The Simple Acid/Base Disorders Dr. Dave Johnson Associate Professor Dept. Physiology UNECOM.
Quantitative Acid-Base Chemistry & Acid-Base Physiology (They are not the same)
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Transcript of Quantitative Acid-Base Chemistry & Acid-Base Physiology (They are not the same)
Quantitative Acid-Base Chemistry & Acid-Base Physiology
(They are not the same)
P Wilkes MD, PhD, FRCPP Wilkes MD, PhD, FRCP
University of Ottawa Heart InstituteUniversity of Ottawa Heart Institute
Guiding Principles
Physiology will not violate the laws of Physiology will not violate the laws of chemistry and physicschemistry and physics
There is a difference between cause and There is a difference between cause and effect vs. correlation (physics vs. math)effect vs. correlation (physics vs. math)
In many ways – you do not know anything In many ways – you do not know anything unless you know everythingunless you know everything
(If the devil is in the details – so is salvation)(If the devil is in the details – so is salvation)
Learning Objectives
The ChemistryThe Chemistry The PhysiologyThe Physiology The Clinical ApproachThe Clinical Approach
The ‘Chemistry’ in the Text BooksThe ‘Chemistry’ in the Text Books‘The CO‘The CO22 Hydration Reaction’ Hydration Reaction’
Chemical Description:Chemical Description:
COCO22 + H + H220 H0 H22COCO3 3 HCO HCO33-- + H + H++
Mathematical Description, the Mathematical Description, the Henderson-Hasselbalch equation:Henderson-Hasselbalch equation:
pH = pH = pK + log [HCOpK + log [HCO33--]]
((..PCOPCO22))
7.0 7.4 7.6 7.0 7.4 7.6
3030
2424
1414
PCOPCO22=40=40[HCO[HCO33
--]]PCOPCO22=60=60
PCOPCO22=30=30
Resp acidosisResp acidosis
Resp alkalosisResp alkalosis
Metab acidosisMetab acidosis
Metab alkalosisMetab alkalosis
7.0 7.4 7.6 7.0 7.4 7.6
3030
2424
1414
PCOPCO22=40=40
[HCO[HCO33--]]
PCOPCO22=60=60
PCOPCO22=30=30HgHg
pHpH
7.0 7.4 7.6 7.0 7.4 7.6
3030
2424
1414
PCOPCO22=40=40
[HCO[HCO33--]]
BaseBase ExcessExcess
MeasuredMeasured DecreaseDecrease in [HCOin [HCO33
--]]
BE = (1-0.014Hgb)(HCO3-24)+(1.43Hgb +7.7)(pH-7.4)BE = (1-0.014Hgb)(HCO3-24)+(1.43Hgb +7.7)(pH-7.4)
pHpH
PCOPCO22=30=30
COCO22 Hydration Reaction Hydration Reaction What is the pH of water equilibrated to PCOWhat is the pH of water equilibrated to PCO22 = =
40 mmHg?40 mmHg? ? ? 7.47.4 ? ? HigherHigher ? ? LowerLower
AnswerAnswer 4.44.4 What is the HCOWhat is the HCO33??
? 24 mEq/L? 24 mEq/L? Higher? Higher? Lower? Lower
Answer < 1 mEq/LAnswer < 1 mEq/L
What is Going On?What is Going On?
The Problem:The Problem:The COThe CO22 hydration reaction does hydration reaction does
not predict what we measure in blood!!not predict what we measure in blood!!Nevertheless, the Henderson-Nevertheless, the Henderson-
Hasselbalch equation does??Hasselbalch equation does?? The Answer:The Answer:
The COThe CO22 hydration reaction is hydration reaction is
either wrong or incompleteeither wrong or incomplete
Modern Quantitative Acid – Base Chemistry(The Real Chemistry)
Peter A Stewart
Can J Physio Pharmacol 61: 1444 - 1461, 1983
Physical Chemistry of COPhysical Chemistry of CO22 and Water and Water Condition: Condition: PCOPCO22 = = 440 mmHg0 mmHg
Equations:Equations: [H[H++][OH][OH--] = K’w] = K’w
COCO22(K(K11) = [HCO) = [HCO33--][H][H++]]
[HCO[HCO33--](K](K22) = [CO) = [CO33
==][H][H++]]
[H[H++] - [HCO] - [HCO33--] - [CO] - [CO33
==] - [OH] - [OH--] = 0] = 0
SolvingSolving: : [H[H++] = ] = 0.00.036 mEq/L, pH = 4.44 !!36 mEq/L, pH = 4.44 !![HCO[HCO33
--] = 0.036 mEq/L] = 0.036 mEq/L
[CO[CO33==] and [OH] and [OH--] < 1 x 10] < 1 x 10-6-6 mEq/L mEq/L
[H[H++], [HCO], [HCO33--], [CO], [CO33
==], [OH], [OH--] are determined ONLY by PCO] are determined ONLY by PCO22 & &
Dissociation ConstantsDissociation Constants
At pH 7.4At pH 7.4
(i.e. [H(i.e. [H++] = 4x10] = 4x10-5-5 mEq/L) mEq/L)
HCOHCO33 = 24 mEq/L = 24 mEq/L
How do you accountHow do you account
for electrical neutrality?for electrical neutrality?
