1 Chapter 16 Ionic Equilibria: Acids and Bases. 2 Chapter Goals 1. A Review of Strong Electrolytes...

1

Chapter 16

Ionic Equilibria: Acids and Bases

2

Chapter Goals

1. A Review of Strong Electrolytes

2. The Autoionization of Water

3. The pH and pOH Scales

4. Ionization Constants for Weak Monoprotic Acids and Bases

5. Polyprotic Acids

6. Solvolysis

7. Salts of Strong Bases and Strong Acids

3

Chapter Goals

8. Salts of Strong Bases and Weak Acids

9. Salts of Weak Bases and Strong Acids

10. Salts of Weak Bases and Weak Acids

11. Salts That Contain Small, Highly Charged Cations

4

A Review of Strong Electrolytes

This chapter details the equilibria of weak acids and bases. We must distinguish weak acids and bases from strong

electrolytes. Weak acids and bases ionize or dissociate

partially, much less than 100%. In this chapter we will see that it is often less than 10%!

Strong electrolytes ionize or dissociate completely. Strong electrolytes approach 100% dissociation in

aqueous solutions.

5

A Review of Strong Electrolytes There are three classes of strong electrolytes.1 Strong Water Soluble Acids

Remember the list of strong acids from Chapter 4.

3(aq)(aq)%100

)3(

3(aq)(aq)3100%

)(2)3(

NOHHNO

or

NOOH OHHNO

6


3(aq)(aq)%100

)3(

3(aq)(aq)3100%

)(2)3(

NOHHNO

or

NOOH OHHNO

7


2 Strong Water Soluble BasesThe entire list of these bases was also introduced in

Chapter 4.

-(aq)

2(aq)

100% OH2(s)

-(aq)(aq)

100% OH(s)

OH 2Sr Sr(OH)

OHKKOH

2

2

8


3 Most Water Soluble SaltsThe solubility guidelines from Chapter 4 will help you remember

these salts.

3(aq)2(aq)

100% OHs23

-(aq)(aq)

100% OH(s)

NO 2Ca )Ca(NO

ClNaNaCl

2

2

9

A Review of Strong Electrolytes The calculation of ion concentrations in solutions of

strong electrolytes is easy. Example 18-1: Calculate the concentrations of ions

in 0.050 M nitric acid, HNO3.

0.050 0.050 050.0

NOOHOHHNO 3(aq)(aq)3100%

)(2)3(

MMM

10


Example 18-2: Calculate the concentrations of ions in 0.020 M strontium hydroxide, Sr(OH)2, solution.

You do it!

M

MMM

0.040

0.0202 0.020 020.0

OH 2SrSr(OH) -(aq)

2(aq)

OH2(s)

2

11

The Autoionization of Water

Pure water ionizes very slightly. The concentration of the ionized water is less than one-

millionth molar at room temperature.

12


We can write the autoionization of water as a dissociation reaction similar to those previously done in this chapter.

Because the activity of pure water is 1, the equilibrium constant for this reaction is:

-(aq)(aq)3)(2)(2 OHOHOH OH

K H O OHc 3+

13


Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 25oC. Note that this is at 25oC, not every temperature!

We can determine the value of Kc from this information.

K H O OH

1.0 x 10 1.0 x 10

1.0 x10

c 3+

-7 -7

14

14


This particular equilibrium constant is called the ion-product for water and given the symbol Kw. Kw is one of the recurring expressions for the remainder of

this chapter and Chapters 19 and 20.

K H O OH

1.0 x10

w 3+

14

15


Example 18-3: Calculate the concentrations of H3O+ and OH- in 0.050 M HCl.

].[OH calculate tous allow willK and OH The

.0.050OH theThus

0.050 0.050 050.0

Cl OH OH + HCl

-w

+3

+3

+32

M

MMM

16


Use the [H3O+] and Kw to determine the [OH-].

You do it!

M13

2

14

+3

14

14+3

100.2OH

100.5

100.1

OH

100.1OH

100.1OHOH

17


The increase in [H3O+] from HCl shifts the equilibrium and decreases the [OH-]. Remember from Chapter 17, increasing the product

concentration, [H3O+], causes the equilibrium to shift to the reactant side.

This will decrease the [OH-] because it is a product!

. 0.050 10 2.0 0.050 ]O[H overall The

.102.0 is K from ]O[H The

OH OH OH OH

0.050 is HCl from ]O[H The

13-3

13-w3

-322

3

MM

M

M

18


Now that we know the [H3O+] we can calculate the [OH-].

You do it!You do it!

M

M

M

13-

14--

3

14--

3

-143w

100.2]OH[

] [0.050

101]OH[

]O[H

101]OH[

. 0.050 ]O[H Since

101][OH-]O[HK

19

The pH and pOH scales

A convenient way to express the acidity and basicity of a solution is the pH and pOH scales.

The pH of an aqueous solution is defined as:

pH = -log H O3+

20


In general, a lower case p before a symbol is read as the ‘negative logarithm of’ the symbol.

