1

10b 1.1x10K 0.20y

2

y = 1.0 x 10-5 M [OH-] = 1.0 x 10-5 M

10b 1.1x10K 0.20y

3

y = 1.0 x 10-5 M [OH-] = 1.0 x 10-5 M Hence pOH = 5.0

10b 1.1x10K 0.20y

4

y = 1.0 x 10-5 M [OH-] = 1.0 x 10-5 M Hence pOH = 5.0

Since pOH + pH = 14.0, therefore pH = 9.0

10b 1.1x10K 0.20y

5

Acid-base titrations: The impact of hydrolysis

6

Acid-base titrations: The impact of hydrolysis

Salt hydrolysis has an important effect on the pH profile of acid-base titrations.

7

Acid-base titrations: The impact of hydrolysis

Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7).

8

Acid-base titrations: The impact of hydrolysis

Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7).

For the titration of a strong acid and a strong base, the equivalence point should be at pH = 7.

9

Acid-base titrations: The impact of hydrolysis

Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7).

For the titration of a strong acid and a strong base, the equivalence point should be at pH = 7.

Example: HCl(aq) + NaOH(aq) NaCl(aq) + H2O

10

11

12

13

14

Indicators

15

Indicators Indicators are often very weak organic acids. We

will represent an indicator as HIn.

16

Indicators Indicators are often very weak organic acids. We

will represent an indicator as HIn. During a titration such as (where we assume NaOH

is being added)

17

Indicators Indicators are often very weak organic acids. We

will represent an indicator as HIn. During a titration such as (where we assume NaOH

is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O

18

Indicators Indicators are often very weak organic acids. We

will represent an indicator as HIn. During a titration such as (where we assume NaOH

is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O

The first drop of excess NaOH then reacts with the indicator that is present:

19

Indicators Indicators are often very weak organic acids. We

will represent an indicator as HIn. During a titration such as (where we assume NaOH

is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O

The first drop of excess NaOH then reacts with the indicator that is present:

HIn(aq) + OH-(aq) H2O + In-

(aq)

20

Indicators Indicators are often very weak organic acids. We

will represent an indicator as HIn. During a titration such as (where we assume NaOH

is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O

The first drop of excess NaOH then reacts with the indicator that is present:

HIn(aq) + OH-(aq) H2O + In-

(aq)

Now HIn and In- have different colors, so we can detect that the acid-base reaction is complete.

21

For the equilibrium: HIn(aq) H+(aq) + In-

(aq)

22

For the equilibrium: HIn(aq) H+(aq) + In-

(aq)

] [HIn]][In[HK

-

In

23

For the equilibrium: HIn(aq) H+(aq) + In-

(aq)

Midway in the transition of the indicator color change: [HIn] = [In-], and hence KIn = [H+] (midway point).

] [HIn]][In[HK

-

In

24

For the equilibrium: HIn(aq) H+(aq) + In-

(aq)

Midway in the transition of the indicator color change: [HIn] = [In-], and hence KIn = [H+] (midway point). Take the log of both sides of this relationship, leads to

pKIn = pH (midway point).

] [HIn]][In[HK

-

In

25

26

IONIC EQUILIBRIUM

27

IONIC EQUILIBRIUM

Buffers

28

Buffers

Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base.

29

Buffers

Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base.

A buffer is a combination of species in solution that maintains an approximately constant pH by virtue of a pair of chemical reactions.

30

Buffers

Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base.

A buffer is a combination of species in solution that maintains an approximately constant pH by virtue of a pair of chemical reactions. One reaction describes a reaction of a buffer component with added acid, the other reaction describes the reaction of a buffer component with added base.

31

Example: acetic acid/sodium acetate buffer

32

Example: acetic acid/sodium acetate buffer A solution containing these two substances has the

ability to neutralize both added acid and added base.

33

Example: acetic acid/sodium acetate buffer A solution containing these two substances has the

ability to neutralize both added acid and added base.

If base is added to the buffer, it will react with the acid component:

CH3CO2H(aq) + OH-(aq) CH3CO2

-(aq) + H2O

34

Example: acetic acid/sodium acetate buffer A solution containing these two substances has the

ability to neutralize both added acid and added base.

If base is added to the buffer, it will react with the acid component:

CH3CO2H(aq) + OH-(aq) CH3CO2

-(aq) + H2O

If acid is added to the buffer, it will react with the base component:

CH3CO2-(aq) + H+

(aq) CH3CO2H(aq)

35

Example: acetic acid/sodium acetate buffer A solution containing these two substances has the

ability to neutralize both added acid and added base.

If base is added to the buffer, it will react with the acid component:

CH3CO2H(aq) + OH-(aq) CH3CO2

-(aq) + H2O

If acid is added to the buffer, it will react with the base component:

CH3CO2-(aq) + H+

(aq) CH3CO2H(aq)

Note that the Na+ is not directly involved in the buffer chemistry.

