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.
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
Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
Check the result: levels fall by half every half-life, and steady state takes about five half-lives.
2Practice problems
1.Lab values & electrolytesMedium
During a red cell transfusion, when are serious acute reactions most likely, so the nurse stays with the patient?
AIn the first 15 minutes
BAfter the last drop is given
CThe next morning
DAfter about 2 hours
2.Lab values & electrolytesHard
A patient with SIADH is on fluid restriction. Which task can the RN delegate to unlicensed assistive personnel (UAP), such as a nursing or healthcare assistant?
AAssessing the patient for new confusion
BDeciding when to recheck the sodium level
CTeaching the reasons for fluid restriction
DMeasuring and recording intake and output
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 22 mmHg (2.9 kPa). Use the rule that in acute respiratory alkalosis HCO₃⁻ falls by 2 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows pH 7.34, PaCO₂ 31 mmHg (4.1 kPa) and HCO₃⁻ 16 mmol/L, with Na⁺ 135 and Cl⁻ 85 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.
Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 10 mg/L at 1 h and 5.7 mg/L at 3 h. Assume first-order elimination. Find the half-life t½.
C1 = 10 mg/L (1 h)C2 = 5.7 mg/L (3 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 93 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.035 L/kg/h. Each dose is 300 mg of aminophylline. Find the average steady-state level Css,avg.
Which equipment must always be kept at the bedside of a patient with a tracheostomy?
AA peak flow meter and a nebuliser
BA blood glucose meter and a spare urinary catheter of the same size
CA blood pressure cuff and a spare nasal cannula and oxygen mask only
DSpare tracheostomy tubes, including one a size smaller, and working suction
8.Clinical nursingHard
A patient with stress incontinence is taught self-care. Which statement shows a need for further teaching?
A"I'll cut down on caffeine."
B"I'll try to lose some weight if I can."
C"I'll do my pelvic floor exercises every day."
D"I'll drink as little as possible so I don't leak."
3Solutions
1.Lab values & electrolytes
Answer: A · In the first 15 minutes
Why this is the answer
Acute hemolytic and anaphylactic reactions usually appear early, so the transfusion is started slowly and the nurse stays with the patient for about the first 15 minutes, checking vital signs before, during and after according to policy.
Answer
A · In the first 15 minutes
2.Lab values & electrolytes
Answer: D · Measuring and recording intake and output
Why this is the answer
UAP can measure and record intake and output and weigh the patient, then report the values to the RN. Assessing mental status, teaching and making decisions about laboratory monitoring need nursing judgment and remain with the RN.
Answer
D · Measuring and recording intake and output
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 20.4 mmol/L
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.
Change in PaCO₂
|22 − 40| = 18 mmHg → 1.8 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1.8 = 20.4 mmol/L
Answer
HCO₃⁻ = 20.4 mmol/L
4.Arterial blood gas interpretation
Answer: added metabolic alkalosis
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.
Anion gap
AG = 135 − (85 + 16) = 34
Delta ratio
(34 − 12)/(24 − 16) = 2.75
added metabolic alkalosis
Answer
added metabolic alkalosis
5.Clinical pharmacokinetics
Answer: t½ = 2.466 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.
Elimination rate constant
k = ln(10/5.7)/2 = 0.5621/2 = 0.2811 h⁻¹
Half-life
t½ = 0.693/0.2811 = 2.466 h
Answer
t½ = 2.466 h
6.Clinical pharmacokinetics
Answer: Css,avg = 9.217 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.
Answer: D · Spare tracheostomy tubes, including one a size smaller, and working suction
Why this is the answer
Emergency equipment such as spare tubes (same size and one smaller), suction, oxygen and tracheal dilators must be at the bedside. This allows rapid management of a blocked or displaced tube.
Answer
D · Spare tracheostomy tubes, including one a size smaller, and working suction
8.Clinical nursing
Answer: D · "I'll drink as little as possible so I don't leak."
Why this is the answer
Restricting fluids concentrates urine, irritates the bladder and risks dehydration and infection. Pelvic floor exercises, weight loss and reducing caffeine are recommended measures.
Answer
D · "I'll drink as little as possible so I don't leak."