NaOH = 5
0
10
20
30
40
0 10 20 30 40 50 60
PCO2 (mmHg)
HC
O3 m
mol
/LHow to Make Bicarb ?
NaOH = 20
NaOH = 40
Necessary ConditionNecessary Condition
[fixed + charge][fixed + charge]
&&
PCOPCO22
The Fixed Positive Charge and Other Acids
Plasma [NaPlasma [Na++] + [K] + [K++] – [Cl] – [Cl--]] 140 mEq/L + 4 mEq/L – 100 mEq/L 140 mEq/L + 4 mEq/L – 100 mEq/L
~ 44 mEq/L positive charge~ 44 mEq/L positive charge This is the Strong Ion Difference (SID)This is the Strong Ion Difference (SID) Other Acids ?Other Acids ?
AlbuminAlbumin PhosphatePhosphate Albumin + Phosphate = [Atot]Albumin + Phosphate = [Atot] ~ Total Protein (g/L) x 0.25 = Atot (mEq/L)~ Total Protein (g/L) x 0.25 = Atot (mEq/L)
The Chemically Complete The Chemically Complete SolutionSolution
Multiple componentsMultiple components
COCO22 hydration reaction hydration reaction
Weak Acids (Blood is thicker than water) Weak Acids (Blood is thicker than water)
Electrical NeutralityElectrical Neutrality Systems Approach:Systems Approach:
Solve All Equations SimultaneouslySolve All Equations SimultaneouslyIndependent vs. Dependent VariablesIndependent vs. Dependent Variables
Water dissociation equilibrium
[H+] x [OH-] = K’wWeak acid dissociation equilibrium
[H+] x [A-] = KA x [HA]Conservation of mass for “A”
[HA] + [A-] = [ATOT]Bicarbonate ion formation equilibrium
[H+] x [HCO3-] = KC x pCO2
Carbonate ion formation equilibrium
[H+] x [CO32-] = K3 x [HCO3
-] Electrical neutrality
[SID] + [H+] - [HCO3-] - [A-] - [CO3
2-] - [OH-] = 0SID – HCOSID – HCO33 – [A – [A--] ~0] ~0
Na + K – Cl – HCONa + K – Cl – HCO33 ~ [A ~ [A--] (Is this familiar?)] (Is this familiar?)
[A[A--] is the anion gap] is the anion gap
• Only the whole set of six equations can explain the quantitative properties of the system, and permit evaluation of any of the dependent variables
• No one of the six equations is more, or less, important than the others. All six are essential
[H+]4 + {KA+[SID]} [H+]3 + {KA([SID]-[Atot])-(K1
.CO2+K’w)} [H+]2 - {KA(K1
.CO2+K’w)+K2.K1
.CO2} [H+] - KA
.K2.K1
.CO2 = 0
A[X]4 + B[X]3 + C[X]2 + D[X] + E = 0
Solve for [H+] and get a big wet kiss:
Please put away all calculators, you have 30 seconds beginning now
A Fourth Order Polynomial !?!
DUOH !?!Now what do I do??
