APPROACH TO BLOOD GAS
ANALYSIS
Dr. MANDAR HAVAL
D.C.H D.N.B
How does the kidney do it?
• The kidney does it in three ways:
– Total reabsorption of filtered bicarbonate (proximal).
– ...
FILTRATE

TUBULAR CELL

BLOOD
FILTRATE

TUBULAR CELL

H2O + CO2
CA II

H2CO3
H+ + HCO3-

BLOOD
FILTRATE

TUBULAR CELL

H2O + CO2
CA II

H2CO3
H+ + HCO3-

BLOOD
FILTRATE

TUBULAR CELL

BLOOD

H2O + CO2
CA II

H2CO3

H+ + HCO3Na+

Na

+
Na

Na K
ATPase

K
TUBULAR CELL

FILTRATE

BLOOD

H2O + CO2
CA II

H2CO3
H+ATPase

Na+

Na+ / H+
Antiporter

+
HH+ + HCO33-HCO
Na
Na+

Na K
A...
TUBULAR CELL

FILTRATE

BLOOD

H2O + CO2
CA II

H2CO3
Na / K

H+

HCO3-

H+ATPase
Na+ / H+
Antiporter

Na+

Na
Na+

Na K
A...
TUBULAR CELL

FILTRATE

H2O + CO2

BLOOD

H2O + CO2
CA II

H2CO3
CA IV

H2CO3

HCO3-

Na / K

H+

HCO3-

H+ATPase
Na+ / H+...
FILTRATE

H2O

TUBULAR CELL

BLOOD

H2O + CO2
CA II
CA IV

H+

HCO3-
COLLECTING
TUBULE CELL

BLOOD

FILTRATE

H2O + CO2
CA II

H2CO3

Cl- / HCO3-

Cl-

Exchanger

HCO3-

H+

H+
ATPase
COLLECTING
TUBULE CELL

BLOOD

FILTRATE

H2O + CO2
CA II

H2CO3

HCO3-

H+
ATPase
Cl- / HCO3-

Exchanger

Cl-

H+
COLLECTING
TUBULE CELL

FILTRATE

BLOOD

HPO4=

H+
ATPase

H+
COLLECTING
TUBULE CELL

FILTRATE

BLOOD

H+
ATPase

H+

HPO4= H2PO4
COLLECTING
TUBULE CELL

FILTRATE

BLOOD

SO4=

H+
ATPase

H+

HSO4-
COLLECTING
TUBULE CELL

FILTRATE

BLOOD

NH3

NH3
H+
ATPase

NH4+
H+
Evaluation of
Systemic Acid Base Disorders
1. Comprehensive history and physical examination.

2. Evaluate simultaneously ...
Step 3:

Identification of the dominant disorder
Primary
disorder
Metabolic
acidosis

pH

Initial
change

Compensatory
cha...
Step 3:

Identification of the dominant disorder
pH

Initial
change

Compensatory
change

Metabolic
acidosis

↓

↓ HCO3

↓...
Step 3:

Identification of the dominant disorder
pH

Initial
change

Compensatory
change

Metabolic
acidosis

↓

↓ HCO3

↓...
• WHERE THE PROBLEM START
Calculation of compensation
Mean "whole body" response equations for simple acid-base disturbances.

Disorder

pH

Primary...
Simple compensation
Disorder

pH

Primary problem

Compensation

Metabolic acidosis

↓

↓ in HCO3-

PaCO2
=1.5xHCO3+8(+/-2...
Calculate the “gaps”
Anion gap

=

Na+ − [Cl− + HCO3−]

Δ AG

=

Anion gap − 12

Δ HCO3

=

24 − HCO3

Δ AG = Δ HCO3 −, th...
SOME FORMULA

•THAT YOU SHOULD
KNOW
CALCULATION OF H+

H   24  HCO


PaCO2

3

20 – 7.70
30 – 7.50
40(H+) – 7.40 (PH)
50 – 7.30
65 – 7.20
pH

H+

pH

H+

6.70

200

7.40

40

6.75

178

7.45

35

6.80

158

7.50

32

6.85

141

7.55

28

6.90

126

7.60

25

6...
CAO2= directly reflects the total
number of oxygen molecules in
arterial blood, both bound and
unbound to hemoglobin
• CaO...
Which patient is more hypoxemic, and why?

