1 acid-base equilibria. 2 solutions of a weak acid or base the simplest acid-base equilibria are...

1

Acid-Base Equilibria

2

Solutions of a Weak Acid or Base

• The simplest acid-base equilibria are those in which a single acid or base solute reacts with water.

– In this chapter, we will first look at solutions of weak acids and bases.

– We must also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water.

3

Acid-Ionization Equilibria

• Acid ionization (or acid dissociation) is the

• (See Animation: Acid Ionization Equilibrium)– When acetic acid is added to water it reacts as follows.

4

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations of ions in solution are determined by the

– Consider the generic monoprotic acid, HA.

5

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).

– The corresponding equilibrium expression is:

6

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).

– Since the concentration of water remains relatively constant, we rearrange the equation to get:

7

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).

– Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.

8

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).

– Table 17.1 lists acid-ionization constants for various weak acids.

9

Experimental Determination of Ka

• The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions.

10

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

11

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

StartingChange

Equilibrium

– Let x be the moles per liter of product formed.

)aq(Nic)aq(OH )l(OH)aq(HNic 32

12

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

– The equilibrium-constant expression is:

13

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

– Substituting the expressions for the equilibrium concentrations, we get

14

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

– We can obtain the value of x from the given pH.

)pHlog(anti]OH[x 3

)39.3log(antix

00041.0101.4x 4

15

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

– Substitute this value of x

– Note that

012.001159.0)00041.0012.0()x012.0( the concentration of unionized acid remains virtually unchanged.

16

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

– Substitute this value of x

522

a 104.1)012.0()00041.0(

)x012.0(x

K

17

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

– To obtain the degree of dissociation:

18

Calculations With Ka

• Once you know the value of Ka, you can calculate the equilibrium concentrations of species HA, A-, and H3O+ for solutions of different molarities.

19

Calculations With Ka

• Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid.

20

Calculations With Ka

• How do you know when you can use this simplifying assumption?

21

Calculations With Ka

• How do you know when you can use this simplifying assumption?

22

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

23

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

24

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.– Note that

25

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

26

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

StartingChange

Equilibrium

– These data are summarized below.

27

– The equilibrium constant expression is

A Problem To Consider• What is the pH at 25°C of a solution

obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

28

– If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

29

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

– Rearranging the preceding equation to put it in the form ax2 + bx + c = 0, we get

– You can solve this equation exactly by using the quadratic formula.

30

– Now substitute into the quadratic formula.

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

a2ac4bb

x2

31

– Now substitute into the quadratic formula.

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

32

– Taking the upper sign, we get

A Problem To Consider

• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

– Now we can calculate the pH.

33

Polyprotic Acids

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

– Sulfuric acid, for example, can lose two protons in aqueous solution.

34

Polyprotic Acids

– For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered.

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

35

– Each equilibrium has an associated acid-ionization constant.

Polyprotic Acids

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

36

– Each equilibrium has an associated acid-ionization constant.

Polyprotic Acids

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

37

Polyprotic Acids

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

38

– However, reasonable assumptions can be made that simplify these calculations as we show in the next example.

Polyprotic Acids

– When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.

• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

Polyprotic Acids

40

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

41

– The pH can be determined by simply solving the equilibrium problem posed by the first ionization.

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

42

– If we abbreviate the formula for ascorbic acid as H2Asc, then the first ionization is:

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

43

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

(aq)AscH(aq)OH )l(OH)aq(AscH 322

StartingChange

Equilibrium

44

– The equilibrium constant expression is

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

45

– Substituting into the equilibrium expression

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

46

– Assuming that x is much smaller than 0.10, you get

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

47

– The hydronium ion concentration is 0.0028 M, so

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

48

– The ascorbate ion, Asc2-, which we will call y, is produced only in the second ionization of H2Asc.

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

49

– Assume the starting concentrations for HAsc- and H3O+ to be those from the first equilibrium.

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

50

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

StartingChange

Equilibrium

51

– The equilibrium constant expression is

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

52

– Substituting into the equilibrium expression

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

53

– Assuming y is much smaller than 0.0028, the equation simplifies to

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

54

– Hence,

A Problem To Consider

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

122 106.1]Asc[y – The concentration of the ascorbate ion equals Ka2.

55

Base-Ionization Equilibria

• Equilibria involving weak bases are treated similarly to those for weak acids.– Ammonia, for example, ionizes in water as

follows.