I know – Use a computer ! ?Insert: SID = Na+K-Cl = 140 + 4 -100
= 44 mEq/LPCO2 = 40 mmHgTprot = 80 g/L = 20 mEq/L
[H+]4 + {KA+ [44] } [H+]3 + {KA( [44] - [20] )-(K1.40+K’w)} [H+]2 - {KA(K1.40+K’w)+K2
.K1.40} [H+] - KA
.K2.K1
.40 = 0
[H[H++] = 3.98 x 10] = 3.98 x 10–8–8
-Log 3.98 x 10-Log 3.98 x 10-8-8 = 7.4 = 7.4
Water dissociation equilibrium
[H+] x [OH-] = K’wWeak acid dissociation equilibrium
[H+] x [A-] = KA x [HA]Conservation of mass for “A”
[HA] + [A-] = [ATOT] (20 mEq/L)Bicarbonate ion formation equilibrium
[H+] x [HCO3-] = KC x pCO2 (40 mmHg)
Carbonate ion formation equilibrium
[H+] x [CO32-] = K3 x [HCO3
-] Electrical neutrality
[SID] (44 mEq/L) + [H+] - [HCO3-] - [A-] - [CO3
2-] - [OH-] = 0[H[H++] is determined by SID, PCO] is determined by SID, PCO22 and Atot and Atot
[H[H++] = 3.98x10] = 3.98x10-8-8 Eq/L (pH=7.4) in all 6 equations only if Eq/L (pH=7.4) in all 6 equations only if
SID=44 mEq/L, PCOSID=44 mEq/L, PCO22=40mmHg and Atot=20 mEq/L=40mmHg and Atot=20 mEq/L
Therefore [HCOTherefore [HCO33] must be ~ 24 mM/L] must be ~ 24 mM/L
(math not chemistry)(math not chemistry)
Why Does the Henderson-Hasselbalch EquationWhy Does the Henderson-Hasselbalch Equation
Seem to Work?Seem to Work?
Paradigm Shift[H[H++] = 3.98 x 10] = 3.98 x 10-8-8 Eq/L (i.e. pH=7.4) Eq/L (i.e. pH=7.4) ANDAND HCO HCO33 = 24 mmol/L because: = 24 mmol/L because:
SID = 44 mEq/LSID = 44 mEq/LPCOPCO22 = 40 mmHg = 40 mmHgTprot = 80 g/L (~20 mEq/L)Tprot = 80 g/L (~20 mEq/L)
1) [H1) [H++] and [HCO] and [HCO33] DO NOT INFLUENCE EACH OTHER] DO NOT INFLUENCE EACH OTHER
2) The CO2) The CO22 hydration reaction: hydration reaction:
a) incomplete chemical description of the systema) incomplete chemical description of the system
b) predicts acid-base but does not offer a cause/effect explanationb) predicts acid-base but does not offer a cause/effect explanation
3) You must understand normal acid-base before you can understand 3) You must understand normal acid-base before you can understand abnormal acid-baseabnormal acid-base
Does the Equation Work?Does the Equation Work?
y = 1.1101x - 0.8447
R2 = 0.8092
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.2 7.3 7.4 7.5 7.6 7.7
pH Measured
pH C
alcu
late
d
(Wilkes AJP:1998)(Wilkes AJP:1998)
Independent vs. DependentIndependent vs. Dependent
DependentDependent
H, HCOH, HCO33, CO, CO33, Prot, ProtH, OH , Prot, ProtH, OH
(not functions of dIn, dOut or Vd)(not functions of dIn, dOut or Vd)
COCO22 NaNa++, Cl, Cl--
ProteinProtein
IndependentIndependent
PCOPCO22 = = ƒ(VCOƒ(VCO22, VA), VA)
SID = ƒ(dIn/dOut, Vd), Na, ClSID = ƒ(dIn/dOut, Vd), Na, ClAtot = ƒ(dIn, dOut, Vd)Atot = ƒ(dIn, dOut, Vd)
In VitroIn Vitro In VivoIn Vivo
SummarySummary
PCOPCO22
[Tprot][Tprot]
[SID][SID][H[H++] = ] = ƒƒ([SID], PCO([SID], PCO22, [Tprot]), [Tprot])
A(X)A(X)4 4 + B(X)+ B(X)3 3 + C(X)+ C(X)22 + D(X) + E = 0 + D(X) + E = 0
[H[H++]]
[OH[OH--]]
[HCO[HCO33--]]
[CO[CO33==]]
[Prot.H][Prot.H]
[Prot[Prot--]]
Cause and EffectCause and Effect
Conceptual Consequences The model predicts what we measureThe model predicts what we measure HH++ and HCO and HCO33 are both determined by are both determined by
chemistry chemistry IN PLASMAIN PLASMA, NOT physiology, NOT physiology HH++ and HCO and HCO33 are not determined by intake are not determined by intake
or loss – the kidney is irrelevant!!or loss – the kidney is irrelevant!! What does physiology control?What does physiology control?
