Stewart and Buffer Base

8/8/2019 Stewart and Buffer Base.

1/69

Stewart & conventionalStewart & conventional

approach in clinicalapproach in clinicalpracticepractice


2/69

The disadvantage of men not

knowing the past is that they do not

know the present.

G. K. Chesterton


3/69

ASAM BASA..ASAM BASA..

pHpH

[H[H++

]]


4/69

Analytic tools used in acid

base chemistry CO

2-bicarbonate (Boston) approach

Schwartz, Brackett et al

H-H equation The Base deficit/excess (Copenhagen) approach

1948 Singer-Hasting, Buffer Base (BB)

1958 Siggaard-Andersen Base Deficit/Excess

(BDE) 1960, Hb into calculation, modified Standard Base

Deficit/Excess (SBE)

1977 Van Slyke equation to computed SBE

Has been validated by Schlitic and Morgan


5/69

Analytic tools used in acid

base chemistry

1977, Anion Gap approach

Emmet and Narins

To address the limitation of Boston and Copenhagen

1978, Stewart introduced the physical-chemicalapproach

3 independent variable; PCO2, SID and weak acid

1993, Stewart-Fencl approach

1998, Anion Gap Corrected

Fencl and Figge

2004, simplified Stewart-Fencl approach

Story DA, Morimatsu et al


6/69

Brief historical perspective


7/69

19081908

Henderson

[H+] = K' [HA] / [A]


8/69

Acid Base

Notasi pH diciptakan oleh seorang ahli kimia dariNotasi pH diciptakan oleh seorang ahli kimia dariDenmark yaitu Soren Peter Sorensen, yang berarti logDenmark yaitu Soren Peter Sorensen, yang berarti log

negatif dari konsentrasi ion hidrogen. Dalam bahasanegatif dari konsentrasi ion hidrogen. Dalam bahasa

Jerman disebutJerman disebutWasserstoffionenexponentWasserstoffionenexponent(eksponen ion(eksponen ion

hidrogen) dan diberi simbol pH yang berarti: hidrogen) dan diberi simbol pH yang berarti: ppotenzotenz

(power) of(power) ofHHydrogen.ydrogen.

19091909


9/69

pH = -log[H+]

defined by Sorensen

[H+] pH

1 x 10-6 6.0

1 x 10-7

7.0

8 x 10-8

7.1

4 x 10

-8

7.42 x 10

-87.7

1 x 10-8

8.0


10/69

pH = 6.1 + logpH = 6.1 + log [ HCO[ HCO33--

] = LUNG] = LUNG

0.03 x0.03 x PcoPco22 = KIDNEY= KIDNEY

Hasselbalch KA. Die Berechnung der Wasserstoffzahl des Blutes aus der freienHasselbalch KA. Die Berechnung der Wasserstoffzahl des Blutes aus der freien

und gebundenen Kohlensaure desselben und die Sauerstoffbindung des Blutesund gebundenen Kohlensaure desselben und die Sauerstoffbindung des Blutes

als Funktion der Wasserstoffzahl. Biochemische Zeitschrift 1916; 78: 11244.als Funktion der Wasserstoffzahl. Biochemische Zeitschrift 1916; 78: 11244.

19161916

Hasselbach:Hasselbach:Used Sorensen's terminology for Henderson'sUsed Sorensen's terminology for Henderson's

equationequation in logarithmic form:in logarithmic form:

Hendersen-Hasselbach equation (H-H)Hendersen-Hasselbach equation (H-H)


11/69

Regulasi asam basa diatur melalui proses di:Regulasi asam basa diatur melalui proses di:

1.1. Ginjal dengan cara mempertahankan [HCOGinjal dengan cara mempertahankan [HCO33--]]sebesar 24 mM dansebesar 24 mM dan

2.2. Mekanisme respirasi dengan cara mempertahankanMekanisme respirasi dengan cara mempertahankan

tekanan parsial COtekanan parsial CO22 arteri (PaCOarteri (PaCO22) sebesar) sebesar

40mmHg.40mmHg.

