Interpretation of Arterial Blood Gases

HA H+ + A- v 1 = k 1 [HA] v 2 = k 2 [H+] [A-] at equilibrium: k 1 [HA] = k 2 [H+] [A-] If k 1 /k 2 = ka then ka [HA] = [H+] [A-] Solving for [H+]: [H+] = ka [ HA ] [ A- ] Taking the -logarithm of both sides: -log [H+] = - log ka - log [HA]/[A-] if: pH = - log [H+], pka = - log ka and + log [A-] / [HA] for - log [HA] / [A-] then: pH = pka + log [ A- ] [ HA ] THE LAW OF MASS ACTION AND THE HENDERSON- HASSELBACH EQUATION

THE HENDERSON- HASSELBACH EQUATION

BICARBONATE/CARBON DIOXIDE BUFFER SYSTEM Carbonic acid can dissociate into H+ and HCO 3 - CO 2 CO 2 + H 2 O H 2 CO 3 H+ + HCO 3 - gas aqueous phase All gases partially dissolve in water. The degree to which this occurs is proportional to the partial pressure of the gas in solution. In humans, the partial pressure of CO 2 = 40 mmHg and [CO 2 ] dis = P CO 2 = 0.03 x 40 = 1.2 mmol/L where 0.03 is the solubility constant for CO 2 in the plasma

The equation can be simplified to: [CO 2 ] dis + H2O H+ + HCO 3 - The law of mass action for this reaction is: Ka = [ H+ ] [HCO 3 - ] [CO 2 ] dis [ H 2 O ] Since the concentration of water is constant, ( Ka x [ H 2 O ]) can be replaced by: K’a = [ H+ ] [HCO 3 - ] [CO 2 ] dis [ H+ ] = K’a [ CO 2 ] or pH = pK’a + log [ CO 2 ] [ HCO 3 - ] [ HCO 3 - ]

In plasma at 37 o C, K’a = 800 nanomol/L and [ CO 2 ] dis = 0.03 x P CO 2 Thus: [ H+ ] = 24 P CO 2 [HCO 3 - ] Since the normal H+ concentration is 40 nanomol/L and the P CO 2 is 40 mmHg, the normal [HCO 3 - ] can be calculated 40 = 24 x. 40. [HCO 3 - ] [ HCO 3 -] = 24 mmol/L

RELATIONSHIP OF pH TO H+ CONCENTRATION pH H (nM) AcidificationAlkalinization (Acidemia) (Alkalemia) H+ doubles for eachH+ halves for each 0.3 unit fall in pH0.3 unit rise in pH

Approach to the Interpretation of the Arterial Blood Gas 1.Get a good history and physical examination. 2.Get your ABG. 3.Determine the serum electrolytes at the same time. 4.Know the primary disorder. 5.Compute for the range of compensation. 6.For METABOLIC ACIDOSIS: GET THE ANION GAP 1. For high anion gap metabolic acidosis: Compare the Δ AG and compare it with Δ HCO3- Interpretation: Δ AG = Δ HCO3- : Pure high AG metabolic acidosis Δ AG > Δ HCO3- : High AG metabolic acidosis with metabolic alkalosis Δ AG < Δ HCO3- : High AG metabolic acidosis with normal AG metabolic acidosis

Approach to the Interpretation of the Arterial Blood Gas… 2.For normal anion gap metabolic acidosis: Get Δ Cl- and compare it with Δ HCO3- Interpretation: Δ Cl- = Δ HCO3- : Pure normal AG metabolic acidosis Δ Cl- > Δ HCO3- : Normal AG metabolic acidosis with metabolic alkalosis Δ Cl- < Δ HCO3- : Normal AG metabolic acidosis with high AG metabolic acidosis

K+ in METABOLIC ACIDOSIS K+ H+ Organic anions SO 4 - PO 4 = K+ OrganicInorganicacidosis

