3. 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.
10. 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
18. 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
24. 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-]
25. 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.”
29. 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
30. 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%
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31. 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
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32. 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%
33. 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
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34. 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.
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35. 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
37. 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
38.
39. a/A Ratio
• Pao2/PAo2 NAORMAL LEVEL IS >0.75
• <0.60 IS INCOMPATIBLE WITH SPONTANIOUS
BREATHING
40. 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
44. RELATION OF ALBUMIN IN ABG
AG corrected = AG + 2.5[4 – albumin]
(AG= Anion gap)
45. 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
47. 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)?
48. 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
49. 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
50. 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)?
51. 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
52. 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
53. 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
54. 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.
55. 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.
56. 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.
57. 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)?
58. 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
59. 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.
60. 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)?
61. 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.
62. 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
63. 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
64. 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
65. 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
66. 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
67. 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
68. 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
69. 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.
70. 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
71. 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
72. 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
73. 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
74. 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
75. 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
76. 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
77. 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
78. 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.
Notes de l'éditeur
Must memorize how to calculate the delta gapJust read off the slide