• Patient A:
pH 7.48
PaCO2 34 mm Hg
PaO2 85 mm Hg
SaO2 95%
Hemoglobin 7 gm%

• ...
ANS CONT…..
• Patient A: Arterial oxygen content = .95 x 7 x
1.34 = 8.9 ml O2/dl
• Patient B: Arterial oxygen content = .8...
PaO2

• Factors affecting the PaO2 include alveolar
ventilation, FIO2, altitude, age, and the
oxyhemoglobin dissociation c...
True or False:
The pO2 in a cup of water open to the
atmosphere is always higher than the arterial
pO2 in a healthy person...
ANS
• The PO2 in the cup of water is always higher. This is for several
reasons. First, there is no barrier to oxygen diff...
A-a Gradient
• Determines the degree of lung function
impairment
• The A-a gradient is the partial pressure of
alveolar ox...
A-a Gradient
• [(713*FIO2)-(PaCO2/0.8)] – PaO2

INTERPRETATION
NORMAL – 10-20
(>30 is SINGNIFICANT)
Seen in – Shunt
Low V/...
A-a Gradient
• PAO2-PaO2 of 20-30mm Hg on room air
indicates mild pulmonary dysfunction, and
greater than 50mm Hg on room ...
a/A Ratio
• Pao2/PAo2 NAORMAL LEVEL IS >0.75
• <0.60 IS INCOMPATIBLE WITH SPONTANIOUS
BREATHING
PaO2/FIO2 Ratio

• To estimate the impairment of oxygenation, calculate
the PaO2/FIO2 ratio
• Normally, this ratio is 500-...
OXYGEN INDEX

• OI =MAP X FIO2
POST DUCTAL
PAO2

x 100
INTERPRETATION
• OI >40 that is unresponsive to iNO predict a
high mortality rate (>80%) and are indications
for ECMO.
VENTILATORY INDEX

• VI =PIP X PCO2 X RR
1000

VI > 65% INDICATE
PREDICTIVE DEATH
IN ARDS
RELATION OF ALBUMIN IN ABG
 AG corrected = AG + 2.5[4 – albumin]
(AG= Anion gap)
DELTA GAP






Delta gap = (actual AG – 12) + HCO3
Adjusted HCO3 should be 24 (+_ 6) {18-30}
If delta gap > 30 -> ad...
SOME CASE DISCUSSION
Case 1
• A 15 yr old juvenile diabetic presents with abdominal
pain, vomiting, fever & tiredness for 1 day. He had
stopped...
Case 1: Solution
• Dominant disorder is Metabolic Acidosis
• Compensation formula:
Δ PaCO2
= 1.2 × Δ HCO3
= 1.2 × 12
= 14....
Case 1: Solution
• Δ AG

= 28 – 12
= 16

• Δ HCO3
• Δ AG > Δ

= 24 – 12
= 12
HCO3

pH
PaCO2
HCO3
PaO2

7.31
26
12
92

Na
K...
Case 2
• A 14 yr old boy presents with continuous vomiting of
3 days duration, mental confusion, giddiness, and
tiredness ...
Case 2: Solution
• Dominant disorder is Metabolic Alkalosis
• Compensation formula:
Δ PaCO2
= 0.7 × Δ HCO3
= 0.7 × 8
= 5.6...
Case 2: Solution
• Δ AG

= 22 – 12
= 10

• High AG metabolic acidosis
• Final Diagnosis:

pH
PaCO2
HCO3
PaO2

7.50
48
32
9...
Case 3: Varieties of Metabolic Acidosis
Patient
ECF volume
Glucose
pH
Na
Cl
HCO3
AG
Ketones

A
Low
600
7.20
140
103
10
27
...
Renal handling of Hydrogen in
Metabolic Acidosis
• In the setting of metabolic acidosis, normal kidneys try to
increase H+...
Urine Electrolytes in Metabolic Acidosis
Patient
U. Na
U. K
U. Cl
Urine AG

A

Dx:

B
10
14
74
–50
Diarrhea

C
50
47
28
+6...
Case 4
• A 17 yr old boy presented with history of
progressive dyspnoea with wheezing for 4 days.
• He also had fever, cou...
Case 4: Laboratory data
• ABG:
pH
PaCO
HCO
PaO

2

3

2

7.30
60 mmHg
28 mEq/L
68 mm Hg

• Serum Electrolytes:
Na
136 mEq/...
Case 4: Solution
• Dominant disorder is Respiratory Acidosis
• Compensation formula:
Δ HCO3
= 0.3 × Δ PaCO2
= 0.3 × 20
=6
...
Case 5
• 12 year old girl presented with complaints of
difficulty in breathing and upper abdominal
discomfort for the past...
Case 5: Laboratory data
• ABG:
pH
PaCO
HCO
PaO