CHARACTERISTICS OF PRIMARY ACID-BASE DISTURBANCES PrimaryCompensatory DisorderpH [H]+disturbanceresponse Metabolic acidosis [HCO 3 -] PCO 2 Metabolic alkalosis [HCO 3 -] PCO 2 Respiratory acidosis PCO 2 [HCO 3 -] Respiratory alkalosis PCO 2 [HCO 3 -]

COMPENSATORY RESPONSES IN SIMPLE ACID-BASE DISORDERS PrimarySecondary DisorderAbnormalityResponse Respiratory AcidosisHypoventilationHCO 3 - generation AlkalosisHyperventilationHCO 3 - consumption Metabolic AcidosisLoss of HCO 3 -Increase in or gain of H+ventilation AlkalosisGain of HCO 3 -Decrease in or Loss of H+ventilation

RENAL AND RESPIRATORY COMPENSATIONS TO PRIMARY ACID-BASE DISTURBANCES IN HUMANS DisorderPrimaryCompensatory Response Metabolic [HCO 3 -]1.2 mmHg decrease in PCO 2 Acidosisfor every 1 meq/L fall in [HCO 3 -] Metabolic [HCO 3 -]0.7 mmHg elevation in PCO 2 for Alkalosisevery 1 meq/L rise in [HCO 3 -] Respiratory PCO 2 1 meq/L increase in [HCO 3 -] for Acidosis (acute)every 10 mmHg rise in PCO 2 Respiratory PCO 2 2 meq /L reduction in [HCO 3 -] for Alkalosis (acute)every 10 mmHg fall in PCO 2

Arterial Measurements in Hypothetical Acid-Base Disorder [HCO3-]PCO2pH Acid base status meq/L mmHg Normal Pure metabolic acidosis Primary metabolic and secondary respiratory acidosis Combined metabolic acidosis and respiratory alkalosis

Approach to the Interpretation of the Arterial Blood Gas 1.Get a good history and physical examination. 2.Get your ABG. 3.Determine the serum electrolytes at the same time. 4.Know the primary disorder. 5.Compute for the range of compensation. 6.For METABOLIC ACIDOSIS: GET THE ANION GAP 1. For high anion gap metabolic acidosis: Compare the Δ AG and compare it with Δ HCO3- Interpretation: Δ AG = Δ HCO3- : Pure high AG metabolic acidosis Δ AG > Δ HCO3- : High AG metabolic acidosis with metabolic alkalosis Δ AG < Δ HCO3- : High AG metabolic acidosis with normal AG metabolic acidosis

Approach to the Interpretation of the Arterial Blood Gas… 2.For normal anion gap metabolic acidosis: Get Δ Cl- and compare it with Δ HCO3- Interpretation: Δ Cl- = Δ HCO3- : Pure normal AG metabolic acidosis Δ Cl- > Δ HCO3- : Normal AG metabolic acidosis with metabolic alkalosis Δ Cl- < Δ HCO3- : Normal AG metabolic acidosis with high AG metabolic acidosis

HCO 3 - Cl- Serum Anion Gap Na+ K+ Ca++ Mg++ Pr- SO 4 = PO 4 = The total number of the cations and the anions in the ECF should be equal

HCO 3 - Cl- Serum Anion Gap Na+ Anion gap : The difference between the concentrations of measured cations and measured anions

HCO 3 - Cl- Serum Anion Gap Na+ Anion gap = Na+ - ( Cl- + HCO 3 -) Normal anion gap = 12 ± - 2

A.Normal ion distribution B.Metabolic acidosis; high anion gap C.Metabolic acidosis; normal anion gap

METABOLIC ACIDOSIS AND THE ANION GAP H+ X-Na+ HCO 3 - Na+ X- H 2 CO 2 H 2 O + CO 2 Normal High Normal AG Acidosis AG Acidosis Na Cl HCO AG Δ HCO Δ AG Lact Δ Lact 0+10