Thus we can write the following notations.

.quantitiesother for forth so and

Ag-log=pAg

OH-log=pOH+

-

21


If either the [H3O+] or [OH-] is known, the pH and pOH can be calculated.

Example 18-4: Calculate the pH of a solution in which the [H3O+] =0.030 M.

52.1pH

100.3logpH

OH-log=pH2

+3

22


Example 18-5: The pH of a solution is 4.597. What is the concentration of H3O+?


M53

4.597-3

3

3

3

1053.2]O[H

10]O[H

-4.597]Olog[H

]O-log[H4.597

]O-log[HpH

23


A convenient relationship between pH and pOH may be derived for all dilute aqueous solutions at 250C.

Taking the logarithm of both sides of this equation gives:

143 100.1]][OHO[H

00.14OHlogOHlog 3

24


Multiplying both sides of this equation by -1 gives:

Which can be rearranged to this form:

00.14OHlogOHlog- 3

14.00 pOH pH

25


Remember these two expressions!! They are key to the next three chapters!

14.00pOH pH

100.1OH OH 143

26


The usual range for the pH scale is 0 to 14.

And for pOH the scale is also 0 to 14 but inverted from pH. pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0.

14.00pH to0pH

100.1OH to 0.1OH 1433

MM

0pOH 00.14pOH

0.1OH toup 100.1OH 14

MM

27


28


Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3 solution. Is HNO3 a weak or strong acid?

What is the [H3O+] ?

70.1pH

100.2-logpH

100.2OH

0.020 0.020 020.0

NOOHOHHNO

2

23

-33

100%23

M

M

MMM

29


Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3 solution.

30.12100.5logpOH

100.5100.2

100.1

OH

100.1OH

100.1OHOHK

13

132

14

3

14

143w

M

30

The pH and pOH scales To help develop familiarity with the pH and pOH scale we can

look at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14 M.

[H3O+] [OH-] pH pOH

1.0 M 1.0 x 10-14 M 0.00 14.00

1.0 x 10-3 M 1.0 x 10-11 M 3.00 11.00

1.0 x 10-7 M 1.0 x 10-7 M 7.00 7.00

2.0 x 10-12 M 5.0 x 10-3 M 11.70 2.30

1.0 x 10-14 M 1.0 M 14.00 0.00

32

Ionization Constants for Weak Monoprotic Acids and Bases Let’s look at the dissolution of acetic acid, a weak

acid, in water as an example. The equation for the ionization of acetic acid is:

The equilibrium constant for this ionization is expressed as:

-3323 COOCHOHOH COOHCH

OH COOHCH

COOCH OHK

23

33c

33

Ionization Constants for Weak Monoprotic Acids and Bases The water concentration in dilute aqueous solutions

is very high. 1 L of water contains 55.5 moles of water. Thus in dilute aqueous solutions:

M5.55OH2

34

Ionization Constants for Weak Monoprotic Acids and Bases The water concentration is many orders of

magnitude greater than the ion concentrations. Thus the water concentration is essentially that of

pure water. Recall that the activity of pure water is 1.

COOHCH

COOCH OH K

COOHCH

COOCH OHOHK

3

33

3

332c

35

Ionization Constants for Weak Monoprotic Acids and Bases We can define a new equilibrium constant for weak

acid equilibria that uses the previous definition. This equilibrium constant is called the acid ionization

constant. The symbol for the ionization constant is Ka.

acid aceticfor

108.1COOHCH

COOCH OHK 5

3

33a

36

Ionization Constants for Weak Monoprotic Acids and Bases In simplified formsimplified form the dissociation equation and

acid ionization expression are written as:

5

3

3a

-33

108.1COOHCH

COOCH HK

COOCHHCOOHCH

37

Ionization Constants for Weak Monoprotic Acids and Bases The ionization constant values for several acids are

given below. Which acid is the strongest?

Acid Formula Ka value

Acetic CH3COOH 1.8 x 10-5

Nitrous HNO2 4.5 x 10-4

Hydrofluoric HF 7.2 x 10-4

Hypochlorous HClO 3.5 x 10-8

Hydrocyanic HCN 4.0 x 10-10

38

Ionization Constants for Weak Monoprotic Acids and Bases From the above table we see that the order of

increasing acid strength for these weak acids is:

The order of increasing base strength of the anions (conjugate bases) of these acids is:

HCN>HClO>COOHCH>HNO>HF 32

---3

-2

- CN<ClO<COOCH<NO<F

39

Ionization Constants for Weak Monoprotic Acids and Bases Example 18-8: Write the equation for the ionization

of the weak acid HCN and the expression for its ionization constant.

10--

a

-

10 x 4.0HCN

CN HK

CN H HCN

40

Ionization Constants for Weak Monoprotic Acids and Bases Example 18-9: In a 0.12 M solution of a weak

monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid.


HY H + Y

KH Y

HY

+ -

a

+ -

41

Ionization Constants for Weak Monoprotic Acids and Bases Since the weak acid is 5.0% ionized, it is also 95% unionized. Calculate the concentration of all species in solution.