2

3

2

7.50
25 mmHg
21 mEq/L
100 mm Hg

• Serum Electrolytes:
Na
137 mEq...
Case 5: Solution
• Dominant disorder is Respiratory Alkalosis
• Compensation formula:
pH
7.50
Δ HCO3
= 0.2 × Δ PaCO2
PaCO2...
Case 7
• Explain the acid-base status of a 18-year-old boy
with history of chronic renal failure treated with high
dose di...
Case 7: Solution
• Dominant disorder is Respiratory Alkalosis
• Compensation formula:
pH
Δ HCO3
= 0.2 × Δ PaCO2
PaCO2
HCO3...
Case 7: Solution
• Δ AG

= 26 – 12
= 14

• Δ HCO3

= 24 – 21
=3

• Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis

• F...
Case 8
• The following values are found in a 65-year-old
patient. Evaluate this patient's acid-base status?
ABG
pH 7.51

S...
Case 8: Solution
• Dominant disorder is Metabolic Alkalosis
• Compensation formula:
Δ PaCO2
= 0.7 × Δ HCO3
= 0.7 × 16
= 11...
Case 8: Solution
• Δ AG

= 25 – 12
= 13

• High AG metabolic acidosis
• Final Diagnosis:
Metabolic Alkalosis +
High AG Met...
Case 9
• A 52-year-old woman has been mechanically ventilated for
two days following a drug overdose. Her arterial blood g...
Case 9: Solution
• Dominant disorder is Chronic Respiratory Alkalosis
• Compensation formula:
pH
7.45
Δ HCO3
= 0.5 × Δ PaC...
Case 9: Solution
• Δ AG

= 24 – 12
= 12

• Δ HCO3

= 24 –18
=6

• Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis

• Fi...
Case 11
• A 21 year old male with progressive renal insufficiency is
admitted with abdominal cramping. He had congenital
o...
Case 11: Solution
• Dominant disorder is Metabolic Acidosis
• Compensation formula:
Δ PaCO2
= 1.2 × Δ HCO3
= 1.2 × 14
= 16...
Case 11: Solution
• Δ AG

= 20 – 12
=8

• Δ HCO3

= 24 –10
= 14

• Δ AG < Δ HCO3-

pH
PaCO2
HCO3

7.20
24
10

Na
K
Cl

140...
Case 12
• A 15 year old female with
hypertension was treated with
low salt diet and diuretics. BP
135/85.
Otherwise normal...
Case 12: Solution
• Initally, dominant disorder is Metabolic Alkalosis

• Compensation formula:
Δ PaCO2
= 0.7 × Δ HCO3
= 0...
Case 12: Solution
• Subsequently, she has developed
pH

HCO3

PaCO2

↓

↓

↓

pH
PaCO2
HCO3

7.51 
47 
35 

7.42
39
25
...
Case 12: Solution
• Subsequently, she has developed
pH

HCO3

PaCO2

↓

↓

↓

Metabolic acidosis

The decrease in bicarbon...
Case 13
• A patient with salicylate overdose.
pH
=
7.45
PCO2
=
20 mmHg
HCO3
=
13 mEq/L

