Chapter 16 Acids and Bases Dr. Peter Warburton [email protected]

Chapter 16Acids and Bases

Dr. Peter [email protected]://www.chem.mun.ca/zcourses/1051.php

2

Arrhenius theory of acids and bases

The Arrhenius concept: acids dissociate in water (aqueous solution) to produce hydrogen ions H+, and bases dissociate in water to give hydroxide ions, OH-.

Arrhenius acid: HA (aq) H+ (aq) + A- (aq)Arrhenius base: MOH (aq) M+ (aq) + OH- (aq)

3

Neutralization reactions

Arrhenius acids and bases react with each other to form water and aqueous salts in neutralization reactions.

H+ (aq) + A- (aq) + M+ (aq) + OH- (aq) H2O (l) + M+ (aq) + A- (aq)

The net ionic equation is

H+ (aq) + OH- (aq) H2O (l)

4

Aqueous salts

The aqueous salt in the reaction comes from the spectator ions in the reaction. These ions are present to balance the proton positive charge or the hydroxide ion negative charge in the acid or base.

If we evaporate all the water, we are left with an ionic solid called a salt e.g. NaCl

5

Arrhenius theory of acids and bases

Many substances that do not contain OH- act like bases!

The key to the Arrhenius description is that we need water to act as a solvent to

promote the dissociation of the acid or base.

6

Brønsted-Lowry theory of acids and bases

Many substances (like NH3) that do not contain OH- act like bases in water!

The Brønsted-Lowry Theory: an acid is any substance that donates protons (H+) while a base is any substance that can accept protons.

This means that Brønsted-Lowry acid-base reactions are proton transfer reactions.

7

Proton transfer reactions

Pairs of compounds are related to each other through Brønsted-Lowry acid-base reactions. These are conjugate acid-base pairs.

Generally, an acid HA has a conjugate base A- (a proton has transferred away from the acid). Conversely, a base B has a conjugate acid BH+ (a proton has transferred toward the base).

8

Water as an acid in BL reactions

When a Brønsted-Lowry base is placed in water, it reacts with the water (which acts as a Brønsted-Lowry acid) and establishes an acid-base equilibrium.

9

BL base strength

The strength of a Brønsted-Lowry base can be quantified by the equilibrium constant as it relates to completeness of reaction with water. For the reaction of ammonia with water

C25at 10 x 1.8

NH

OHNHK

5-

3

-4

b

10

BL base strength

We give the equilibrium constant a special name for reactions like these – the base dissociation (or ionization) constant Kb. Since the value of the constant is less than one, the ammonia does not dissociate to a great extent – it is a weak base!

C25at 10 x 1.8

NH

OHNHK

5-

3

-4

b

11

Water as a base in BL reactions

When a Brønsted-Lowry acid is placed in water, it reacts with the water (which acts as a Brønsted-Lowry base) and establishes an acid-base equilibrium.

12

BL acid strength

The strength of a Brønsted-Lowry acid can be quantified by the equilibrium constant as it relates to completeness of reaction with water. For the reaction of acetic acid with water

C25at 10 x 1.8

COOHCH

OHCOOCHK

5-

3

33a

13

BL acid strength

We give the equilibrium constant a special name for reactions like these – the acid dissociation (or ionization) constant Ka. Since the value of the constant is less than one, the acetic acid does not dissociate to a great extent – it is a weak acid!

C25at 10 x 1.8

COOHCH

OHCOOCHK

5-

3

33a

14

Strong BL acids and bases

Strong BL acids and bases have the exact same reaction with water as do weak acids and bases, they just are much more complete. This is reflected in the acid or base dissociation constant – it is much larger than one!Hydrochloric acid is a strong acid.

C25at 10 x 1

HCl

OHClK

6

3a

15

Hydrated protons and hydronium ions

What is the strongest Brønsted-Lowry acid there is? The strongest Brønsted-

Lowry acid is H+.

The ultimate proton-donor is a proton itself!

In water there is no such thing as H+.

large! very K OHOHH 32

16

Hydrated protons and hydronium ions

Often more than one water molecule will crowd around the hydronium ion (H3O+) to give hydrates

with the formula [H3O(H2O)n]+ where n is 1 to 4.