MM

M

MM

11.0)12.0(95.0HY

100.6YH

0060.0)12.0(05.0YH3+

+

42

Ionization Constants for Weak Monoprotic Acids and Bases

Use the concentrations that were just determined in the ionization constant expression to get the value of Ka.

4

a

33

a

a

103.3K

11.0

100.6 100.6K

HY

Y HK

43

Ionization Constants for Weak Monoprotic Acids and Bases Example 18-10: The pH of a 0.10 M solution of a

weak monoprotic acid, HA, is found to be 2.97. What is the value for its ionization constant?

pH = 2.97 so [H+]= 10-pH

M3

97.2

101.1H

10H

44

Ionization Constants for Weak Monoprotic Acids and Bases Use the [H3O+] and the ionization reaction to

determine concentrations of all species.

HA H A

Equil. []'s 0.10 -1.1 10 1.1 10 1.1 10

0.10

+ -

-3 -3 -3

45

Ionization Constants for Weak Monoprotic Acids and Bases Calculate the ionization constant from this

information.

5a

-3-3

a

102.1K

0.10

101.1101.1

HA

AHK

46

Ionization Constants for Weak Monoprotic Acids and Bases Example 18-11: Calculate the concentrations of the

various species in 0.15 M acetic acid, CH3COOH, solution.

It is always a good idea to write down the ionization reaction and the ionization constant expression.

5

3

-33

a

-3323

108.1COOHCH

COOCHOHK

COOCHOH OHCOOHCH

47

Ionization Constants for Weak Monoprotic Acids and Bases Next, combine the basic chemical concepts with

some algebra to solve the problem.

M0.15 [] Initial

COOCH OH OHCOOHCH -3323

48

Ionization Constants for Weak Monoprotic Acids and Bases Next we combine the basic chemical concepts with

some algebra to solve the problem

xMxMxM

M

- Change

0.15 [] Initial


49

Ionization Constants for Weak Monoprotic Acids and Bases Next we combine the basic chemical concepts with

some algebra to solve the problem

xMxM-x)M.(

xMxMxM

M

150 [] mEquilibriu

- Change

0.15 [] Initial


50

Substitute these algebraic quantities into the ionization expression.

5

3

33a

108.115.0

COOHCH

COOCH OHK

x

xx


51

Ionization Constants for Weak Monoprotic Acids and Bases Solve the algebraic equation, using a simplifying assumption that is

appropriate for all weak acid and base ionizations.

52

52

108.115.0

108.115.0

xx

x

x

52

Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations.


52

3a

52

52

108.115.0

[]. tocompared ignore enough to small is

.assumption thismake then 10 K If

108.115.0

108.115.0

x

x

xx

x

x

53

Complete the algebra and solve for the concentrations of the species.


MM

Mx

x

15.0106.115.0COOHCH

COOCHOH106.1

107.2

33

333

62

54

Ionization Constants for Weak Monoprotic Acids and Bases Note that the properly applied simplifying assumption gives

the same result as solving the quadratic equation does.

2a

4acbb

c b a

0107.2108.1

108.115.0

2

652

5

x

xx

X

xx

55


3-3

6255

101.6- and 106.1

12

107.214108.1108.1

x

x

56

Ionization Constants for Weak Monoprotic Acids and Bases Let us now calculate the percent ionization for the

0.15 M acetic acid. From Example 18-11, we know the concentration of CH3COOH that ionizes in this solution. The percent ionization of acetic acid is

%1.1%10015.0

106.1ionization %

%100COOHCH

COOHCH= ionization %

3

original3

ionized3

M

M

57

Ionization Constants for Weak Monoprotic Acids and Bases Example 18-12: Calculate the concentrations

of the species in 0.15 M hydrocyanic acid, HCN, solution.

Ka= 4.0 x 10-10 for HCN


58


MMx

Mx

x

x

xx

MxMxMx

MxMxMx

M

15.0 15.0HCN

CNH107.7

100.6

100.415.0

HCN

CN HK

-0.15 mEquilibriu

+ + - Change

0.15 Initial

CN OH OH HCN

6

112

10a

-32

59

Ionization Constants for Weak Monoprotic Acids and Bases The percent ionization of 0.15 M HCN solution is

calculated as in the previous example.

%0051.0%10015.0

107.7ionization %

%100HCN

HCN = ionization %

6

original

ionized

M

M

60

Ionization Constants for Weak Monoprotic Acids and Bases Let’s look at the percent ionization of two weak acids

as a function of their ionization constants. Examples 18-11 and 18-12 will suffice.

Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

Solution Ka [H+] pH % ionization

0.15 M acetic acid

1.8 x 10-5 1.6 x 10-3 2.80 1.1

0.15 M

HCN

4.0 x 10-10 7.7 x 10-6 5.11 0.0051

61

Ionization Constants for Weak Monoprotic Acids and Bases All of the calculations and understanding we have at

present can be applied to weak acids and weak bases!

