Dr. Anand Bhabhor
Senior Intensivist
Jaslok Hospital & Research Centre
 ABG analysis is an essential component of diagnosing

and managing critically ill patients in the ICU.

 A step wise approach to the evaluation of these

disorders helps delineate underlying causes,
compensatory mechanism, and the correct approach
to management.
Range

For calculation

pH

7.35 – 7.45

7.4

PaCO2

35 – 45 mmHg

40 mmHg

PaO2

> 80 mmHg

> 95 mmHg

HCO3-

22 – 26 mEq/L

24 mEq/L
 PaO2 and SaO2

PaO2 / FiO2 ratio
e.g. PaO2 :84 mmHg with 0.21 FiO2 [room air]
PaO2 / FiO2 = 84 / 0.21 = 400
 Normal = 300 – 500 mmHg
 < 300 = acute lung injury [previous definition]
 < 200 = ARDS [previous definition]
Berlin definition:
 200 – 300 [with PEEP/CPAP > 5] = mild ARDS
 < 200 [with PEEP > 5]
= moderate ARDS
 <100 [with PEEP > 5]
= severe ARDS
 Step 1: Acidemic, alkalemic, or normal?
 Step 2: Is the primary disturbance respiratory or metabolic?
 Step 3: For a primary respiratory disturbance, is it acute or chronic?

 Step 4: For a metabolic disturbance, is the respiratory system

compensating OK?

 Step 5: For a metabolic acidosis, is there an increased anion gap?
 Step 6: For an increased anion gap metabolic acidosis, are there other

derangements?
 The pH is the –log [H]. So by altering either the PCO2

or the HCO3-, [H] will change, and so will pH.
 An acidemia(low pH) can result from either a low

HCO3- or a high CO2
 An alkalemia (high pH) can result from either a high

HCO3 or a low CO2
 If pH and PaCO2 changes in same direction, the

primary disorder is metabolic and if they change in
opposite direction ,the primary disorder is respiratory.
Chemical
change

Primary
disorder

Compensation

pH

Low HCO3-

Metabolic
acidosis

Respiratory
alkalosis

low pH

High HCO3-

Metabolic
alkalosis

Respiratory
acidosis

High pH

High PaCO2

Respiratory
acidosis

Metabolic
alkalosis

Low pH

Low PaCO2

Respiratory
alkalosis

Metabolic
acidosis

High pH
 Respiratory acidosis is due to a primary rise in CO2
 Hypercapnia almost always results from alveolar

hypoventilation due to one of the following causes:
Respiratory center depression
2. Neuromuscular disorder
3. Upper airway obstruction
4. Pulmonary disease
1.
 A respiratory alkalosis is due to decrease in PaCO2.
 It results from hyperventilation leading to decrease in

CO2.
 Causes of respiratory alkalosis:

Hypoxemia from any causes
2. Respiratory center stimulation
3. Mechanical hyperventilation
4. Sepsis, pain
1.
 Normal pH is 7.4

 Calculate the change in pH (from 7.4)

A.

in acute respiratory disorder (acidosis / alkalosis)
change in pH = 0.008 X [PaCO2 -40]
expected pH = 7.4 +/-change in pH
B. in chronic respiratory disorder (acidosis/alkalosis)
change in pH = 0.003 X [PaCO2 -40 ]
expected pH = 7.4 +/- change in pH
Compare the pH on ABG
if pH on ABG is close to A, it is acute disorder
if pH on ABG is close to B, it is chronic disorder
 M/60 yrs, k/c/o C.O.P.D. admitted with U.T.I.
 ABG: 7.26 / 84 / 74 / 37 / 94%
 [A]. For Acute change in pH;

change in pH = 0.008 X [84 – 40 ] =0.008 X [44] =0.35
Expected pH = 7.4 – 0.35 = 7.05
 [B]. For chronic change in pH’

Change in pH = 0.003 X [84 – 40 ] = 0.003 X [44] = 0.13
Expected pH =7.4 -0.13 = 7.27
So B is near to the patient’s ABG which is 7.26; so primary disorder
is chronic respiratory acidosis.
Primary
disorder