• Dominant disorder: Respiratory a...
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    1. 1. APPROACH TO BLOOD GAS ANALYSIS Dr. MANDAR HAVAL D.C.H D.N.B
    2. 2. How does the kidney do it? • The kidney does it in three ways: – Total reabsorption of filtered bicarbonate (proximal). – Controlled secretion of H+ into filtrate (distal). – Judicious use of urinary buffers.
    3. 3. FILTRATE TUBULAR CELL BLOOD
    4. 4. FILTRATE TUBULAR CELL H2O + CO2 CA II H2CO3 H+ + HCO3- BLOOD
    5. 5. FILTRATE TUBULAR CELL H2O + CO2 CA II H2CO3 H+ + HCO3- BLOOD
    6. 6. FILTRATE TUBULAR CELL BLOOD H2O + CO2 CA II H2CO3 H+ + HCO3Na+ Na + Na Na K ATPase K
    7. 7. TUBULAR CELL FILTRATE BLOOD H2O + CO2 CA II H2CO3 H+ATPase Na+ Na+ / H+ Antiporter + HH+ + HCO33-HCO Na Na+ Na K ATPase K
    8. 8. TUBULAR CELL FILTRATE BLOOD H2O + CO2 CA II H2CO3 Na / K H+ HCO3- H+ATPase Na+ / H+ Antiporter Na+ Na Na+ Na K ATPase K
    9. 9. TUBULAR CELL FILTRATE H2O + CO2 BLOOD H2O + CO2 CA II H2CO3 CA IV H2CO3 HCO3- Na / K H+ HCO3- H+ATPase Na+ / H+ Antiporter Na+ Na Na+ Na K ATPase K
    10. 10. FILTRATE H2O TUBULAR CELL BLOOD H2O + CO2 CA II CA IV H+ HCO3-
    11. 11. COLLECTING TUBULE CELL BLOOD FILTRATE H2O + CO2 CA II H2CO3 Cl- / HCO3- Cl- Exchanger HCO3- H+ H+ ATPase
    12. 12. COLLECTING TUBULE CELL BLOOD FILTRATE H2O + CO2 CA II H2CO3 HCO3- H+ ATPase Cl- / HCO3- Exchanger Cl- H+
    13. 13. COLLECTING TUBULE CELL FILTRATE BLOOD HPO4= H+ ATPase H+
    14. 14. COLLECTING TUBULE CELL FILTRATE BLOOD H+ ATPase H+ HPO4= H2PO4
    15. 15. COLLECTING TUBULE CELL FILTRATE BLOOD SO4= H+ ATPase H+ HSO4-
    16. 16. COLLECTING TUBULE CELL FILTRATE BLOOD NH3 NH3 H+ ATPase NH4+ H+
    17. 17. Evaluation of Systemic Acid Base Disorders 1. Comprehensive history and physical examination. 2. Evaluate simultaneously performed ABG & serum electrolytes. 3. Identification of the dominant disorder. 4. Calculation of compensation. 5. Calculate the anion gap and the Δ. 1. Anion Gap 2. Δ AG 3. Δ Bicarbonate
    18. 18. Step 3: Identification of the dominant disorder Primary disorder Metabolic acidosis pH Initial change Compensatory change ↓ ↓ HCO3 ↓ PCO2
    19. 19. Step 3: Identification of the dominant disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2 Primary disorder
    20. 20. Step 3: Identification of the dominant disorder pH Initial change Compensatory change Metabolic acidosis ↓ ↓ HCO3 ↓ PCO2 Metabolic alkalosis ↑ ↑ HCO3 ↑ PCO2 Respiratory acidosis ↓ ↑ PCO2 ↑ HCO3 Respiratory alkalosis ↑ ↓ PCO2 ↓ HCO3 Primary disorder
    21. 21. • WHERE THE PROBLEM START
    22. 22. Calculation of compensation Mean "whole body" response equations for simple acid-base disturbances. Disorder pH Primary change Compensatory Response Equation Metabolic Acidosis   [HCO3-]  PCO2 ΔPCO2  1.2  ΔHCO3 Metabolic Alkalosis   [HCO3-]  PCO2 ΔPCO2  0.7  ΔHCO3 Respiratory Acidosis   PCO2  [HCO3-] Acute: ΔHCO3-  0.1  ΔPCO2 Chronic: ΔHCO3-  0.3  ΔPCO2 Respiratory Alkalosis   PCO2  [HCO3-] Acute: ΔHCO3-  0.2  ΔPCO2 Chronic: ΔHCO3-  0.5  ΔPCO2 Note: The formula calculates the change in the compensatory parameter.
    23. 23. Simple compensation Disorder pH Primary problem Compensation Metabolic acidosis ↓ ↓ in HCO3- PaCO2 =1.5xHCO3+8(+/-2) Metabolic alkalosis ↑ 10↑ in HCO3- 7↑ in PaCO2 Respiratory acidosis ↓ ACUTE -10↑ in PaCO2 CHRONIC -10↑ in PaCO2 1↑ in [HCO3-] 3.5↑ in [HCO3-] Respiratory alkalosis ↑ ACUTE-10↓ in PaCO2 CHRONIC-10↓ in PaCO2 2↓ in [HCO3-] 4↓ in [HCO3-]