One example of a weak base ionization is ammonia ionizing in water.

62

Ionization Constants for Weak Monoprotic Acids and Bases All of the calculations and understanding we have at

present can be applied to weak acids and weak bases!

Example 18-13: Calculate the concentrations of the various species in 0.15 M aqueous ammonia.

xMxM-x)M.(

xMxMxM

M

150 [] mEquilibriu

- Change

0.15 [] Initial

OH NH OH NH -423

63


MM

M

Mxx

x

xx

x

xx

x

xx

xMxM-x)M.(

xMxMxM

M

15.0101.615.0NH

101.6OHNH

101.6 and 107.2

108.1)15.0(

108.115.0

15.0

0.15 x - 0.15 thus0.15x

valid.is assumption gsimplifyin The

108.115.0

NH

OH NHK

150 [] mEquilibriu

- Change

0.15 [] Initial

OH NH OH NH

3-3

3--4

3-62

52

5

5

3

-4

b

-423

64

Ionization Constants for Weak Monoprotic Acids and Bases The percent ionization for weak bases is calculated

exactly as for weak acids.

%1.1

%10015.0

106.1

%100NH

NHionization %

3

original3

ionized3

M

M

65

Ionization Constants for Weak Monoprotic Acids and Bases Example 18-14: The pH of an aqueous

ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution.


66


M

M3

4

363.2pOH-

103.2NH

103.21010OH

2.63 11.37-14.00pH - 14.00=pOH

pOH. thederivecan we14.00, = pOH + pH From

11.37=pH

67

Ionization Constants for Weak Monoprotic Acids and Bases Use the ionization equation and some algebra to get

the equilibrium concentration.

3-3-3-

3-3-3-

-423

102.3+ 102.3+ 102.3- m[]Equilibriu

102.3+ 102.3+ 102.3- Change

Initial[]

OH NH OH NH

Mx

xM

68

Ionization Constants for Weak Monoprotic Acids and Bases Substitute these values into the ionization constant

expression.

3

335

5

3

4b

103.2

103.2 103.2108.1

108.1NH

OH NHK

x

69

Ionization Constants for Weak Monoprotic Acids and Bases Examination of the last equation suggests that our

simplifying assumption can be applied. In other words (x-2.3x10-3) x.

Making this assumption simplifies the calculation.

3

5

23

NH 30.0

108.1103.2

Mxx

70

Polyprotic Acids Many weak acids contain two or more acidic hydrogens.

Examples include H3PO4 and H3AsO4.

The calculation of equilibria for polyprotic acids is done in a stepwise fashion. There is an ionization constant for each step.

Consider arsenic acid, H3AsO4, which has three ionization constants.

1 Ka1 = 2.5 x 10-4

2 Ka2 = 5.6 x 10-8

3 Ka3 = 3.0 x 10-13

71

Polyprotic Acids

The first ionization step for arsenic acid is:

4

43

421

-4243

105.2AsOH

AsOH HKa

AsOHHAsOH

72

Polyprotic Acids

The second ionization step for arsenic acid is:

8-1

42

24

a2

-24

-142

106.5AsOH

HAsO HK

HAsOHAsOH

73

Polyprotic Acids

The third ionization step for arsenic acid is:

13-2

4

34

a3

-34

-24

100.3HAsO

AsOHK

AsO HHAsO

74

Polyprotic Acids

Notice that the ionization constants vary in the following fashion:

This is a general relationship. For weak polyprotic acids the Ka1 is always > Ka2, etc.

a3a2a1 KKK

75

Polyprotic Acids

Example 18-15: Calculate the concentration of all species in 0.100 M arsenic acid, H3AsO4, solution.

1 Write the first ionization step and represent the concentrations.Approach this problem exactly as previously done.

xMxMMx 100.0

AsOHHAsOH 4243

76

Polyprotic Acids

2 Substitute the algebraic quantities into the expression for Ka1.

apply.not does assumption gsimplifyin thecase, In this

0105.2105.2

105.210.0

K

105.2AsOH

AsOH HK

542

4a1

4

43

42a1

xx

x

xx

77

Polyprotic Acids

Use the quadratic equation to solve for x, and obtain both values of x.

MMx

MxM

MxMx

x

095.0100.0AsOH

109.4AsOHH

109.4 and 101.5

12

105.214105.2105.2

43

342

33

5244

78

Polyprotic Acids

4 Next, write the equation for the second step ionization and represent the concentrations.

yMyMMy

M

)10(4.9lly algebraica

)10(4.9 step1st ]from [

HAsO + H AsOH

3-

3-

-24

+-42

79

Polyprotic Acids

5 Substitute the algebraic expressions into the second step ionization expression.

.assumption apply thecan westep For this

104.9

104.9=K

106.5AsOH

HAsO OH=K

3-

3-

a2

8

42

243

a2

y

yy

80

Polyprotic Acids

2nd1st

242nd

8

83-

3-

a2

3-3-

3-

3-

3-

a2

8

42

243

a2

HH that Note

HAsOH106.5

106.5104.9

104.9=K

104.9104.9 Thus,

104.9 y

applied. becan assumption thecase In this

104.9

104.9=K

106.5AsOH

HAsOOH=K

My

y

y

y

yy

81

Polyprotic Acids

6 Finally, repeat the entire procedure for the third ionization step.

MzMzMz

MM

105.6 changes ] [ of tionsrepresenta algebraic

105.6104.9 105.6 sionization 2 and 1 from s]' [

AsO H HAsO

8-

8-3-8-ndst

-34

24

82

Polyprotic Acids

7. Substitute the algebraic representations into the third ionization expression.

.105.6 z applied, becan assumption The

106.5

106.5109.4=K

100.3HAsO

AsOOH=K

8-

8

83

a3

1324

343

a3

z

zz

83

Polyprotic Acids

Use Kw to calculate the [OH-] in the 0.100 M H3AsO4 solution.

M12

3

1414

14

100.2OH

109.4

100.1

H

100.1OH

100.1OHH

343rd

18

138

3

AsOH104.3

100.3106.5

109.4

Mz

z

84

Polyprotic Acids A comparison of the various species in 0.100 M

H3AsO4 solution follows.

Species Concentration

H3AsO4 0.095 M

H+ 0.0049 M

H2AsO4- 0.0049 M

HAsO42- 5.6 x 10-8 M

AsO43- 3.4 x 10-18 M

OH- 2.0 x 10-12 M

85

Solvolysis This reaction process is the most difficult concept in this chapter. Solvolysis is the reaction of a substance with the solvent in which

it is dissolved. Hydrolysis refers to the reaction of a substance with water or its

ions. Combination of the anion of a weak acid with H3O+ ions from

water to form nonionized weak acid molecules.

86

Solvolysis Hydrolysis refers to the reaction of a

substance with water or its ions. Hydrolysis is solvolysis in aqueous solutions.

The combination of a weak acid’s anion with H3O+ ions, from water, to form nonionized weak acid molecules is a form of hydrolysis.

A H O HA H O

recall H O + H O H O OH

-3 2

2 2 3

87

Solvolysis

The reaction of the anion of a weak monoprotic acid with water is commonly represented as:

mequilibriu water the

upsets OH of removal The

OH HA OH A+

3

-2

-

88

Solvolysis

Recall that at 25oC in neutral solutions:

[H3O+] = 1.0 x 10-7 M = [OH-]

in basic solutions:[H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M

in acidic solutions:

[OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M

89

Solvolysis

Remember from BrØnsted-Lowry acid-base theory: The conjugate base of a strong acid is a very weak base. The conjugate base of a weak acid is a stronger base. Hydrochloric acid, a typical strong acid, is essentially

completely ionized in dilute aqueous solutions.

HCl H O H O Cl2 3 ~100%

90

Solvolysis The conjugate base of HCl, the Cl- ion, is a very weak base.

The chloride ion is such a weak base that it will not react with the hydronium ion.

This fact is true for all strong acids and their anions.

Cl H O No rxn. in dilute aqueous solutions3

91

Solvolysis HF, a weak acid, is only slightly ionized in dilute aqueous solutions. Its conjugate base, the F- ion, is a much stronger base than the Cl-

ion. The F- ions combine with H3O+ ions to form nonionized HF.

Two competing equilibria are established.

HF + H O H O F

only slightly

F + H O HF + H O

nearly completely

2 3+ -

-3

+2

92

Solvolysis Dilute aqueous solutions of salts that

contain no free acid or base come in four types:

1. Salts of Strong Bases and Strong Acids

2. Salts of Strong Bases and Weak Acids

3. Salts of Weak Bases and Strong Acids

4. Salts of Weak Bases and Weak Acids

93

Salts of Strong Bases and Weak Acids Salts made from strong acids and strong soluble

bases form neutral aqueous solutions. An example is potassium nitrate, KNO3, made from

nitric acid and potassium hydroxide.

neutral. issolution theThus

OHOHupset oreaction t no is There

amounts. equalin present are HNO and KOH The

solution in are that ions The

OH OH OHOH

NOK KNO

-+3

3

HNOKOH

3-

22

3+OHin %100~

)(3

3

2

s

94

Salts of Strong Bases and Weak Acids Salts made from strong soluble bases and weak acids hydrolyze to

form basic solutions.

Anions of weak acids (strong conjugate bases) react with water to form hydroxide ions.

An example is sodium hypochlorite, NaClO, made from sodium hydroxide and hypochlorous acid.

base?or acidstronger theisWhich

solution in ions Notice

OH + OH OH + OH

ClONaNaClO

HClONaOH

3-

22

-OHin %100~)(

2

s

95

Salts of Strong Bases and Weak Acids

Na ClO Na ClO

H O + H O OH + H O

ClO H O HClO H O

+ - in H O -

2 2-

3+

-3 2

2( )

~s

100%

We can combine these last two equations into one single equation that represents the total reaction.

