Ionization of water, pHAcids and basesAcid Dissociation, pKaThe Henderson-Hasselbalch EquationTitration Buffers

Prof. Ramune Morkuniene, Biochemistry dept., LUHS

Water is a weak electrolyte: among 555 x l06

its molecules only 1 molecule undergoes ionization

The ionization is represented by the equation:2H2O ↔ H3O+ + OH-

hydronium ion hydroxide ion

The process is usually simplified:H2O ↔ H+ + OH-

Ionization of water

[H2O], [H+], [OH-] – concentration in moles per liter (M)

H2O H+ + OH-

O][H]][OH[HK

2eq

−+

=

The degree of ionisation of water:Keq=1/ 555 x l0-6 =1.8×10-16M

[H+][OH-] = 1.8×10-16×55.5 = 1×10-14 M2

[H+] = [OH-]

-log[H+] = pH; -log[OH-] = pOH; pH + pOH=14

100018[H2O] = ≈ 55.5 M

[H+] = [OH-] = 10-7 M Concentrations of H+

and OH- are expressed as negative log of 10-7M

(-log10-7 = 7):-log[H+] = pH, this is

the index of acidity.

[H+][OH-] = Keq×[H2O]

The equilibrium constant (Keq):

pH scalepH is defined as negative logarithm of the concentration of H+

• Is used to indicate the acidity of a solution.• Has values that usually range from 0 to 14.• Indicates an acidic solution when the values are less than 7.• Indicates a neutral solution with a pH of 7.• Indicates a basic solution when the values are greater than 7.

Calculating pHExample 1:What is the pH of a solution of 10 mM H+?

Example 3:The [H+] of lemon juice is 1.0×10-3M. Whatis the [OH-] of the solution?

Solution:Kw=[H+][OH-]=1×10-14 M101

101101][OH 11

3

14−

−

−− ×=

××

=

Example 2:What is the [H+] of a solution of pH 6?

Solution:10 mM = 10 mmoles/litre = 0.01 moles/litrelog (0.01) = (0.01= 10-2) = 2pH = 2

Solution:pH 6 = 10-6 M (or 1 μM)

• Acids are compounds, which dissociate in protons H+ and anions in water solutions:

HA ↔ H+ + A-

• Bases are compounds, which dissociate in cations and hydroxide anions OH- in water solutions:

KOH ↔ K+ + OH-

Acids and basesaccording to the theory of electrolytic dissociation

The Arrhenius Theory of acids and bases

Protolytic (Bronsted-Lowry)theory of acids and bases

• Acids are hydrogen ion (H+) donors.• Bases are hydrogen ion (H+) acceptors.

• An acid that donates H+ forms a conjugate base• A base that accepts a H+ forms a conjugate acidAn acid cannot exist without its conjugate base and

vice versa

• In an acid-base reaction, there are two conjugateacid-base pairs.

• If an acid is strong, its conjugate base is weak.• If a base is strong, its conjugate acid is weak.

Conjugate acid-base pairs

Example:Identify conjugate acid-base pairs:1. HNO2; NO2

-

2. H2CO3; CO32-

3. HCl; ClO4-

4. HS-; H2S5. NH3; NH+

46. HBr; Br-

Solution: 1; 4; 5; 6

Acid Dissociation• Strong acids and bases dissociate

completely in water

HCl Cl- + H+

• Weak acids and bases do not dissociate completely in H2O

Acid strenght

HA A- + H+

aeq K[HA]

]][A[HK ==−+

The strenght of an acid is a measure of its tendency to release protons.

Can be indicated from its dissociation constant Ka (acidity constant)

-logKa=pKa

The pKa is a measure of acid strength.The larger Ka, the stronger the acidThe smaler the pKa, the stronger the acid

The Henderson-Hasselbalch Equation

Defines:

(1) The pH of a solution

(2) The pKa of the acid

(3) Concentrations of the acid (HA) and conjugate base (A-).

When the solution contains [HA]=[A-], the pH = pKa

[HA][A-][H+] = Ka×

[H+][A-][HA]Ka =

HA A- + H+

Determination of pKa and acid (or base) concentration

• By titration - the technique of volumetric analysis.

• It is based on the measurement of the volume of one reactant required to react with a measured volume of another reactant in the chemical reaction.

• Titrant (reactant of known concentration) is added drop by drop to the measured volume of the solution of unknown concentration until the reaction is complete - when all the particles of the substance have reacted with the particles of the titrant. This is called reaction endpoint or equivalence point.

Acid - Base Titration• An acid-base titration is the determination of the concentration of an

acid or base by exactly neutralizing the unknown concentration of acid/base with an acid or base of known concentration.

• This allows to determine the concentration of an unknown acid or base solution.