Initial Compens Compensa
chemic a-tory
al
response tory
change
mechanis
s
m

Respiratory
acidosis

Expected level of compensation

↑PCO2 ↑HCO3-

Acute

Bufferingrule of 1

↑[HCO3-] = 1 mEq/L for every
10 mmHg delta PCO2

Chronic

Generation
of new
HCO3 –
rule of 3

[HCO-] = 3 mEq/L for every
10mmHg delta PCO2
Primary
disorder

Initial
chemic
al
change
s

Compen
sa-tory
respons
e

Compensatory
mechanism

Expected level of
compensation

Respiratory
alkalosis

↓PCO2 ↓HCO3-

Acute

buffering- rule
of 2

↓[HCO-] = 2 mEq/L for
every 10 mmHg delta PCO2

chronic

Decreased
reabsorption
of HCO3- rule
or 4

↓[HCO3-] = 4 mEq/L for every
10 mmHg delta PCO2
 Metabolic acidosis results from a primary decrease in

plasma [HCO3-]

 Check the respiratory compensation by winter’s

formula:
Primary
disorder

Initial
chemica
l
changes

Compens Compensaa-tory
tory
response mechanism

Expected level of
compensation

Metabolic
acidosis

↓HCO3-

↓PCO2

Hyperventilation

Metabolic
alkalosis

↑HCO3-

↑PCO2

Hypoventilation PCO2 = 0.7 X [HCO3] +
21+/- 2

PCO2 = 1.5 X [HCO3-]+
8 +/-2
 7.23 / 34 /88 /17 [ metabolic acidosis]

 Winter ‘s formula:
 Expected PaCO2 = 1.5 X [HCO3-] + 8 +/- 2

= 1.5 X [17] + 8 +/ -2
= 33.5 + / -2 = 31.5 – 35.5 mmHg
 7.45 / 43 / 95 / 30 [ metabolic alkalosis]
 Winter ‘s formula:
 Expected PaCO2= 0.7 X [HCO3-] + 21 + / - 2

= 0.7 X [30] + 21 + / - 2
=42 + / -2 = 40 – 44 mmHg
 Metabolic alkalosis reflects an increase in plasma

[HCO3-]

 It can be classified into saline responsive or

nonresponsive.
 More than 20 mEq/L urinary chloride is saline

unresponsive and less than 20 mEq/L is saline
responsive.
Causes of Urine CL > 20 mEq/L

Causes of Urine CL < 20 mEq/L

Mineralocorticoid excess

Vomiting, nasogastric suctioning

K + and Mg ++ deficiency

chloride-wasting diarrhea

Liddle’s syndrome

Villous adenoma of colon

Barter’s syndrome

Posthypercapnia

Hypercalcemia with secondary
hyperparathyroidism

Poorly reabsorbed anions like
carbenicillin

Milk – alkali syndrome

Diuretic therapy
 Calculate AG in case of metabolic acidosis
 High denotes raised AG metabolic acidosis, and

normal or narrow denotes non-AG acidosis.
 It is used to determine if a metabolic acidosis is due to

an accumulation of non- volatile acids [e.g. lactic
acidosis] or a net loss of bicarbonate [e.g. diarrhea]
 Na + UC = [ Cl + HCO3 ] + UA
 UA – UC [ Anion gap] = Na –[ Cl + HCO3- ]
 AG= Na –[Cl + HCO3]; normal AG is 12+/-2 mEq/L
Unmeasured Anions

Unmeasured Cation

Albumin: 15 mEq/L

Calcium: 5 mEq/L

Organic Acids: 5 mEq/L

Potassium: 4.5 mEq/L

Phosphate: 2 mEq/L

Magnesium: 1.5 mEq/L

Sulfate: 1 mEq/L
Total UA: 23 mEq/L

Total UC: 11 mEq/L

Anion AG = UA – UC = 12 mEq/L

Adjusted AG = calculated AG + 2.5 X [4 – S.albumin gm%]
 7.23 / 34 /88 /17 : Metabolic Acidosis
 Na : 135 / Cl: 99 / K: 3.5

 AG = Na - [ Cl + HCO3-] = 135 – [ 99 + 17] = 19
 High AG
Pneumonic

Causes

M

Methanol

U

Uremia

D

Diabetic ketoacidosis

P

Paraldehyde

I

Isoniazid / iron

L

Lactate

E

Ethanol, ethylene glycol

R

Rhabdomyolysis / renal failure

S

Salicylate / sepsis
Pneumonic

Causes

H

Hyper alimentation

A

Acetazolamide

R

Renal tubular acidosis

D

Diarrhea

U

Uremia (acute)