    24. 24. Calculate the “gaps” Anion gap = Na+ − [Cl− + HCO3−] Δ AG = Anion gap − 12 Δ HCO3 = 24 − HCO3 Δ AG = Δ HCO3 −, then Pure high AG Met. Acidosis Δ AG > Δ HCO3 −, then High AG Met Acidosis + Met. Alkalosis Δ AG < Δ HCO3 −, then High AG Met Acidosis + Normal AG Met A Note:  Add Δ AG to measured HCO3− to obtain bicarbonate level  Delta _ AG    Pr e _ existing _ Bicarb  that would have existed IF the high AG metabolic acidosis Current _ Bicarb   were to be absent, i.e., “Pre-existing Bicarbonate.”
    25. 25. SOME FORMULA •THAT YOU SHOULD KNOW
    26. 26. CALCULATION OF H+ H   24  HCO  PaCO2  3 20 – 7.70 30 – 7.50 40(H+) – 7.40 (PH) 50 – 7.30 65 – 7.20
    27. 27. pH H+ pH H+ 6.70 200 7.40 40 6.75 178 7.45 35 6.80 158 7.50 32 6.85 141 7.55 28 6.90 126 7.60 25 6.95 112 7.65 22 7.00 100 7.70 20 7.05 89 7.75 18 7.10 79 7.80 16 7.15 71 7.85 14 7.20 63 7.90 13 7.25 56 7.95 11 7.30 50 8.00 10 7.35 45
    28. 28. CAO2= directly reflects the total number of oxygen molecules in arterial blood, both bound and unbound to hemoglobin • CaO2 = (1.34 x HB x SPO2) +(0.003 x PaO2) Normal CaO2 ranges from 16 to 22 ml O2/dl
    29. 29. Which patient is more hypoxemic, and why? • Patient A: pH 7.48 PaCO2 34 mm Hg PaO2 85 mm Hg SaO2 95% Hemoglobin 7 gm% • Patient B: pH 7.32 PaCO2 74 mm Hg PaO255 mm Hg SaO2 85% Hemoglobin 15 gm% www.dnbpediatrics.com
    30. 30. ANS CONT….. • Patient A: Arterial oxygen content = .95 x 7 x 1.34 = 8.9 ml O2/dl • Patient B: Arterial oxygen content = .85 x 15 x 1.34 = 17.1 ml O2/dl • Patient A, with the higher PaO2 but the lower hemoglobin content, is more hypoxemic www.dnbpediatrics.com
    31. 31. PaO2 • Factors affecting the PaO2 include alveolar ventilation, FIO2, altitude, age, and the oxyhemoglobin dissociation curve • Relation between PaO2 and SaO2: PaO2 60mm Hg 50mm Hg 40mm Hg 30mm Hg corresponds to SaO2 90% 80% 70% 60%
    32. 32. True or False: The pO2 in a cup of water open to the atmosphere is always higher than the arterial pO2 in a healthy person (breathing room air) who is holding the cup www.dnbpediatrics.com
    33. 33. ANS • The PO2 in the cup of water is always higher. This is for several reasons. First, there is no barrier to oxygen diffusing into the water; thus the PO2 in the cup will be the same as the atmosphere, at sea level approximately 160 mm Hg. • Second, there is no CO2 coming from the cup to dilute the oxygen, as there is in people. • Third, there is no V-Q inequality or shunt; even healthy people have a difference between alveolar PO2 and arterial PO2 for this reason. Thus a healthy person and a cup of water exposed to the atmosphere at sea level would have PO2 values of about 100 mm Hg and 160 mm Hg, respectively. www.dnbpediatrics.com
    34. 34. A-a Gradient • Determines the degree of lung function impairment • The A-a gradient is the partial pressure of alveolar oxygen minus the partial pressure of arterial oxygen (PAO2-PaO2) • Normal is 2-10mm Hg or 10 plus one tenth the person’s age
    35. 35. A-a Gradient • [(713*FIO2)-(PaCO2/0.8)] – PaO2 INTERPRETATION NORMAL – 10-20 (>30 is SINGNIFICANT) Seen in – Shunt Low V/Q Hypoventilation
    36. 36. A-a Gradient • PAO2-PaO2 of 20-30mm Hg on room air indicates mild pulmonary dysfunction, and greater than 50mm Hg on room air indicates severe pulmonary dysfunction • The causes of increased gradient include intrapulmonary shunt, intracardiac shunt, and diffusion abnormalities