1) The concentrations of Na, K, Cl 1) The concentrations of Na, K, Cl
and thus the strong ion differenceand thus the strong ion difference
2) PCO2) PCO22
3) Total weak acid concentration3) Total weak acid concentration
Influence of [SID]Influence of [SID]
[Na[Na++] = 137 mEq/L] = 137 mEq/L [Na[Na++] = 143 mEq/L ] = 143 mEq/L
[K[K++] = 4 mEq/L] = 4 mEq/L [K[K++] = 4 mEq/L ] = 4 mEq/L
[Cl[Cl--] = 105 mEq/L] = 105 mEq/L [Cl[Cl--] = 96 mEq/L ] = 96 mEq/L
[SID] = 36 mEq/L[SID] = 36 mEq/L [SID] = 51 mEq/L[SID] = 51 mEq/L
PCOPCO22 = 40 mmHg = 40 mmHg PCOPCO22 = 40 mmHg = 40 mmHg
[Tprot] = 80 g/L[Tprot] = 80 g/L [Tprot] = 80 g/L[Tprot] = 80 g/L
pH = 7.29pH = 7.29 pH = 7.51pH = 7.51
[HCO[HCO33--] = 20 mEq/L] = 20 mEq/L [HCO[HCO33
--] = 34 mEq/L] = 34 mEq/L
[Prot[Prot--] = 16 mEq/L] = 16 mEq/L [Prot[Prot--] = 17 mEq/L] = 17 mEq/L
‘‘Electrolyte’ AcidosisElectrolyte’ Acidosis
Usually iatrogenicUsually iatrogenic Assumed to be ‘benign’Assumed to be ‘benign’ However:However:
Gut hypoperfusion/Post Op N/VGut hypoperfusion/Post Op N/V(Gan, Anesth.1999; Williams, Anesth Analg. 1999)(Gan, Anesth.1999; Williams, Anesth Analg. 1999)
Impaired Renal Blood Flow, GFRImpaired Renal Blood Flow, GFR(Wilcox, J Clin Invest, 1983)(Wilcox, J Clin Invest, 1983)
Difficulty diagnosing Lactacidosis?Difficulty diagnosing Lactacidosis?
First Patient 74 yo male, 48 hrs post MI, on BIPAP74 yo male, 48 hrs post MI, on BIPAP S. aureus infection, osteomyolytis (by S. aureus infection, osteomyolytis (by
MRI for neck pain) vs. endocarditis (by MRI for neck pain) vs. endocarditis (by ECHO)ECHO)
Mod MR, EF 50%, no inotropes, stable Mod MR, EF 50%, no inotropes, stable BP (No Swan)BP (No Swan)
COPD, DM, Creat 173COPD, DM, Creat 173 7.29/42/80/20/-6.7 7.29/42/80/20/-6.7 AG = 15AG = 15 Are you worried ?Are you worried ?
First Patient: Approach Electrolytes:Electrolytes:
Na = 132, K = 4.9, Cl = 102Na = 132, K = 4.9, Cl = 102SID = 34 mEq/LSID = 34 mEq/L
Total Protein:Total Protein:64 g/L ~ 15 mEq/L64 g/L ~ 15 mEq/L
Electrical Neutrality:Electrical Neutrality:SID – HCO3 – Prot- ~ 0SID – HCO3 – Prot- ~ 0
34 - 20 - 15 = -134 - 20 - 15 = -1 Lactate = 1.5 mEq/LLactate = 1.5 mEq/L Electrolyte based acidosis, low SIDElectrolyte based acidosis, low SID
SID and Acid-Base
0
20
40
60
80
100
120
140
160
Cations Anions
[Na+]
[Cl-]
[HCO3-]
[Prot-]
0
20
40
60
80
100
120
140
160
Cations Anions
[Na+]
[Cl-]
[HCO3-]
[Prot-]
Very common cause of met acidosisVery common cause of met acidosis
Iatrogenic hyperchloremic metabolic acidosis Iatrogenic hyperchloremic metabolic acidosis
How does this occur?How does this occur?
