Ionic composition of body fluids and acid-base balance Eva Samcová Petr Tůma.

Ionic composition of body fluids and acid-base balance

Eva Samcová

Petr Tůma

Life

• Life possesses the properties of replication, catalysis, and mutability (N. Horowitz)

• Biochemistry is the study of life on the molecular level.

Life in aqueous environments

• Nearly all biological molecules assume their shapes (by that their function) in response to the physical and chemical properties of the surrounding water.

• Reactants and products of metabolic reactions, nutrients as well as waste products, depend on water for transport within and between cells.

• The reactivity of many functional groups on biological molecules depend on the relative concentrations of H and OH in the surrounding medium

Water is a polar molecule

• Water molecules form hydrogen bonds.• The angular geometry of the water molecule has

enormous implications for living system.• Van der Waals distance – the distance of closest

approach between two nonbonded atoms.• Hydrogen bonds and other weak interactions

influence biological molecules.• The strength of bonds and weak interactions: ionic

interactions (two charge groups)►hydrogen bonds►van der Waals forces - arise from electrostatic interactions among permanent or induced dipoles (two permanent dipoles, dipole - induced dipole, London dispersion forces)

Life has evolved in water and is still dependent on it.

• electrical dipole – hydrogen bonds– Important in bonds: H-F, H-O, H-N

H2O CH4

Molar mass, g/mol 18 16

Dipole moment, 10-30 C.m 6,2 0

Boiling point, °C 100 -162

Heat of vaporization, kJ/mol

41 8

Dissolving of inorganic substances in water• electrolytic dissociation and hydration

• NaCl → Na+ + Cl-

• KNO3 → K+ + NO3-

• Na2SO4 → 2Na+ + SO42-

Insoluble salts• BaSO4 (s) ↔ Ba2+ (aq) + SO4

2- (aq)• Product of solubility KS

• KS = [Ba2+].[SO42-]

salt KS

BaSO4 1,3.10-10

AgCl 1,8.10-10

Ca3(PO4)2 1.10-26

Dissolving of organic substances in water• Formation of hydrogen

bonds– Alcohols– Monosaccharides

– Carboxylic acids– Length of chain

• Amphipathic substances– Soaps and detergents

• Hydrophobic substances– Oil in water

Composition of Body Fluids

Organic substances• Glucose• Amino Acids• Urea

Constituent Plasma Interstitial fluid

Intracellular fluid

Na+ 142 145 12

K+ 4 4.1 150

Ca2+ 1-2 1-2 <10-6 mol/l*

Mg2+ 0.75-1.5 2

Cl- 103 113 4

HCO3- 25 27 12

Protein 60 25

The normal composition of the major body fluid compartments is approximately as follows (mmol/l, except Ca2+)* Free ionic Ca2+ is very low inside cells, total calcium may be much higher (1-2mmol/l).

Acids and Bases Acid-Base Balance

Brönsted´s theory: Acid – substance which is able to donate H+

(proton) to another substance (proton donors) Base – substance which is able to accept H+ from another substance (proton acceptors)

Acid ↔ conjugate base + H+

Conjugate pairs:HCl ↔ Cl- + H+

NH3 + H+ ↔ NH4+

Relation to water:

HCl + H2O ↔ Cl- + H3O+

NH3 + H2O ↔ NH4+ + OH-

Disociation of water and pH concept• Autoprotolysis of water – ionic product of water

[H3O+] [OH-] = 1.10-14 při 25 °CpH + pOH = 14

• pH = -log[H3O+]

• Acidic solution: [H3O+] > [OH-] pH < 7• Neutral solution: [H3O+] = [OH-] pH = 7• Basic solution: [H3O+] < [OH-] pH > 7

Solutions of strong acids and strong bases

• Strong acids: HCl, HNO3, H2SO4, HClO4

• Strong bases: NaOH, KOH, Ca(OH)2, Ba(OH)2

X log10X

10 000 = 104 4

1 000 = 103 3

100 = 102 2

10 = 101 1

1 = 100 0

0,1 = 10-1 -1

0,01 = 10-2 -2

0,001 = 10-3 -3

0,0001 = 10-4 -4

pH calculation of solutions of strong acids and bases

• Strong acids: HCl, HNO3, H2SO4, HClO4

• Strong bases: NaOH, KOH, Ca(OH)2, Ba(OH)2

• Calculate the pH of 0,001M HCl. What are concentrations of H3O+ and OH- in this solution?

Calculate the pH of 0,01M KOH. What are concentrations of H3O+ a OH- in this solution?

Weak acids and dissociation constant

• electrolytic dissociation of acid

HA + H2O ↔ H3O+ + A- • Dissociation constant of acid KA

pH weak acid solution• pH = ½ pKA - ½ log [HA]

Calculate pH of 0,01M solution of acetic acid KA= 1,75 10-5

AA

A

KpK

HA

AOHK

log

3

Power of acids

Conjugate pair KA pKA α, [%]

HClO4 ClO4- 95

HNO3 NO3- 90-95

HCl Cl- 90-95

H2SO4 HSO4- 60

H3PO4 H2PO4- 7,5 .10-3 2,12 12

HF F- 6,6 .10-4 3,18 10

CH3COOH CH3COO- 1,75 .10-5 4,76 1,3

H2CO3 HCO3- 4,47 .10-7 6,35 0,12

H2S HS- 1,07 .10-7 6,97

H2PO42- HPO4

- 6,20 .10-8 7,21

HCN CN- 6,17 .10-10 9,21 0,07

HCO3- CO3

2- 4,68 .10-11 10,33

HPO42- PO4

3- 4,27 .10-13 12,37

Po

we

r o

f a

cid

s

Po

we

r o

f b

ase

s

BasesPower of bases

• strong bases: NaOH, KOH, CsOH, Ca(OH)2,

Ba(OH)2

• weak bases: NH4OH, pyridine, aniline, purine

Dissociation constant of bases KB

B + H2O ↔ BH+ + OH- BOHBH

KB

Báze KB

NaOH

KOH

Ca(OH)2

NH3 1,80 . 10-5

pyridin 1,62 . 10-9

anilin 4,07 . 10-11

Solutions which buffer deviationof pH

Mixture of weak acid and its conjugate salt:• CH3COOH + NaOH → CH3COONa + H2O• CH3COONa + HCl → CH3COOH + NaCl

Hendersonova-Haselbalch equation kyseliny

solipKpH A log

What is result pH of buffer which was prepared by mixing of 200mL 0,1M CH3COOH with 200mL 0,2M CH3COONa? KA=1,75 10-5

Buffers

pH

Physiological buffers

Phosphate buffer • H2PO4

- + H2O ↔ HPO4-2 + H3O+ pK = 7,2

• Buffers intracellulary, in urine, 1% of plasma• Inorganic phosphate, AMP, ADP, ATPBicarbonate buffer• H2CO3 + H2O ↔ HCO3

- + H3O+

• Buffering of blood plasmaProteins• Side chains of amino acids – His, Lys, Arg, Glu, Asp• The most important hemoglobin 30% of buffer capacity

of blood, other proteins only 6%

Dissociation of phosphoric acid

24

334

33342

24

42

324

2324242

43

3421342243

HPO

OHPOKOHPOOHHPO

POH

OHHPOKOHHPOOHPOH

POH

OHPOHKOHPOHOHPOH