)aq(OH)aq(NH )l(OH)aq(NH 423

– The corresponding equilibrium constant is:

]OH][NH[]OH][NH[

K23

4c

56

Base-Ionization Equilibria

• Equilibria involving weak bases are treated similarly to those for weak acids.– Ammonia, for example, ionizes in water as

follows.

– The concentration of water is nearly constant.

57

Base-Ionization Equilibria

• Equilibria involving weak bases are treated similarly to those for weak acids.– In general, a weak base B with the base ionization

has a base ionization constant equal to

Table 17.2 lists ionization constants for some weak bases.

58

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.

1. Write the equation and make a table of concentrations.

2. Set up the equilibrium constant expression.

3. Solve for x = [OH-].

– As before, we will follow the three steps in solving an equilibrium.

59

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Pyridine ionizes by picking up a proton from water

(as ammonia does).

StartingChange

Equilibrium

60

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Note that

61

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– The equilibrium expression is

62

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– If we substitute the equilibrium

concentrations and the Kb into the equilibrium constant expression, we get

63

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Using our simplifying assumption that the x

in the denominator is negligible, we get

64

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Solving for x we get

65

A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Solving for pOH

66

Acid-Base Properties of a Salt Solution

• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.– Consider a solution of sodium cyanide, NaCN.

67

• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.

Acid-Base Properties of a Salt Solution

68

• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.

Acid-Base Properties of a Salt Solution

69

• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.– You can also see that OH- ion is a product, so you

would expect

– The reaction of the CN- ion with water is referred to as the hydrolysis of CN-.

Acid-Base Properties of a Salt Solution

70

• The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

)aq(OH)aq(HCN )l(OH)aq(CN 2

Acid-Base Properties of a Salt Solution

71)aq(OH)aq(HCN )l(OH)aq(CN 2

Acid-Base Properties of a Salt Solution

• The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

72)aq(OH)aq(NH )l(OH)aq(NH 3324

Acid-Base Properties of a Salt Solution

• The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

73)aq(OH)aq(NH )l(OH)aq(NH 3324

Acid-Base Properties of a Salt Solution

• The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

74

Predicting Whether a Salt is Acidic, Basic, or Neutral

• How can you predict whether a particular salt will be acidic, basic, or neutral?– The Brønsted-Lowry concept illustrates the

inverse relationship in the strengths of conjugate acid-base pairs.

75

Predicting Whether a Salt is Acidic, Basic, or Neutral

• How can you predict whether a particular salt will be acidic, basic, or neutral?

reaction no)l(OH)aq(Cl 2

76

Predicting Whether a Salt is Acidic, Basic, or Neutral

• How can you predict whether a particular salt will be acidic, basic, or neutral?

reaction no)l(OH)aq(Na 2

77

Predicting Whether a Salt is Acidic, Basic, or Neutral

• To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.– Consider potassium acetate, KC2H3O2.

The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

reaction no)l(OH)aq(K 2

78

Predicting Whether a Salt is Acidic, Basic, or Neutral• To predict the acidity or basicity of

a salt, you must examine the acidity or basicity of the ions composing the salt.– Consider potassium acetate, KC2H3O2.

OHOHCH )l(OH)aq(OHC 2322232

79

Predicting Whether a Salt is Acidic, Basic, or Neutral

• These rules apply to normal salts (those in which the anion has no acidic hydrogen)

1. A salt of a strong base and a strong acid.

80

Predicting Whether a Salt is Acidic, Basic, or Neutral

• These rules apply to normal salts (those in which the anion has no acidic hydrogen)

2. A salt of a strong base and a weak acid.

81

Predicting Whether a Salt is Acidic, Basic, or Neutral

• These rules apply to normal salts (those in which the anion has no acidic hydrogen)

3. A salt of a weak base and a strong acid.

82

Predicting Whether a Salt is Acidic, Basic, or Neutral

• These rules apply to normal salts (those in which the anion has no acidic hydrogen)

4. A salt of a weak base and a weak acid.

83

The pH of a Salt Solution

• To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion. (see Figure 17.8)

– The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species.

– Thus the Kb for CN- is related to the Ka for HCN.

84

The pH of a Salt Solution

• To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

)aq(CN)aq(OH )l(OH)aq(HCN 32

)aq(OH)aq(HCN )l(OH)aq(CN 2

Ka

Kb

– When these two reactions are added you get the ionization of water.

)aq(OH)aq(OH )l(OH2 32 Kw

85

The pH of a Salt Solution

• To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

)aq(CN)aq(OH )l(OH)aq(HCN 32

)aq(OH)aq(HCN )l(OH)aq(CN 2

Ka

Kb

– When two reactions are added, their equilibrium constants are multiplied.

