To calculate the new pH, use the Henderson- Hasselbalch equation: 1141

1

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

H]CO[CH

] CO[CHlogpKpH23

-23a

2

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

= 4.6

H]CO[CH

] CO[CHlogpKpH23

-23a

M]1.10M0.90log4.7

3

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

= 4.6

Note in this example that the [H+] changed from [H+] = 10-pH = 10-4.7 = 2.0 x 10-5 M to [H+] = 10-pH = 10-4.6 = 2.5 x 10-5 M

H]CO[CH

] CO[CHlogpKpH23

-23a

M]1.10M0.90log4.7

4

To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.

5

To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.

The initial [H+] = 1.0 x 10-7 M (from self-dissociation of water)

6

To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.

The initial [H+] = 1.0 x 10-7 M (from self-dissociation of water)

After the HCl is added, [H+] = 1.0 x 10-1 M , so there is a million fold increase in [H+]!

7

To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.

The initial [H+] = 1.0 x 10-7 M (from self-dissociation of water)

After the HCl is added, [H+] = 1.0 x 10-1 M , so there is a million fold increase in [H+]!

Whereas for the buffer, [H+] changes from 2.0 x 10-5 M to 2.5 x 10-5 M.

8

Distribution Curve

9

Distribution Curve Distribution curve (for a buffer): Gives the fraction

of the acid component and the base component present in solution as a function of the pH.

10

Distribution Curve Distribution curve (for a buffer): Gives the fraction

of the acid component and the base component present in solution as a function of the pH.

Example: Consider the acetic acid/acetate buffer.

11

Distribution Curve Distribution curve (for a buffer): Gives the fraction

of the acid component and the base component present in solution as a function of the pH.

Example: Consider the acetic acid/acetate buffer. At low pH: H+ + CH3CO2

- CH3CO2H

12

Distribution Curve Distribution curve (for a buffer): Gives the fraction

of the acid component and the base component present in solution as a function of the pH.

Example: Consider the acetic acid/acetate buffer. At low pH: H+ + CH3CO2

- CH3CO2H

At high pH: OH- + CH3CO2H CH3CO2-

13

Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:

14

Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:

1pKpH a

15

Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:

For the acetic acid/acetate ion buffer (pKa =4.7) the

buffer range is pH = 3.7 – 5.7 .

1pKpH a

16

Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:

For the acetic acid/acetate ion buffer (pKa =4.7) the

buffer range is pH = 3.7 – 5.7 . Note that the buffer functions best at pH = 4.7, i.e. when

[CH3CO2H] = [CH3CO2-]

1pKpH a

17

18

19

Exercise: Comment on the following as possible buffer systems.

(a) NaCl(aq)/HCl(aq)

(b) NH3(aq)/NH4Cl(aq)

(c) H2PO4-(aq)/ HPO4

2-(aq)

(d) NaHCO3(aq)

20

(a) NaCl(aq)/HCl(aq)

To be a buffer, it needs to be able to react with both added H+ and added OH-. If H+ is added, the following reaction does not occur to any significant extent: H+

(aq) + Cl-

(aq) HCl(aq)

(because HCl is a strong acid).

That means that (a) cannot be a buffer system.

21

(b) NH3(aq)/NH4Cl(aq)

Addition of H+: H+(aq) + NH3(aq) NH4

+(aq)

Addition of OH-: OH-

(aq) + NH4+

(aq) NH3(aq) + H2O

Hence (b) is a buffer system.

22

(b) NH3(aq)/NH4Cl(aq)

Addition of H+: H+(aq) + NH3(aq) NH4

+(aq)

Addition of OH-: OH-

(aq) + NH4+

(aq) NH3(aq) + H2O

Hence (b) is a buffer system.

(c) H2PO4-(aq)/ HPO4

2-(aq)

Addition of H+: H+(aq) + HPO4

2-(aq) H2PO4

-(aq)

Addition of OH-: OH-

(aq) + H2PO4-(aq) HPO4

2-(aq) + H2O

Hence (c) is a buffer system.

23

(d) NaHCO3(aq)

24

(d) NaHCO3(aq)

(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)

25

(d) NaHCO3(aq)

(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)

Addition of H+: H+(aq) + HCO3

-(aq) H2CO3(aq)

26

(d) NaHCO3(aq)

(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)

Addition of H+: H+(aq) + HCO3

-(aq) H2CO3(aq)

Addition of OH-: OH-

(aq) + HCO3-(aq) CO3

2-(aq) + H2O

27

(d) NaHCO3(aq)

(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)

Addition of H+: H+(aq) + HCO3

-(aq) H2CO3(aq)

Addition of OH-: OH-

(aq) + HCO3-(aq) CO3

2-(aq) + H2O

Hence (d) is a buffer system.

28

(d) NaHCO3(aq)

(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)

Addition of H+: H+(aq) + HCO3

-(aq) H2CO3(aq)

Addition of OH-: OH-

(aq) + HCO3-(aq) CO3

2-(aq) + H2O

Hence (d) is a buffer system. Note that in this example, the HCO3

- ion functions as both the acid and the base.