P

Post ventilation hypocapnia
 Check urinary AG in non-AG metabolic acidosis
 U Na + U K – U Cl
 Normal :

negative

 Non-renal loss of bicarbonate [diarrhea] : negative
 Renal loss of bicarbonate[ RTA /

H+ excretion]
: positive
 In less obvious cases, the coexistence of two metabolic

acid-base disorders may be apparent by calculating the
difference between the change in AG [delta AG] and
the change in serum HCO3- [delta HCO3-].
 e.g. Diabetic ketoacidosis

 This is called the Delta gap or gap –gap.
 Delta gap = delta AG – delta HCO3 Where delta AG = patient’s AG – 12 mEq/L
 Delta HCO3- = 24 mEq/L – patient’s HCO3 Normally the delta gap is zero :
 AG acidosis

 A positive delta gap of more than 6 mEq/L :
 metabolic alkalosis and/or HCO3- retention.

 The delta gap of less than 6 mEq/L :
 Hypercholremic acidosis and/or HCO3- excretion.
 7.23 / 34 /88 /17 : Metabolic Acidosis

 Na : 138 / Cl: 99 / K: 3.5
 AG = Na - [ Cl + HCO3-] = 138 – [ 99 + 17] = 22

Next step is to calculate the Delta Gap.
 Delta AG = patient’s AG -12 = 22 – 12 = 10
 Delta HCO3- = 24 – patient’s HCO3- = 24 – 17 = 7
 Delta gap = Delta AG- Delta HCO3- = 10 – 7 = 3
 Additional metabolic alkalosis is also present with high AG

metabolic acidosis.
 Step 1: Acidemic, alkalemic, or normal?
 Step 2: Is the primary disturbance respiratory or metabolic?
 Step 3: For a primary respiratory disturbance, is it acute or chronic?

 Step 4: For a metabolic disturbance, is the respiratory system

compensating OK?

 Step 5: For a metabolic acidosis, is there an increased anion gap?
 Step 6: For an increased anion gap metabolic acidosis, are there other

derangements?
 65 years old male with CKD presenting with nausea,

diarrhea and acute respiratory distress
 ABG: 7.23/17/235/7 with 50% FiO2 on V.M.
 Electrolytes: Na: 123 mEq/L, Cl: 97 mEq/L, S.K 3.5
 Renal function: S. Creat: 5.1 mg%, BUN: 119
 Acidemic ?
 Alkalemic ?
 Normal ?
 Respiratory / metabolic ?
 ABG: 7.23/17/235/7 with 50% FiO2 on V.M.
 pH and PaCO2 goes in same direction; so it is

primarily metabolic disorder.
 ABG: 7.23/17/235/7 with 50% FiO2 on V.M.
 Winter’s formula:
 Expected PaCO2 = 1.5 X [7] + 8 +/- 2 = 18.5 +/-2
 ABG: 7.23/17/235/7 with 50% FiO2 on V.M.
 Electrolytes: Na: 123 mEq/L, Cl: 97 mEq/L, S.K 3.5
 AG = Na – [Cl +HCO3-] = 123 – [97 + 7] = 19
 High AG metabolic acidosis
 Delta gap = Delta AG – Delta HCO3-

= [19 - 12 ] – [24 – 7 ]

=7–7=0
 Non –anion gap metabolic acidosis

 High AG metabolic acidosis with non-anion metabolic

acidosis
 High AG metabolic acidosis is due to BUN 119
 Non anion metabolic acidosis is due to diarrhea
 Very sick 56 year old man being evaluated for a





possible double lung transplant
Dyspnoea on minimal exertion
On home oxygen therapy
(nasal prongs, 2 lpm)
Numerous pulmonary medications
ABG: 7.30/65/88/31.1
 Acidemia ?
 Alkalemia ?
 Normal ?
 pH and PaCO2 goes in opposite direction; so primary

disorder is respiratory.
 For Acute disorder;
 Change in pH = 0.008 X [65 – 40 ]=0.2
 Expected pH = 7.40 – 0.2 = 7.20