Hendersen-Hasselbalch


12/69

pH = 6.1 + logH = 6.1 + log [ HCO[ HCO33--]]

0.030.03xx

1. Change in1. Change in

Metabolic disturbanceMetabolic disturbance

2. Change after2. Change afterRenal compensation forRenal compensation forRespiratory disturbanceRespiratory disturbance

1. Change in1. Change inRespiratory disturbanceRespiratory disturbance

2. Change after2. Change afterRespiratory compensation forRespiratory compensation for

Renal disturbanceRenal disturbance

pCO2pCO2


13/69

pHpH = 6.1 + log= 6.1 + log

[HCO[HCO33--]]

pCOpCO22

GINJALGINJAL

PARUPARU

BASABASA

ASAMASAM CO2

HCO3HCO3

CO2

KompensasiKompensasi

NormalNormal

NormalNormal


14/69

Diagram Davenport

[HCO

3-

]

PCO2 = 80 40

20

pH

7.0 7.2 7.4 7.6 7.8

10

20

30

40

50

pH = 6.1 + Ginjal

ParuBB

AA

CC

7.4 / 40 / 247.4 / 40 / 24

7.2 / 80 / 307.2 / 80 / 30

7.6 / 20 / 187.6 / 20 / 18NormalNormal


15/69

Low

pH

Alkalosis Metabolik

PCO

2

HCO3-

normal

High

Low

pH

Alkalosis Respiratori

PCO

2

HCO3 -

normal

High

High

pH

PCO2

HCO3

-

Asidosis Respiratori

norm

al

Low

High

pH

PCO2

HCO3

-

Asidosis Metabolik

norm

al

Low

Gangguan asam-basa primerGangguan asam-basa primer


16/69

Diagnosis menggunakan nilai asam basa serum:Davenport Diagram

[HC

O3

-]

PCO2 = 80 40

20

pH7.0 7.2 7.4 7.6 7.8

10

20

30

40

50

Henderson- Hasselbalch:

pH = pK + log [HCO3-]

s PCO2AsidosisAsidosis

RespiratoriRespiratori AlkalosisAlkalosis

MetabolikMetabolik

AlkalosisAlkalosis

RespiratoriRespiratori

AsidosisAsidosisMetabolikMetabolik

pH = 6.1 + Ginjal

Paru

atau,

Normal


17/69

RESPON KOMPENSASIRESPON KOMPENSASI


18/69

Alkalosis Respiratori

[HC

O3

-]

PCO2 = 80 40

20

pH7.0 7.2 7.4 7.6 7.8

10

20

30

40

50

AlkalosisAlkalosis


terkompensasiterkompensasi

Penyebab:

1) Nyeri

2) Histerik

3) Hipoksia

AlkalosisAlkalosis


Normal

kompensasi = [HCO3-]


19/69

Asidosis Respiratori

[HC

O3

- ]

PCO2 = 80 40

20

pH7.0 7.2 7.4 7.6 7.8

10

20

30

40

50

AsidosisAsidosis


kompensasi = [HCO3-]Penyebab:

1) PPOK, Gagal jantung

kronik, bbrp pnykt

paru

2) Obat anestesi

AsidosisAsidosis




20/69

Metabolic Alkalosis

[HC

O3

-]

PCO2 = 80 40

20

pH7.0 7.2 7.4 7.6 7.8

10

20

30

40

50

AlkalosisAlkalosis

MetabolikMetabolik

kompensasi =PCO2Penyebab:

1) Intake basa >>

2) Kehilangan asam

(Muntah,

penyedotan lambung)

AlkalosisAlkalosis

MetabolikMetabolik



21/69

Metabolic Asidosis

[HC

O3

-]

PCO2 = 80 40

20

pH7.0 7.2 7.4 7.6 7.8

10

20

30

40

50

AsidosisAsidosis

MetabolikMetabolik

kompensasi = PCO2

Penyebab:

