Interpretation Of Arterial 2

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Interpretation Of Arterial 2

  1. 1. Interpretation of arterial blood gases Sarah Ramsay Dept of Anaesthesia & Intensive Care The Chinese University of Hong Kong Prince of Wales Hospital Modified by Charles Gomersall Version 1.1 June 2004
  2. 2. Disclaimer Although considerable care has been taken in the preparation of this tutorial, the author, the Prince of Wales Hospital and The Chinese University of Hong Kong take no responsibility for any adverse event resulting from its use.
  3. 3. ContentsBasic physiologyInterpretation of ABGsMixed disturbancesExamples
  4. 4. Acid-baseH20 + CO2 H2CO3 HCO3- + H+
  5. 5. H20 + CO2 H2CO3 HCO3- + H+ Normal [H+] = 40 nmol/l pH = - log [H+] = 7.4
  6. 6. H20 + CO2 H2CO3 HCO3- + H+ Normal PaCO2 = 5.3 kPa
  7. 7. ALVEOLAR VENTILATIONH20 + CO2 H2CO3 HCO3- + H+ Normal PaCO2 = 5.3 kPa
  8. 8. Normal HCO3- = 22-26 mmol/lH20 + CO2 H2CO3 HCO3- + H+
  9. 9. ALVEOLAR VENTILATION Normal HCO3- = 22-26 mmol/lH20 + CO2 H2CO3 HCO3- + H+ RENAL HCO3- HANDLING Click here to continue tutorial
  10. 10. Interpretation of arterialblood gases pH • Oxygenation PaCO2 PaO2 • Ventilation HCO3- • Acid base status Base excess Saturation
  11. 11. Interpretation of arterialblood gases pH • Oxygenation PaCO2 • Ventilation PaO 2 HCO3- • Acid base status Base excess Saturation
  12. 12. Interpretation of arterialblood gases pH • Oxygenation PaCO2 • Ventilation PaO2 HCO3- • Acid base status Base excess Saturation
  13. 13. Oxygenation• What is the PaO2? pH• Is this is adequate PaCO2 for the amount of PaO 2 inspired oxygen? HCO3-• Does the ABG result Base excess agree with the Saturation saturation probe?
  14. 14. Oxygenation• Normal PaO2 breathing air (FiO2 = 21%) is 12- 13.3 kPa ; small reduction with age• Lower values constitute hypoxaemia• PaO2 <6.7 kPa on room air = respiratory failure• PaO2 should go up with increasing FiO2• A PaO2 of 13.3 kPa breathing 60% O2 is not normal• You need to know the FiO2 to interpret the ABG
  15. 15. Oxygenation• Correlate the ABG result with the saturation probe result• If there is a discrepancy: – Is there a problem with the probe (poor perfusion? etc) – Is there a problem with the blood gas (is it a venous sample?)
  16. 16. Oxygenation• Is the PO2 is lower than expected?• Calculate the A-a gradient to assess if the low PO2 is due to: – Low alveolar PO2 – Structural lung problems causing failure of oxygen transfer
  17. 17. OxygenationThe alveolar gas equation: PAO2 = [94.8 x FIO2] – [PaCO2 x 1.25]The alveolar-arterial oxygen difference (A-a) PO2 = PAO2 - PaO2
  18. 18. OxygenationThe alveolar gas equation: PAO2 = [94.8 x FIO2] – [PaCO2 x 1.25]The alveolar-arterial oxygen difference (A-a) PO2 = PAO2 - PaO2
  19. 19. Oxygenation The alveolar gas equation: PAO2 = [94.8 x FIO2] – [PaCO2 x 1.25] The alveolar-arterial oxygen difference (A-a) PO2 = PAO2 - PaO2In the normal state there is only a small gradient between thealveolus and the arterial blood (1.33kPa). As CO2 accumulates in thealveolus due to HYPOVENTILATION there is less room for oxygen. Ifthe lung is otherwise normal this oxygen can pass into blood asnormal. There just is not enough passing. If there are problems thatlimit oxygen diffusion the gradient will get bigger.
  20. 20. OxygenationThe alveolar gas equation: PAO2 = [94.8 x FIO2] – [PaCO2 x 1.25]The alveolar-arterial oxygen difference (A-a) PO2 = PAO2 - PaO2 Continue tutorial Examples
  21. 21. Acid base problems Is there acidaemia or alkalaemia? Normal pH = 7.38 – 7.42Acidaemia < 7.38 Alkalaemia > 7.42
  22. 22. Acid base problems Is the primary problem respiratory or metabolic? Look at the PaCO2 Normal PaCO2 = 5.3 kPa
  23. 23. Acid base problems Is the primary problem respiratory or metabolic? Look at the [HCO3-] Normal [HCO3-] = 24 mmol/l
  24. 24. Is there Is the PaCO2 Is the HCO3- It isAcidaemia High Normal/high Respiratory acidosisAcidaemia Low Low Metabolic acidosisAlkalaemia Low Normal/low Respiratory alkalosisAlkalaemia High High Metabolic alkalosis Click to continue with tutorial
  25. 25. Is there Is the PaCO2 Is the HCO3- It isAcidaemia High Normal/high Respiratory ( > 6 kPa) ( 24 mmol/l) acidosisAcidaemia Low Low Metabolic acidosisAlkalaemia Low Normal/low Respiratory alkalosisAlkalaemia High High Metabolic alkalosisH20 + CO2 H2CO3 HCO3- + H+
  26. 26. Is there Is the PaCO2 Is the HCO3- It isAcidaemia High Normal/high Respiratory acidosisAcidaemia Low Low Metabolic ( < 4.5 kPa) ( 23 mmol/l) acidosisAlkalaemia Low Normal/low Respiratory alkalosisAlkalaemia High High Metabolic alkalosisH20 + CO2 H2CO3 HCO3- + H+
  27. 27. Is there Is the PaCO2 Is the HCO3- It isAcidaemia High Normal/high Respiratory acidosisAcidaemia Low Low Metabolic acidosisAlkalaemia Low Normal/low Respiratory ( < 4.5 kPa) ( 23 mmol/l) alkalosisAlkalaemia High High Metabolic alkalosisH20 + CO2 H2CO3 HCO3- + H+
  28. 28. Is there Is the PaCO2 Is the HCO3- It isAcidaemia High Normal/high Respiratory acidosisAcidaemia Low Low Metabolic acidosisAlkalaemia Low Normal/low Respiratory alkalosisAlkalaemia High High Metabolic ( > 6 kPa) ( 24 mmol/l) alkalosisH20 + CO2 H2CO3 HCO3- + H+
  29. 29. If there is a respiratoryproblem…• Is there an acidosis or an alkalosis?• Is it acute or chronic?• Is there renal compensation?• Does the pH change as much as expected?• What is the bicarbonate?