 For chronic disorder;
 Change n pH= 0.003 X [65 – 40 ] = 0.07
 Expected pH = 7.40 – 0.07 = 7.33
 So its chronic respiratory acidosis.
 delta PaCO2 = 25
 So rise in HCO3- will be by 7.5 mEq/L, which should

be 24 + 7.5 = 31.5 mEq/L
 Which is expected HCO3-
 Primary chronic respiratory acidosis.
 Patient has a long standing pulmonary disease so

bicarbonate is in the compensation.
 A 44 year old moderately dehydrated man was

admitted with a two day history of acute severe
diarrhea.
 ABG: 7.31 / 33 / 88 /16 / 95%
 Elect: Na: 136 mEq/L, Cl: 103 mEq/L, K: 2.9 mEq/L
 Acidemia ?
 Alkalemia ?
 Normal ?
 pH and PaCO2 goes in the same direction; so primary

disorder is metabolic.
 Winter’s formula
 Expected PaCO2= 1.5 X [16] + 8 +/ -2


= 32 -/+ 2 = 30 – 34 mEq/L

 So it is fully compensated metabolic disorder.
 AG = Na – [Cl + HCO3-] = 134 – [104 + 16 ] = 14
 Non AG metabolic acidosis
 Patient lost bicarbonate in diarrhea leading to non

anion gap metabolic acidosis.
 A 38-year-old woman is 12 weeks pregnant. For the last 10

days she has had worsening nausea and vomiting. When
seen by her physician, she is dehydrated and has shallow
respirations. Arterial blood gas data is as follow
 ABG: 7.56/54/110/45
 Acidemia ?
 Alkalemia ?
 Normal ?
 pH and PaCO2 goes in same direction so the primary

disorder is metabolic alkalosis.
 Expected PaCO2 = 0.7 X [ HCO3-] + 21 +/- 2

= 0.7 X [ 45 ] + 21 +/ - 2
= 52.5 +/-2
= 50.5 – 54.5
 Metabolic alkalosis
 Patient has lost lot of gastric juice and hydrochloric

acid in vomit leading to metabolic alkalosis.
 Step 1: Acidemic, alkalemic, or normal?
 Step 2: Is the primary disturbance respiratory or metabolic?
 Step 3: For a primary respiratory disturbance, is it acute or chronic?

 Step 4: For a metabolic disturbance, is the respiratory system

compensating OK?

 Step 5: For a metabolic acidosis, is there an increased anion gap?
 Step 6: For an increased anion gap metabolic acidosis, are there other

derangements?
THANK YOU ! !!