Influence of Plasma ProteinInfluence of Plasma Protein
Normal Protein = 80 g.LNormal Protein = 80 g.L-1-1 SID = 44 mEq.LSID = 44 mEq.L-1-1 PCOPCO22 = 40 mmHg = 40 mmHg
pH = 7.42 pH = 7.42 [HCO[HCO33
--] = 27 mmol.L] = 27 mmol.L-1-1
[Prot[Prot--] = 16.8 mEq.L] = 16.8 mEq.L-1-1
Low Protein = 40 g.LLow Protein = 40 g.L-1-1 SID = 44 mEq.LSID = 44 mEq.L-1-1 PCOPCO22 = 40 mmHg = 40 mmHg
ppH = 7.533H = 7.533 [HCO[HCO33
--] = 35 mEq.L] = 35 mEq.L-1-1
[Prot[Prot--] = 8.7 mEq.L] = 8.7 mEq.L-1-1
Hypoproteinemic Metabolic Hypoproteinemic Metabolic AlkalosisAlkalosis
0
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Cations Anions
0
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140
160
Cations Anions
NaNa ClCl
HCO3Prot-
HCO3
Prot-
NaNa ClCl
HypoproteinemiaHypoproteinemia
Decrease in COP Decrease in COP (Schupbach et al., Vox Sang (Schupbach et al., Vox Sang 35:332, 1978; Zabala Ann Ital Chir LXIV 4:387, 1993)35:332, 1978; Zabala Ann Ital Chir LXIV 4:387, 1993)
Fluid Shifts/RetentionFluid Shifts/Retention Glomerular Filtration PressureGlomerular Filtration Pressure
Metabolic Alkalosis Metabolic Alkalosis (McAuliffe et al., Am J Med (McAuliffe et al., Am J Med 81:86, 1986)81:86, 1986)
Consequences of Consequences of Hypoproteinemic Metabolic Hypoproteinemic Metabolic
AlkalosisAlkalosis Hypoventilation Hypoventilation (McAuliffe AJM, 1986)(McAuliffe AJM, 1986)
Hyperventilation ! Hyperventilation ! (Rossing, J Lab Clin Med, 1988)(Rossing, J Lab Clin Med, 1988)
Difficulty assessing acid-base statusDifficulty assessing acid-base status Anion gap = (Na + K – Cl – HCOAnion gap = (Na + K – Cl – HCO33
--) = Prot) = Prot--
Hides lactacidosisHides lactacidosis
Case #3: POD #1 2200 hr, Emergency CABG
LV III, Inotropic support, IABPLV III, Inotropic support, IABP 105/60, 38/20, CI = 2.0, SVR = 1200105/60, 38/20, CI = 2.0, SVR = 1200 ABG = 7.45/40/155/29/-2.2ABG = 7.45/40/155/29/-2.2 Anion gap = Na + K – Cl – HCO3 = 17Anion gap = Na + K – Cl – HCO3 = 17 Are you concerned ?Are you concerned ?
Third Patient: ApproachThird Patient: Approach
Step 1: Electrolytes and SIDStep 1: Electrolytes and SIDNa = 139Na = 139K = 3.5K = 3.5Cl = 96Cl = 96SID = 46 mEq/LSID = 46 mEq/L
Step 2: Total Protein = 40 g/LStep 2: Total Protein = 40 g/L Convert to mEq/L: 40 g/L x 0.25 = 10 mEq/LConvert to mEq/L: 40 g/L x 0.25 = 10 mEq/L
Third Patient: Approach
Step 3: Electrical NeutralityStep 3: Electrical NeutralitySID – HCO3 – ProtSID – HCO3 – Prot-- ~ 0 ~ 0 ++ 5 mEq/L 5 mEq/L
46.5 – 29 – 10 = 7.5 mEq/L46.5 – 29 – 10 = 7.5 mEq/L
Either:Either: Lab variation (~ 2%)Lab variation (~ 2%)Unmeasured acidUnmeasured acid
Lactate = 8 mEq/LLactate = 8 mEq/L (Type A) (Type A)
Dead gut on laparotomyDead gut on laparotomy
The Danger of Hypoproteinemic Alkalosis
0
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40
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120
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Cations Anions
[Na+][Cl-]
[HCO3-]
[Prot-]
0
20
40
60
80
100
120
140
160
Cations Anions
[Na+] [Cl-]
[HCO3-]
[Lact-]
[Prot-]
How does Stewart approach affect
concept of Anion Gap ?
AG = Na+K-Cl-AG = Na+K-Cl-HC0HC033 = [Prot = [Prot--] ]
= [A= [A--]]
from [HA]Ka = [A from [HA]Ka = [A--][H][H++]] [Atot] = [HA] + [Atot] = [HA] +
[A[A--]] [A [A--] = ] = KaAtotKaAtot Ka+[H Ka+[H++]]
Therefore: AG can be Therefore: AG can be measured directlymeasured directly
0
20
40
60
80
100
120
140
160
Cations Anions
NaNa ClCl
HCO3HCO3
ProtProtSIDSID
AGAG
How does the Stewart approach affect concept of BE?