  30. 30. pH and HCO3- changes pH [HCO 3-]Acute respiratory Falls 0.06 Rises 0.8 mmol for every 1 kPa riseacidosis (up to 30 mmol/l) in PaCO 2Acute respiratory Rises 0.06 Falls 1.5 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2Chronic respiratory Falls 0.02 Rises 3.0 mmol for every 1 kPa riseacidosis (up to 36 mmol/l) in PaCO 2Chronic respiratory Rises 0.02 Falls 3.8 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2
  31. 31. For acute respiratory conditions pH [HCO 3-]Acute respiratory Falls 0.06 Rises 0.8 mmol for every 1 kPa riseacidosis (up to 30 mmol/l) in PaCO 2Acute respiratory Rises 0.06 Falls 1.5 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2Chronic respiratory Falls 0.02 Rises 3.0 mmol for every 1 kPa riseacidosis (up to 36 mmol/l) in PaCO 2Chronic respiratory Rises 0.02 Falls 3.8 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2
  32. 32. Early renal compensation for respiratory conditions pH [HCO 3-]Acute respiratory Falls 0.06 Rises 0.8 mmol for every 1 kPa riseacidosis (up to 30 mmol/l) in PaCO 2Acute respiratory Rises 0.06 Falls 1.5 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2Chronic respiratory Falls 0.02 Rises 3.0 mmol for every 1 kPa riseacidosis (up to 36 mmol/l) in PaCO 2Chronic respiratory Rises 0.02 Falls 3.8 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2
  33. 33. pH [HCO 3-]Acute respiratory Falls 0.06 Rises 0.8 mmol for every 1 kPa riseacidosis (up to 30 mmol/l) in PaCO 2Acute respiratory Rises 0.06 Falls 1.5 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2Chronic respiratory Falls 0.02 Rises 3.0 mmol for every 1 kPa riseacidosis (up to 36 mmol/l) in PaCO 2Chronic respiratory Rises 0.02 Falls 3.8 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2 Late renal compensation for respiratory conditions
  34. 34. pH and HCO3- changes pH [HCO 3-] Acute respiratory Falls 0.06 Rises 0.8 mmol for every 1 kPa rise acidosis (up to 30 mmol/l) in PaCO 2 Acute respiratory Rises 0.06 Falls 1.5 mmol for every 1 kPa fall in alkalosis (down to 18 mmol/l) PaCO 2 Chronic respiratory Falls 0.02 Rises 3.0 mmol for every 1 kPa rise acidosis (up to 36 mmol/l) in PaCO 2 Chronic respiratory Rises 0.02 Falls 3.8 mmol for every 1 kPa fall in alkalosis (down to 18 mmol/l) PaCO 2Take time to review the table then click to continue
  35. 35. Causes of respiratory disturbances RESPIRATORY ACIDOSIS RESPIRATORY ALKALOSIS Click to return to tutorial
  36. 36. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  37. 37. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  38. 38. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  39. 39. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  40. 40. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  41. 41. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  42. 42. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  43. 43. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  44. 44. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  45. 45. Respiratory acidosis Brainstem Spinal cord Airway Nerve root Lung Nerve Pleura Neuromuscular Chest wall junction Respiratory muscle
  46. 46. Respiratory alkalosis• Catastrophic CNS event (CNS haemorrhage)• Drugs (salicylates, progesterone)• Pregnancy (especially the 3rd trimester)• Decreased lung compliance (interstitial lung disease)• Liver cirrhosis• Anxiety Click to return to causes
  47. 47. If there is a metabolicproblem…• Is it an acidosis or an alkalosis?• What can we find out about a metabolic acidosis?• Information about base excess/deficit• Is there respiratory compensation?• Is there anything else going on?