17

Requirements of Brønsted-Lowry bases

For a molecule or ion to accept a proton (to act as a base) requires it to have an unshared pair of

electrons which can then be used to create a bond to the H+.

All Brønsted-Lowry bases have at least one lone pair of electrons.

In the previous reactions we’ve seen NH3 has a lone pair and can act as a base. Also, water has two lone pairs, and can act as a base.

18

Amphiprotic substances

Some substances, like water, have protons that can be donated (BL acid),

and lone pairs of electrons that can accept protons (BL base). This is why it

can act like an acid AND a base.

Such substances are said to be

amphiprotic.

19

Problem

Write a balanced equation for the dissociation of each of the following Brønsted-Lowry acids in water:

a) H2SO4 b) HSO4

-

c) H3O+ d) NH4

+

20

Problem

What is the conjugate acid of each of the following Brønsted- Lowry bases?

a) HCO3-

b) CO32-

c) OH-

d) H2PO4-

21

Problem

Of the following species, one is acidic, one is basic, and one is amphiprotic in their reactions with water: HNO2, PO4

3-, HCO3-.

Write the four equations needed to represent these facts.

OH COH OH HCO CO OH OH HCO

OH HPO OH PO NO OH OH HNO

322-

3

2

332-

3

-242

-342322

22

Problem

For each of the following reactions, identify the acids and bases in both the forward and reverse directions:

Cl OHHC HCl OHC

NH SO NH HSO

OH F OH HF

232-

232

42

43-

4

32

Acid Base Base Acid

Acid Base Base Acid

Base Acid Acid Base

23

A 2nd look at acid and base strength

Acid-base equilibria are competitions!

The equilibrium is a tug-of war between the two bases in the system as they fight for protons given

away by the two acids.

The acid that is “better at giving away protons” (or the base that is “better at taking protons”) will be found in lesser amounts at equilibrium

than the other acid (or base).

24

Strong acids in water

A strong acid (HA) is one that almost completely dissociates in water (which acts as a weak base). The conjugate base of the strong acid (A-) will be a very weak base.

At equilibrium, there will be very little to no HA present in the system, and the concentration of A- will essentially be the same as the initial concentration of HA.

acidweaker

3basevery weak baseweak

2acid strong

OHAOHHA

25

Strong bases in water

A strong base (A-) is one that almost completely dissociates in water (which acts as an acid). The conjugate acid of the strong base (HA) will be a very weak acid.

At equilibrium, there will be very little to no A- present in the system, and the concentration of HA will essentially be the same as the initial concentration of A-.

baseweaker acidvery weak acidweak

2base strong

- OHHAOHA

26

Relationship of acid/base strengths

The stronger the acid, the weaker its conjugate base.

The stronger the base, the weaker its conjugate acid.

We’ll look at this a bit more in depth later.

27

28

Differentiating strong acids

Very strong acids like HClO4 and HCl dissociate in water so completely it is almost impossible to find accurate Ka values, and therefore determine which is the stronger acid. We must place the acids in a solvent that is a much weaker base than water (poorer at taking protons). Whichever reaction in the new solvent has a higher K value tells us which is really the stronger acid.

29

Differentiating strong acids

HClO4 must be a stronger acid than HCl because it forces the very weak base diethyl ether to accept protons much more readily (reaction is essentially complete – large K) than does HCl, which establishes an equilibrium – smaller K.

30

Self-ionization of water

Water can act as an acid or a base because the molecule has both protons and lone pairs available – its amphiprotic!It is possible for one water molecule to act as an acid

while another water molecule acts as a base at the same time. This leads to the auto-dissociation (or self-

ionization) of water equilibrium reaction:

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

The equilibrium constant for this reaction is called the

ion-product constant for water, Kw.

Kw = [H3O+] [OH-]

31

Autoionization of water

32

At 25 °C, Kw = 1.0 x 10-14

so [H3O+] = [OH-] = 1.0 x 10-7 mol/L

Relatively few water molecules are dissociated at equilibrium at room temperature!

