental status: knows own limits → 0
Add
Morse total = 0 + 0 + 30 + 0 + 10 + 0 = 40
Answer
MFS = 40
2.Clinical assessment scores
Answer: DBP = 40 mmHg
Blood pressure values: pulse pressure PP = SBP − DBP. Mean arterial pressure MAP ≈ (SBP + 2·DBP)/3 = DBP + PP/3, an estimate based on the heart spending about two thirds of each beat in diastole at a normal heart rate (less accurate when the heart rate is fast); a MAP of at least about 65 mmHg is a common target for organ perfusion. The shock index SI = heart rate/SBP is normally about 0.5–0.7; a value above about 1 is a warning sign of shock.
Rearrange
DBP = (3·MAP − SBP)/2
Substitute
DBP = (3·60 − 100)/2 = 80/2 = 40 mmHg
Answer
DBP = 40 mmHg
3.Arterial blood gas interpretation
Answer: uncompensated
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 7.20: acidaemia (below 7.35)
PaCO₂ 40 mmHg: normal (35–45)
HCO₃⁻ 15 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 28.5–32.5 mmHg, so the measured 40 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: A–a gradient = 21.98 mmHg
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.
Answer: A · To prevent lumpy fatty swelling that affects absorption
Why this is the answer
Injecting in the same place causes lipohypertrophy, which makes absorption unpredictable. Patients rotate within an area and check sites for lumps.
Answer
A · To prevent lumpy fatty swelling that affects absorption
6.Nursing fundamentals
Answer: A · From the sampling port, using aseptic technique
Why this is the answer
Samples are taken from the needle-free sampling port after cleaning it. Urine in the bag is stale and contaminated, and breaking the system raises infection risk.
Answer
A · From the sampling port, using aseptic technique
7.Nursing math & unit conversions
Answer: T = 98.4 °F
Temperature: °C = (°F − 32) × 5/9 and °F = °C × 9/5 + 32. Subtract 32 before you multiply when going from °F, and add 32 after you multiply when going from °C. A quick check: 37 °C is 98.6 °F, and every 1 °C is 1.8 °F.
Multiply by 9/5
36.9·9/5 = 66.42
Add 32
T = 66.42 + 32 = 98.42 ≈ 98.4 °F
Answer
T = 98.4 °F
8.Nursing math & unit conversions
Answer: V = 557.5 mL
Household measures and weight: nursing exams use rounded equivalents: 1 teaspoon (tsp) = 5 mL, 1 tablespoon (tbsp) = 15 mL = 3 tsp, 1 fluid ounce (fl oz) = 30 mL = 2 tbsp, and 1 cup = 8 fl oz = 240 mL. For weight, 1 kg = 2.2 lb and 1 lb = 16 oz; for length, 1 inch = 2.54 cm. Set the conversion up so that the unit you do not want cancels out.