Monoprotic Acid-Monoprotic Acid-Base EquilibriaBase Equilibria

Kw = [H+] [HO-] = 1.0 x 10-14

-log Kw = pH + pOH = 14.00 at 25oC

So what is the pH of 1.0 x 10-8 M KOH?

[H+] = Kw / [HO-] =(1.0 x 10-14)(1.0 x 10-8)

=1.0 x 10-6

pH = 6.00Incorrect !! KOH is a base so pH >7.

What’s wrong ?

Water is an acid. It ionizes to form H+ and HO-

In pure water: [HO-] = 1.0 x 10-7 M

Hence the reactions in the solution are:

KOH K+ + HO-

H2O H+ + HO-

The chemical species in the solution are K+, HO- and H+

The charge balance is :

[K+] + [H+] = [HO-] (1)

The mass balance is :

[K+] = 1.0 x 10-8 M (2)

The equilibrium constant expression :

Kw = [H+] [HO-] (3)

From (1) : [HO-]= [K+] + [H+]

= 1.0 x 10-8 + [H+] (4)

Putting (4) into (3) :

[H+](1.0 x 10-8 + [H+]) = 1.0 x 10-14

[H+] = 9.6 x 10-8 M or –1.1 x 10-7 M

Rejecting the negative value, we have :

When concentration is low ( 10-8 M), the pH is 7.00. There is insufficient acid or base to significantly affect the pH of water itself

At intermediate concentration (~ 10-6

to 10-8 M), the effects of water ionization and the added acid or base is comparable

[H+] = 9.6 x 10-8 M

pH = -log [H+] = 7.02

When concentration is high ( 10-6 M), the pH has the value we would calculate by just considering the added H+ or HO-

Weak Acids and Weak Acids and BasesBasesA weak acid is one that not

completely dissociated

Examples of weak acids : carboxylic acids, alcohols

HA + H2O H3O+ + A-

Acid-dissociation constant, Ka =

Similarly, a weak base is one whose dissociation does not go to completion

Examples of weak base: primary and secondary amines

B + H2O BH+ + HO-

Base-dissociation constant, Kb =

[H+][A-]

[HA]

[BH+][HO-]

[B]

Salts of Weak Acids Salts of Weak Acids and Basesand Bases

Recall : In the Brønsted and Lowry classification: the products of a reaction between an acid and a base are also classified as acids and bases. Example :

acetic acid methylamine acetate ion methyl-

ammonium ion

acidconjuga

te base

CH3 C

O

O H

+ CH3 NH

H

CH3 C

O

O -+ CH3 N

HH

H

+

base conjugate acid

Hence the anion of a weak acid, eg -

OAc, is a Brønsted base, which will accept protons :

-OAc + H2O HOAc + -

OH

The stronger the conjugate base the more strongly the salt will combine with a proton, as from water, to shift the ionization to the right.

Equilibrium constant:

KH = Kb =[HOAc][HO-]

[AcO-]

hydrolysis constant

Multiply numerator and denominator by [H+]

Kb =

=

= Kw = KaKb

[HOAc][HO-]

[AcO-]

[H+][H+]

[HOAc] Kw

[AcO-][H+]

Kw

Ka

Similar equation can be derived for the cations of salts of weak bases, ie :

BH+ + H2O B + H3O+

Ka = =

[B][H3O+]

[BH+]

Kw

Kb

Example :

Consider a weak acid, HA with Ka =10-4

The conjugate base, A- has Kb = Kw/Ka = 10-10

the conjugate base of a weak acid is a weak base

If Ka was 10-5, then Kb = Kw/Ka = 10-9

as HA becomes a weaker acid, A- becomes a stronger base (but never a strong base)

Weak Is Conjugate to Weak Is Conjugate to WeakWeak

Weak Acid EquilibriaWeak Acid Equilibria

How does one find the pH of a solution of a 0.0500 M weak acid, HA ? (Given: Ka = 1.07 x 10-3)