We will always assume that [H3O+] [OH-] = 1.0 x 10-14 at 25 °C.

[H3O+] > 1.0 x 10-7 is acidic ([OH-] < 1.0 x 10-7)

[H3O+] < 1.0 x 10-7 is basic ([OH-] > 1.0 x 10-7)

[H3O+] = 1.0 x 10-7 is neutral ([OH-] = 1.0 x 10-7)

33

We also find, since

[H3O+] [OH-] = 1.0 x 10-14 = Kw

then

[H3O+] = 1.0 x 10-14 / [OH-]

and

[OH-] = 1.0 x 10-14 / [H3O+]

at 25 °C

34

35

Problem

The concentration of OH- in a sample of seawater is 5.0 x 10-6 molL-1. Calculate the concentration of H3O+ ions, and classify the solution as acidic, neutral, or basic.Answer: [H3O+] = 2.0 x 10-9. Solution is basic.

36

Problem

At 50 °C the value of Kw is 5.5 x 10-14. What are the [H3O+] and [OH-]in a neutral solution at 50 °C?

Answer: Both are 2.3 x 10-7 molL-1

37

The pH scale

Because [H3O+] in water solutions can range from very small (strongly basic) to very large (strongly acidic) it is sometimes easier to use a logarithmic (power of 10) scale to express [H3O+].

Additionally, since using a negative number is sometime awkward, we actually use a negative logarithmic scale to express the hydronium ion concentration with a term we call the pH of a solution.

pH = - log [H3O+]

38

pH and acidity

For [H3O+] = 1.0 x 10-7 molL-1 (neutral), pH = 7.00

For [H3O+] = 1.0 x 10-4 molL-1 (acidic), pH = 4.00

For [H3O+] = 1.0 x 10-11 molL-1 (basic), pH = 11.00

In general,

pH > 7 is basic

pH < 7 is acidic

pH = 7 is neutral

39

pOH and acidity

We occasionally see the pOH mentioned, which is

pOH = - log [OH-] or

[OH-] = 10-pOH

pOH < 7 is basic

pOH > 7 is acidic

pOH = 7 is neutral

40

Kw = 1.0 x 10-14 = [H3O+] [OH-]

14.00 = pH + pOH at 25 C

41

Problem

Calculate the pH of each of the following solutions:

a) A sample of seawater that has an

OH- concentration of 1.58 x 10-6 molL-1

b) A sample of acid rain that has an

H3O+ concentration of 6.0 x 10-5 molL-1

Answers: a) pH = 8.19 b) pH = 4.22

42

Problem

Calculate [H3O+] and [OH-] in each of the following solutions:

a) Human blood (pH 7.40)

b) A cola beverage (pH 2.8)Answers:

a) [H3O+] = 4.0 x 10-8 M and [OH-] = 2.5 x 10-7 M

b) [H3O+] = 1.6 x 10-3 M and [OH-] = 6.3 x 10-12 M

43

Problem

The pH of a solution of HCl in water is found to be 2.50. What volume of water would you add to 1.00 L of this solution to raise the pH to 3.10?

Answer: You would add 3.0 L of water

44

Strong acids and strong bases

Most of the strong acids are monoprotic acids, that are capable of donating only one proton.

Sulphuric acid (H2SO4), which is a diprotic acid capable of donating two protons, is also a strong acid (for the first proton!).

Strong monoprotic acids (HA) essentially dissociate 100% in water to give H3O+ and A-, leaving virtually no HA in solution at equilibrium.

If we know the initial concentration of HA, then the equilibrium concentration of H3O+ will be the same, and we can calculate the pH.

45

pH and strong bases

Strong bases, such as the alkali metal (Li, Na, K, etc.) hydroxides MOH completely dissociate in water to give metal ions and hydroxide ions. H2O

MOH (s) M+ (aq) + OH- (aq)

Again, calculating the pH is straight-forward, as the final concentration of OH- will be the same as the initial concentration of MOH. Of course, if we know [OH-], we can calculate [H3O+], and the pH.

46

pH and strong bases

Alkaline earth (Ca, Mg, etc) metal hydroxides M(OH)2 are strong bases and will completely dissociate in water up to the point of their low solubility.

