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 Braden scale: sensory perception: slightly limited; moisture: very moist; activity: chairfast; mobility: very limited; nutrition: adequate; friction and shear: no apparent problem. Find the Braden total B.
6–23
B =
2.Clinical assessment scoresHard
Practice exercise (made-up values, not clinical guidance). An adult weighs 91 kg and is 156 cm tall. Calculate the BMI and click the WHO weight category.
91 kg156 cm
1underweight
2healthy weight
3overweight
4obesity class I
5obesity class II
6obesity class III
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.54, PaCO₂ 29 mmHg (3.9 kPa) and HCO₃⁻ 24 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.
pH 7.54PaCO₂ 29 mmHg (3.9 kPa)HCO₃⁻ 24 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 31 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 31 mmol/L
PaCO₂ = mmHg
5.Nursing fundamentalsMedium
How is a 24-hour urine collection started?
AOnly urine passed at night is collected
BA sample is taken every four hours and the rest discarded
CThe first urine of the morning is discarded and the time noted, then all urine is kept
DThe first urine of the morning is kept and the rest of the day's urine is discarded
6.Nursing fundamentalsHard
A pulse oximeter reads 82% on a patient with cold hands who is talking in full sentences with a normal respiratory rate. What is the best first action?
ACheck the patient and probe, then warm the hand or move the probe
BRecord the reading and recheck it at the next routine set of observations
CIgnore the reading because the patient looks well
DRemove any oxygen because the reading must be wrong
7.Nursing math & unit conversionsMedium
Practice exercise (made-up values, not clinical guidance). A patient is 4 ft 9 in tall. Using 1 ft = 12 in and 1 in = 2.54 cm, find the height h in centimetres, to the nearest whole centimetre.
4 ft 9 in1 ft = 12 in1 in = 2.54 cm
h = cm
8.Nursing math & unit conversionsHard
Practice exercise (made-up values, not clinical guidance). An infusion delivers 6 mcg/min. Express this dose rate D in mg/h.
6 mcg/min1 h = 60 min1 mg = 1000 mcg
D = mg/h
3Solutions
1.Clinical assessment scores
Answer: B = 15
Braden scale: six subscales for pressure-injury risk. Sensory perception, moisture, activity, mobility and nutrition score 1 to 4, from the worst level (1) to no problem (4); friction and shear scores 1 to 3. The total runs from 6 to 23, and a lower total means a higher risk. Common bands: 19–23 not at risk, 15–18 mild risk, 13–14 moderate risk, 10–12 high risk and 9 or below very high risk.
Score each subscale
sensory perception: slightly limited → 3
moisture: very moist → 2
activity: chairfast → 2
mobility: very limited → 2
nutrition: adequate → 3
friction and shear: no apparent problem → 3
Add
Braden total = 3 + 2 + 2 + 2 + 3 + 3 = 15
Answer
B = 15
2.Clinical assessment scores
Answer: obesity class II
Body mass index: BMI = weight (kg)/height (m)², with the height converted to metres first. With pounds and inches, BMI = 703 × weight (lb)/height (in)². WHO adult categories: below 18.5 underweight, 18.5–24.9 healthy weight, 25.0–29.9 overweight, 30.0–34.9 obesity class I, 35.0–39.9 obesity class II and 40 or more obesity class III.
BMI
BMI = 91/1.56² = 37.39
Category
obesity class II
Answer
obesity class II
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 7.54: alkalaemia (above 7.45)
PaCO₂ 29 mmHg: low (35–45)
HCO₃⁻ 24 mmol/L: normal (22–26)
Compensation
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: PaCO₂ = 42.7 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.
Expected PaCO₂
0.7·31 + 21 = 42.7 mmHg
Answer
PaCO₂ = 42.7 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Nursing fundamentals
Answer: C · The first urine of the morning is discarded and the time noted, then all urine is kept
Why this is the answer
The patient empties the bladder at the start and that urine is discarded. All urine for the next 24 hours is kept, including a final void at the end time.
Answer
C · The first urine of the morning is discarded and the time noted, then all urine is kept
6.Nursing fundamentals
Answer: A · Check the patient and probe, then warm the hand or move the probe
Why this is the answer
Cold peripheries, movement and nail varnish can give falsely low readings. The nurse assesses the patient first, then checks the waveform and probe site; a low reading is never ignored until it is explained.
Answer
A · Check the patient and probe, then warm the hand or move the probe
7.Nursing math & unit conversions
Answer: h = 145 cm
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.
Height in inches
4·12 + 9 = 57 in
Inches to centimetres
h = 57·2.54 = 144.8 ≈ 145 cm
Answer
h = 145 cm
8.Nursing math & unit conversions
Answer: D = 0.36 mg/h
Dimensional analysis: write the given quantity, then multiply by conversion factors written as fractions (for example 60 min/1 h or 1 mg/1000 mcg) so that each unwanted unit appears once on top and once underneath and cancels. When only the wanted unit is left, calculate. Use the equivalents the question gives and round only the final answer.