HA + H2O H3O+ + A-

-A + H2O HA + -OH

2H2O H3O+ + -OH

Charge balance: [H3O+ ] = [A-] + [-OH] (1)

Mass balance: CHA= [A-] + [HA] (2)

= 0.0500 M

Equilibria: Ka = (3)

[H3O+][A-][HA]

In weak acid, the concentration of H3O+ due to acid dissociation >>the H3O+ from water dissociation. Thus,

[A-] >> [HO-]

Thus (1) becomes: [H3O+ ] [A-]

From (3) :

1.07 x 10-3 =

[H3O+ ] = 6.80 x 10-3 M

= [A-]

pH = -log (6.80 x 10-

3 )= 2.17

[H3O+][H3O+]

0.0500 - [H3O+]

Kw = [H3O+ ][-OH] (4)

There are 4 chemical species ([A-], [-OH], [HA] and [H3O+ ]) and 4 quations

[HA] = 0.0500 - [H3O+ ]

= 0.0432 M

[-OH] = Kw /[H3O+ ]

= 1.47 x 10-12 M

[H3O+ ] from H2O dissociation = 1.47 x 10-12

M

[H3O+ ] from HA dissociation = 6.80 x 10-3 M

The assumption that [H3O+ ] is derived

mainly from HA is acceptable

Fraction of Fraction of DissociationDissociation

The fraction of dissociation, , is defined as the fraction of the acid HA that forms A-

= =

=

For a solution of a 0.0500 M weak acid, HA, with Ka = 1.07 x 10-3, the fraction of dissociation is :

=

= 0.136

= 13.6%

[A-]

[A-] + [HA]

[A-]

[A-] + (CHA - [A-])

[A-]

CHA

(6.80 x 10-3)0.0500

Weak-Base EquilibriaWeak-Base Equilibria

The treatment of weak bases is almost the same as that of weak acids

B + H2O BH+ + HO-

2H2O H3O+ + -OH

Kb =

Charge balance : [BH+] + [H3O+] = [HO-]

Mass balance : CB = [BH+] + [B]

Assuming that nearly all of the HO- comes from the reaction of B + H2O [BH+] [HO-]

Kb =

If CB = 0.0372 M and Kb =2.6 x 10-6, then

[BH+][HO-]

[B]

[BH+] [BH+]CB – [BH+]

[BH+] = 3.1 x 10-4 M

= [HO-] from base hydrolysis

From : CB = [BH+] + [B]

[B] = 0.0372 – (3.1 x 10-4)

= 0.03689 M

From : [H3O+] = Kw /[HO-]

= (1.0 x 10-14)/(3.1 x 10-4)

= 3.2 x 10-11 M

= [HO-] from water hydrolysis

pH = -log (3.2 x 10-11)

= 10.49

Fraction of association =

= 0.0083 =0.83%

[BH+][BH+] + [B]

BuffersBuffers

A buffer :

- resists changes in pH when acids or bases are added or when dilution occurs

- is a composed of a weak-acid and its salt (conjugate base) or a weak-base and its salt (conjugate acid)

If A moles of a weak acid is mixed with B moles of its conjugate base, the moles of acid remain close to A and the moles of base remain close to B. Why ?

Consider an acid, 0.10 M solution of HA, with pKa = 4.00 and its conjugate base,

0.10 mol A-, with pKb = 10.00

HA H+ + A- pKa = 4.00

Let x be the concentration of acid that dissociates

[H+][A-]

[HA]x2

CHA - xKa = =

= = 1.0 x 10-4

x =3.1 x 10-3

Fraction of dissociation, = = 0.031

This means that only 3.1% of the acid has dissociated

HA dissociates very little. Adding extra A- to the solution will make HA dissociate even less.

What about the reaction of A- with water ?

x2

0.10 - x

x

CHA

A- + H2O HA + HO- pKb = 10.00

Let y be the concentration of A- that reacts with water

Kb = = = 1.0 x 10-10

y = 3.2 x 10-6

Fraction of association = = 3.2 x 10-5

The degree of association is 3.2 x 10-3 %

A- does not react very much with water and the presence of HA makes A- react even less with water.