ClO H O HClO OH-2

96

Salts of Strong Bases and Weak Acids

K =

HClO OH

ClOb

-

-

The equilibrium constant for this reaction, called the hydrolysis constant, is written as:

97

Salts of Strong Bases and Weak Acids Algebraic manipulation of the previous expression

give us a very useful form of the expression. Multiply the expression by one written as [H+]/ [H+].

H+/H+ = 1

K =HClO OH

ClO

H

H

K =HClO

H ClO

H OH

b

-

-

b -

-

1

98

Salts of Strong Bases and Weak Acids Which can be rewritten as:

K =

HClO

H ClO

H OH

K =1

KK

b -

-

ba for HClO

w

1

99

Salts of Strong Bases and Weak Acids Which can be used to calculate the hydrolysis

constant for the hypochlorite ion:

K =1

KK

K =K

K=

1 10

3.5 10

K =HClO OH

ClO

ba for HClO

w

bw

a for HClO

-14

-8

b

2 9 10 7.

100

Salts of Strong Bases and Weak Acids This same method can be applied to the anion of

any weak monoprotic acid.

HAfor a

Wb

-2

K

K

A

OHHA=K

OHHAOHA

101

Salts of Strong Bases and Weak Acids Example 18-16: Calculate the hydrolysis constants

for the following anions of weak acids.

1. The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4.

F H O HF OH

KHF OH

F

KK

K

2

bw

a for HF

b

10 10

7 2 1014 10

14

411.

..

102

Salts of Strong Bases and Weak Acids The cyanide ion, CN-, the anion of hydrocyanic acid, HCN.

For HCN, Ka = 4.0 x 10-10.


510

14

b

HCNfor a

wb

-2

105.2100.4

100.1K

K

K

CN

OHHCNK

OH+ HCNOHCN

103

Salts of Strong Bases and Weak Acids Example 18-17: Calculate [OH-], pH and percent

hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.

MMM 0.10 0.10 0.10

ClO Na NaClO OHin 100%~(s)

2

104

Salts of Strong Bases and Weak Acids Set up the equation for the hydrolysis and the

algebraic representations of the equilibrium concentrations.

ClO + H O HClO + OH

Initial: 0.10

Change: - + +

At equil: 0.10 -

-2

-M M M

xM xM xM

x M xM xM

0 0

105

Salts of Strong Bases and Weak Acids Substitute the algebraic expressions into the

hydrolysis constant expression.

7b 109.2

ClO

OH HClOK

Kb

x xx010

2 9 10 7

..

106

Salts of Strong Bases and Weak Acids Substitute the algebraic expressions into the

hydrolysis constant expression.

10.23. pH get the we14.00 pOH pH From

3.77 pOH get the we][OH theFrom

OHClO107.1 becomesWhich

109.2 toreducesequation The

0.10 -0.10 and 0.10

case. in this made becan assumption gsimplifyin The

-

4

82

Mx

x

xx

107

Salts of Strong Bases and Weak Acids The percent hydrolysis for the hypochlorite ion may

be represented as:

%17.0%1000.10

101.7=hydrolysis %

%100ClO

ClO=hydrolysis %

4-

original-

hydrolyzed-

M

M

108

Salts of Strong Bases and Weak Acids If a similar calculation is performed for 0.10 M NaF

solution and the results from 0.10 M sodium fluoride and 0.10 M sodium hypochlorite compared, the following table can be constructed.

Solution Ka Kb [OH-] (M) pH%

hydrolysis

NaF 7.2 x 10-4 1.4 x 10-11 1.2 x 10-6 8.08 0.0012

NaClO 3.5 x 10-8 2.9 x 10-7 1.7 x 10-4 10.23 0.17

109

Salts of Weak Bases and Strong Acids Salts made from weak bases and strong acids form

acidic aqueous solutions. An example is ammonium bromide, NH4Br, made

from ammonia and hydrobromic acid.

base?or acidstronger theisWhich

aresolution in Ions

OH OH OHOH

Br NH BrNH

HBrOHNH

3-

22

-4

100%~OHs4

4

2

110

Salts of Weak Bases and Strong Acids

OH excess generates

OHNHOHNH

OH excess leavingsolution fromit

removingion OH with thereacts

,NH acid, strong relatively The

3

23-

4

3

-

4

The reaction may be more simply represented as:

OHNH OHNH 3324

111

Salts of Weak Bases and Strong Acids

The hydrolysis constant expression for this process is:

Or even more simply as:

4

3a

4

33a NH

H NHKor

NH

OH NHK

HNH NH 34

112

Salts of Weak Bases and Strong Acids Multiplication of the hydrolysis constant expression

by [OH-]/ [OH-] gives:

1

OH OH

OH NH

NHK

OH

OH

NH

OH NHK

-3

-4

3a

-

-

4

33a

113

Salts of Weak Bases and Strong Acids Which we recognize as:

K

KK1

KK

K

ab NH

w w

b NH

a

3 3

1

10 10

18 105 6 10

14

510.

..