36

Quantitative treatment of Buffers:The Henderson-Hasselbalch Equation

37

Quantitative treatment of Buffers:The Henderson-Hasselbalch Equation

Consider the equilibrium: CH3CO2H(aq) CH3CO2

-(aq) + H+

(aq)

38

Quantitative treatment of Buffers:The Henderson-Hasselbalch Equation

Consider the equilibrium: CH3CO2H(aq) CH3CO2

-(aq) + H+

(aq)

H]CO[CH

] -CO][CH[HK23

23a

39

Quantitative treatment of Buffers:The Henderson-Hasselbalch Equation

Consider the equilibrium: CH3CO2H(aq) CH3CO2

-(aq) + H+

(aq)

Now take the log of both sides of the preceding

equation, to obtain

H]CO[CH] -CO][CH[HK

23

23a

40

Quantitative treatment of Buffers:The Henderson-Hasselbalch Equation

Consider the equilibrium: CH3CO2H(aq) CH3CO2

-(aq) + H+

(aq)

Now take the log of both sides of the preceding

equation, to obtain

H]CO[CH] -CO][CH[HK

23

23a

H]CO[CH

] -CO[CH][HloglogK23

23a

41

That is,

]log[HlogKa

H]CO[CH] CO[CHlog

23

-23

42

That is,

]log[HlogKa

H]CO[CH

] CO[CHloglogK23

-23a ]log[H

H]CO[CH

] CO[CHlog23

-23

43

That is,

]log[HlogKa

H]CO[CH

] CO[CHloglogK23

-23a

H]CO[CH

] CO[CHlogpKpH23

-23a

]log[H

H]CO[CH

] CO[CHlog23

-23

44

That is,

This is the Henderson-Hasselbalch equation for the acetic acid system.

]log[HlogKa

H]CO[CH

] CO[CHloglogK23

-23a

H]CO[CH

] CO[CHlogpKpH23

-23a

]log[H

H]CO[CH

] CO[CHlog23

-23

45

We could repeat the previous approach for the weak acid HA to obtain:

46

We could repeat the previous approach for the weak acid HA to obtain:

[HA]

][AlogpKpH-

a

47

We could repeat the previous approach for the weak acid HA to obtain:

In a more general form it would be:

[HA]

][AlogpKpH-

a

acid] [conjugatebase] [conjugatelogpKpH a

48

We could repeat the previous approach for the weak acid HA to obtain:

In a more general form it would be:

Either of the preceding two equations are called the Henderson-Hasselbalch equation.

[HA]

][AlogpKpH-

a

acid] [conjugatebase] [conjugatelogpKpH a

49

Example: Calculate the pH of a buffer system containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The Ka for acetic acid is 1.8 x 10-5.

50

Example: Calculate the pH of a buffer system containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The Ka for acetic acid is 1.8 x 10-5.

Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that

[CH3CO2H] = 1.0 M

51

Example: Calculate the pH of a buffer system containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The Ka for acetic acid is 1.8 x 10-5.

Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that

[CH3CO2H] = 1.0 M

It is also important to keep in mind that there is a lot of acetate ion present, and this will suppress the dissociation of the acetic acid (Le Châtelier’s Principle).

52

A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that

[CH3CO2-] = 1.0 M

53

A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that

[CH3CO2-] = 1.0 M

Now Ka = 1.8 x 10-5 so that pKa = 4.7

54

A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that

[CH3CO2-] = 1.0 M

Now Ka = 1.8 x 10-5 so that pKa = 4.7

Hence, from the Henderson-Hasselbalch equation:

= 4.7

H]CO[CH

] CO[CHlogpKpH23

-23a

M1.0M1.0log4.7

55

Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs:

56

Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs:

CH3CO2- + H+ CH3CO2H

57

Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs:

CH3CO2- + H+ CH3CO2H

0.10 mols 0.10 mols 0.10 mols (from the HCl)

58

Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs:

CH3CO2- + H+ CH3CO2H

0.10 mols 0.10 mols 0.10 mols (from the HCl) At equilibrium, the concentrations of the buffer

components are (keep in mind the total volume is 1.0 liter):

59

Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs:

CH3CO2- + H+ CH3CO2H

0.10 mols 0.10 mols 0.10 mols (from the HCl) At equilibrium, the concentrations of the buffer

components are (keep in mind the total volume is 1.0 liter):

[CH3CO2H] = 1.0 + 0.10 = 1.10 M

[CH3CO2-] = 1.0 - 0.10 = 0.90 M

60

To calculate the new pH, use the Henderson-Hasselbalch equation: