Acid/Base Chemical EquilibriaAcid/Base Chemical Equilibria

The BrThe Brøønsted Definitionsnsted Definitions

Brønsted Acid proton donor

Brønsted Base proton acceptor

Conjugate acid - base pair an acid and its conjugate base or a base and its conjugate acid

Example Acid-Base ReactionsExample Acid-Base Reactions

Look at acetic acid dissociatingCH3COOH(aq) CH3COO-(aq) + H+(aq)

Brønsted acid Conjugate base

Look at NH3(aq) in waterNH3(aq) + H2O(l) NH4

+(aq) + OH-(aq) Brønsted base conjugate acid

Representing Protons in Aqueous Representing Protons in Aqueous SolutionSolution

CH3COOH(aq) CH3COO-(aq) + H+(aq)

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

HCl (aq) Cl-(aq) + H+(aq)

HCl(aq) + H2O(l) Cl-(aq) + H3O+(aq)

What is HWhat is H++ (aq)? (aq)?

+H O

H

HH3O+

HO

H

H +O

H

H

H5O2+

+H

HO

OH2H2O

H2O

+

H

H

H+

H9O4+

Representing ProtonsRepresenting Protons

Both representations of the proton are equivalent.

H5O2+ (aq), H7O3

+ (aq), H9O4+ (aq) have

been observed.

We will use H+(aq)!

The Autoionization of Water The Autoionization of Water

Water autoionizes (self-dissociates) to a small extent

2H2O(l) H3O+(aq) + OH-(aq)

H2O(l) H+(aq) + OH-(aq) These are both equivalent definitions of the

autoionization (self ionization) reaction. Water is acting as a base and an acid in the above reaction water is amphoteric.

The Autoionization The Autoionization EquilibriumEquilibrium

From the equilibrium chapter

)(

)( )(

)(

)( )(

2

3

2

OH

OHOH or

OH

OHH = Kc

But we know (H2O) is 1.00 (pure liquids are not included in an equilibrium expression)!

The Defination of KThe Defination of Kww

Kw = (H+) (OH-)

Ion product constant for water, Kw, is the product of the activities of the H+ and OH-

ions in pure water at a temperature of 298.15 K

Kw = (H+) (OH-) = 1.0x10-14 at 298.2 K

The pH scaleThe pH scale

Attributed to Sørenson in 1909

We should define the pH of the solution in terms of the hydrogen ion concentration in solution

pH -log (H+)

Determination of pHDetermination of pH

What are we really measuring when we measure the pH?

pH -log (H+)

(H+) is the best approximation to the hydrogen ion activity (physical chemistry term) in solution.

How do we measure (H+)?

Equilibria in Aqueous Solutions Equilibria in Aqueous Solutions of Weak Acids/ Weak Basesof Weak Acids/ Weak Bases

By definition, a weak acid or a weak base does not ionize completely in water ( <<100%). How would we calculate the pH of a solution of a weak acid or a weak base in water?

To obtain the pH of a weak acid solution, we must apply the principles of chemical equilibrium.

Equilibria of Weak Acids in Equilibria of Weak Acids in WaterWater: The K: The Kaa Value Value

Define the acid dissociation constant Ka

For a general weak acid reaction

HA (aq) H+ (aq) + A- (aq)

HA

AHKa

Equilibria of Weak Acids in Equilibria of Weak Acids in WaterWater

For the dissolution of HF(aq) in water.

HF (aq) H+ (aq) + F- (aq)

The small value of Ka indicates that this acid is only ionized to a small extent at equilibrium.

41017

xHF

FHKa .

Equilibria of Weak Bases in Equilibria of Weak Bases in WaterWater

To calculate the percent dissociation of a weak base in water (and the pH of the solutions)

CH3NH2 (aq) + H2O CH3NH3+(aq) + OH- (aq)

We approach the problem as in the case of the weak acid above, i.e., from the chemical equilibrium viewpoint.

The KThe Kbb Value Value

Define the base dissociation constant Kb For a general weak base reaction with water

B (aq) + H2O (aq) B+ (aq) + OH- (aq)

BOHB

K b

Examples of Acid-Base Examples of Acid-Base CalculationsCalculations

Determining the pH of a strong acid (or base solution).

Determining the pH of a weak acid (or base solution).

Calculating the pH of Solutions Calculating the pH of Solutions of Strong Acidsof Strong Acids

For the dissolution of HCl, HI, or any of the other seven strong acids in water

HCl (aq) H+ (aq) + Cl- (aq) HI (aq) H+ (aq) + I- (aq) % eq = 100%

The pH of these solutions can be estimated from the molarity of the dissolved acid

pH -log [H+]

For the dissolution of NaOH, Ba(OH)2, or any of the other strong bases in water

NaOH (aq) Na+ (aq) + OH- (aq) Ba(OH)2 (aq) Ba2+ (aq) + 2OH- (aq)

% eq = 100%

Calculating the pH of Solution of Calculating the pH of Solution of Strong BasesStrong Bases

Calculating the pH of Solution of Calculating the pH of Solution of Strong BasesStrong Bases

The pH of these solutions is obtained by first estimating the pOH from the molarity of the dissolved base

pOH -log [OH-]

pOH -log[Ba(OH)2]

pH = 14.00 - pOH

The Definition of a Buffer The Definition of a Buffer

Buffer a reasonably concentrated solution of a weak acid and its conjugate base that resists changes in the pH when an additional amount of strong acid or strong base is added to the solutions.