1) Kehilangan basa

(eg. diare)

2) Akumulasi asam

(diabetes, gagal ginjal)

3) Asidosis Tubular Ginjal

AsidosisAsidosis

MetabolikMetabolik

terkompensasterkompensas

ii


22/69

Kompensasi ginjal terhadap asidosis resp. kronik

Kompensasi ginjal & paru terhadap asidosis non ginjal

PPOK

Keto/Laktatasidosis


23/69

DISORDER pH PRIMER RESPON

KOMPENSASI

ASIDOSISASIDOSIS

METABOLIKMETABOLIK

HCO3- pCO2 ALKALOSISALKALOSIS

METABOLIKMETABOLIK

HCO3- pCO2ASIDOSISASIDOSIS

RESPIRATORIRESPIRATORI

pCO2 HCO3-

ALKALOSISALKALOSISRESPIRATORIRESPIRATORI

pCO2 HCO3-

RANGKUMAN GANGGUANRANGKUMAN GANGGUANKESEIMBANGAN ASAM BASAKESEIMBANGAN ASAM BASA

TRADISIONALTRADISIONAL


24/69

BDE (Base DeficitBDE (Base DeficitExcess)Excess)

&&SBE (Standard BaseSBE (Standard Base

Excess)Excess)


25/69

Singer and Hastings, 1948,Singer and Hastings, 1948,

Whole blood buffer base (BB), defined as the sum ofWhole blood buffer base (BB), defined as the sum of

bicarbonate [HCObicarbonate [HCO33--] & nonvolatile buffer anions (A] & nonvolatile buffer anions (A--))

The change in BB from "normal" was called delta BBThe change in BB from "normal" was called delta BB

(BB). This change in BB was an expression of the(BB). This change in BB was an expression of the

metabolic component of an acid-base disturbancemetabolic component of an acid-base disturbance

BE (Base Excess)BE (Base Excess)

Singer RB, Hastings AB: An improved clinical method for the estimation of disturbances of theSinger RB, Hastings AB: An improved clinical method for the estimation of disturbances of the

acid-base balance of human blood.acid-base balance of human blood. Medicine (Baltimore)Medicine (Baltimore) 1948; 27:223-2421948; 27:223-242

Introduced the concept of buffer base [BB]Introduced the concept of buffer base [BB]

[BB] = [HCO[BB] = [HCO33--] + [A] + [A--]]


26/69

Modified Base Excess

ECF includes plasma, red cells, and thesurrounding interstitial fluid. Its where the action

takes place in the body regarding acid-basemovement.

Blood-gas machines calculate SBE as:

Standard Base Excess

SBE = (1 - 0.014Hgb) (HCOSBE = (1 - 0.014Hgb) (HCO33 24 + (1.43Hgb + 7.7) (pH - 24 + (1.43Hgb + 7.7) (pH -

7.4)`7.4)`


27/69

[HC

O3

-]

pCO2 = 80 40

20

pH

7.0 7.2 7.4 7.6 7.8

10

20

30

40

50

AsidosisAsidosis

MetabolikMetabolik

Base deficit;Base deficit;

Kekurangan Basa;Kekurangan Basa;

Jumlah Basa (HCOJumlah Basa (HCO33) yg) yg

harus ditambahkan agarharus ditambahkan agarpH normalpH normal

AlkalosisAlkalosis

MetabolikMetabolik

Base excess;Base excess;

Kelebihan Basa; JumlahKelebihan Basa; Jumlah

Basa (HCOBasa (HCO33) yang harus) yang harus

dikurangi agar pHdikurangi agar pHnormalnormal

Base Excess/Base Excess/

Base DeficitBase Deficit

Normal

Base

Defisit

BaseExcess


28/69

Unmeassured anion

Limitasi dari H-H tidak dapat

mengukur metabolic acidosis yangdisebabkan unmeassured anionseperti laktat dan keton