  48. 48. Metabolic acidosisWhat is the anion gap?What is the base excess/deficitIs there any respiratory compensation? Click to return to tutorial
  49. 49. Anion Gap Anion Gap = [Na+] – [Cl-] - [HCO3-] • The anion gap is an artificial difference between the commonly measured anions and cations. • In reality there is electrochemical neutrality[Na+] + [unmeasured cations] = [Cl-] + [HCO3-] + [unmeasured anions][unmeasured anions] - [unmeasured cations] = [Na+] - ([Cl-] + [HCO3-])
  50. 50. Anion Gap Cations Anions Na+ HCO3- K+ Chloride- Ca2+ Protein (albumin) Mg2+ Organic acids Phosphates Sulphates[unmeasured anions] - [unmeasured cations] = [Na+] - ( [Cl-] + [HCO3-] ) Anion Gap = [Na+] – ( [Cl-] + [HCO3-] )
  51. 51. Anion Gap normal anion gap = 12 mmol/l[Na+] – ( [Cl-] + [HCO3-] ) = Anion Gap Na+ 144 – ( 108 + 24 ) = 12
  52. 52. Anion gap acidosis If there is accumulation of an organic acid not normally present in serum (eg lactic acid, ketones etc) these will replace HCO3- A fall in [HCO3-] will widen the anion gap Link to causes of anion gap acidosis Continue with tutorial
  53. 53. Normal anion gap acidosis If there is loss of HCO3- (GI tract or renal) there will be an increase in [chloride] and the anion gap will not change If there is administration of exogenous chloride, [HCO3-] will fall but the anion gap will not change. Link to causes of non-anion gap acidosis
  54. 54. Anion gap acidosis• Lactic acidosis – shock – severe hypoxaemia – generalized convulsions – severe sepsis• Ketoacidosis – diabetic, alcoholic• Alcohol poisons or drug intoxication – methanol, ethylene glycol, paraldehyde, salicylates• Renal failure (late stage) Non-anion gap acidosis
  55. 55. Normal anion gap acidosis• GI loss of HCO3- – Diarrhoea – Pancreatic/biliary drainage – Urinary diversion• Renal loss of HCO3- - Compensation for respiratory alkalosis - Renal tubular acidosis - Renal hypoperfusion - Carbonic anhydrase inhibitor (acetazolamide)• Other causes: HCl or NH4Cl infusion, Cl gas inhalation Return to tutorial
  56. 56. Base excess / deficit• Amount or acid or base that needs to be added to 1 litre of blood to return pH to normal, assuming standard conditions (temp 37oC, PaCO2 = 5.3 kPa, pressure 1 atm) – A measure of the metabolic component of a disturbance – Normal: -2 to +2 mmol/l – BE is POSITIVE in metabolic alkalosis (or compensation for a respiratory acidosis) – BE is NEGATIVE in metabolic acidosis (or compensation for a respiratory alkalosis) – Useful to follow the TREND to assess treatment. Return to tutorial
  57. 57. Metabolic acidosis• Is there any respiratory compensation? – Occurs rapidly after the change in pH – Predictable for metabolic acidosis by Winter’s formula – PaCO2 outside the predicted range suggest additional respiratory disturbances
  58. 58. Winter’s formula: Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 Link to examples
  59. 59. Using Winter’s formula: A patient with a metabolic acidosis has a [HCO3-] of 10 mmol/l. Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 = [ (1.5 x 10) + (8 ± 2) ] x 0.133 = 2.8 – 3.3A value outside this range suggests an additional respiratory disturbance Click to continue
  60. 60. Using Winter’s formula: A patient with a metabolic acidosis has a [HCO3-] of 10 mmol/l. Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 = [ (1.5 x 10) + (8 ± 2) ] x 0.133 = 2.8 – 3.3A value out side this range suggests an additional respiratory disturbance• If the actual PaCO2 is less than 2.8 kPa there is also RESPIRATORY ALKALOSIS Click to continue
  61. 61. Using Winter’s formula: A patient with a metabolic acidosis has a [HCO3-] of 10 mmol/l. Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 = [ (1.5 x 10) + (8 ± 2) ] x 0.133 = 2.8 – 3.3A value out side this range suggests an additional respiratory disturbance • If the actual PaCO2 is more than 3.3 kPa there is also RESPIRATORY ACIDOSIS Return to tutorial
  62. 62. Metabolic alkalosis• Is there respiratory compensation? – Occurs rapidly after the change in pH – Not complete or easily predictable for metabolic alkalosis; – Rarely achieve PaCO2 > 7 kPa – A suggested formula: Expected PaCO2 = 0.8 kPa per 10 mmol/l in HCO3-
  63. 63. Causes of metabolic disturbances METABOLIC ACIDOSIS METABOLIC ALKALOSIS Click to continue tutorial
  64. 64. Anion gap acidosis• Lactic acidosis – shock – severe hypoxaemia – generalized convulsions – severe sepsis• Ketoacidosis – diabetic, alcoholic• Alcohol poisons or drug intoxication – methanol, ethylene glycol, paraldehyde, salicylates• Renal failure (late stage) Non-anion gap acidosis
  65. 65. Normal anion gap acidosis• GI loss of HCO3- – Diarrhoea – Pancreatic/biliary drainage – Urinary diversion• Renal loss of HCO3- - Compensation for respiratory alkalosis - Renal tubular acidosis - Renal hypoperfusion - Carbonic anhydrase inhibitor (acetazolamide)• Other causes: HCl or NH4Cl infusion, Cl gas inhalation Return to causes
  66. 66. Metabolic alkalosis• Volume contraction (vomiting, overdiuresis, ascites)• Hypokalemia• Alkali ingestion (bicarbonate)• Excess gluco- or mineralocorticoids• Bartters syndrome Return to causes
  67. 67. Mixed disturbances These are difficult to interpret Expected corrections Acid base nomogram Examples
  68. 68. Primary change Compensatory changeRespiratory acidosis Rise in PaCO2 Rise in [HCO3-] 1. pH change consistent with PaCO2 2. Calculate expected rise in [HCO3-]Respiratory alkalosis Fall in PaCO2 Fall in [HCO3-] 1. pH change consistent with PaCO2 2. Calculate expected fall in [HCO3-]Metabolic acidosis Fall in [HCO3-] Fall in PaCO2 1. Winter’s formula for expected PaCO2 2. Corrected [HCO3-] in anion-gap acidosisMetabolic alkalosis Rise in [HCO3-] Rise in PaCO2 1. Difficult to predict; use suggested formula If the correction is NOT as expected there is another disturbance.
