Buffers - ufjf.br

Buffers

A buffered solution resists changes in pH when small amounts of

acids or bases are added or when dilution occurs.

The buffer consists of a mixture of an acid and its conjugate base.

Example: acetic acid/sodium acetate

The Henderson-Hasselbalch equation

][

][log

][

][loglog]log[

][

][log]log[

][

]][[loglog

][

]][[

HA

ApKpH

HA

AKH

HA

AH

HA

AHK

HA

AHK

a

a

a

a

pH pKa

HA(aq) ⇌ H+(aq) + A-(aq)

For base

][

][log

][

][loglog]log[

][

][log]log[

][

]][[loglog

][

]][[

B

BHpKpOH

B

BHKOH

B

BHOH

B

BHOHK

B

BHOHK

b

b

b

b

When [A-] = [HA], pH = pKa

Regardless of how complex a solution may be, whenever pH = pKa, [A-] must

equal [HA].

a

a

a

pKpH

HK

A

AH

HA

AHK

]log[log

][

]][[

][

]][[

Application of Henderson-

Hasselbalch equation

Sodium hypochlorite (NaOCl, the active ingredient of bleach) was dissolved in a

solution buffered to pH 6.20. Find the ratio [OCl-]/[HOCl] in this solution.

pKa = 7.53 for HOCl

HOCl(aq) + H2O(l) ⇌ H3O+(aq) + OCl-(aq)



pKa = 7.53 for HOCl



][

][log33.1

][

][log53.720.6

][

][log

HOCl

OCl

HOCl

OCl

HOCl

OClpKpH a

][

][047.0

1010 ][

][log

33.1

HOCl

OCl

HOCl

OCl

HOCl(aq) + H2O(l) ⇌ H3O+(aq) + OCl-(aq)





pKa = 7.53 for HOCl

Problems

1. What is the pH of a buffered prepared by 0.12 mol L-1 lactic acid

(CH3CH(OH)COOH) and 0.10 mol L-1 sodium lactate (CH3CH(OH)COONa)

solution? Ka = 1.4 x 10-4.

2. Find the pH of a buffered solution prepared by 0.040 mol L-1 NH4Cl(aq) and 0.030

mol L-1 NH3(aq) solution. Kb = 1.8 x 10-5.

3. What weight of NH4Cl (M. W. = 53.5 g mol-1) should be added to 100 mL of 0.1

mol L-1 NH4OH solution to give a pH of 9? Kb = 1.8 x 10-5.

4. A solution of equal concentrations of lactic acid and sodium lactate has pH = 3.08.

(a) What are the pKa and Ka values of lactic acid? (b) What would be the pH if the

acid concentration was twice the salt concentration?

Effect of adding acid to a buffer

If we add 100 mL of 0.1 mol L-1 HCl solution to a one liter of 0.1 mol L-1 acetic acid and

0.1 mol L-1 sodium acetate solution (Ka = 1.75x10-5), what will the pH be?




pKa = 4.76CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)





Initial:

CH3COO- n = C V = 0.1 x 1.0 = 0.10 mol

CH3COOH n = C V = 0.1 x 1.0 = 0.10 mol

H3O+ n = C V = 0.1 x 0.1 = 0.01 mol





Initial:

CH3COO- n = C V = 0.1 x 1.0 = 0.10 mol

CH3COOH n = C V = 0.1 x 1.0 = 0.10 mol

H3O+ n = C V = 0.1 x 0.1 = 0.01 mol

Final:

CH3COO- n = 0.10 – 0.01 = 0.09mol

CH3COOH n = 0.10 + 0.01 = 0.11mol

H3O+ totally reacted

Disregarding the change in volume

67.409.076.411.0

09.0log76.4 pH

The preceding example illustrates that the pH of a buffer does not change very

much when a limited amount of a strong acid or base is added.