Emmet dan Narins

Anion Gap AG= Na+ - (Cl- + HCO

3-)

Normal Gap 10-15

N lN l


29/69

Increased anion gap acidosisIncreased anion gap acidosisNormal anion gap acidosisNormal anion gap acidosis

Na

K Cl

AG

HCO-3

AG = 10-15AG = 10-15

2525

105105145145

NormalNormal+

Na

K Cl

HCO-3

AG

1515

115115 145145

= 15 (normal)= 15 (normal)

Na

K

Cl

HCO-3

AG/Other

anion

= 25= 25 1515

105105

(normal)(normal)

145145

HCO3- decreases and replaced by

Cl- so there is a Cl- shift :Eg.Diarrhea or simple gain of H+

HCO3- decreases and replaced by

anions other than Cl- so no Cl-shift: Eg.renal failure anddiabetic keto-acidosis

Metabolic acidosis

HCO3-


30/69

Stewart and traditional

approach(Stewart-Fencl approach)


31/69

NaNa++

140140

KK++ 44CaCa++++MgMg++++

ClCl--

102102

HCOHCO33

A-

Buffer BaseBuffer Base = SID = HCO3- +

A-

SID & BASE EXCESSSID & BASE EXCESS

(expected if pH = 7.4 and pCO2 =

40)Any deviation in [NaAny deviation in [Na++], [Cl], [Cl--] or [Alb] or [Alb--] from normal values will] from normal values will

produce aproduce a base excess or deficitbase excess or deficit

ClCl--

102102

A-

HCOHCO33--

BECl (-)

BufferBasea

Siggaard-Andersen, Base excess or buffer base (strong ion difference) as a measure of a

non-respiratory acid-base disturbance. Acta Anaesthesiol Scand, 1995

Buffer BaseBuffer Baseaa -- Expected Buffer BaseExpected Buffer BaseBEBEClCl ==


32/69

SID = Buffer BaseSID = Buffer Base

0

20

40

60

80

100

120

140

160

Cations Anions

Na+ Cl-

K+, Mg++ , Ca++

mE

q/L

Lactate,Other

anions

Cl-

A-

HCO3-

SID BB


33/69


0

20

40

60

80

100

120

140

160

Cations Anions

Na+ Cl-

K+, Mg++ , Ca++

mE

q/L

Cl-

A-

HCO3-

SID BB

BaseDeficit


34/69


The standard base

excess corresponds to

the change in SID

required to restore theplasma (in vivo) to pH

7.40 with pCO2 of 40 mm

Hg

RR22=0.9527=0.9527

-10-10

-8-8

-6-6

-4-4

-2-2

00

22

44

66

-8-8 -6-6 -4-4 -2-2 00 22 44

A/V SIDeA/V SIDe

A/VS

BE

A/VSBE

Kellum et al. J Crit Care 1997; 12: 7-12


35/69

SBE = SIDexSBE = SIDex

[SIDex] mEq/L

[S

BE]m

Eq/L

-30

-20

-10

0

10

20

-30 -20 -10 0 10 20

Schlichtig R. Adv Exper Med Bio. 1997;411:91-95


36/69

BEBE from the Blood Gas Machine

SID effectSID effect, mEq/l = A + B

A. Free Water effect on Na+

= 0.3 x ([Na+

] 140)B. Corrected Cl- effect

= 102 ([Cl-] x 140/[Na+]) Total weak acids effectTotal weak acids effect, mEq/l

= 0.123 x pH - 0.6310 x (42 - [Albumin])UA effect = BE efBE ef SID efSID ef AATotTot efef

BASE EXCESS DAN STEWARTBASE EXCESS DAN STEWART

Stewart-Fencl, 2000


37/69

Steps to combine base-excessSteps to combine base-excess

with the Stewart approachwith the Stewart approach


38/69

Efek SIDEfek SID

Blood Gas Analysis

Step 1; Base-excess effect of the Na & Cl

[(efek Na + efek Cl)[(efek Na + efek Cl)+ efek Alb]+ efek Alb]BEBEUAUA ==BE BE

Step 2; Base-excess of the albumin

BEUA effect is the residual difference

If there were no abnormalities inIf there were no abnormalities in

Na,Cl,Alb, or UA anions, then BENa,Cl,Alb, or UA anions, then BE

would be equivalent to BEwould be equivalent to BEUAUA ..