Abg analysis

  • 1.
    Dr. Anand Bhabhor SeniorIntensivist Jaslok Hospital & Research Centre
  • 2.
     ABG analysisis an essential component of diagnosing and managing critically ill patients in the ICU.  A step wise approach to the evaluation of these disorders helps delineate underlying causes, compensatory mechanism, and the correct approach to management.
  • 4.
    Range For calculation pH 7.35 –7.45 7.4 PaCO2 35 – 45 mmHg 40 mmHg PaO2 > 80 mmHg > 95 mmHg HCO3- 22 – 26 mEq/L 24 mEq/L
  • 5.
     PaO2 andSaO2 PaO2 / FiO2 ratio e.g. PaO2 :84 mmHg with 0.21 FiO2 [room air] PaO2 / FiO2 = 84 / 0.21 = 400  Normal = 300 – 500 mmHg  < 300 = acute lung injury [previous definition]  < 200 = ARDS [previous definition] Berlin definition:  200 – 300 [with PEEP/CPAP > 5] = mild ARDS  < 200 [with PEEP > 5] = moderate ARDS  <100 [with PEEP > 5] = severe ARDS
  • 6.
     Step 1:Acidemic, alkalemic, or normal?  Step 2: Is the primary disturbance respiratory or metabolic?  Step 3: For a primary respiratory disturbance, is it acute or chronic?  Step 4: For a metabolic disturbance, is the respiratory system compensating OK?  Step 5: For a metabolic acidosis, is there an increased anion gap?  Step 6: For an increased anion gap metabolic acidosis, are there other derangements?
  • 7.
     The pHis the –log [H]. So by altering either the PCO2 or the HCO3-, [H] will change, and so will pH.  An acidemia(low pH) can result from either a low HCO3- or a high CO2  An alkalemia (high pH) can result from either a high HCO3 or a low CO2
  • 8.
     If pHand PaCO2 changes in same direction, the primary disorder is metabolic and if they change in opposite direction ,the primary disorder is respiratory. Chemical change Primary disorder Compensation pH Low HCO3- Metabolic acidosis Respiratory alkalosis low pH High HCO3- Metabolic alkalosis Respiratory acidosis High pH High PaCO2 Respiratory acidosis Metabolic alkalosis Low pH Low PaCO2 Respiratory alkalosis Metabolic acidosis High pH
  • 9.
     Respiratory acidosisis due to a primary rise in CO2  Hypercapnia almost always results from alveolar hypoventilation due to one of the following causes: Respiratory center depression 2. Neuromuscular disorder 3. Upper airway obstruction 4. Pulmonary disease 1.
  • 10.
     A respiratoryalkalosis is due to decrease in PaCO2.  It results from hyperventilation leading to decrease in CO2.  Causes of respiratory alkalosis: Hypoxemia from any causes 2. Respiratory center stimulation 3. Mechanical hyperventilation 4. Sepsis, pain 1.
  • 11.
     Normal pHis 7.4  Calculate the change in pH (from 7.4) A. in acute respiratory disorder (acidosis / alkalosis) change in pH = 0.008 X [PaCO2 -40] expected pH = 7.4 +/-change in pH B. in chronic respiratory disorder (acidosis/alkalosis) change in pH = 0.003 X [PaCO2 -40 ] expected pH = 7.4 +/- change in pH Compare the pH on ABG if pH on ABG is close to A, it is acute disorder if pH on ABG is close to B, it is chronic disorder
  • 12.
     M/60 yrs,k/c/o C.O.P.D. admitted with U.T.I.  ABG: 7.26 / 84 / 74 / 37 / 94%  [A]. For Acute change in pH; change in pH = 0.008 X [84 – 40 ] =0.008 X [44] =0.35 Expected pH = 7.4 – 0.35 = 7.05  [B]. For chronic change in pH’ Change in pH = 0.003 X [84 – 40 ] = 0.003 X [44] = 0.13 Expected pH =7.4 -0.13 = 7.27 So B is near to the patient’s ABG which is 7.26; so primary disorder is chronic respiratory acidosis.
  • 13.
    Primary disorder Initial Compens Compensa chemica-tory al response tory change mechanis s m Respiratory acidosis Expected level of compensation ↑PCO2 ↑HCO3- Acute Bufferingrule of 1 ↑[HCO3-] = 1 mEq/L for every 10 mmHg delta PCO2 Chronic Generation of new HCO3 – rule of 3 [HCO-] = 3 mEq/L for every 10mmHg delta PCO2
  • 14.
    Primary disorder Initial chemic al change s Compen sa-tory respons e Compensatory mechanism Expected level of compensation Respiratory alkalosis ↓PCO2↓HCO3- Acute buffering- rule of 2 ↓[HCO-] = 2 mEq/L for every 10 mmHg delta PCO2 chronic Decreased reabsorption of HCO3- rule or 4 ↓[HCO3-] = 4 mEq/L for every 10 mmHg delta PCO2