Initial:

CH3COO- n = C V = 0.1 x 1.0 = 0.10 mol

CH3COOH n = C V = 0.1 x 1.0 = 0.10 mol

H3O+ n = C V = 0.1 x 0.1 = 0.01 mol

Final:

CH3COO- n = 0.10 – 0.01 = 0.09mol

CH3COOH n = 0.10 + 0.01 = 0.11mol

H3O+ totally reacted

Disregarding the change in volume

The preceding example illustrates that the pH of a buffer does not change very

much when a limited amount of a strong acid or base is added.

67.409.076.411.0

09.0log76.4 pH

The same volume of HCl added to one liter of

water (pH = 7) would lead to a pH = -log 0.01 =

2 solution, a change of 5 pH units. This shows

how a buffer solution regulates the pH.

Buffer capacity

1a

pKpH

Buffer capacity measure how well a solution resists changes in pH when acid or

base is added. The greater the buffer capacity, the less the pH changes.

Effect of adding 0.01 mol of H+ or

OH- to a buffer containing HA and

A- (total quantity of HA + A- = 1

mol). The minimum change in pH

occurs when the initial pH of the

buffer equals pKa for HA. That is,

the maximum buffer capacity

occurs when pH = pKa.

Problem

1) When it is desired to study reactions of biological interest at pHs close to

neutrality, the buffer formed by the mixture of KH2PO4 and K2HPO4 is often used.

a) Find the pH of 0.2 mols of KH2PO4 and 0.1 mols of K2HPO4 dissolved in 1L

solution.

b) If you add 0.05 mols of nitric acid to a solution of item a, what will the new pH

be? Consider that the volume variation is negligible.

c) If you add 0.05 mols of NaOH to a solution of item a, what will the new pH be?

Consider that the volume variation is negligible.

Ka1 = 5.9x10-3; Ka2 = 6.2x10-8; Ka3 = 4.8x10-13

Problem

2) Two buffers were prepared:

a) 0.1 mol L-1 HAc + 0.1 mol L-1 NaAc

b) 0.1 mol L-1 HAc + 0.5 mol L-1 NaAc

25 mL of each solution was added in a flask a and b and on them was added 5

mL of 0.1 mol L-1 NaOH solution.

25 mL of each solution was added in a flask a and b and on them was added 5

mL of 0.1 mol L-1 HCl solution.

In which flask there was greater variation of pH when NaOH was added? And

HCl? Ka = 1.8 x 10-5

Problem

3) Which of the buffer systems below would be a good choice to prepare a buffer

with pH = 5? And pH = 10?

Composition pKa

CH3COOH / CH3COO- 4.75

HNO2 / NO2- 3.37

HClO2 / ClO2- 2.00

NH4+ / NH3 9.25

(CH3)3NH+ / (CH3)3N 9.81

H2PO4- / HPO4

2- 7.21

Mix

Two weak acids: (1) HA and (2) HL.

KHA>>>KHL

H2O ⇌ OH-(Aq.) + H+

(Aq.)

HA ⇌ H+(Aq.) + A-

(Aq.)

HL ⇌ H+(Aq.) + L-

(Aq.)

[H+] = [OH-] + [A-] + [L-]

Each species contribution:

From water: ][

][

H

KOH w

][

][][

H

HAKA HA

][

][][

H

HLKL HLand

Equilibrium of proton

][][][

][][][

][

][

][

][

][][

2

HLKHAKKH

HLKHAKKH

H

HLK

H

HAK

H

KH

HLHAw

HLHAw

HLHAw

HA

HAKH

HCHA

AHA

][

][][

][][cHA

HL

HLKH

HCHL

LHL

][

][][

][][cHL

Mix

By replacing the expressions of [HA] and [HL] in the equilibrium expression of the

próton, a fifth order equation is obtained in relation to the H+ ion. This is a complex

formulation, requiring calculations of successive approximation. A possible

simplification is to assume that: CHA [HA] e CHL [HL], thus a first approximation

of the hydrogen ion concentration can be obtained:

HLHLHAHAw

HLHLHAHAw

CKCKKH

CKCKKH

0

2

0

][

][

HA

HAKH

HCHA

0

01

][

][][

HL

HLKH

HCHL

0

01

][

][][

Mix

11

11

2

][][][

][][][

HLKHAKKH

HLKHAKKH

HLHAw

HLHAw

The approximation test is performed, taking the maximum error value in the

calculation of the hydrogen ion concentration as 1.0% ([HA]1 ≠ [HA]0 = 1.0%)

and restarting the cycle, if the error is greater than this value. It is a calculation

of successive approximation.