)aq(OH)aq(OH )l(OH2 32 Kw

86

The pH of a Salt Solution

• To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

)aq(CN)aq(OH )l(OH)aq(HCN 32

)aq(OH)aq(HCN )l(OH)aq(CN 2

Ka

Kb

Therefore,

)aq(OH)aq(OH )l(OH2 32 Kw

87

The pH of a Salt Solution

• For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases.– The only difference is first obtaining the Ka

or Kb for the ion that hydrolyzes.

88

A Problem To Consider

• What is the pH of a 0.10 M NaCN solution at 25 °C? The Ka for HCN is 4.9 x 10-10.

89

A Problem To Consider

• What is the pH of a 0.10 M NaCN solution at 25 °C? The Ka for HCN is 4.9 x 10-10.

– The CN- ion is acting as a base, so first, we must calculate the Kb for CN-.

90

A Problem To Consider

– Let x = [OH-] = [HCN

• What is the pH of a 0.10 M NaCN solution at 25 °C? The Ka for HCN is 4.9 x 10-10.

91

A Problem To Consider

– This gives

• What is the pH of a 0.10 M NaCN solution at 25 °C? The Ka for HCN is 4.9 x 10-10.

92

A Problem To Consider

– Solving the equation, you find that

• What is the pH of a 0.10 M NaCN solution at 25 °C? The Ka for HCN is 4.9 x 10-10.

93

The Common Ion Effect

• The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.– Consider a solution of acetic acid (HC2H3O2), in

which you have the following equilibrium.

94

The Common Ion Effect

• The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232

95

The Common Ion Effect

• The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232

96

The Common Ion Effect

• The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232

97

A Problem To Consider

• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– Consider the equilibrium below.

(aq)OCH(aq)OH 23 )l(OH)aq(OHCH 22

Starting 0.025 0 0.018Change -x +x +x

Equilibrium

0.025-x x 0.018+x

98

A Problem To Consider

• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– The equilibrium constant expression is:

a2

23 K]OHCH[

]OCH][OH[

99

A Problem To Consider

• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– Substituting into this equation gives:

4107.1)x025.0()x018.0(x

100

A Problem To Consider

• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.

101

A Problem To Consider

• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– The equilibrium equation becomes

102

A Problem To Consider

• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.– Hence,

103

A Problem To Consider

• An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.

Buffers

105

Buffers

• A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

106

Buffers

• A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

107

Buffers

• A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

108

Buffers

• A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

109

The Henderson-Hasselbalch Equation

• How do you prepare a buffer of given pH?

– To illustrate, consider a buffer of a weak acid HA and its conjugate base A-.

The acid ionization equilibrium is:

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The Henderson-Hasselbalch Equation

• How do you prepare a buffer of given pH?

– The acid ionization constant is:

– By rearranging, you get an equation for the H3O+ concentration.

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The Henderson-Hasselbalch Equation

• How do you prepare a buffer of given pH?– Taking the negative logarithm of both sides of the

equation we obtain:

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The Henderson-Hasselbalch Equation

• How do you prepare a buffer of given pH?

– More generally, you can write

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The Henderson-Hasselbalch Equation

• How do you prepare a buffer of given pH?

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• Calculate the pH of a 44ml solution of .202 M acetic acid and 15.5 ml of .185 M NaOH solution.

HC2H3O2 + NaOH NaC2H3O2 + H2O 44.0 ml 15.5 ml

.202 M .185 M

The Henderson-Hasselbalch Equation

115

The Henderson-Hasselbalch Equation

• Calculate the pH of a 44ml solution of .202 M acetic acid and 15.5 ml of .185 M NaOH solution.

Calculating the new molarity in 59.5 ml

HC2H3O2 + H2O <->C2H3O2- + H3O+

1.01 x 10-1M 4.82 x 10-2M x

Titration Curves

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Acid-Ionization Titration Curves

• An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid).

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Titration of a Strong Acid by a Strong Base

Figure 17.12: Curve for the titration of a strong acid by a strong base.

120

• Figure 17.12 shows a curve for the titration of HCl with NaOH.

– At the equivalence point, the pH of the solution is

Titration of a Strong Acid by a Strong Base

121

• Figure 17.12 shows a curve for the titration of HCl with NaOH.