(aq)OH 2(aq)M(aq)M(OH)

(aq)M(OH)(s)M(OH)

2ondissociatifull2

2amountsmalldissolve

2

47

pH and strong bases

Alkaline earth (Ca, Mg, etc) metal oxides MO are stronger bases than the equivalent hydroxides because O2- is a very strong base. In fact, much like bare H+ can’t exist in water, neither can bare O2-.

O2- (aq) + H2O (l) 2 OH- (aq), so H2O

MO (s) M2+ (aq) + 2 OH- (aq)

48

Be careful!

We are assuming the pH of the solution will be determined solely by the initial concentration of the strong acid or strong base. This is true if the initial concentration is large enough that the autoionization of water contributes insignificant amounts of H3O+ and OH-!

What is the pH of 1.0 x 10-8 M HCl?

49

Common strong acids and bases

50

Problem

Calculate the pH of:

a) 0.050 M HClO4

b) 6.0 M HCl

c) 4.0 M KOH

d) 0.010 M Ba(OH)2

Answers:

a) pH = 1.30 b) pH = -0.78

c) pH = 14.60 d) pH = 12.30

51

Problem

If 535 mL of gaseous HCl at 26.5 C and 747 mmHg is dissolved in enough water to prepare 625 mL of solution, what is the pH of this solution?

The gas constant R = 0.08206 LatmK-

1mol-1 and 760 mmHg = 1 atm exactly!

Answer: pH = 1.467

52

Problem

Milk of magnesia is a saturated solution of Mg(OH)2. Its solubility is 9.63 mg Mg(OH)2 per 100.0 mL of solution at 20 C. What is the pH of saturated Mg(OH)2 at 20 C?

The molar mass of Mg(OH)2 is 58.3197 gmol-1

Answer: pH = 11.52

53

Problem

Calculate the pH of an aqueous solution that is 3.00% KOH by mass and has a density of 1.0242 gmL-1.

The molar mass of KOH is 56.1056 gmol-1.

Answer: pH = 13.74

54

pKa and pKb

We can define pKa = - log Ka and pKb = - log Kb

exactly like pH = - log [H3O+]

A very small Ka or Kb value is the same as a large positive pKa or pKb which means the

acid or base is weak (partially dissociated in water).

As Ka (or pKa ) or as Kb (or pKb ) acid strength or base strength increases.

55

pKa and pKb

56

Problem

The pH of 0.10 molL-1 HOCl is 4.27. Calculate Ka for hypochlorous acid

HOCl (aq) + H2O (l) H3O+ (aq) + ClO- (aq)

Answer: 2.9 x 10-8

57

Identifying weak acids

Generally there are three categories of weak acids:

carboxylic acids

oxoacids

miscellaneous acids

58

Carboxylic acids

Carboxylic acids contain the -COOH (carboxyl) group. The hydrogen of this group is the proton that is donated.

59

Carboxylic acids

60

Oxoacids

Oxoacids are generally weak acids with the formula

HmXOn where m = 1 to 3 and n = 1 to 4

This formula is often quite misleading because the structure is actually usually

(HO)mXOn where (m + n) = 1 to 4

61

Strong oxoacids

Nitric acid –

HNO3

Sulfuric acid –

H2SO4

Perchloric acid –

HClO4

62

Some weak oxoacids

Nitrous acid –

HNO2

Phosphoricic acid –

H3PO4

Chlorous acid –

HClO2

63

Miscellaneous weak acids

There are other weak acids that are not carboxylic acids or oxoacids. Some of the more common ones are

Hydrofluoric acid – HF

Hydrocyanic acid – HCN

Hydrazoic acid – HN3

64

Identifying weak bases

Many weak bases are amines, which contain nitrogen. The lone pair on the nitrogen allow the amine to be a proton acceptor (Brønsted-Lowry base).

65

Identifying weak bases

66

Amino acids

Amino acids have both a carboxyl group and an amine group, meaning different parts of the molecule can act as an acid and a base.