ANION GAP IN MAJOR CAUSES OF METABOLIC ACIDOSIS High Anion Gap A. Lactic acidosis: Lactate B. Ketoacidosis: B-hydroxybutyric acid C. Renal failure: Sulfate, phosphate, urate D.Ingestions 1.Salicylate: ketones, lactate, salicylate 2.Methanol or formaldehyde 3.Ethylene glycol: glycolate, oxalate Normal Anion Gap A.Gastrointestinal loss of HCO3- 1. Diarrhea B.Renal HCO3- loss 1. Type I and Type II Renal Tubular Acidosis C.Ingestion: 1. Ammonium Chloride

Urine anion gap in Normal AG metabolic acidosis Urine AG = ([Na+] + [K+]) – [Cl-] N= near zero to positive value NH 4 + is the major unmeasured urinary cation. In metabolic acidosis, the excretion of NH4+ (and of Cl- to maintain electroneutrality) should increase markedly, if renal acidification is intact, resulting in a value that varies from -20 to more than -50

Problem: 1 A 81 year old, diabetic for 15 years, and hypertension for 10 years. Claims to have good control of both medical problems. Initial laboratory data: Na+ = 133 meq/L pH = 7.32 K+ = 5.6 meq/L pCO 2 = 27 mm Hg Cl- = 100 meq/L HCO 3 - = 14.4 meq/L 1. What is the primary disorder? Metabolic Acidosis 2. What is the compensatory response? 24 – 14.4 = 9.6 X 1.2 = = Compensated metabolic acidosis

Prob. 1… Na+ = 133 meq/L pH = 7.32 K+ = 5.6 meq/L pCO 2 = 27 mm Hg Cl- = 100 meq/L HCO 3 - = 14.4 meq/L 3. What is the anion gap?AG =133 – ( ) = Get your Δ AG __?____ Δ HCO _______ ___<___ Final interpretation? High anion gap metabolic acidosis, compensated, with Normal anion gap metabolic acidosis

Problem:2 A 45 year-old woman with peptic ulcer disease complains of persistent vomiting. PE showed a BP of 100/60 mmHg, poor skin turgor, and flat neck veins. Initial laboratory data were: [ Na+] = 140 meq/LpH = 7.53 [ K+ ] = 2.2 meq/LpCO2 = 53 mmHg [ Cl-] = 86 meq/L [ HCO3-] = 42 meq/L 1. What are your diagnostic clues? 2. What is your primary disorder?Metabolic alkalosis 3. What is your compensatory response? 42 – 24 = 18 x.7 = 12.6 ; = What is your final diagnosis? Compensated metabolic alkalosis

Pathophysiology of Metabolic Alkalosis 1.Why do patients become alkalotic? 2.Why do they remain alkalotic, since renal excretion of the excess HCO3- should rapidly restore normal acid-base balance?

Check serum and urine pH 1.Serum and urine pH both increased: loss of acid through the GIT, gain of alkali, exogenous 2.Urine pH low: loss of acid through the kidneys (mineralocorticoid excess)

Causes of Impaired HCO3- excretion that allow metabolic alkalosis to persist Decreased GFR - Effective circulating volume depletion - Renal failure ( usually associated with metabolic acidosis) Increased tubular reabsorption - Effective circulating volume depletion - Chloride depletion - Hypokalemia - Hyperaldosteronism

Problem: 3 A 27 year-old male with insulin-dependent diabetes mellitus has not been taking his insulin and is admitted to the hospital in a semicomatose condition. The following laboratory data are obtained: [ Na+] = 140 meq/LpH = 7.10 [ K+ ] = 7.0 meq/LpCO2 = 20 mmHg [ Cl-] = 105 meq/L [ HCO3-] = 6 meq/L Glucose = 800 mg/L 1. What are your diagnostic clues? 2. What is your primary disorder? Metabolic Acidosis 3. What is the compensatory response? Fall in bicarbonate: 24 – 6 = 18; Expected decrease in PCO2: 18 X 1.2 = 21.6; Expected PCO2 level: 40 – 21.6 = 18.4