114

Salts of Weak Bases and Strong Acids In its simplest form for this hydrolysis:

NH NH H

KNH H

NH

4 3

a3

4

5 6 10 10.

115

Salts of Weak Bases and Strong Acids Example 18-18: Calculate [H+], pH, and percent

hydrolysis for the ammonium ion in 0.10 M ammonium bromide, NH4Br, solution.

1. Write down the hydrolysis reaction and set up the table as we have done before:

NH H O NH H

Initial[]: 0.10 + +

Change: - + +

Equilibrium[]: 0.10

4+

2 3

M xM xM

xM xM xM

x M xM xM

116

Salts of Weak Bases and Strong Acids2. Substitute the algebraic expressions into the

hydrolysis constant.

0.10 - 0.10 thus0.10

.applicable is assumption The

105.6=10.0

K

106.5NH

H NHK

10-a

10

4

3a

xx

x

xx

117

Salts of Weak Bases and Strong Acids3. Complete the algebra and determine the

concentrations and pH.

12.5pH

107.5HNH

107.5=

106.5

106.510.0

6-3

6-

112

10

M

Mx

x

x

xx

118

Salts of Weak Bases and Strong Acids4. The percent hydrolysis of the ammonium ion in

0.10 M NH4Br solution is:

%

%.

%

hydrolysis =NH

NH

hydrolysis =7.5 10

hydrolysis = 0.0075%

4+

hydrolized

4+

original

-6

100%

010100%

MM

119

Salts of Weak Bases and Weak Acids Salts made from weak acids and weak bases can

form neutral, acidic or basic aqueous solutions. The pH of the solution depends on the relative values of

the ionization constant of the weak acids and bases.

1. Salts of weak bases and weak acids for which parent Kbase =Kacid make neutral solutions.

An example is ammonium acetate, NH4CH3COO, made from aqueous ammonia, NH3,and acetic acid, CH3COOH.

Ka for acetic acid = Kb for ammonia = 1.8 x 10-5.

120

Salts of Weak Bases and Weak Acids The ammonium ion hydrolyzes to produce H+ ions.

Its hydrolysis constant is:

NH NH H

KNH H

NH

4+

3

a3

4+

5 6 10 10.

121

Salts of Weak Bases and Weak Acids The acetate ion hydrolyzes to produce OH- ions. Its

hydrolysis constant is:

CH COO H O CH COOH OH

KCH COOH OH

CH COO

3 2 3

b3

3

5 6 10 10.

122

Salts of Weak Bases and Weak Acids Because the hydrolysis constants for both ions are

equal, their aqueous solutions are neutral. Equal numbers of H+ and OH- ions are produced.

solution!in formed are base and acidA weak

aresolution in Ions

OH OH OH OH

COOCHNHCOOCHNH

COOHCHOHNH

3-

22

34100%~OH

34

34

2

123

Salts of Weak Bases and Weak Acids2. Salts of weak bases and weak acids for

which parent Kbase > Kacid make basic solutions.

An example is ammonium hypochlorite, NH4ClO, made from aqueous ammonia, NH3,and hypochlorous acid, HClO.

Kb for NH3 = 1.8 x 10-5 > Ka for HClO = 3.5x10-8

124

Salts of Weak Bases and Weak Acids The ammonium ion hydrolyzes to produce H+ ions.

Its hydrolysis constant is:

10

4

3a

34

106.5NH

H NHK

HNHNH

125

Salts of Weak Bases and Weak Acids The hypochlorite ion hydrolyzes to produce OH-

ions. Its hydrolysis constant is:

Because the Kb for ClO- ions is three orders of magnitude larger than the Ka for NH4

+ ions, OH- ions are produced in excess making the solution basic.

ClO H O HClO OH

KHClO OH

ClO

2

b -

2 9 10 7.

126

Salts of Weak Bases and Weak Acids3. Salts of weak bases and weak acids for which

parent Kbase < Kacid make acidic solutions. An example is trimethylammonium fluoride,

(CH3)3NHF, made from trimethylamine, (CH3)3N,and hydrofluoric acid acid, HF.

Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4

127

Salts of Weak Bases and Weak Acids Both the cation, (CH3)3NH+, and the anion, F-,

hydrolyze.

CH NH F CH NH F3 3H O~100%

3 32

128

Salts of Weak Bases and Weak Acids The trimethylammonium ion hydrolyzes to produce

H+ ions. Its hydrolysis constant is:

(CH ) NH CH N H

KCH N H

(CH ) NH

KK

K

3 3+

3

a3

3 3+

w

b for CH N

a

3

( )

( )

.

..

( )

3

3

14

510

3

10 10

7 4 1014 10

129

Salts of Weak Bases and Weak Acids The fluoride ion hydrolyzes to produce OH- ions. Its

hydrolysis constant is:

Because the Ka for (CH3)3NH+ ions is one order of magnitude larger than the Kb for F- ions, H+ ions are produced in excess making the solution acidic.