67

Equilibrium in solutions of weak acids and bases

We can calculate equilibrium concentrations of reactants and products in acid-base reactions with known values for Ka or Kb.

We need to figure out what is an acid and what is a base in our system. Water will be an acid or base depending on whether we added a base or an acid to it. For example, if we start with 0.10 molL-1 HCN, then HCN is an acid, and water is a base.

HCN (aq) + H2O (l) H3O+ (aq) + CN- (aq)

Ka = 4.9 x 10-10

68


However, since the acid-base reaction we are looking at takes place in water, we must include the autoionization of water reaction as a source of H3O+ and OH-!

For our example reaction (HCN is a weak acid)

HCN (aq) + H2O (l) H3O+ (aq) + CN- (aq)

Ka = 4.9 x 10-10

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

Kw = 1.0 x 10-14

69

The strongest acid or base (larger K value) will dominate a system. We call this equilibrium reaction with the larger K value the principal reaction. Any other reactions are subsidiary reactions.


70

Since all reactions take place in one container, then

[H3O+] = [H3O+] (principal) + [H3O+] (subsidiary)OR

[OH-] = [OH-] (principal) + [OH-] (subsidiary)

If the K of the principal reaction is much greater than K for the subsidiary reactions, then we assume

[H3O+] [H3O+] (principal reaction)OR

[OH-] [OH-] (principal reaction)


71

We can solve this equation using the quadratic formula and get the right answer, but it might be possible to do it more simply.

Let’s assume the initial concentration of the acid and the equilibrium concentration of HCN are essentially the same (that is x << 0.10 in this case).

(all in mol/L) HCN (aq) + H2O (l) H3O+ (aq) + CN- (aq)

Initial conc. 0.10 N/A 0.0 0.0 Conc. change -x N/A +x +x Equil. conc. 0.10 – x N/A +x +x

x)(0.10

(x)(x)10 x 4.9 so

[HCN]

]][CNO[H10 x 4.9K 10310

a


72

4.9 x 10-10 = x2 / 0.10x2 = (4.9 x 10-10)(0.10)

x2 = 4.9 x 10-11

x = 4.9 x 10-11

x = 7.0 x 10-6 molL-1

Based on the assumption we’ve made, at equilibrium [H3O+] = [CN-] = 7.0 x 10-6 molL-1 and [HCN] = 0.10 mol/L.

However, any time we make an assumption, we must check it! (5% rule)


73

We also assumed that [H3O+] [H3O+] (principal reaction) so we must check this.

For the subsidiary reactionH2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

Kw = 1.0 x 10-14

If the assumption we’ve made is good, then [H3O+] = 7.0 x 10-6 molL-1 and so

[OH-]subsidiary = Kw / [H3O+]principal

[OH-]subsidiary = 1.0 x 10-14 / 7.0 x 10-6

[OH-]subsidiary = 1.4 x 10-9 molL-1


74

From the subsidiary balanced equation we see if [OH-]subsidiary = 1.4 x 10-9 molL-1, then

[H3O+]subsidiary = 1.4 x 10-9 molL-1.

Lets check our assumption

The ratio [H3O+]subidiary / [H3O+]principal,

is 1.4 x 10-9 molL-1 / 7.0 x 10-6 molL-1 = 0.02%.

Our assumption is valid (5% rule)!


75

[H3O+] [H3O+] (principal reaction)

= 7.0 x 10-6 molL-1

pH = - log [H3O+]

pH = - log 7.0 x 10-6

pH = 5.15


76

Generally it turns out that if our principal acid-base calculation gives a [H3O+] or

[OH-] (depending on reaction type!) less than 10-6 or so then the auto-dissociation of

water reaction will actually contribute a significant amount of [H3O+] or [OH-] to our system and we must include the subsidiary

reaction contribution to pHSee page 673 (9th ed.) or 682 (8th ed.) of

textbook

77

Problem

Acetic acid CH3COOH is the solute that gives vinegar its characteristic odour and sour taste. Calculate the pH and the concentration of all species present in:

a) 1.00 molL-1 CH3COOH

b) 0.0100 molL-1 CH3COOH

78

Problem a)

(all in mol/L) CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)