[ Na+] = 140 meq/LpH = 7.1 [ K+ ] = 7.0 meq/LpCO2 = 20 mmHg [ Cl-] = 105 meq/L [ HCO3-] = 6 meq/L Glucose = 800 mg/L 4. What is the anion gap? 140 – ( ) = 29 High Anion gap metabolic acidodsis 5. Get your Δ AG __?____ Δ HCO _______ _______ 18 6.What is your final diagnosis? High anion gap metabolic acidosis, compensated Problem 3

Problem 4 A 17 year old male has a history of on and off weakness since 2 years ago. A few hrs before admission, he can’t move his lower extremities [ Na+] = 154 meq/LpH = 7.36 [ K+ ] = 2.3 meq/LpCO2 = 37.5 mmHg [ Cl-] = 115 meq/L [ HCO3-] = 20.9 meq/L Base excess = What are your diagnostic clues? 2. What is your primary disorder?Metabolic Acidosis 3. What is your compensatory response? Decrease in bicarbonate = 24 – 20.9 = 3.1 Expected fall in PCO2 = 3.1 X 1.2 = 3.72 Expected pCO2 level = 40 – 3.72 = 36.27

[ Na+] = 154 meq/LpH = 7.36 [ K+ ] = 2.3 meq/LpCO2 = 37.5 mmHg [ Cl-] = 115 meq/L [ HCO3-] = 20.9 meq/L Base excess = What is the anion gap? ( ) = 18.1High anion gap metabolic acidosis 5. Get your Δ AG ___?____ Δ HCO ___?_____ 24 – ____>____ 3.1 What is your final diagnosis? High anion gap metabolic acidosis with metabolic alkalosis, compensated Problem 4

Base excess: the easiest and most accurate way of determining whether a metabolic disorder is present. Normal : to +2.5 Interpretation: 1.BE > 2.5, the diagnosis is metabolic alkalosis from excess of base or loss of H+ 2. BE < -2.5, the diagnosis is metabolic acidosis from deficit of base or excess of H+ 3. If the number is within normal limits, then the diagnosis is metabolic balance

Look at the pH: 1.If the pH value is in the same direction as the metabolic diagnosis, then it is the primary and the respiratory changes secondary 2. If the pH value is in the opposite direction as the metabolic diagnosis, then it is the compensatory change and the respiratory component is the primary one BE pH

Problem 5 A 54 year old lawyer complained of leg pains and eventually underwent femoral popliteal bypass surgery. Preop renal function and acid-base status were normal. 24 hrs later, he was noted to be oliguric, cold and clammy and drowsy. PE revealed a BP of 90/60 mmHg, HR = 126/min and absent pulses in his R foot. Laboratory results show: [ Na+] = 140 meq/LpH = 7.0 [ K+ ] = 6.4 meq/LpCO2 = 32 mmHg [ Cl-] = 103 meq/L [ HCO3-] = 8.0 meq/L 1. What are your diagnostic clues? 2. What is your primary disorder? Metabolic acidosis 3. What is your compensatory response? Fall in bicarbonate = 24 – 8 = 16; Expected change in PCO2 = 16 X 1.2 = 19.2 Expected PCO2 level = 40 – 19.2 = 20.8 Respiratory acidosis

[ Na+] = 140 meq/LpH = 7.0 [ K+ ] = 6.4 meq/LpCO2 = 32 mmHg [ Cl-] = 103 meq/L [ HCO3-] = 8.0 meq/L 4. What is the anion gap? 140 – ( 8 + !03) = 29 High anion gap 5. Get your Δ AG ___?____ Δ HCO ___?____ _______ What is your final interpretation? High anion gap metabolic acidosis with respiratory acidosis Problem 5