  69. 69. pH and HCO3- changes pH [HCO 3-]Acute respiratory Falls 0.06 Rises 0.8 mmol for every 1 kPa riseacidosis (up to 30 mmol/l) in PaCO 2Acute respiratory Rises 0.06 Falls 1.5 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2Chronic respiratory Falls 0.02 Rises 3.0 mmol for every 1 kPa riseacidosis (up to 36 mmol/l) in PaCO 2Chronic respiratory Rises 0.02 Falls 3.8 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2 Return to expected corrections
  70. 70. Expected corrections Primary change Compensatory changeRespiratory acidosis Rise in PaCO2 Rise in [HCO3-] 1. pH change consistent with PaCO2 2. Calculate expected rise in [HCO3-]Respiratory alkalosis Fall in PaCO2 Fall in [HCO3-] 1. pH change consistent with PaCO2 2. Calculate expected fall in [HCO3-]Metabolic acidosis Fall in [HCO3-] Fall in PaCO2 1. Winter’s formula for expected PaCO2 2. Corrected [HCO3-] in anion-gap acidosisMetabolic alkalosis Rise in [HCO3-] Rise in PaCO2 1. Difficult to predict; use suggested formula If the correction is NOT as expected there is another disturbance.
  71. 71. Expected corrections Primary change Compensatory changeRespiratory acidosis Rise in PaCO2 Rise in [HCO3-] 1. pH change consistent with PaCO2 2. Calculate expected rise in [HCO3-]Respiratory alkalosis Fall in PaCO2 Fall in [HCO3-] 1. pH change consistent with PaCO2 2. Calculate expected fall in [HCO3-]Metabolic acidosis Fall in [HCO3-] Fall in PaCO2 1. Winter’s formula for expected PaCO2 2. Corrected [HCO3-] in anion-gap acidosisMetabolic alkalosis Rise in [HCO3-] Rise in PaCO2 1. Difficult to predict; use suggested formula If the correction is NOT as expected there is another disturbance.
  72. 72. Respiratorycompensation• Not complete or easily predictable for metabolic alkalosis;• Rarely achieve PaCO2 > 7 kPa• A suggested formula: Expected PaCO2 = 0.8 kPa per 10 mmol/l in HCO3- Return to expected corrections
  73. 73. Winter’s formula: Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 Click to continue
  74. 74. Using Winter’s formula: A patient with a metabolic acidosis has a [HCO3-] of 10 mmol/l. Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 = [ (1.5 x 10) + (8 ± 2) ] x 0.133 = 2.8 – 3.3A value out side this range suggests an additional respiratory disturbance Click to continue
  75. 75. Using Winter’s formula: A patient with a metabolic acidosis has a [HCO3-] of 10 mmol/l. Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 = [ (1.5 x 10) + (8 ± 2) ] x 0.133 = 2.8 – 3.3A value out side this range suggests an additional respiratory disturbance• If the actual PaCO2 is less than 2.8 kPa there is also RESPIRATORY ALKALOSIS Click to continue
  76. 76. Using Winter’s formula: A patient with a metabolic acidosis has a [HCO3-] of 10 mmol/l. Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133 = [ (1.5 x 10) + (8 ± 2) ] x 0.133 = 2.8 – 3.3A value out side this range suggests an additional respiratory disturbance • If the actual PaCO2 is more than 3.3 kPa there is also RESPIRATORY ACIDOSIS Click to continue
  77. 77. Corrected bicarbonate• An anion gap acidosis may co-exist with an non-anion gap acidosis or a metabolic alkalosis.• In a simple anion gap acidosis the widened gap is due to absent bicarbonate – adding the actual and the missing bicarbonate adds up to a normal value for bicarbonate Click to continue
  78. 78. Corrected bicarbonate Corrected [HCO3-] = measured [HCO3-] + (anion gap – 12)• An patient with metabolic acidosis and an anion gap of 26 mmol/l has a serum [HCO3-] of 10 mmol/l.• The corrected [HCO3-] = 24 mmol/l Corrected [HCO3-] = 10 + (26 – 12)• No other metabolic disturbance exists Click to continue
  79. 79. Corrected bicarbonate Corrected [HCO3-] = measured [HCO3-] + (anion gap – 12)• An patient with metabolic acidosis and an anion gap of 26 mmol/l has a serum [HCO3-] of 15 mmol/l.• The corrected [HCO3-] = 29 mmol/l Corrected [HCO3-] = 15 + (26 – 12)• There is extra bicarbonate in the system and a metabolic alkalosis co-exists Return to expected corrections
  80. 80. 7.0 100 6 9 12 15 18 21 24 90 27 HCO3 -(mmol/l) 80 30 7.1 33 70 36 7.2 60 39 42 H+ (nmol/l) 45 50 7.3 48 51 40 N 57 7.4 63 69 30 74 7.5 7.6 20 7.7 7.8 10 8.0 8.5 0 PCO2 (kPa)Use two parameters to see if the result falls into expected values
  81. 81. Click for examples.Example 1. Example 6.Example 2. Example 7.Example 3. Example 8.Example 4. Example 9.Example 5. Example 10.