From BE = (1-0.014Hgb)(HCOFrom BE = (1-0.014Hgb)(HCO33-24)+(1.43Hgb +7.7)(pH-7.4) -24)+(1.43Hgb +7.7)(pH-7.4)
BE = 0 when pH = 7.4 and [HCO BE = 0 when pH = 7.4 and [HCO33] = 24] = 24
From Stewart:From Stewart: pH and pH and [HCO[HCO33] are dependent upon prevailing PCO] are dependent upon prevailing PCO22, SID and Atot, SID and Atot
Therefore BE is also a function of PCOTherefore BE is also a function of PCO22, SID and Atot, SID and Atot
Atot and SID can both be altered by pathophysiology such that Atot and SID can both be altered by pathophysiology such that neither pH or HCOneither pH or HCO33 change change
Both Atot and SID should be assessed during a clinical evaluationBoth Atot and SID should be assessed during a clinical evaluation
Information from BEBE = (1-0.014Hgb)(HCO3-24)+(1.43Hgb +7.7)(pH-7.4)BE = (1-0.014Hgb)(HCO3-24)+(1.43Hgb +7.7)(pH-7.4)
HgB = 14HgB = 14
pH = 7.4pH = 7.4
HCO3 = 24HCO3 = 24
PCO2 = 40PCO2 = 40
BE = 0BE = 0
Na = 140Na = 140
K = 4K = 4
Cl = 100Cl = 100
AG = 16AG = 16
pH = 7.3pH = 7.3
HCO3 = 16HCO3 = 16
PCO2 = 30PCO2 = 30
BE = -9.2BE = -9.2
Na = 140Na = 140
K = 4K = 4
Cl = 100Cl = 100
AG = 24 AG = 24
Lact = 10Lact = 10
SID = 44 (-10)SID = 44 (-10)
Tprot = 20 mEq/LTprot = 20 mEq/L
SID-HCO3-TP~0SID-HCO3-TP~0
34 – 16 – 20 = 234 – 16 – 20 = 2
SID = 44SID = 44
Tprot = 20 mEq/LTprot = 20 mEq/L
SID-HCO3-TP~0SID-HCO3-TP~0
44 – 24 – 20 ~ 044 – 24 – 20 ~ 0
pH = 7.38pH = 7.38
HCO3 = 25HCO3 = 25
PCO2 = 40PCO2 = 40
BE = +1.2BE = +1.2
Na = 140Na = 140
K = 4K = 4
Cl = 100Cl = 100
AG = 15AG = 15
SID = 44SID = 44
Tprot = 10 mEq/LTprot = 10 mEq/L
SID-HCO3-TP~10SID-HCO3-TP~10
44 – 25 – 10 = 944 – 25 – 10 = 9
pH = 7.24pH = 7.24
HCO3 = 16HCO3 = 16
PCO2 = 35PCO2 = 35
BE = -10.9BE = -10.9
Na = 130Na = 130
K = 4K = 4
Cl = 105Cl = 105
AG = 9AG = 9
SID = 29SID = 29
Tprot = 15Tprot = 15
SID-HCO3–TP~ 0SID-HCO3–TP~ 0
29 – 16 – 15 = -229 – 16 – 15 = -2
Lact = 0Lact = 0Lact = 10Lact = 10
Model SummaryModel Summary
SID (electrolyte or metabolic) SID (electrolyte or metabolic) acidosisacidosis
SID (electrolyte)SID (electrolyte) alkalosisalkalosis
Weak acid (protein)Weak acid (protein) alkalosis alkalosis ((Weak acid (phosWeak acid (phos--, protein), protein) acidosis)acidosis)
PCOPCO22 acidosis acidosis
PCOPCO22 alkalosisalkalosis
Compensatory Responses Metabolic acidosisMetabolic acidosis HyperventilateHyperventilate
Lower PCOLower PCO22
Metabolic alkalosisMetabolic alkalosis 22ndnd low protein low protein Decrease SID Decrease SID
by increase Clby increase Cl 22ndnd increased SID increased SID HypoventilateHypoventilate
Increase PCOIncrease PCO22
Respiratory acidosisRespiratory acidosis Increase SID byIncrease SID bydecreasing Cldecreasing Cl
Respiratory alkalosisRespiratory alkalosis ?increase protein?increase protein?decrease SID?decrease SID