    37. 37. a/A Ratio • Pao2/PAo2 NAORMAL LEVEL IS >0.75 • <0.60 IS INCOMPATIBLE WITH SPONTANIOUS BREATHING
    38. 38. PaO2/FIO2 Ratio • To estimate the impairment of oxygenation, calculate the PaO2/FIO2 ratio • Normally, this ratio is 500-600 • Below 300 is acute lung injury* • Below 200 is ARDS* *Along with diffuse infiltrates, normal PCWP, and appropriate mechanism
    39. 39. OXYGEN INDEX • OI =MAP X FIO2 POST DUCTAL PAO2 x 100
    40. 40. INTERPRETATION • OI >40 that is unresponsive to iNO predict a high mortality rate (>80%) and are indications for ECMO.
    41. 41. VENTILATORY INDEX • VI =PIP X PCO2 X RR 1000 VI > 65% INDICATE PREDICTIVE DEATH IN ARDS
    42. 42. RELATION OF ALBUMIN IN ABG  AG corrected = AG + 2.5[4 – albumin] (AG= Anion gap)
    43. 43. DELTA GAP      Delta gap = (actual AG – 12) + HCO3 Adjusted HCO3 should be 24 (+_ 6) {18-30} If delta gap > 30 -> additional metabolic alkalosis If delta gap < 18 -> additional non-gap metabolic acidosis If delta gap 18 – 30 -> no additional metabolic disorders
    44. 44. SOME CASE DISCUSSION
    45. 45. Case 1 • A 15 yr old juvenile diabetic presents with abdominal pain, vomiting, fever & tiredness for 1 day. He had stopped taking insulin 3 days ago. Examination revealed tachycardia, BP- 100/60, signs of dehydration. Abdominal examination was normal. • ABG: pH PaCO2 HCO3 PaO2 7.31 26 mmHg 12 mEq/L 92 mm Hg Serum Electrolytes: Na 140 mEq/L K 5.0 mEq/L Cl 100 mEq/L • Evaluate the acid-base disturbance(s)?
    46. 46. Case 1: Solution • Dominant disorder is Metabolic Acidosis • Compensation formula: Δ PaCO2 = 1.2 × Δ HCO3 = 1.2 × 12 = 14.4 PaCO2 = 40 – 14 = 26 Compensation is appropriate. • Anion Gap = 140 – (100 + 12) = 28 AG is high. pH PaCO2 HCO3 PaO2 7.31 26 12 92 Na K Cl 140 5.0 100
    47. 47. Case 1: Solution • Δ AG = 28 – 12 = 16 • Δ HCO3 • Δ AG > Δ = 24 – 12 = 12 HCO3 pH PaCO2 HCO3 PaO2 7.31 26 12 92 Na K Cl 140 5.0 100 • Final Diagnosis: High AG Met. Acidosis + Met. Alkalosis
    48. 48. Case 2 • A 14 yr old boy presents with continuous vomiting of 3 days duration, mental confusion, giddiness, and tiredness for 1 day. • Examination revealed tachycardia, hypotension and dehydration. • ABG pH 7.50 Serum Electrolytes: PaCO2 48 Na 139 HCO3 32 K 3.9 PaO2 90 Cl 85 • Evaluate the acid-base disturbance(s)?
    49. 49. Case 2: Solution • Dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 8 = 5.6 PaCO2 = 40 + 6 = 46 Compensation is appropriate. • Anion Gap = 139 – (85 + 32) = 22 AG is high. pH PaCO2 HCO3 PaO2 7.50 48 32 90 Na K Cl 139 3.9 85
    50. 50. Case 2: Solution • Δ AG = 22 – 12 = 10 • High AG metabolic acidosis • Final Diagnosis: pH PaCO2 HCO3 PaO2 7.50 48 32 90 Na K Cl 139 3.9 85 Metabolic Alkalosis + High AG Met. Acidosis
    51. 51. Case 3: Varieties of Metabolic Acidosis Patient ECF volume Glucose pH Na Cl HCO3 AG Ketones A Low 600 7.20 140 103 10 27 4+ B Low 120 7.20 140 118 10 12 0 C Normal 120 7.20 140 118 10 12 0 High-AG Met. Acidosis Non-AG Met. Acidosis Non-AG Met. Acidosis
    52. 52. Renal handling of Hydrogen in Metabolic Acidosis • In the setting of metabolic acidosis, normal kidneys try to increase H+ excretion by increasing titratable acidity and ammonia. The latter is excreted as NH4+. • When NH4+ is excreted, it also causes increased chloride loss, to maintain electrical neutrality. • Chloride loss, therefore, will be in excess of Na and K. • Urine Anion-Gap = Na + K – Cl • In metabolic acidosis, if Urine anion gap is negative, it suggests that the kidneys are excreting H+ effectively.