  • 15.
     Metabolic acidosisresults from a primary decrease in plasma [HCO3-]  Check the respiratory compensation by winter’s formula:
  • 16.
    Primary disorder Initial chemica l changes Compens Compensaa-tory tory response mechanism Expectedlevel of compensation Metabolic acidosis ↓HCO3- ↓PCO2 Hyperventilation Metabolic alkalosis ↑HCO3- ↑PCO2 Hypoventilation PCO2 = 0.7 X [HCO3] + 21+/- 2 PCO2 = 1.5 X [HCO3-]+ 8 +/-2
  • 17.
     7.23 /34 /88 /17 [ metabolic acidosis]  Winter ‘s formula:  Expected PaCO2 = 1.5 X [HCO3-] + 8 +/- 2 = 1.5 X [17] + 8 +/ -2 = 33.5 + / -2 = 31.5 – 35.5 mmHg  7.45 / 43 / 95 / 30 [ metabolic alkalosis]  Winter ‘s formula:  Expected PaCO2= 0.7 X [HCO3-] + 21 + / - 2 = 0.7 X [30] + 21 + / - 2 =42 + / -2 = 40 – 44 mmHg
  • 18.
     Metabolic alkalosisreflects an increase in plasma [HCO3-]  It can be classified into saline responsive or nonresponsive.  More than 20 mEq/L urinary chloride is saline unresponsive and less than 20 mEq/L is saline responsive.
  • 19.
    Causes of UrineCL > 20 mEq/L Causes of Urine CL < 20 mEq/L Mineralocorticoid excess Vomiting, nasogastric suctioning K + and Mg ++ deficiency chloride-wasting diarrhea Liddle’s syndrome Villous adenoma of colon Barter’s syndrome Posthypercapnia Hypercalcemia with secondary hyperparathyroidism Poorly reabsorbed anions like carbenicillin Milk – alkali syndrome Diuretic therapy
  • 20.
     Calculate AGin case of metabolic acidosis  High denotes raised AG metabolic acidosis, and normal or narrow denotes non-AG acidosis.
  • 21.
     It isused to determine if a metabolic acidosis is due to an accumulation of non- volatile acids [e.g. lactic acidosis] or a net loss of bicarbonate [e.g. diarrhea]  Na + UC = [ Cl + HCO3 ] + UA  UA – UC [ Anion gap] = Na –[ Cl + HCO3- ]  AG= Na –[Cl + HCO3]; normal AG is 12+/-2 mEq/L
  • 22.
    Unmeasured Anions Unmeasured Cation Albumin:15 mEq/L Calcium: 5 mEq/L Organic Acids: 5 mEq/L Potassium: 4.5 mEq/L Phosphate: 2 mEq/L Magnesium: 1.5 mEq/L Sulfate: 1 mEq/L Total UA: 23 mEq/L Total UC: 11 mEq/L Anion AG = UA – UC = 12 mEq/L Adjusted AG = calculated AG + 2.5 X [4 – S.albumin gm%]
  • 23.
     7.23 /34 /88 /17 : Metabolic Acidosis  Na : 135 / Cl: 99 / K: 3.5  AG = Na - [ Cl + HCO3-] = 135 – [ 99 + 17] = 19  High AG
  • 24.
    Pneumonic Causes M Methanol U Uremia D Diabetic ketoacidosis P Paraldehyde I Isoniazid /iron L Lactate E Ethanol, ethylene glycol R Rhabdomyolysis / renal failure S Salicylate / sepsis
  • 25.
    Pneumonic Causes H Hyper alimentation A Acetazolamide R Renal tubularacidosis D Diarrhea U Uremia (acute) P Post ventilation hypocapnia
  • 26.
     Check urinaryAG in non-AG metabolic acidosis  U Na + U K – U Cl  Normal : negative  Non-renal loss of bicarbonate [diarrhea] : negative  Renal loss of bicarbonate[ RTA / H+ excretion] : positive
  • 27.
     In lessobvious cases, the coexistence of two metabolic acid-base disorders may be apparent by calculating the difference between the change in AG [delta AG] and the change in serum HCO3- [delta HCO3-].  e.g. Diabetic ketoacidosis  This is called the Delta gap or gap –gap.
  • 28.
     Delta gap= delta AG – delta HCO3 Where delta AG = patient’s AG – 12 mEq/L  Delta HCO3- = 24 mEq/L – patient’s HCO3 Normally the delta gap is zero :  AG acidosis  A positive delta gap of more than 6 mEq/L :  metabolic alkalosis and/or HCO3- retention.  The delta gap of less than 6 mEq/L :  Hypercholremic acidosis and/or HCO3- excretion.
  • 29.
     7.23 /34 /88 /17 : Metabolic Acidosis  Na : 138 / Cl: 99 / K: 3.5  AG = Na - [ Cl + HCO3-] = 138 – [ 99 + 17] = 22 Next step is to calculate the Delta Gap.  Delta AG = patient’s AG -12 = 22 – 12 = 10  Delta HCO3- = 24 – patient’s HCO3- = 24 – 17 = 7  Delta gap = Delta AG- Delta HCO3- = 10 – 7 = 3  Additional metabolic alkalosis is also present with high AG metabolic acidosis.
  • 30.