Permissive Hypercapnia(Wilkes, unpublished)
0102030405060708090
100
0 5 10 15 20 25 30
Days
PCO2 (mmHg)
[HCO3-] (mEq/L)
[SID] (mEq/L)
Permissive Hypercapnia(Wilkes, unpublished)
0
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160
0 5 10 15 20 25 30
Days
mEq
.L-1
[Na+],(mEq/L)
[Cl-], (mEq/L)
[SID], (mEq/L)
Respiratory Acidosis
Compensatory response to Compensatory response to hypercapnea in severe COPD is to hypercapnea in severe COPD is to increase SID by lowering Cl increase SID by lowering Cl (Alfero 1996)(Alfero 1996)
Influence of [Atot] on [SID](Wilkes AJP 1998)
20
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160
8 10 12 14 16 18 20Atot (mEq.L-1)
[Na
+],
[Cl
- ], [S
ID] (
mEq
.L-1
)
[SID]=0.99[Atot]+24.7
[Cl-]=-0.66[Atot]+110.7
[Na+]=0.18[Atot]+134.6
Influence of [Atot] on [ProtInfluence of [Atot] on [Prot--], [HCO], [HCO33--] ]
and [Hand [H++]]
0
10
20
30
40
50
60
8 10 12 14 16 18 20Atot (mEq.L-1)
[Pro
t- ], [
HC
O3
- ], [
H+]
[H+], nEq.L-1
[HC03-], mEq.L-1
[Prot-], mEq.L-1
The Law of Electrical NeutralityA Law of Physics – it ALWAYS exist
SID SID + H+ H++ - HCO - HCO33-- - Prot - Prot-- - Phos - Phos-- - CO- CO33
== - OH - OH-- =0 =0
SID – HCOSID – HCO33-- - Prot - Prot-- - Phos - Phos-- ~ 0 ~ 0
{HCO{HCO33-- + Prot + Prot-- + Phos + Phos--} ~ SID} ~ SID
{HCO{HCO33-- - Prot - Prot-- - Phos - Phos--} ~ {Na + Mg + Ca + K – Cl – Lact}} ~ {Na + Mg + Ca + K – Cl – Lact}
What if electrical neutrality APPEARS not to exist?What if electrical neutrality APPEARS not to exist?
Influence of Apparent Charge ImbalanceInfluence of Apparent Charge Imbalance[SID] - [HCO[SID] - [HCO33
--] - [Prot] - [Prot--] ~ 0] ~ 0
(Wilkes, AJP, 1998)(Wilkes, AJP, 1998)
y = -0.0146x + 0.0214
R2 = 0.8599
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
-15 -10 -5 0 5 10 15 20
Un-Measured Ions (UMI), (mEq/L)
pH(m
sd-c
alc)
if UMI = 0, dpH = 0.0214if UMI = 0, dpH = 0.0214
Explanations of The Strong Ion Gap
SID SID + H+ H++ - HCO - HCO33-- - Prot - Prot-- - CO- CO33
== - OH - OH-- =0 =0
SID – HCOSID – HCO33-- - Prot - Prot-- = 0 = 0
SID = HCOSID = HCO33-- + Prot + Prot--
If SID = HCOIf SID = HCO33-- + Prot + Prot-- (Strong Ion Gap)? (Strong Ion Gap)?
1.1. Measurement error (Na, Cl Measurement error (Na, Cl ++ 2%) 2%) (Jones 1993)(Jones 1993)
2.2. Error of omission - lactate, ketoacids, Error of omission - lactate, ketoacids, krebs cycle intermediates krebs cycle intermediates (Gilfix 1993, Forni 2005, (Gilfix 1993, Forni 2005,
Moviat 2008)Moviat 2008)
3.3. Ionic concentration vs. activity Ionic concentration vs. activity (Stewart 1978)(Stewart 1978)
Acids – Bases & Salts & Ionic Activity
If NS is equal parts Na and Cl (SID=0) how If NS is equal parts Na and Cl (SID=0) how can you explain the pH = 5.5?can you explain the pH = 5.5?