Hence if 0.10 mol A- and 0.10mol HA are added to water, there will be close to 0.10 mol A- and close to 0.10mol HA in the solution in equilibrium.

This approximation however breaks down for dilute solutions or at extremes of pH

yCA

-

[HA][HO-][A-]

y2

0.10 - y

Henderson-Henderson-Hasselbalch EquationHasselbalch Equation

The Henderson-Hasselbalch equation is :

- a central equation for buffers

- basically a rearranged form of the Ka equilibrium expression

Ka =

log Ka = log

= log[H+] + log

-log [H+] = -log Ka + log

pH = pKa + log

[H+][A-]

[HA]

[H+][A-]

[HA][A-][HA]

[A-][HA]

[A-][HA]

The Henderson-Hasselbalch equation tells us the pH of a solution provided we know the ratio of the concentrations of conjugate and base and the pKa for the acid

If a solution is prepared from a weak base B and its conjugate acid, the equation becomes

pH = pKa + log

where pKa = acid dissociation constant of the weak acid BH+

[B][BH+]

What does the equation tell us:

pH = pKa + log

HA H+ + A-

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

- for every power-of-10 change in the ratio [A-] /[HA] , the pH changes by 1 unit

- as the base concentration ([A-] or [B]) increases, the pH goes up. Similarly, as the acid concentration increases, the pH comes down

[A-][HA]

Example :

What is the ratio of [H2CO3]/[HCO3-] in the

blood buffered to a pH of 7.40 ? Given: Ka

= 4.4 x 10-7 M

H2CO3 HCO3- + H+

Ka = = 4.4 x 10-7 M

=

pH = -log[H+] = 7.40

[H+] = 10-7.40

= = 9.0 x 10-2

[H+][HCO3-]

[H2CO3]

[H2CO3]

[HCO3-] Ka

[H+]

[H2CO3]

[HCO3-]

10-7.40

4.4 x 10-7

Preparing a BufferPreparing a Buffer

What should you do if you would wish to prepare 1.00 l of buffer containing 0.100 M tris at pH 7.60 from solid tris hydrochloride and ~1M NaOH?

Step 1: Weigh out 0.100 mol of tris hydrochloride and dissolve it in a beaker containing about 800 ml of water

Step 2: Place a pH electrode in the solution and monitor he pH

Step 3: Add NaOH solution until the pH is exactly 7.60

Step 4: Transfer the solution to a volumetric flask and wash the beaker a few times. Add the washings to the volumetric flask

Step 5: Dilute to the mark and mix thoroughly

Reason for adding about 800 ml of water is to make the volume reasonably close to the final volume during pH adjustment. Addition of large quantities of solvent will affect the ionic strength affect the pH

Addition of Strong Addition of Strong Acid/Base to a BufferAcid/Base to a Buffer

HA H+ + A-

pH = pKa + log

What happens when a small amount of strong acid/base is added to the buffered solution ?

If a small amount of a strong acid is added :

- it will combine with an equal amount of the A- to convert it to HA.

- the change in the ratio [A-]/[HA] is small and hence the change in pH is small

If a small amount of a strong base is added :

- it will combine with part of the HA to form an equivalent amount of A-

- the change in the ratio [A-]/[HA] is small and hence the change in pH is small

[A-][HA]

Buffer CapacityBuffer Capacity

The buffer capacity, , is a measure of how well a solution resists changes in pH when strong acid or base is added.

where Ca and Cb are the number of moles

of strong acid and strong base per liter needed to produce a unit change in pH.

The larger the value of the more resistant the solution is to pH change

is determined by the concentrations of HA and A-.

The higher the HA and A- concentrations, the more acid or base the solution can tolerate

dCb

dpHdCa

dpH

A buffer is most effective in resisting

changes in pH when pH = pKa

maximum buffer

capacity