  82. 82. Example 1. A 33 male patient with SARS has a pH 7.43 saturation of 91% PaCO2 4.76 on Fi02 0.4 PaO2 8.1 HCO3- 231. Is he hypoxic? Base -0.6 excess2. Is there an acid Saturation 90% base or ventilation problem? Click to continue
  83. 83. Example 1. Is he hypoxic? pH 7.43 PaCO2 4.76 YES. PaO2 8.1 The SpO2 and HCO3- 23 calculated Base excess -0.6 saturation agree Saturation 90% Click to continue
  84. 84. Example 1. Is he hypoxic? YES. pH 7.43 PaCO2 4.76 (A-a) PO2 = 23.9 kPa PaO2 8.1 HCO3- 23 There is major Base -0.6 problem with oxygen excess transfer into the lung Saturation 90% To calculate (A-a) PO2 Click to continue
  85. 85. Example 1. Is there an acid base or ventilation pH 7.43 problem? PaCO2 4.76 NO. PaO2 8.1 pH, PaCO2 and PaCO2 PaCO2 23 are normal Base -0.6 excess This is pure Saturation 90% hypoxaemic respiratory failure Return to examples
  86. 86. Example 2. A patient with in the recovery room has pH 7.08 been found to be PaCO2 10.6 cyanosed, with PaO2 4.9 shallow breathing. HCO3- 26 This is the ABG result Base +2 on room air. excess Saturation 86% Click to continue
  87. 87. Example 2. Is the patient hypoxic due simply because of pH 7.08 hypoventilation as a PaCO2 10.6 result of residual PaO2 4.9 anaesthetic agents or have they also aspirated HCO3- 26 and developed lung Base +3 excess parenchymal problems? Saturation 86% Click to continue
  88. 88. Example 2.Calculate the A-a gradient: pH 7.08 PAO2 = [94.8 x 0.21] – [10.6 x 1.25] = 6.65 kPa PaCO2 10.6 PaO2 4.9 (A-a) PO2 = 6.65 – 4.9 HCO3- 26 = 1.75 kPa Base +3 excess This is a near normal A-a gradient, Saturation 86% and hypoventilation alone can explain the hypoxaemia. Increased ventilation will improve hypercapnia and oxygenation too. To calculate (A-a) PO2 Click to continue
  89. 89. Example 2. Is there an acid base or pH 7.08 ventilation PaCO2 10.6 problem? PaO2 4.9 HCO3- 26 Base +2 excess YES. Saturation 86% Click to continue
  90. 90. Example 2.There is:• Acidosis pH 7.08• PaCO2 is elevated PaCO2 10.6 PaO2 4.9 RESPIRATORY ACIDOSIS HCO3- 28 Base +2 excess Saturation 86% Diagnose disturbance Click to continue
  91. 91. Example 2.There is:• HCO3- = 28 pH 7.08• Expected HCO3- PaCO2 10.6• = 24 + [(10.6 – 5.3) x 0.8] = 28.2 PaO2 4.9 This is the expected [HCO3- ] if HCO3- 28 there has only been a small Base +2 amount of renal compensation excess Saturation 86% ACUTE RESPIRATORY ACIDOSIS Click to continue
  92. 92. Example 2.There is:• pH change: pH 7.08 [10.6 – 5.3] x 0.06 = 0.32 PaCO2 10.6 pH = [7.4 – 0.32] = 7.08 PaO2 4.9 HCO3- 28 CONSISTENT WITH SIMPLE Base +2 ACUTE RESPIRATORY ACIDOSIS; excess NO ADDITIONAL DISTURBANCE Saturation 86% Renal compensation Return to examples
  93. 93. Example 3. A patient has been brought to A&E after pH 7.23 a head injury; he is PaCO2 8.1 deeply unconscious. PaO2 4.9 This is the ABG on HCO3- 26 room air. Base +3 excess Saturation 86% Clearly he is very hypoxic Click to continue
  94. 94. Example 3. Is the patient hypoxic due simply because pH 7.23 of hypoventilation as PaCO2 8.1 a result of CNS PaO2 4.9 depression or have HCO3- 26 they also aspirated Base +3 and developed lung excess parenchymal Saturation 86% problems? Click to continue
  95. 95. Example 3.Calculate the A-a gradient: pH 7.23 PAO2 = [94.8 x 0.21] – [8.1 x 1.25] = 10.1 kPa PaCO2 8.1 PaO2 4.9 (A-a) PO2 = 10.1 – 4.9 HCO3- 26 = 5.2 kPa Base +3 excess The A-a gradient is increased Saturation 86% suggesting that less of the O2 available in the alveolus is able to get into the arterial blood. There is a lung problem; possibly aspiration To calculate (A-a) PO2 Click to continue
  96. 96. Example 3. Is there an acid base or pH 7.23 ventilation PaCO2 8.1 problem? PaO2 4.9 HCO3- 26 Base +3 excess YES. Saturation 86% Click to continue
  97. 97. Example 3.There is• Acidosis pH 7.23• PaO2 is elevated PaCO2 8.1 RESPIRATORY PaO2 4.9 ACIDOSIS HCO3- 26 Base +3 excess Saturation 86% Diagnose disturbance Click to continue
  98. 98. Example 3.There is:• HCO3- = 26 pH 7.23• Expected HCO3- PaCO2 8.1• = 24 + [(8.1 – 5.3) x 0.8] = 26.2 PaO2 4.9 This is the expected [HCO3- ] if HCO3- 26 there has only been a small Base +3 amount of renal compensation excess Saturation 86% ACUTE RESPIRATORY ACIDOSIS Click to continue
  99. 99. Example 3.There is:• pH change: pH 7.23 [8.1 – 5.3] x 0.06 = 0.32 PaCO2 8.1 pH = [7.4 – 0.17] = 7.23 PaO2 4.9 HCO3- 26 CONSISTENT WITH SIMPLE Base +3 ACUTE RESPIRATORY ACIDOSIS; excess NO ADDITIONAL DISTURBANCE Saturation 86% Renal compensation Return to examples
  100. 100. Example 4. A patient with COPD has a ABG pH 7.34 taken in out- PaCO2 8.0 patient clinic to PaO2 7.5 HCO3- 32.1 assess his need Base +8 for home oxygen. excess He is breathing Saturation 86% room air. Click to continue