Fencl V, Leith DE: Stewarts quantitative acid-baseFencl V, Leith DE: Stewarts quantitative acid-base

chemistry: Applications in biology and medicine.chemistry: Applications in biology and medicine. RespResp

PhysiolPhysiol 1993, 91:1--16.1993, 91:1--16.


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BASE EXCESS DANBASE EXCESS DAN

STEWARTSTEWART

(a) BE of Free water(a) BE of Free water

0.3 x (Na-140)0.3 x (Na-140)

(b) BE of Chloride effect(b) BE of Chloride effect

102-(Cl x 140/Na)102-(Cl x 140/Na)

(c) BE of Albumin effect(c) BE of Albumin effect

(0.123 x pH - 0.631) (42-[alb])(0.123 x pH - 0.631) (42-[alb])

BEBEUAUA = BE= BEastrupastrup [(a) + (b) + (c)] mEq/L [(a) + (b) + (c)] mEq/L


40/69

Kasus 1:Kasus 1:

7.25 / 30 / -10 / 147.25 / 30 / -10 / 14

Na 140; Cl 112; Alb 4.0Na 140; Cl 112; Alb 4.0

(a) Free water effect:(a) Free water effect:

0.3 x (140-140) = 00.3 x (140-140) = 0

(b) Chloride effect(b) Chloride effect 102-(112) x 140/140) = - 10102-(112) x 140/140) = - 10

(c) Albumin effect(c) Albumin effect

(0.123 x 7.25 - 0.631) (42-[40]) = 0.5(0.123 x 7.25 - 0.631) (42-[40]) = 0.5

UA = - 10 [(0) + (-10) + (0.5)] mEq/L = - 0.5UA = - 10 [(0) + (-10) + (0.5)] mEq/L = - 0.5


41/69

Alb


140

150

102

112BE akibat pe Cl- -10

HCO3-

Na+ Cl-

Alb

7.25 / 30 / -10 / 147.25 / 30 / -10 / 14

BECl = Buffer Base expected buffer

base

(expected if pH = 7.4 and pCO2 = 40)


42/69

Kasus 2:

7.48 / 45 / +10 / 34Na 150; Cl 102; Alb 4.0

(a) Free water effect:

0.3 x (150-140) = 3

(b) Chloride effect 102-(102 x 140/150) = 6.8

(c) Albumin effect

(0.123 x 7.25 - 0.631) (42-[40]) = 0.5

UA = 10 [(3) + (6.8) + (0.5)] mEq/L = - 0.3


43/69

Alb


140

150

Na+ Cl-

HCO3-

Alb

BENa = Buffer Base expected buffer base


BE akibat Na+ +10


44/69

Kasus 3:

7.48 / 45 / + 9 / 34Na 140; Cl 93; Alb 4.2


0.3 x (140-140) = 0

(b) Chloride effect 102-(93 x 140/140) = 9

(c) Albumin effect

(0.123 x 7.25 - 0.631) (42-[42]) = 0

UA = 9 [(0) + (9) + (0)] mEq/L = 0


45/69

Alb


140

Na+ Cl-

HCO3

-

BECl = Buffer Base expected buffer base


BE akibat Cl- +9

Alb


46/69

Kasus 4:

7.48 / 45 / +7 / 34Na 140; Cl 102; Alb 2.0


0.3 x (140-140) = 0

(b) Chloride effect 102-(102 x 140/140) = 0

(c) Albumin effect

(0.123 x 7.48 - 0.631) (42-[20]) = 6.4

UA = 7 [(0) + (0) + (6.4)] mEq/L = 0.6


47/69

Alb


140

Na+ Cl-

HCO3-

Hipoalb

BEAlb = Buffer Base expected buffer base


BE akibat Alb + 5

UA- (4.8)