How would we calculate the pH of a buffer solution?

pH of a BufferpH of a Buffer

HCOOH

HCOOHKa

HCOOH

HCOOHKpK aa

loglog

pH of a BufferpH of a Buffer

HCOOHHCOOHpK a logloglog

note pH = -log (H+)

Define pKa = -log (Ka )

The Buffer EquationThe Buffer Equation

HCOOH

HCOOpHpKa

)(log

Substituting and rearranging

HCOOH

HCOOpKpH a

)(log

The Generalized Buffer The Generalized Buffer EquationEquation

The pH of the solution determined by the ratio of the weak acid to the conjugate base. This equation (the Henderson-Hasselbalch equation) is often used by chemists, biochemists, and biologists for calculating the pH of a solution of a weak acid and its conjugate base!

acid

weak

baseconjpKpH a

).(log

The Generalized Buffer EquationThe Generalized Buffer Equation

Note: The Henderson-Hasselbalch equation is really only valid for pH ranges near the pKa of the weak acid!

The Generalized Buffer EquationThe Generalized Buffer Equation

Buffer CH3COONa (aq) and CH3COOH (aq))

CH3COOH (aq) ⇄ CH3COO- (aq) + H+ (aq)

The Equilibrium Data Table

n(CH3COOH) n(H+) n(CH3COO-)

I A 0 B

C -x + x +x

E (A-x) (x) (B+x)

The pH of the solution will be almost entirely due to the original molarities of acid and base!!

][

logA

BpKpH a

This ratio will be practically unchanged in the presence of a small amount of added strong acid or base

The pH of the solution changes very little after adding strong acid or base (i.e., it is buffered)

How does the pH change after the addition of strong acid or base?

Example of Buffer Example of Buffer CalculationsCalculations

How do we calculate the pH of a buffer solution?

The pH of a Buffer SolutionThe pH of a Buffer Solution

The major task in almost all buffer calculations is to obtain the ratio of the concentrations of conjugate base to weak acid!

Using the Ka of the appropriate acid, the pH of the solution is obtained from the Henderson-Hasselbalch equation.

Adding Strong Acid or Base to Adding Strong Acid or Base to Buffer Solutions Buffer Solutions

To obtain the pH after the addition of a strong acid or base, we must calculate the new amount of weak acid and conjugate base from the reaction of the strong acid (or base) to the buffer system.

The pH of the solution may again be calculated with the Henderson-Hasselbalch equation.

Buffer CapacityBuffer Capacity

Buffer solutions do not have unlimited capacity to keep the pH relatively constant. When the weak acid of a buffer is used up, no more buffering action toward additional base is possible.

The buffer capacity is the amount of acid or base that can be added to a certain volume of a buffer solution before the Ph begins to change significantly.

Buffer CapacityBuffer Capacity

If you need to make a buffer solution up to a certain pH, use the Henderson-Hasselbalch Equation.

pH = pKa + log

A- = conjugate base (salt) HA = acid Generally easier to use than solving quadratic.

Calculating the pH of a Buffer Calculating the pH of a Buffer Using the H-H Eq.Using the H-H Eq.

What is the pH of a buffer solution that is 0.10 M chloroacetic acid and 0.15 M sodium chloroacetate? Ka = 1.4 X10-3

pH = pKa + log

ExampleExample

pH = pKa + log

pH = -log(1.4 X10-3) + log[0.15M]/[0.10M]

pH = 3.03

Calculating the pH of a Buffer from Calculating the pH of a Buffer from Given Volumes of SolutionGiven Volumes of Solution

What is the pH of a buffer that contains 60 mL of 0.100 M NH3 with 40 mL of 0.100 M NH4Cl?

NH3 + H2O NH4+ + OH-

The problem is to obtain the concentration of OH- in this mixture. We must, then, calculate the new concentrations of the NH3 and NH4

+.

ExampleExample

NH3 M = 0.001 moles/L = X/0.060L

X = 0.0060 mols NH4

+ M= 0.100 moles/L = X/0.40L

X = 0.0040 mols NH3 molarity = 0.0060 mols/0.100L

= 0.060M NH4

+ M = 0.0040 mols/0.100L = 0.040M

ExampleExample

Must use Kb since a weak base. Create table.

NH3 + H20 NH4+ + OH-

Initial 0.060 0.040 0

Change -X X X

Equili 0.060 - X 0.040+ X X

First Way to CompleteFirst Way to Complete

Kb =

1.8 X 10-5 = (0.040 + X) (X)/(0.060 – X) Assume X is small and delete 1.8 X 10-5 = 0.040 X/0.060 X = 2.7 X 10-5 = [OH-] pOH = -log[2.7 X 10-5 ] = 4.56 pH = 14 - 4.56 = 9.43

Second Way to Complete Using the Second Way to Complete Using the Henderson-HasslebachHenderson-Hasslebach

pOH = pKb + log[salt]/[acid]

pOH = -log[1.8 X 10-5] + log 0.040/0.060 pOH = 4.74 + -0.176 pOH = 4.56 pH = 9.43