Problem 6 This is a 50 year old male who underwent nephrolithotomy for R staghorm calculus. There was no ff-up after discharge. A few days prior to admission, he complained of weakness, nausea and vomiting. [ Na+] = 143 meq/LpH = 7.25 [ K+ ] = 6.2 meq/LpCO2 = 24.5 mmHg [ Cl-] = 102 meq/L [ HCO3-] = 10.8 meq/L 1. What are your clinical clues? 2. What is the primary disorder?Metabolic acidosis 3. Compensatory response? Fall in bicarbonate = 24 – 10.8 = 13.2 Expected PCO2 fall = 13.2 X 1.2 = Expected PCO2 level = 40 – = Compensated

[ Na+] = 143 meq/LpH = 7.25 [ K+ ] = 6.2 meq/LpCO2 = 24.5 mmHg [ Cl-] = 102 meq/L [ HCO3-] = 10.8 meq/L 4.Anion Gap 143 – ( ) = 30.2 High anion gap 5.Get your Δ AG ___?____ Δ HCO3 – ____?___ ___>_____ Interpretation? High anion gap metabolic acidosis, with metabolic alkalosis Problem 6

Problem 7 A 74 year old male was admitted for pneumonia. He presented with the ff. labs. [ Na+] = 144 meq/LpH = 7.36 [ K+ ] = 4.2 meq/LpCO2 = 54.0 mmHg [ Cl-] = 105 meq/L [ HCO3-] = 30.6 meq/L 1. Clinical clues? 2. Primary disorderRespiratory Acidosis 3. Compensatory response? Increase in PCO2 = 54 – 40 = 14 Expected increase in bicarbonate: (14/10) X 1 = 1.4 Expected bicarbonate value = = Final answer? Acute respiratory acidosis with metabolic alkalosis

Problem 8 A 25 year old patient with epilepsy suffered a grand mal seizure. Immediately after the seizure, the ff labs were obtained: [ Na+] = 140 meq/LpH = 7.14 [ K+ ] = 4.2 meq/LpCO2 = 45.0 mmHg [ Cl-] = 98 meq/L [ HCO3-] = 14 meq/L 1. What are your diagnostic clues? 2. What is the primary disorder? a. Respiratory acidosis? 45 – 40 = 5 /40 =.125 b. Metabolic acidosis? 24 – 14 = 10 / 24 = Compensatory response? 4. Anion gap 140 – ( ) = 28

[ Na+] = 140 meq/LpH = 7.14 [ K+ ] = 4.2 meq/LpCO2 = 45.0 mmHg [ Cl-] = 98 meq/L [ HCO3-] = 14 meq/L 5.Get your Δ AG ___?____ Δ HCO3 – ____?___ ___>_____ Final answer? High anion gap metabolic acidosis, with metabolic alkalosis and respiratory acidosis Problem 8

Problem Application A patient with salicylate poisoning has the following ABG values: pH = 7.32PCO2 = 30 mmHgHCO3- = 15 meq/L What is your interpretation? Metabolic Acidosis Compensatory response? 24 – 15 = 9 x 1.2 =10.8; Diagnosis: Compensated metabolic acidosis

An important aspect of salicylate poisoning is to alkalinize the blood which will decrease the concentration of salicylate In the tissues.If the aim is to raise the pH to 7.45, to what level does the plasma HCO3- concentration have to be increased to achieve this goal. (Assume that PCO2 remains constant) pH = 7.32PCO2 = 30 mmHgHCO3- = 15 meq/L [H+] at pH = 7.40 = 40 nmol/L [H+] at pH = 7.70 = 20 nmol/L [H+] at pH = 7.45 = ? 36.6 [ H+ ] = 24 x P CO 2 [HCO 3 - ] [ 36.6 ] = 24 x 30 [HCO 3 - ] [HCO3-] = 19.6 mmol/L

How much bicarbonate are you going to give your patient? Actual = 15 meq/L Desired = 19.6 meq/L Formula: Bicarbonate needed = (Desired – Actual) x 60% BW = (4.6) 42 ( Let’s say that pt weighs 70 kg) = 193.2

Thank you!!!