Mix

Acid-base equilibrium in non-

aqueous media

In order to treat the acid-base equilibrium in non-aqueous or partially aqueous

médium, two fundamental characteristics of the solvents must be considered:

the autoprotolysis constant and the dieletric constant.

The autoprotolysis constant reflects the amphiprotic or aprotic behaviour of the

solvent.

The dieletric constant reflects its ability to separate the charges, that is, to avoid

the formation of pair between oppositely charged ion.

Solvents can be divided into 4 different classes. (in order to facilitate the

discussion)

Protic solvents are polar and have as main characteristic the ability to

make a hydrogen bond, for example water. Aprotic solvents do not have

this characteristic. Ex: hexane.

First class

Polar amphiprotic solvents have a water-like behaviour and include methanol,

formic acid, water-alcohol mixtures, etc.. When an HA acid is dissolved in a polar

amphiprotic solvent HS, the following equilibrium reaction occurs:

where H2S+ is the solvated proton. The formal equilibrium treatment given to this

class of solvents is similar to that given to water.

HA + HS ⇌ H2S+ + A-

HA

ASH

aa

aaK

.

2

Acid pKa (H2O) pKa (MeOH) pKa (EtOH)

Acetic 4.75 9.7 10.4

Benzoic 4.20 9.4 10.1

Phenol 9.97 14.2 15.3

Second class

It comprises apolar amphiprotic solvents, the typical example is glacial acetic

acid. An HA acid, when dissolved in this type of solvent, ionizes and dissociates

according to the equations:

HA + HS H2S+A-

HA

ASH

ia

aK

2

H2S+A- H2S

+ + A-

ASH

ASH

da

aaK

2

2

.

As in amphiprotic solvent with low dielectric constant the ion pair formation is

inevitable, the ionization constant is the one that expresses exactly the force of

the acid or base.

The ionization is the acidity / basicity function of the solvent.

The dissociation is a function of the dielectric constant.

In solvents with a dielectric constant greater than 30-40, the ion pair formation is

negligible and the treatment is done as in water.

Acid pKa Base pKa

Percloric 4.87 Tribenzylamine 5.38

Sulfuric (K1) 7.24 Pyridine 6.10

Hydrochloric 8.55 2 5-dichloroaniline 9.48

Third class

It is the one that includes the aprotic polar, that are better known as aprotic

dipolar. They are relatively inert and can be subdivided into protophilic and

protophobic.

Protophilics: DMSO, DMF, Py - solved cations and can accept hydrogen bonds.

Protophobics: ACN, nitromethane - have little capacity to solvate cations and

form hydrogen bonds.

Acid pKa (ACN) pKa (DMSO)

Acetic 22.3 12.6

Benzoic 20.7 11.1

Salicylic 16.7 6.8

Fourth class

Aprotic apolar - inert

This class includes benzene, toluene, n-hexane, etc.

They have low dielectric constant (usually less than 10) and play no role in

neutralization reactions.

The equilibrium treatment in non-aqueous medium, as many be noted, is not as

simple as in water, and therefore, much care must be taken.

TITRATIONS IN ANALYTICAL CHEMISTRY

Titration

A titration is a procedure in which increments of the known reagent

solution – the tritant – are added to analyte until the reaction is

complete.

Titrant: is usually delivered from a buret.

Analyte: is contained in a flask.

Each increment of titrant should be completely and quickly consumed by

reaction with analyte until the analyte is used up.

Common titrations are based on acid-base, oxidation-reduction, complex

formation, or precipitation reactions.

Equivalence and end point

The equivalence point occurs when the quantity of titrant added is the

exact amount necessary for stoichiometric reaction with the analyte.