     Step 1:Acidemic, alkalemic, or normal?  Step 2: Is the primary disturbance respiratory or metabolic?  Step 3: For a primary respiratory disturbance, is it acute or chronic?  Step 4: For a metabolic disturbance, is the respiratory system compensating OK?  Step 5: For a metabolic acidosis, is there an increased anion gap?  Step 6: For an increased anion gap metabolic acidosis, are there other derangements?
  • 31.
     65 yearsold male with CKD presenting with nausea, diarrhea and acute respiratory distress  ABG: 7.23/17/235/7 with 50% FiO2 on V.M.  Electrolytes: Na: 123 mEq/L, Cl: 97 mEq/L, S.K 3.5  Renal function: S. Creat: 5.1 mg%, BUN: 119
  • 32.
     Acidemic ? Alkalemic ?  Normal ?
  • 33.
     Respiratory /metabolic ?  ABG: 7.23/17/235/7 with 50% FiO2 on V.M.  pH and PaCO2 goes in same direction; so it is primarily metabolic disorder.
  • 34.
     ABG: 7.23/17/235/7with 50% FiO2 on V.M.  Winter’s formula:  Expected PaCO2 = 1.5 X [7] + 8 +/- 2 = 18.5 +/-2
  • 35.
     ABG: 7.23/17/235/7with 50% FiO2 on V.M.  Electrolytes: Na: 123 mEq/L, Cl: 97 mEq/L, S.K 3.5  AG = Na – [Cl +HCO3-] = 123 – [97 + 7] = 19  High AG metabolic acidosis  Delta gap = Delta AG – Delta HCO3- = [19 - 12 ] – [24 – 7 ]  =7–7=0  Non –anion gap metabolic acidosis 
  • 36.
     High AGmetabolic acidosis with non-anion metabolic acidosis
  • 37.
     High AGmetabolic acidosis is due to BUN 119  Non anion metabolic acidosis is due to diarrhea
  • 38.
     Very sick56 year old man being evaluated for a     possible double lung transplant Dyspnoea on minimal exertion On home oxygen therapy (nasal prongs, 2 lpm) Numerous pulmonary medications ABG: 7.30/65/88/31.1
  • 39.
     Acidemia ? Alkalemia ?  Normal ?
  • 40.
     pH andPaCO2 goes in opposite direction; so primary disorder is respiratory.
  • 41.
     For Acutedisorder;  Change in pH = 0.008 X [65 – 40 ]=0.2  Expected pH = 7.40 – 0.2 = 7.20  For chronic disorder;  Change n pH= 0.003 X [65 – 40 ] = 0.07  Expected pH = 7.40 – 0.07 = 7.33  So its chronic respiratory acidosis.
  • 42.
     delta PaCO2= 25  So rise in HCO3- will be by 7.5 mEq/L, which should be 24 + 7.5 = 31.5 mEq/L  Which is expected HCO3-
  • 43.
     Primary chronicrespiratory acidosis.
  • 44.
     Patient hasa long standing pulmonary disease so bicarbonate is in the compensation.
  • 45.
     A 44year old moderately dehydrated man was admitted with a two day history of acute severe diarrhea.  ABG: 7.31 / 33 / 88 /16 / 95%  Elect: Na: 136 mEq/L, Cl: 103 mEq/L, K: 2.9 mEq/L
  • 46.
     Acidemia ? Alkalemia ?  Normal ?
  • 47.
     pH andPaCO2 goes in the same direction; so primary disorder is metabolic.
  • 48.
     Winter’s formula Expected PaCO2= 1.5 X [16] + 8 +/ -2  = 32 -/+ 2 = 30 – 34 mEq/L  So it is fully compensated metabolic disorder.
  • 49.
     AG =Na – [Cl + HCO3-] = 134 – [104 + 16 ] = 14
  • 50.
     Non AGmetabolic acidosis
  • 51.
     Patient lostbicarbonate in diarrhea leading to non anion gap metabolic acidosis.
  • 52.
     A 38-year-oldwoman is 12 weeks pregnant. For the last 10 days she has had worsening nausea and vomiting. When seen by her physician, she is dehydrated and has shallow respirations. Arterial blood gas data is as follow  ABG: 7.56/54/110/45
  • 53.
     Acidemia ? Alkalemia ?  Normal ?
  • 54.
     pH andPaCO2 goes in same direction so the primary disorder is metabolic alkalosis.
  • 55.
     Expected PaCO2= 0.7 X [ HCO3-] + 21 +/- 2 = 0.7 X [ 45 ] + 21 +/ - 2 = 52.5 +/-2 = 50.5 – 54.5
  • 56.
  • 57.
     Patient haslost lot of gastric juice and hydrochloric acid in vomit leading to metabolic alkalosis.
  • 58.
     Step 1:Acidemic, alkalemic, or normal?  Step 2: Is the primary disturbance respiratory or metabolic?  Step 3: For a primary respiratory disturbance, is it acute or chronic?  Step 4: For a metabolic disturbance, is the respiratory system compensating OK?  Step 5: For a metabolic acidosis, is there an increased anion gap?  Step 6: For an increased anion gap metabolic acidosis, are there other derangements?
  • 60.