F H O HF OH

KHF OH

F

KK

K

2

b -w

a for HF

b

10 10

7 2 1014 10

14

411.

..

130

Salts of Weak Bases and Weak Acids Summary of the major points of

hydrolysis up to now.1 The reactions of anions of weak monoprotic

acids (from a salt) with water to form free molecular acids and OH-.

A + H O HA + OH

KK

K

-2

-

bw

a HA

131

Salts of Weak Bases and Weak Acids2. The reactions of anions of weak monoprotic acids

(from a salt) with water to form free molecular acids and OH-.

BH + H O B + H O

KK

K B = weak base

+2 3

+

aw

b B

132

Salts of Weak Bases and Weak Acids Aqueous solutions of salts of strong acids

and strong bases are neutral. Aqueous solutions of salts of strong bases

and weak acids are basic. Aqueous solutions of salts of weak bases and

strong acids are acidic. Aqueous solutions of salts of weak bases and

weak acids can be neutral, basic or acidic.The values of Ka and Kb determine the pH.

133

Hydrolysis of Small Highly-Charged Cations Cations of insoluble bases (metal hydroxides) become

hydrated in solution. An example is a solution of Be(NO3)3. Be2+ ions are thought to be tetrahydrated and sp3 hybridized.

Be Be(OH )

1s 2s 2p

Be

aq2+

2 4

4

2

1s 2s 2p

Be

Be(OH )

e pairs on coordinated water molecules

2

form sp hybrids

2 4

-

3

2 xx xx xx xx

134

Hydrolysis of Small Highly-Charged Cations In condensed form it is represented as:

or, even more simply as:

Be(OH ) H O Be OH OH H O2 4 2 2 3 32

Be H O Be(OH) H2+2

135

Hydrolysis of Small Highly-Charged Cations The hydrolysis constant expression for [Be(OH2)4]2+

and its value are:

or, more simply

K

Be(OH ) (OH) H O

Be(OH )a2 3

+3

2 42

10 10 5.

K

Be(OH) H

Bea

+

2+

10 10 5.

136

Hydrolysis of Small Highly-Charged Cations Example 18-19: Calculate the pH and percent

hydrolysis in 0.10 M aqueous Be(NO3)2 solution.

1. The equation for the hydrolysis reaction and representations of concentrations of various species are:

xMxMMx 10.0

H Be(OH) OH Be 22

137

Hydrolysis of Small Highly-Charged Cations2. Algebraic substitution of the expressions into the

hydrolysis constant:

00.3pH

100.1BeH

100.1

100.1

100.110.0

10.0 - 0.10 and 0.10

applies. assumption gsimplifyin The

100.110.0

3hydrolyzed

2

3

62

5

5

M

Mx

x

xx

xx

x

xx

138

Hydrolysis of Small Highly-Charged Cations3 Calculate the percent hydrolysis of Be2+.

% hydrolyzed =Be

Be

% hydrolyzed =1.0 10

0.10

2+

hydrolyzed2+

original

-3

100%

100% 10%.

139

Hydrolysis of Small Highly-Charged Cations This table is a comparison of 0.10 M Be(NO3 )2

solution and 0.10 M CH3COOH solution.

Solution [H3O+] pH

% hydrolysis or

% ionization0.10 M Be(NO3)2 1.0 x 10-3 M 3.00 1.0%

0.10 M CH3COOH 1.3 x 10-3 M 2.89 1.3%

Notice that the Be solution is almost as acidic as the acetic acid solution.

140

Synthesis Question Rain water is slightly acidic because it

absorbs carbon dioxide from the atmosphere as it falls from the clouds. (Acid rain is even more acidic because it absorbs acidic anhydride pollutants like NO2 and SO3 as it falls to earth.) If the pH of a stream is 6.5 and all of the acidity comes from CO2, how many CO2 molecules did a drop of rain having a diameter of 6.0 mm absorb in its fall to earth?

141

Synthesis Question

Lmol77

32

7

32

77

a

7

32

3a

-332

3222

-75.6

104.2104.2COH

102.4COH

102.3102.3K

102.4COH

HCOHK

HCOHCOH

COH OH CO

102.310H6.5pH

M

M

142

Synthesis Question

molecules CO 106.1

mol

molecules106.022mol 106.2

104.2L101.1 molecules CO ofnumber

L101.1cm 11.0

cm 0.3 r droplet water of volume

213

2311

Lmol74

2

41000cm

1L3

33

433

4

3

143

Group Question

A common food preservative in citrus flavored drinks is sodium benzoate, the sodium salt of benzoic acid. How does this chemical compound behave in solution so that it preserves the flavor of citrus drinks?

144

End of Chapter 18

Weak aqueous acid-base mixtures are called buffers. They are the subject of Chapter 19.

1 Chapter 16 Ionic Equilibria: Acids and Bases. 2 Chapter Goals 1. A Review of Strong Electrolytes...

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Transcript of 1 Chapter 16 Ionic Equilibria: Acids and Bases. 2 Chapter Goals 1. A Review of Strong Electrolytes...