  101. 101. Example 4.Is he hypoxic?YES. pH 7.34 PaCO2 8.0The (A-a) PO2 = 2.4 kPa PaO2 7.5The (A-a) gradient is HCO3- 32.1increased, and home Base +8 excessoxygen might be Saturation 86%appropriate To calculate (A-a) PO2 Click to continue
  102. 102. Example 4.Is there an acid baseor ventilation pH 7.34problem? PaCO2 8.0 PaO2 7.5 HCO3- 32.1YES. Base excess +8 Saturation 86% Click to continue
  103. 103. Example 4.There is:• Mild acidosis pH 7.34• PaCO2 is elevated PaCO2 8.0 PaO2 7.5 RESPIRATORY ACIDOSIS HCO3- 32.1 Base +8 excess Saturation 86% Diagnose disturbance Click to continue
  104. 104. Example 4.There is:• HCO3- = 32.1 pH 7.34• Expected HCO3- PaCO2 8.0• = 24 + [(8.0 – 5.3) x 3.0] = 33.9 PaO2 7.5 This is the expected [HCO3- ] if HCO3- 32.1 there has been significant renal Base +8 compensation over a long period; excess in addition the base excess has Saturation 86% increased. CHRONIC RESPIRATORY ACIDOSIS Click to continue
  105. 105. Example 4.There is:• pH change: pH 7.34 [8.0 – 5.3] x 0.02 = 0.054 PaCO2 8.0 pH = [7.4 – 0.054] = 7.35 PaO2 7.5 HCO3- 32.1 CONSISTENT WITH SIMPLE Base +8 CHRONIC RESPIRATORY ACIDOSIS; excess NO ADDITIONAL DISTURBANCE Saturation 86% Renal compensation Return to examples
  106. 106. Example 5. A 35 year old woman with a pH 7.54 history of anxiety PaCO2 2.9 attacks presents PaO2 12.1 HCO3- 22 to A&E with Base +2 palpitations. excess Saturation 100% Click to continue
  107. 107. Example 5. Is she hypoxic? pH 7.54 PaCO2 2.9 NO. PaO2 12.1 This is a normal HCO3- 22 PaO2 for room air Base excess +2 Saturation 100% Click to continue
  108. 108. Example 5. Is there an acid base or pH 7.54 ventilation PaCO2 2.9 problem? PaO2 12.1 HCO3- 22 Base +2 excess YES. Saturation 100% Click to continue
  109. 109. Example 5. There is:• Alkalosis pH 7.54• PaCO2 is decreased PaCO2 2.9 PaO2 12.1 RESPIRATORY ALKALOSIS HCO3- 22 Base +2 excess Saturation 100% Diagnose disturbance Click to continue
  110. 110. Example 5.There is:• HCO3- = 20 pH 7.54• Expected HCO3- PaCO2 2.9• = 24 - [(5.3 – 2.9) x 1.5] = 20.4 This is the expected [HCO3- ] if PaO2 12.1 there has only been a small HCO3- 20 amount of renal Base +2 excess compensation Saturation 100% ACUTE RESPIRATORY ALKALOSIS Click to continue
  111. 111. Example 5.There is:• pH change: pH 7.54 [5.3-2.9] x 0.06 = 0.144 PaCO2 2.9 pH = [7.4 + 0.144] = 7.54 PaO2 12.1 HCO3- 22 CONSISTENT WITH SIMPLE Base +2 ACUTE RESPIRATORY ALKALOSIS; excess NO ADDITIONAL DISTURBANCE Saturation 100% Renal compensation Return to examples
  112. 112. Example 6. A 42 year old diabetic woman pH 7.23 present with UTI PaCO2 3.3 PaO2 29.9 symptoms; she HCO3- 12 has deep sighing Base -10 respiration. This is excess Saturation 100% the ABG on FiO2 0.4 Click to continue
  113. 113. Example 6. Is she hypoxic? pH 7.23 PaCO2 3.3 NO. PaO2 29.9 This PaO2 is HCO3- 12 adequate for an Base -10 excess FiO2 of 0.4 Saturation 100% Click to continue
  114. 114. Example 6. Is there an acid base or ventilation pH 7.23 problem? PaCO2 3.3 PaO2 29.9 HCO3- 12 YES. Base excess -10 Saturation 100% Click to continue
  115. 115. Example 6. There is:• Acidosis pH 7.23• PaCO2 is decreased PaCO2 3.3• NOT respiratory acidosis PaO2 29.9 HCO3- 12 Look at [HCO3-] Base -10 excess• [HCO3-] is reduced Saturation 100%• Base excess is negative METABOLIC ACIDOSIS Diagnose disturbance Click to continue
  116. 116. Example 6.Using Winter’s formula:Expected PaCO2 pH 7.23= [ (1.5 x 12) + (8 ± 2) ] x 0.133 PaCO2 3.3 PaO2 29.9= 3.2 – 3.7 kPa HCO3- 12The PaCO2 falls within this range Base -10 excess SIMPLE METABOLIC ACIDOSIS Saturation 100% Respiratory compensation Click to continue
  117. 117. Example 6.What is the anion gap?= [Na+] – ( [Cl-] + [HCO3-] ) pH 7.23= [135] – ( 99 + 12 ) Na PaCO2 3.3 PaO2 29.9= 24 mmol/l HCO3- 12 Base -10• There is an anion gap excess acidosis due to Na+ 135 accumulation of organic Cl- 99 acids caused by diabetic ketoacidosis Click to continue
  118. 118. Example 6. Corrected bicarbonate = 24 mmol/l pH 7.23 PaCO2 3.3 The PaCO2 falls within the PaO2 29.9 expected range HCO3- 12 Base -10 excess SIMPLE METABOLIC ACIDOSIS; Na+ 135 NO OTHER DISTURBANCE Cl- 99More on metabolic acidosis Return to examples
  119. 119. Example 7. A 70 year old man presents with a 3 pH 7.5 day history of PaCO2 6.2 severe vomiting. PaO2 10.6 HCO3- 38 Here is his ABG on Base +8 room air. excess Saturation 96% Click to continue
  120. 120. Example 7. Is he hypoxic? pH 7.5 PaCO2 6.2 NO. This is a PaO2 10.6 normal PaO2 for a HCO3- 38 patient this age Base excess +8 breathing room air Saturation 96% Click to continue