• It makes use of the neutralization reaction that occurs between acids and bases and the knowledge of how acids and bases will react if their formulas are known.

Acid + Base → Salt + H2OHCl + NaOH → NaCl + H2O

Indicators -

• An acid-base indicator is a weak organic acid (H-Ind) that has a different colour than itsconjugate base (Ind-) with the colour changeoccuring over a specific and relatively narrow pH range.

H-Ind [H+] + [Ind-]

Ind][H][Ind logpKpH ind −

+=−

To assess the pH range at which indicator work the Henderson -Hasselbalch equation is used:

Indicators are compounds which dissociation dependson the pH of solution.

Indicator Color Range of color changing

Methyl orangeIonized (Ind-)Non-ionized (H-Ind)

RedYelow

3.2-4.4

PhenolphtaleinIonized (Ind-)Non-ionized (H-Ind)

PinkColorless

8.2-10

A typical acid-base titration requires a:

• Burette • Pipette • Indicator – changes color

based on pH. • Flask• Analyte – “unknown” solution

to be titrated • Titrant – solution of known

concentration

Titration curve of acetic acid CH3COOH

• Titration curves are used to determine pKa values

CH3COOH + NaOH ↔ CH3COO- Na+ + H2O

• The middle point of thecurve indicates the pKvalue.

• At the middle point, theconcentration of acid (or a proton donor) is equal to theconcentration of conjugatebase (or proton acceptor).

Calculating molarity in acid-basetitrations

Example 1:What is the molarity of an HCl solution if 18 mL of a 0.25 M NaOH are required to neutralize 10 mL of the acid?Solution:HCl + NaOH → NaCl + H2O0.018L NaOH × 0.25 M = 0.0045 moles NaOH1 mole of HCl reacts with 1 mole NaOH, therefore: 0.0045 moles HCl.0.0045 moles HCl0.01 L HCl

Example 2:Calculate the L of 2 M H2SO4 required to neutralize 50 mL of 1 M KOH.Solution:H2SO4 + 2KOH → K2SO4 + 2H2O0.05 L KOH × 1 M = 0.05 mol KOH1 mole of H2SO4 reacts with 2 moles of KOH, therefore 0.05 × 1/2=0.025 moles H2SO4. 2 mol H2SO4 - in 1 L; 0.025 - in x. x=0.025/2=0.0125 L

= 0.45 M HCl

Buffers

• When an acid or base is added to water, the pH changes drastically.

• A buffer solution resists a change in pH uponaddition of an acid or a base or upon dilution.

• Contain a combination of acid-base conjugatepairs, usually a weak acid and its conjugatebase.

A solution capable to maintain its own acidity almost constant is considered as a buffer.

Buffering capacity• Buffering capacity - the ability of buffer to resist against pH changing:

• Buffering capacity is expressed as the as quantity of an acid or a base which should be added into 1 L of buffer solution to induce pH change by 1 unit.

• The buffer capacity depends on the nature and concentration of the buffer components as well as on the ratio of these concentrations:

- The buffer capacity increases when the concentration of buffer components increases;

- The capacity for both acid and base is highest in a buffer where ratio of concentrations equals 1.

01 pHpHCB−

=C - as quantity of acid or base added;pH0 - initial pH of the buffer solution;pH1 - pH of the buffer after acid or base addition

[weak acid] = [conjugate base]- best buffering occurs at pKa range +/- 1

[A-][HA]

pH = pKa +log

• The pKa of acetic acid is 4.74, • So acetic acid and acetate ion buffer

functions well over pH range 3.74 –4.74

Buffer action (1)• The acetic acid/acetate buffer contains:

– acetic acid (CH3COOH) – sodium acetate (CH3COONa).

• The salt dissociates completely and producessodium and acetate ions:CH3COONa → CH3COO- + Na+

• Dissociation of the weak acid is diminisheddue to the presence of ions of the same type (CH3COO-)

CH3COOH CH3COO- + H+

• Therefore, large amount of both CH3COO-

and CH3COOH is present in the solution.

Buffer action (2)

• The weak acid in a buffer neutralizes base:CH3COOH + OH- → CH3COO- + H2O

• The conjugate base in the buffer neutralizesacid:CH3COO- + H+ → CH3COOH

• The pH of the solution is maintained.

The Henderson-Hasselbalchequation describes the pH of a

buffer solution:

[A-][HA]

pH = pKa +log

pH calculations with buffers: Example

What is the pH of a H2CO3 buffer that is 0.2 M H2CO3 and 0.1 M HCO3

-?Ka(H2CO3) = 4.3 × 10-7.

Solution:

]CO[H][HCOlogpKpH

32

3a

−

+=

6.070.3)(6.370.20.1log]10log[4.3pH 7 =−+=+×−= −

• Reactions of metabolism produce about 22 000 meq of acids per day. If we place such amount of acids in water, whose volume is equal to the volume of body fluids, this water acidity would be equal to 1 unit of pH.