1) Electrical neutrality refers to ionic 1) Electrical neutrality refers to ionic activity (functional concentration) not activity (functional concentration) not ionic concentration (actual ionic concentration (actual concentration)concentration)
2) NS is also equal parts HCl and NaOH 2) NS is also equal parts HCl and NaOH and HCl is a stronger acid than NaOH is and HCl is a stronger acid than NaOH is base base (Wilkes, 2009)(Wilkes, 2009)
Strong Ion Gap (SIG) In the Critically Ill (Kellum 1998)
Increased SIG in pts with:Increased SIG in pts with: Sepsis Sepsis (Reeves, 1983, Mehta 1986)(Reeves, 1983, Mehta 1986)
Liver disease Liver disease (Kellum 1995, Kirschbaum 1997)(Kellum 1995, Kirschbaum 1997)
Critical illness Critical illness (Mehta 1986)(Mehta 1986)
Animal studies with endotoxemia Animal studies with endotoxemia (Kellum (Kellum
1995)1995)
Critically ill pts with ARF Critically ill pts with ARF (Rocktaeschel 2003)(Rocktaeschel 2003)
May be a predictor of mortality May be a predictor of mortality (Kellum 2004)(Kellum 2004)
Clinical Outcome (Gunnerson CC: 2005)
Definitions of Acid-Base Disturbances Consistent definitions are a necessity to Consistent definitions are a necessity to
assess incidence and outcomeassess incidence and outcome Respiratory disturbances are straight Respiratory disturbances are straight
forwardforward Metabolic acidosis secondary to high Metabolic acidosis secondary to high
chloride is not equivalent to high lactatechloride is not equivalent to high lactate Organic vs. electrolyteOrganic vs. electrolyte
Metabolic alkalosis from elevated SID or Metabolic alkalosis from elevated SID or hypoproteinemiahypoproteinemia Organic vs. electrolyteOrganic vs. electrolyte
SID and the ICU Need to define acid-base disorders Need to define acid-base disorders
based on physical principles, not HCObased on physical principles, not HCO33 and Hand H++
Clinical outcome may be a function of Clinical outcome may be a function of underlying physical principle more so underlying physical principle more so than the actual pHthan the actual pH
Need to define what is normal for a Need to define what is normal for a given populationgiven population
We still do not have a complete We still do not have a complete understanding of the physical chemistryunderstanding of the physical chemistry
Basic Clinical Approach at 02:00
pH=7.05, HCOpH=7.05, HCO33=12, BE=-12, PCO=12, BE=-12, PCO22=30=30 ETT, O2, #14 iv and volume, inotropes, Sx and/or antibxETT, O2, #14 iv and volume, inotropes, Sx and/or antibx Most other casesMost other cases Measure electrolytes – calculate SID Measure electrolytes – calculate SID
Normal is 40-44 mEq/LNormal is 40-44 mEq/L Measure Measure TPrt x 0.25 = Atot (mEq/L)TPrt x 0.25 = Atot (mEq/L) Estimate electrical Estimate electrical neutrality fromneutrality from [SID] – Atot – [SID] – Atot – [HCO[HCO33] ] ++ 5 5 If >5 then check If >5 then check lactatelactate If lactate normal then If lactate normal then +SIG (equally bad)+SIG (equally bad)
Case 2: POD # 0, Elective CABG
PMHx:PMHx:79 yo male with peripheral, cerebral 79 yo male with peripheral, cerebral
vascular diseasevascular disease Intra Op:Intra Op:
SVG x 2, off CPB with mil/levoSVG x 2, off CPB with mil/levoincreased abdominal girth increased abdominal girth
CI=1.8, SVR=1477, BP=115/55, Ppa 50/20CI=1.8, SVR=1477, BP=115/55, Ppa 50/207.35/36/104/20/-5.4 AG = 9.27.35/36/104/20/-5.4 AG = 9.2
Are you worried ?Are you worried ?
Second Patient: Approach1) Electrolytes:1) Electrolytes:
Na = 136Na = 136K = 5.2K = 5.2Cl = 112Cl = 112SID = 29.2 mEq/LSID = 29.2 mEq/L
2) Total Protein = 33 g/L ~ 7.6 mEq/L2) Total Protein = 33 g/L ~ 7.6 mEq/L
3) Electrical Neutrality:3) Electrical Neutrality:29.2 – 20 – 7.6 = 1.6 mEq/l29.2 – 20 – 7.6 = 1.6 mEq/l
4) Lactate = 2 mEq/L4) Lactate = 2 mEq/L
5) Compensated hypoproteinemic alkalosis 5) Compensated hypoproteinemic alkalosis (Wilkes, AJP, 1998)(Wilkes, AJP, 1998)
Questions