    53. 53. Urine Electrolytes in Metabolic Acidosis Patient U. Na U. K U. Cl Urine AG A Dx: B 10 14 74 –50 Diarrhea C 50 47 28 +69 RTA Urine Anion Gap = (U. Na + U. K – U. Cl) In Normal anion gap Metabolic Acidosis, Positive Urine AG suggests distal Renal Tubular Acidosis Negative Urine AG suggests non-renal cause for Metabolic Acidosis.
    54. 54. Case 4 • A 17 yr old boy presented with history of progressive dyspnoea with wheezing for 4 days. • He also had fever, cough with yellowish expectoration. • He had increased sleepiness for 1 day. • On examination, he was tachypnoeic, pulse100/min bounding, BP-160/96, central cyanosis +, drowsy, asterixis +, RS – B/L extensive wheezing +. • CXR- hyperinflated lung fields with tubular heart.
    55. 55. Case 4: Laboratory data • ABG: pH PaCO HCO PaO 2 3 2 7.30 60 mmHg 28 mEq/L 68 mm Hg • Serum Electrolytes: Na 136 mEq/L K 4.5 mEq/L Cl 98 mEq/L • Evaluate the acid-base disturbance(s)?
    56. 56. Case 4: Solution • Dominant disorder is Respiratory Acidosis • Compensation formula: Δ HCO3 = 0.3 × Δ PaCO2 = 0.3 × 20 =6 HCO3 = 24 + 6 = 30 Compensation is appropriate. • Anion Gap = 136 – (98 + 28) = 10 AG is normal. pH PaCO2 HCO3 PaO2 7.30 60 28 68 Na K Cl 136 4.5 98
    57. 57. Case 5 • 12 year old girl presented with complaints of difficulty in breathing and upper abdominal discomfort for the past 1 hr. • On examination, vitals normal, patient hyperventilating, RS – normal, Abdomen – normal.
    58. 58. Case 5: Laboratory data • ABG: pH PaCO HCO PaO 2 3 2 7.50 25 mmHg 21 mEq/L 100 mm Hg • Serum Electrolytes: Na 137 mEq/L K 3.9 mEq/L Cl 99 mEq/L Calcium 9.0 mEq/L • Evaluate the acid-base disturbance(s)?
    59. 59. Case 5: Solution • Dominant disorder is Respiratory Alkalosis • Compensation formula: pH 7.50 Δ HCO3 = 0.2 × Δ PaCO2 PaCO2 25 HCO3 21 = 0.2 × 15 PaO2 100 =3 HCO3 = 24 – 3 = 21 Na 137 K 3.9 Compensation is appropriate. Cl 99 Calcium 9.0 • Anion Gap = 137 – (99 + 21) = 17 AG is slightly high which can be seen in respiratory alkalosis.
    60. 60. Case 7 • Explain the acid-base status of a 18-year-old boy with history of chronic renal failure treated with high dose diuretics admitted to hospital with pneumonia and the following lab values: ABG pH 7.52 PaCO2 30 mm Hg PaO2 62 mm Hg Serum Electrolytes Na+ 145 mEq/L K+ 2.9 mEq/L Cl 98 mEq/L - HCO3 21 mEq/L
    61. 61. Case 7: Solution • Dominant disorder is Respiratory Alkalosis • Compensation formula: pH Δ HCO3 = 0.2 × Δ PaCO2 PaCO2 HCO3 = 0.2 × 10 PaO2 =2 HCO3 = 24 – 2 = 22 Na K Compensation is appropriate. Cl • Anion Gap = 145 – (98 + 21) = 26 AG is very high suggestive of metabolic acidosis. 7.52 30 21 62 145 2.9 98
    62. 62. Case 7: Solution • Δ AG = 26 – 12 = 14 • Δ HCO3 = 24 – 21 =3 • Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis • Final Diagnosis: Respiratory Alkalosis + High AG Metabolic Acidosis + Metabolic Alkalosis pH PaCO2 HCO3 PaO2 7.52 30 21 62 Na K Cl 145 2.9 98
    63. 63. Case 8 • The following values are found in a 65-year-old patient. Evaluate this patient's acid-base status? ABG pH 7.51 Serum Chemistry Na + 155 mEq/L PaCO2 50 mm Hg HCO3- 40 mEq/L K+ 5.5 mEq/L Cl- 90 mEq/L CO2 40 mEq/L BUN 121 mg/dl Glucose 77 mg/dl
    64. 64. Case 8: Solution • Dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 16 = 11.2 PaCO2 = 40 + 11 = 51 Compensation is appropriate. • Anion Gap AG is high. = 155 – (90 + 40) = 25 pH PaCO2 HCO3 PaO2 7.51 50 40 62 Na K Cl BUN 155 5.5 90 121