  121. 121. Example 7. Is there an acid base or ventilation pH 7.5 problem? PaCO2 6.2 PaO2 10.6 HCO3- 38 YES. Base excess +8 Saturation 96% Click to continue
  122. 122. Example 7.There is:• Alkalosis pH 7.5• PaCO2 is elevated PaCO2 6.2• NOT respiratory PaO2 10.6 alkalosis HCO3- 38 Base +8 Look at [HCO3-] excess• [HCO3-] is increased Saturation 96%• Base excess is positive METABOLIC ALKALOSIS Diagnose disturbance Click to continue
  123. 123. Example 7. Is there respiratory compensation? pH 7.5 PaCO2 6.3Expected PaCO2 PaO2 10.6= 0.8 kPa per 10 mmol/l in HCO3- 38 HCO3- Base +8 excess= 5.3 + (0.8 x ([ 38 – 24 ]/10)) Saturation 96%= 6.4 CONSISTENT WITH SIMPLE METABOLIC ALKALOSIS Respiratory compensation Return to examples
  124. 124. Example 8. A 54 year old woman has multiple organ pH 7.07 failure due to intra- PaCO2 8.63 abdominal sepsis. She PaO2 11.8 has ARDS, renal HCO3- 17.9 failure and requires Base excess -6.5 inotropic support. Saturation 95% This is her ABG on FiO2 1.0 Click to continue
  125. 125. Example 8. Is she hypoxic? pH 7.07 PaCO2 8.63 YES. This PaO2 is PaO2 11.8 very low for an HCO3- 17.9 FiO2 of 1.0 Base excess -6.5 Saturation 95% Click to continue
  126. 126. Example 8. Is there an acid base or pH 7.07 ventilation PaCO2 8.63 problem? PaO2 11.8 HCO3- 17.9 Base -6.5 excess YES. Saturation 95% Click to continue
  127. 127. Example 8.There is• Acidosis pH 7.07• PaO2 is elevated PaCO2 8.63 RESPIRATORY PaO2 11.8 ACIDOSIS HCO3- 17.9 Base -6.5 excess Saturation 95% Diagnose disturbance Click to continue
  128. 128. Example 8.Expected pH= 7.4 – ([8.63-5.3] x 0.03) pH 7.07= 7.2 PaCO2 8.63 Observed pH is lower PaO2 11.8 HCO3- 17.9Expected bicarbonate Base -6.5= 24 + ([8.63-5.3] x 0.8) excess Saturation 95%= 26.7 mmol/l Observed bicarbonate is too low Renal compensation Click to continue
  129. 129. Example 8.Lower pHLower bicarbonate pH 7.07Base deficit negative PaCO2 8.63 ADDITIONAL METABOLIC PaO2 11.8 ACIDOSIS HCO3- 17.9 Base -6.5 excess Severe ARDS leads to Saturation 95% hypoxia & hypercapnia with respiratory acidosis; renal failure and poor perfusion lead to metabolic acidosis Return to examples
  130. 130. Example 9. A 77 year old man presents with a 3 pH 7.23 day history of PaCO2 3.3 severe diarrhoea. PaO2 10.6 HCO3- 8 Here is his ABG on Base -10 room air. excess Saturation 96% Click to continue
  131. 131. Example 9. Is he hypoxic? pH pH 7.5 7.23 PPCO2 2 a aCO 6.2 3.3 NO. This is a PPO2 2 a aO 10.6 10.6 normal PaO2 for a HCO3- - HCO3 38 8 patient this age Base Base excess excess +8 -10 breathing room air Saturation 96% Saturation 96% Click to continue
  132. 132. Example 9. Is there an acid base or ventilation pH 7.23 problem? PaCO2 3.3 PaO2 10.6 HCO3- 8 YES. Base excess -10 Saturation 96% Click to continue
  133. 133. Example 9. There is:• Acidosis pH 7.23• PaCO2 is decreased PaCO2 3.3• NOT respiratory acidosis PaO2 10.6 29.9 HCO3- 8 12 Look at [HCO3-] Base -10 excess• [HCO3-] is reduced Saturation 96% 100% METABOLIC ACIDOSIS Diagnose disturbance Click to continue
  134. 134. Example 9.Using Winter’s formula:Expected PaCO2 pH 7.23= [ (1.5 x 12) + (8 ± 2) ] x 0.133 PaCO2 3.3 PaO2 10.6= 3.2 – 3.7 kPa HCO3- 8The PaCO2 falls within this range Base -10 excess SIMPLE METABOLIC ACIDOSIS Saturation 96% Respiratory compensation Click to continue
  135. 135. Example 9.What is the anion gap?= [Na+] – ( [Cl-] + [HCO3-] ) pH 7.23= [135] – ( 115 + 8 ) Na PaCO2 3.3 PaO2 10.6= 12 mmol/l HCO3- 8 Base -10• There is an non-anion excess gap acidosis due to loss Na+ 135 of bicarbonate in Cl- 115 diarrhoea. Dehydration concentrates [Cl-] Return to examples
  136. 136. Example 10. A 43 year old man presents with an pH 7.37 overdose of PaCO2 2.3 aspirin. This is his PaO2 12 HCO3- 10 ABG on air. Base -7.4 excess Saturation 97% Click to continue
  137. 137. Example 10. Is he hypoxic? pH 7.37 PaCO2 2.3 NO. This is a PaO2 12 normal PaO2 for a HCO3- 10 patient this age Base excess -7.4 breathing room Saturation 97% air Click to continue
  138. 138. Example 10. Is there an acid base or ventilation pH 7.37 problem? PaCO2 2.3 PaO2 12 HCO3- 10 NO. Or is there? Base excess -7.4 Saturation 97% Click to continue
  139. 139. Example 10.PaCO2 is low• Respiratory alkalosis? pH 7.37• Metabolic acidosis? PaCO2 2.3HCO3- is low PaO2 12 HCO3- 10Negative base deficit Base -7.4• Metabolic acidosis? excess Saturation 97% Diagnose disturbance Click to continue
  140. 140. Example 10.Expected PaCO2 by Winter’s formula pH 7.37 =2.8 – 3.3 kPa PaCO2 2.3Observed PaCO2 is out of PaO2 12 this range HCO3- 10 Base -7.4 MIXED DISTURBANCE: excess Saturation 97% RESPIRATORY ALKALOSIS AND METABOLIC ACIDOSIS Respiratory compensation Click to continue