48/69

Kasus 5 :Kasus 5 :

7.42 / 35 / 100 / -2 / 21 ;7.42 / 35 / 100 / -2 / 21 ;

(a) Free water(a) Free water

0.3 x (140-140) = 00.3 x (140-140) = 0

(b) Chloride effect(b) Chloride effect

102-(102 x 140/140) = 0102-(102 x 140/140) = 0

(c) Albumin effect(c) Albumin effect

(0.123 x 7.42 - 0.631) (42-[18]) = 6.7(0.123 x 7.42 - 0.631) (42-[18]) = 6.7

UA = - 2 [(0) + (0) + (6.7)] mEq/L = - 8.7UA = - 2 [(0) + (0) + (6.7)] mEq/L = - 8.7

Na 140; Cl 102; Alb 1.8Na 140; Cl 102; Alb 1.8

Menurut H-HMenurut H-H normalnormal


49/69


140

102

HCO3

-

24

Alb

Na+ Cl-

hipoalbumin

HCO3

-

30.7

SID normalBE astrup = - 8.7 + 6.7 = - 2

HCO3-

22

BE akibat UA -8.7

SID BE astrup = - 8.7 + 0 = - 8.7

UA = - 8.7

UA =- 8.7

Alb normal

BE akibat hipoalb +

6.7


50/69


51/69

BDE adjustment for serum albumin


52/69

Strong ions, weak acids and base excess:Strong ions, weak acids and base excess:

aasimplified FenclStewart approachsimplified FenclStewart approach totoclinical acidbase disordersclinical acidbase disorders

Story, Morimatsu, BellomoStory, Morimatsu, Bellomo (2004),(2004), British Journal of Anaesthesia. Vol. 92,British Journal of Anaesthesia. Vol. 92,

SBESBE = from a blood gas machine == from a blood gas machine =

NaCl effectNaCl effect = [Na= [Na++][Cl][Cl]38 =...]38 =...

Albumin effectAlbumin effect = 0.25 x [42Alb(g/l)] == 0.25 x [42Alb(g/l)] =

UAUA = SBE (NaCl)effect Albumin effect == SBE (NaCl)effect Albumin effect =


53/69

RESPON KOMPENSASIRESPON KOMPENSASI

Kompensasi terhadap kronikKompensasi terhadap kronik


54/69

PCO2

PPOK

pH

NHNH44ClCl Hipoalbumin..?Hipoalbumin..?

Sintesis Alb


55/69

PaCO2 pH

NHNH44ClClHipoalbuminHipoalbumin

Laktat- / keto-

SIDSID

Sintesis Alb


56/69

WORKSHOP ACIDBASESTEWART PERDICI 2006

Group 1 Group 2 Group

3

paCOpaCO22 < 40< 40 paCOpaCO22 40-5040-50 paCOpaCO22 > 50> 50

pH

SID


57/69



58/69


Respon kompensasiespon kompensasi


59/69


60/69

Conclusions: Critically ill patients may present severehyperlactatemia with normal values of pH, [HCO3], and

[BE] as a result of associated hypochloremic alkalosis.

CONCLUSION


61/69

SID

PCO2

Weak acid

pH

Renal Failure

Lactic acidosisKeto-acidosisVomiting

Diare

Heart Failure

Lung diseaseHyperventilationHypoventilation

Nephrotic SyndromeDehydrationMalnutrition

Charge BalanceDissociation of:WaterProteinCarbonic acid

CONCLUSION


62/69

SBESBE = -10= -10

NaCl effectNaCl effect = [Na= [Na++

][Cl][Cl

]38 = 14011238 = -10]38 = 14011238 = -10 Albumin effectAlbumin effect = 0.25 x [4240(g/l)] = 0.5= 0.25 x [4240(g/l)] = 0.5