• Blood pH of a healthy individual is 7,35-7,45.• Intracellular medium has pH 6,9-7,4• Exstracellular medium pH 6,8-7,8.• ↓ or ↑ by > 1/10 unit can be fatal.• Blood, cerebrospinal fluid, extracellular fluid and

intracellular fluid are capable to resist against acidic orbasic compounds due to presence of buffer systems.

Acidity of fluids of humanorganisms

The main buffers systems in our body

• 1. Bicarbonate-carbonic acid buffer system; it acts in blood and extracellular medium.

• 2. Phosphate buffer; it acts intracellullarly.• 3. Hemoglobin buffer system efficient in RBC.• 4. Protein buffer system is efficient in

intracellular medium and in blood plasma.

Carbonic acid - bicarbonatebuffer system

• The principal buffer system in extracellularfluid and in blood.

• It is composed of:– H2CO3 (proton donor), weak acid– HCO3

- (proton acceptor), conjugate base

• In normal conditions metabolic reactions of human body produce 0.5-1.0 kg of CO2.

• CO2 combines with water producing carbonic acid:

CO2+ H2O → H2CO3 ↔ HCO3- + H+

Carbonic anhydrase

Carbonic acid -proton donor (weak acid)

Bicarbonate ion – proton acceptor (conjugate base)

Dissociation

• pKa of carbonic acid is 6.4; therefore at normal blood acidity (7.4) almost all molecules of this acid dissociate into ions.

Mechanism of Carbonic acid -bicarbonate buffer action

• Binding of acids (not neutralization in chemical terms)

• H+ + HCO3-↔ H2CO3 ↔CO2↑+ H2O

• Neutralization of bases• OH- + H2CO3 ↔H2O+ HCO3

-↑

Carbonic acid - bicarbonatebuffer system

][CO][HCO log6.4pH

2

3−

+=

2CO

3

P0.03][HCO log6.4pH

×+=

−

blood in CO of pressure partialP 2CO2−

CO2↑ - pH↓; CO2 ↓ - pH ↑

In body fluids, there is more dissolved CO2 than H2CO3. Amount of dissolved CO2 is expressed in pressure units as partial pressure of CO2 - pCO2 in mmHg.

Effective buffer in pH ranges 6.4 ± 1 [5.4-7.4]

• Changes in carbonic acid concentration can be effected within seconds through increased or decreased respiration.

• Changes in hydrogen carbonate ion concentration, however, require hours through the relatively slow elimination through the kidneys

The concentrations of HCO3- and of H2CO3 are controlled by two

independent physiological systems:1. H2CO3 concentration is controlled by respiration (through the

lungs). Carbonic acid is in equilibrium with dissolved CO2:H2CO3 ↔ CO2 + H2O

2. The concentration of HCO3- is controlled through the

kidneys. Excess hydrogen carbonate ions are excreted in the urine.

Phosphate buffer systemActs in intracellular mediumIt is composed of a mixture of two phosphate salts:

– H2PO4- (proton donor)

– HPO42- (proton acceptor)

• H3PO4 ↔ H+ + H2PO4− (pKa 2.12)

• H2PO4− ↔ H+ + HPO4

2− (pKa 7.21) Dominant!• HPO4

2− ↔ H+ + PO43− (pKa 12.67) unlikely

Response of phosphate buffersystem

]PO[H][HPO log7pH -

42

24−

+= 2.

H2PO4- + OH- ↔ HPO4

2- + H2O

HPO42- + H+ ↔ H2PO4

2-

Effective buffer in pH ranges 7.2 ± 1 [6.2-8.2]

• Important in ICF and ECF• Amino acid buffers and plasma protein

buffers:~ are slower than other chemical buffers;~ remove either excess H+ or excess OH-

depending on pH;~ if pH ↑, amino acid acts like an acid;~ if pH ↓, amino acid acts like a base.

Protein buffer systems

Abnormalities of acid-base balance:Acidosis and alkalosis

• Acidosis - pH < 7.35 – excess of acid in body or deficiency of alkali

• Alkalosis –pH > 7.45 excess of alkali in blood or deficiency of acid.

• Metabolic reactions causes is production of acids such as acetoacetic, lactic and pyruvic. When production of those acids exceeds capacidyof buffers, pH of body fluids decreases. This is metabolic acidosis.

• Intoxication by alkaline compounds, vomiting, hyperaldosteronism can lead to metabolic alkalosis.

• Respiratory diseases cause blood to loose CO2, therefore pH ↑. This is respiratory alkalosis.

• Some respiration diseases (obstruction) in which respiration rate decreases, so that CO2 release diminishes. This leads to respiratory acidosis.