– To detect the equivalence point, you need

Titration of a Strong Acid by a Strong Base

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A Problem To Consider

• Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

NaOH + HCl NaCl + H2O

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A Problem To Consider

• Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

NaOH + HCl NaCl + H2O– We get the amounts of reactants by multiplying

the volume of each (in liters) by their respective molarities.

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A Problem To Consider

• Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

NaOH + HCl NaCl + H2O

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A Problem To Consider

• Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

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A Problem To Consider

• Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

– Hence,

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• The titration of a weak acid by a strong base gives a somewhat different curve.

Titration of a Weak Acid by a Strong Base

Figure 17.13: Curve for the titration of a weak acid by a strong base.

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A Problem To Consider

• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

HC2H3O2 + NaOH NaC2H3O2 + H2O– At the equivalence point,.

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A Problem To Consider

• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

HC2H3O2 + NaOH NaC2H3O2 + H2O

– First, calculate the concentration of the acetate ion.

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A Problem To Consider• Calculate the pH of the solution at the equivalence point when

25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

HC2H3O2 + NaOH NaC2H3O2 + H2O 2.5 x 10-3 mol

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A Problem To Consider

• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

– The total volume of the solution is 50.0 mL. Hence,

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• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

HC2H3O2 + NaOH NaC2H3O2 + H2O

.025 M .025 M - .025 M - .025 M +.025 M 0 0 .025 M

A Problem To Consider

134

• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

NaC2H3O2 Na+ + C2H3O2-

.025 M .025 M .025 M

A Problem To Consider

135

A Problem To Consider

• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

• The Kb for the acetate ion is 5.9 x 10-10

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• The titration of a weak base with a strong acid is a reflection of our previous example.– Figure 17.14 shows the titration of NH3 with HCl.

– In this case, the pH declines slowly at first, then falls abruptly from about pH 7 to pH 3.

– Methyl red, which changes color from yellow at pH 6 to red at pH 4.8, is a possible indicator.

Titration of a Strong Acid by a Weak Base

Figure 17.14: Curve for the titration of a weak base by a strong acid.

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Just a Reminder…

139

Building Equations with K values

• Weak acids and bases are assigned a Ka or Kb values based on the degree to which they ionize in water.

• Larger K values indicate a greater degree of ionization (strength).

• Ka and Kb, along with other K values that we will study later (Ksp, KD, Kf) are all

manipulated in the same manner.

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• When equations are added K values are multiplied.

• MnS ↔ Mn+2 + S-2 K= 5.1 x 10-15

• S-2 + H2O ↔HS- + OH- K= 1.0 x 10-19

• 2H+ + HS- OH- ↔ H2S +H2O K= 1.0 x 10-7

• MnS + 2H+ ↔ Mn+2+ H2K=

Building Equations with K values

141

• When equations are reversed the K values are reciprocated.

• • Al(OH)3 ↔ Al+3 + 3OH- K = 1.9 x 10-33

• 3OH- + Al+3 ↔ Al(OH)3 K = 1/1.9 x 10-33 =

Building Equations with K values

142

• When equations are multiplied the K values are raised to the power.

• NH3 + H2O ↔ NH4++ OH- K = 1.8 x 10-5

• 2NH3 + 2H2O ↔ 2NH4++ 2OH- K = (1.8 x 10-5)2

=

Building Equations with K values

143

• When equations are divided the root of the K values are taken.

• 2HPO4-2 ↔ 2H+ + 2PO4

3- K = 1.3 x 10-25

• HPO4-2 ↔ H+ + PO4

3- K = √1.3 x 10-25 =

Building Equations with K values

144

Operational Skills

• Determining Ka (or Kb) from the solution pH

• Calculating the concentration of a species in a weak acid solution using Ka

• Calculating the concentration of a species in a weak base solution using Kb

• Predicting whether a salt solution is acidic, basic, or neutral

• Obtaining Ka from Kb or Kb from Ka

• Calculating concentrations of species in a salt solution

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Operational Skills

• Calculating the common-ion effect on acid ionization

• Calculating the pH of a buffer from given volumes of solution

• Calculating the pH of a solution of a strong acid and a strong base

• Calculating the pH at the equivalence point in the titration of a weak cid with a strong base

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Animation: Acid Ionization Equilibrium

Return to slide 3

(Click here to open QuickTime animation)

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Figure 17.3: Variation of percent ionization of a weak acid with concentration.

Return to slide 19

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Figure 17.8: A pH meter reading NH4Cl. Photo courtesy of American Color.

Return to slide 82

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Animation: Adding an Acid to a Buffer

Return to slide 104

(Click here to open QuickTime animation)