The equivalence point is the ideal result we seek in a titration. What we

actually measure is the end point, which is marked by a sudden change in

a physical property of the solution.

The end point cannot exactly equal the equivalence point. The difference

between the end point and the equivalence point is an inescapable titration

error.

endeequivalenc VVE

Requirements of a reaction employed

in a titration

✓Simple reaction with known stoichiometry

✓Quick reaction

✓Present chemical or physical changes (pH, temperature,

conductivity), mainly at the point of equivalence.

Methods

➢Visual indicator

Color change – is caused by the disappearance of analyte

or the appearance of excess titrant.

➢Instrumental methods

Change in voltage or current between a pair of electrodes

Monitoring the absorbance of light by species in the

reaction

Primary standard

A primary standard is a reagent that is pure enough to be

weighed out and used directly to provide a known number

of moles.

Ex: sodium carbonate

Requirements – primary standard

❖High purity. Established methods for confirming purity

should be available.

❖Stability toward air.

❖Absence of hydrate water so that the composition of the

solid does not change with variations in relative humidity.

❖Ready availability at modest cost.

❖Reasonable solubility in the titration medium.

❖Reasonably large molar mass so that the relative error

associated with weighing the standard is minimized.

Standardization

❖99.9% pure or better

❖It should not decompose under ordinary storage

❖It should be stable when dried by heating or vacum

We say that a solution whose concentration is known is a

standard solution.

Secondary standard

It is a compound which allows to prepare a titrant solution,

but its concentration is determined by comparison

(standardization) against a primary standard.

Ex: NaOH

Standard solutions

The ideal standard solution for a titrimetric method will

1. be sufficiently stable so that it is only necessary to determine its

concentration once;

2. react rapidly with the analyte so that the time required between

additions of reagente is minimized;

3. react more or less completely with the analyte so that satisfactory end

points are realized;

4. undergo a selective reaction with the analyte that can be described by

a simple balanced equation.

Types

Direct titration

Titrant is added to analyte until the end point is observed.

analyte + titrant product

Indirect titration

The reagent to be titrated is generated in the solution.

Back titration

A known excess of a standard reagent is added to the analyte. Then a second

standard reagent is used to titrate the excesso of the first reagent.

analyte + reagent 1 product + excess reagent 1

excess reagent 1 + reagent 2 product

Neutralization titration

Standard solutions of strong acids and strong bases are use extensively

for determing analytes that are themselves acids or bases or analytes that

can be converted to such species by chemical treatment.

The standard solutions employed in neutralization titrations are Strong

acids or strong bases because these substances react more completely

with na analyte than do their weaker counterparts and thus yield sharper

end points.

Strong base: NaOH, KOH

Strong acid: HCl, H2SO4, HClO4

OBS: HNO3 is rarely used due to its high oxidizing power.

Acid/Base indicators

An acid/base indicator is a weak organic acid or weak organic base whose

undissociated form differs in color from its conjugate base or its conjugate

acid form.

For example, the behaviour of an acid-type indicator, HIn, is described by

the equilibrium

HIn (acid color) + H2O ⇌ H3O+ + In- (base color)

Here, internal structural changes accompany dissociation and cause the

color change. The equilibrium for a base-type indicator, In, is

In (base color) + H2O ⇌ InH+ (acid color) + OH-


aa xKIn

HInOH

HIn

InOHK

][

][][

][

]][[3

3

The human eye is not very sensitive to color diferences in a solution containing

a mixture of In- and HIn, particularly when the ratio [In-]/[HIn] is greater than

about 10 or smaller than about 0.1.

1)10

1log()(

10

1][

10

1

][

][

1)10log()(10][1

10

][

][

3

3

aaa

aaa

pKKcolorbasicpHKOHIn

HIn

pKKcoloracidpHKOHIn

HIn

1 apKpHIndicator range:

PHENOLPHTHALEIN

transaction range: 8.3-10.0


PHENOL RED



RED

REDYELLOW

METHYL ORANGE



RED

YELLOW

Curve Titration

Sigmoidal curve

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Buffers - ufjf.br