29

(d) NaHCO3(aq)

(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)

Addition of H+: H+(aq) + HCO3

-(aq) H2CO3(aq)

Addition of OH-: OH-

(aq) + HCO3-(aq) CO3

2-(aq) + H2O

Hence (d) is a buffer system. Note that in this example, the HCO3

- ion functions as both the acid and the base.

Various anions of multi-proton acids can function as buffer systems with a single species present.

30

IONIC EQUILIBRIUMSolubility Products

31

IONIC EQUILIBRIUMSolubility Products

A saturated solution of an insoluble salt is a heterogeneous equilibrium.

32

IONIC EQUILIBRIUMSolubility Products

A saturated solution of an insoluble salt is a heterogeneous equilibrium.

Example: In a saturated solution of AgCl solution, the following equilibrium is present:

33

IONIC EQUILIBRIUMSolubility Products

A saturated solution of an insoluble salt is a heterogeneous equilibrium.

Example: In a saturated solution of AgCl solution, the following equilibrium is present:

AgCl(s) Ag+(aq) + Cl-

(aq)

34

IONIC EQUILIBRIUMSolubility Products

A saturated solution of an insoluble salt is a heterogeneous equilibrium.

Example: In a saturated solution of AgCl solution, the following equilibrium is present:

AgCl(s) Ag+(aq) + Cl-

(aq)

The equilibrium constant is:

] [AgCl]-][Cl[AgK

35

Now

]][Cl[Ag] K[AgCl -

36

Now

but [AgCl] is a constant, recall

]][Cl[Ag] K[AgCl -

37

Now

but [AgCl] is a constant, recall

]][Cl[Ag] K[AgCl -

AgClmass molarAgCldensity[AgCl]

38

Now

but [AgCl] is a constant, recall

We set where the subscript sp stands for solubility product.

]][Cl[Ag] K[AgCl -

AgClmass molarAgCldensity[AgCl]

] K[AgClKsp

39

Now

but [AgCl] is a constant, recall

We set where the subscript sp stands for solubility product. Hence,

]][Cl[Ag] K[AgCl -

AgClmass molarAgCldensity[AgCl]

] K[AgClKsp

]][Cl[AgK -sp

40

Examples: MgF2(s) Mg2+(aq) + 2 F-

(aq)

41

Examples: MgF2(s) Mg2+(aq) + 2 F-

(aq)

2-2sp ]][F[MgK

42

Examples: MgF2(s) Mg2+(aq) + 2 F-

(aq)

Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO4

3-(aq)

2-2sp ]][F[MgK

43

Examples: MgF2(s) Mg2+(aq) + 2 F-

(aq)

Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO4

3-(aq)

2-2sp ]][F[MgK

2-34

32sp ][PO][CaK

44

Examples: MgF2(s) Mg2+(aq) + 2 F-

(aq)

Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO4

3-(aq)

Note that the pure solids do not occur in the expression for Ksp.

2-2sp ]][F[MgK

2-34

32sp ][PO][CaK

45

Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.

46

Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.

In a solution containing Ag+(aq) and Cl-

(aq) at 25 oC, we have one of the following situations:

47

Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.

In a solution containing Ag+(aq) and Cl-

(aq) at 25 oC, we have one of the following situations:

unsaturated solution

10- 10x1.6]][Cl[Ag

48

Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.

In a solution containing Ag+(aq) and Cl-

(aq) at 25 oC, we have one of the following situations:

unsaturated solution saturated solution

10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag

49

Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.

In a solution containing Ag+(aq) and Cl-

(aq) at 25 oC, we have one of the following situations:

unsaturated solution saturated solution supersaturated solution

10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag

50

Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.

In a solution containing Ag+(aq) and Cl-

(aq) at 25 oC, we have one of the following situations:

unsaturated solution saturated solution supersaturated solution Some AgCl precipitate will form until the product of

the ionic concentrations is equal to 1.6 x 10-10.

10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag

51

Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3

2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.

52

Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3

2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.

Ag2CO3(s) 2 Ag+(aq) + CO3

2-(aq)

53

Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3

2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.

Ag2CO3(s) 2 Ag+(aq) + CO3

2-(aq)

= (2.54 x 10-4)2 (1.27 x 10-4) = 8.19 x 10-12

][CO][AgK -23

2sp

54

Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3

2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.

Ag2CO3(s) 2 Ag+(aq) + CO3

2-(aq)

= (2.54 x 10-4)2 (1.27 x 10-4) = 8.19 x 10-12 The concentration of CO3

2-(aq) is equal to the number

of moles of Ag2CO3(s) that have dissolved.

][CO][AgK -23

2sp

55

molar mass of Ag2CO3 = 275.8 g/mol

The solubility of Ag2CO3

= 1.27 x 10-4 mol l-1 275.8 g mol-1

= 0.0350 g/l

56

Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.

57

Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.

PbF2(s) Pb2+(aq) + 2 F-

(aq)

58

Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.

PbF2(s) Pb2+(aq) + 2 F-

(aq)

The abbreviated ICE table for this problem looks like

Pb2+ F-

59

Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.

PbF2(s) Pb2+(aq) + 2 F-

(aq)

The abbreviated ICE table for this problem looks like

Pb2+ F-

y 2y

60

Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.

PbF2(s) Pb2+(aq) + 2 F-

(aq)

The abbreviated ICE table for this problem looks like

Pb2+ F-

y 2y

228sp ]-][F[Pb10x4.1K