  141. 141. Example 10.Aspirin overdosecharacteristically causes a pH 7.37metabolic acidosis duethe effect of salicylic acid PaCO2 2.3and a respiratory alkalosis PaO2 12due to hyperventilation HCO3- 10 Base -7.4 excess Saturation 97% Return to examples
  142. 142. A patient with in therecovery room has pH 7.08been found to be PaCO2 10.6cyanosed, with PaO2 4.9shallow breathing. HCO3- 26This is the ABG result Base +3on room air. excess Saturation 86% Click to continue
  143. 143. Is the patient hypoxicdue simply because pH 7.08of hypoventilation as PaCO2 10.6a result of residual PaO2 4.9anaesthetic agents or HCO3- 26have they also Base +3aspirated and excessdeveloped lung Saturation 86%parenchymalproblems? Click to continue
  144. 144. Calculate the A-a gradient: pH 7.08 PAO2 = [94.8 x 0.21] – [10.6 x 1.25] = 6.65 kPa PaCO2 10.6 PaO2 4.9 (A-a) PO2 = 6.65 – 4.9 HCO3- 26 = 1.75 kPa Base +3 excess This is a near normal A-a gradient, Saturation 86% and hypoventilation alone can explain the hypoxaemia. Increased ventilation will improve hypercapnia and oxygenation too. Click to continue
  145. 145. A patient has beenbrought to A&E after pH 7.23a head injury; they PaCO2 8.1are deeply PaO2 4.9unconscious. This is HCO3- 26the ABG on room air Base +3 excess Saturation 86% Click to continue
  146. 146. Is the patient hypoxicdue simply because pH 7.23of hypoventilation as PaCO2 8.1a result of CNS PaO2 4.9depression or have HCO3- 26they also aspirated Base +3and developed lung excessparenchymal Saturation 86%problems? Click to continue
  147. 147. Calculate the A-a gradient: pH 7.23 PAO2 = [94.8 x 0.21] – [8.1 x 1.25] = 10.1 kPa PaCO2 8.1 PaO2 4.9 (A-a) PO2 = 10.1 – 4.9 HCO3- 26 = 5.2 kPa Base +3 excess The A-a gradient is increased Saturation 86% suggesting that less of the O2 available in the alveolus is able to get into the arterial blood. There is a lung problem; possibly aspiration Return to tutorial
  148. 148. Oxygenation The alveolar gas equation: PAO2 = [94.8 x FIO2] – [PaCO2 x 1.25] The alveolar-arterial oxygen difference (A-a) PO2 = PAO2 - PaO2As CO2 accumulates in the alveolus due to HYPOVENTILATION thereis less room for oxygen. If the lung is otherwise normal this oxygencan pass into blood as normal. There just is not enough passing. Inthe normal state there is only a small gradient between the alveolusand the arterial blood (1.33kPa). If there are problems that limitoxygen diffusion the gradient will get bigger. Return to example
  149. 149. Oxygenation The alveolar gas equation: PAO2 = [94.8 x FIO2] – [PaCO2 x 1.25] The alveolar-arterial oxygen difference (A-a) PO2 = PAO2 - PaO2As CO2 accumulates in the alveolus due to HYPOVENTILATION thereis less room for oxygen. If the lung is otherwise normal this oxygencan pass into blood as normal. There just is not enough passing. Inthe normal state there is only a small gradient between the alveolusand the arterial blood (1.33kPa). If there are problems that limitoxygen diffusion the gradient will get bigger. Return to example
  150. 150. RespiratorycompensationFor metabolic acidosis Winter’s formula: Expected PaCO2 = [ (1.5 x HCO3-) + (8 ± 2) ] x 0.133For metabolic alkalosis: Expected PaCO2 = 0.8 kPa per 10 mmol/l in HCO3- Return to example
  151. 151. Metabolic acidosis Anion Gap Anion Gap = [Na+] – [Cl-] - [HCO3-] Correcting bicarbonate Corrected [HCO3-] = measured [HCO3-] + (anion gap – 12) Return to example
  152. 152. pH and HCO3- changes pH [HCO 3-]Acute respiratory Falls 0.06 Rises 0.8 mmol for every 1 kPa riseacidosis (up to 30 mmol/l) in PaCO 2Acute respiratory Rises 0.06 Falls 1.5 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2Chronic respiratory Falls 0.02 Rises 3.0 mmol for every 1 kPa riseacidosis (up to 36 mmol/l) in PaCO 2Chronic respiratory Rises 0.02 Falls 3.8 mmol for every 1 kPa fall inalkalosis (down to 18 mmol/l) PaCO 2 Return to example
  153. 153. Is there Is the PaCO2 Is the HCO3- It isAcidosis High Normal/high Respiratory acidosisAcidosis Low Low Metabolic acidosisAlkalosis Low Normal/low Respiratory alkalosisAlkalosis High High Metabolic alkalosis Return to example
  154. 154. Oxygen dissociation curve10050 3.5 13.3 PO2 kPa
  155. 155. Oxygen dissociation curve1007550 3.5 5.3 13.3 PO2 kPa
  156. 156. Oxygen dissociation curve100887550 3.5 5.3 6.7 13.3 PO2 kPa
  157. 157. Oxygen cascadeDry atmospheric gas: 21 kPa Humidified tracheal gas: 19.8 kPa Alveolar gas: 14 kPa Arterial blood: 13.3 kPa Capillary blood: 6-7 kPa Mitochondria: 1-5 kPaClick for acid base physiology Venous blood: 5.3 kPa

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