UAUA = -10 (-10) 0.5 = -0.5= -10 (-10) 0.5 = -0.5

Kasus 1:Kasus 1:

7.25 / 30 / -10 / 147.25 / 30 / -10 / 14

Na 140; Cl 112; Alb 4.0Na 140; Cl 112; Alb 4.0



Albumin effectAlbumin effect = 0.25 x [42Alb(g/l)] == 0.25 x [42Alb(g/l)] = UAUA = SBE (NaCl)effect Albumin effect == SBE (NaCl)effect Albumin effect =


63/69

Alb


140

150

102

112BE akibat pe Cl- -10

HCO3-

Na+ Cl-

Alb

7.25 / 30 / -10 / 147.25 / 30 / -10 / 14

WD/: Asidosis metabolik karena hiperkloremiaCausal: - Pemberian Lar NaCl berlebihan

- Gagal ginjal akutTh/: Batasi NaCl

HD/CRRT

K 2K 2


64/69




Kasus 2:Kasus 2:

7.48 / 50 / + 9 / 347.48 / 50 / + 9 / 34

Na 140; Cl 93; Alb 4.2Na 140; Cl 93; Alb 4.2

SBESBE = +9= +9


][Cl][Cl

]38 = 1409338 = 9]38 = 1409338 = 9 Albumin effectAlbumin effect = 0.25 x [4242(g/l)] = 0= 0.25 x [4242(g/l)] = 0

UAUA = 9 9 0 = 0= 9 9 0 = 0


65/69

Alb


140

Na+ Cl-

HCO3-

BE akibat Cl- +9

Alb

7.48 / 45 / + 9 / 347.48 / 45 / + 9 / 34

WD/: Alkalosis metabolik karena hipokloremiaCausal:

- Diuretik Lasik- Muntah, Enterokutan Fistula

Th: NaCl 0.9%, kurangi furosemide, cairan fisteldimasukkan lagi

K 3Kasus 3:


66/69




Kasus 3:Kasus 3:

7.30 / 27 / -7 / 187.30 / 27 / -7 / 18

Na 128; Cl 100; Alb 3.0Na 128; Cl 100; Alb 3.0

SBESBE = -7= -7


][Cl][Cl

]38 = 12810038 = -10]38 = 12810038 = -10 Albumin effectAlbumin effect = 0.25 x [4230(g/l)] = 3= 0.25 x [4230(g/l)] = 3

UAUA = -7 + 10 3 = 0= -7 + 10 3 = 0


67/69

Alb


140

Na+ Cl-

BE akibat Na -7

7.30 / 27 / -7 / 187.30 / 27 / -7 / 18

WD/: Acidosis metabolik karena hiponatremia

Causal:- hemodilusi- Overload cairan, fase awal shock oligouri

Th: perbaiki shock, inotropik, HD/CRRT

128

Kasus 4 :Kasus 4 :


68/69

Kasus 4 :

7.42 / 35 / 100 / -2 / 21 ;7.42 / 35 / 100 / -2 / 21 ;

Na 140; Cl 102; Alb 1.8Na 140; Cl 102; Alb 1.8

Menurut H-HMenurut H-H normalnormal




SBESBE = -2= -2


][Cl][Cl

]38 = 14010238 = 0]38 = 14010238 = 0 Albumin effectAlbumin effect = 0.25 x [4218(g/l)] = 6= 0.25 x [4218(g/l)] = 6

UAUA = -2 0 6 = -8= -2 0 6 = -8


69/69


140

102

HCO3

-

24

Alb

Na+ Cl-

hipoalbumin

HCO3

-

30.7

SID normalBE astrup = - 8 + 6 = - 2

HCO3-

22

BE akibat lact -8

UA = - 8BE akibat hipoalb +

6

7.42 / 35 / 100 / -2 / 217.42 / 35 / 100 / -2 / 21

Lactic Asidosis metabolik maskingoleh hipoalbumin

Documents

Stewart and Buffer Base