    65. 65. Case 8: Solution • Δ AG = 25 – 12 = 13 • High AG metabolic acidosis • Final Diagnosis: Metabolic Alkalosis + High AG Metabolic Acidosis pH PaCO2 HCO3 PaO2 7.51 50 40 62 Na K Cl BUN 155 5.5 90 121
    66. 66. Case 9 • A 52-year-old woman has been mechanically ventilated for two days following a drug overdose. Her arterial blood gas values and electrolytes, stable for the past 12 hours, show: ABG pH 7.45 PaCO2 25 mm Hg Serum Chemistry Na + 142 mEq/L K+ 4.0 mEq/L Cl- 100 mEq/L HCO3- 18 mEq/L
    67. 67. Case 9: Solution • Dominant disorder is Chronic Respiratory Alkalosis • Compensation formula: pH 7.45 Δ HCO3 = 0.5 × Δ PaCO2 PaCO2 25 = 0.5 × 15 HCO3 18 = 7.5 HCO3 = 24 – 8 = 16 Na 142 K 4.0 Compensation is appropriate. Cl 100 • Anion Gap = 142 – (100 + 18) = 24 AG is very high suggestive of metabolic acidosis.
    68. 68. Case 9: Solution • Δ AG = 24 – 12 = 12 • Δ HCO3 = 24 –18 =6 • Δ AG > Δ HCO3High AG Met Acidosis + Met. Alkalosis • Final Diagnosis: Chronic Respiratory Alkalosis + High AG Metabolic Acidosis + ? Metabolic Alkalosis
    69. 69. Case 11 • A 21 year old male with progressive renal insufficiency is admitted with abdominal cramping. He had congenital obstructive uropathy with creation of ileal loop for diversion. On admission, ABG pH 7.20 PaCO2 24 mm Hg Serum Chemistry Na + 140 mEq/L K+ 5.6 mEq/L Cl- 110 mEq/L HCO3- 10 mEq/L
    70. 70. Case 11: Solution • Dominant disorder is Metabolic Acidosis • Compensation formula: Δ PaCO2 = 1.2 × Δ HCO3 = 1.2 × 14 = 16.8 PaCO2 = 40 – 17 = 23 Compensation is appropriate. • Anion Gap = 140 – (110 + 10) = 20 High anion-gap metabolic acidosis. pH PaCO2 HCO3 7.20 24 10 Na K Cl 140 5.6 110
    71. 71. Case 11: Solution • Δ AG = 20 – 12 =8 • Δ HCO3 = 24 –10 = 14 • Δ AG < Δ HCO3- pH PaCO2 HCO3 7.20 24 10 Na K Cl 140 5.6 110 High AG Met Acidosis + Normal-AG Met. Acidosis • Final Diagnosis: Mixed Metabolic Acidosis
    72. 72. Case 12 • A 15 year old female with hypertension was treated with low salt diet and diuretics. BP 135/85. Otherwise normal. See initial lab values. • She developed profound watery diarrhea, nausea and weakness. • On exam, HR = 96, T=100.6 F, BP 115/70. Abdominal tenderness with guarding on palpation. Parameter Initial Subseq uent Na 137 138 K+ 3.1 2.8 Cl- 90 102 HCO3 35 25 pH 7.51 7.42 PaCO2 47 39
    73. 73. Case 12: Solution • Initally, dominant disorder is Metabolic Alkalosis • Compensation formula: Δ PaCO2 = 0.7 × Δ HCO3 = 0.7 × 11 = 7.7 PaCO2 = 40 + 8 = 48 Compensation is appropriate. • Anion Gap AG is normal. = 137 – (90 + 35) = 12 pH PaCO2 HCO3 7.51 47 35 Na K Cl 137 3.1 90
    74. 74. Case 12: Solution • Subsequently, she has developed pH HCO3 PaCO2 ↓ ↓ ↓ pH PaCO2 HCO3 7.51  47  35  7.42 39 25 Na K Cl 137  3.1  90  138 2.8 102
    75. 75. Case 12: Solution • Subsequently, she has developed pH HCO3 PaCO2 ↓ ↓ ↓ Metabolic acidosis The decrease in bicarbonate is almost same as the rise in chloride. • Final Diagnosis: Metabolic Alkalosis + Hyperchloremic (non-AG) Metabolic Acidosis
    76. 76. Case 13 • A patient with salicylate overdose. pH = 7.45 PCO2 = 20 mmHg HCO3 = 13 mEq/L • Dominant disorder: Respiratory alkalosis • Appropriate Compensation would have been HCO3 of 20 (24 – 4) • Lower than expected HCO3 suggests presence of metabolic acidosis as well.

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