Initial conc. 1.00 N/A 0.0 0.0 Conc. change -x N/A +x +x Equil. conc. 1.00 – x N/A +x +x

x)(1.00

(x)(x)10 x 1.8 so

COOH][CH

]COO][CHO[H10 x 1.8K 5

3

335a

Let’s assume that x << 1.00 and so1.8 x 10-5 = x2 / 1.00x2 = (1.8 x 10-5)(1.00)

x = 1.8 x 10-5

x = 4.2 x 10-3 molL-1 (must be +ve value since x = [H3O+])

CHECK ASSUMPTION

79

We also assumed that [H3O+] [H3O+] (principal reaction) so we must check this.

For the subsidiary reaction

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

Kw = 1.0 x 10-14

If the assumption we’ve made is good, then [H3O+] = 4.2 x 10-3 molL-1 and so

[OH-]subsidiary = Kw / [H3O+]principal

[OH-]subsidiary = 1.0 x 10-14 / 4.2 x 10-3

[OH-]subsidiary = 2.4 x 10-12 molL-1

Problem a)

80

From the subsidiary balanced equation we see if

[OH-]subsidiary = 2.4 x 10-12 molL-1, then

[H3O+]subsidiary = 2.4 x 10-12 molL-1.

Lets check our assumption

The ratio [H3O+]subidiary / [H3O+]principal,

is 2.4 x 10-12 molL-1 / 4.2 x 10-3 molL-1

= 0.00000006%.

Our assumption is valid (5% rule)!

Problem a)

81

Problem a)

pH = - log [H3O+]

pH = - log 4.2 x 10-3

pH = 2.38

82

Let’s assume that x << 0.00100 and so1.8 x 10-5 = x2 / 0.00100x2 = (1.8 x 10-5)(0.00100)

x = 1.8 x 10-8

x = 1.3 x 10-4 molL-1 (must be +ve value since x = [H3O+])

CHECK ASSUMPTION

Problem b)

x)(0.00100

(x)(x)10 x 1.8 so

COOH][CH


3

335a

83

Problem b)

We can not assume that x << 0.00100 and so

0x]x10 x 1.8[]10 x [1.8

0xx)](0.0010010 x [1.8258

25

x)(0.00100

(x)(x)10 x 1.8 so

COOH][CH


3

335a

84

Problem b)

pH = - log [H3O+]pH = - log 1.25 x 10-4

pH = 3.90

mol/L10 x 25.1xormol/L10 x 43.1xso2

10 x 1.7xor

2

10 x 8.2x

2

10 x 2.610 x 1.8xor

2

10 x 2.610 x 1.8x

2

10 x 7.210 x 3.210 x 1.8xor

2

10 x 7.210 x 3.210 x 1.8x

1)2(

)10 x 1)(1.84()10 x 1.8()10 x 1.8(xso

2a

4acbbx

445

94

7

48

54

85

8104

58104

5

82552

85

Problem

Piperidine (C5H11N; molar mass = 85.149 gmol-1) is a base found in small amounts in black pepper. What is the pH of 315 mL of an aqueous solution containing 114 mg of piperidine if Kb is 1.6 x 10-3?

Answer: pH = 11.29

86

Percent ionization of weak acids

The pH of a solution of a weak acid like acetic acid will depend on the initial concentration of the weak acid. Therefore, we can define a second measure of the strength of a weak acid by looking of the percent ionization (or dissociation) of the acid.

%ionized = [H3O+]eqm / [HA]initial x 100%

87

Percent ionization

We saw on slide 78 that an acetic acid solution with initial concentration of 1.00 molL-1 at equilibrium had

[H3O+] = 4.2 x 10-3 molL-1


%ionized = 4.2 x 10-3 molL-1 / 1.00 molL-1 x 100%

%ionized = 0.42%

88

Percent ionization

We saw on slide 84 that an acetic acid solution with initial concentration of 0.0010 molL-1 at equilibrium had

[H3O+] = 1.25 x 10-4 molL-1


%ionized = 1.25 x 10-4 molL-1 / 0.0010 molL-1 x 100%

%ionized = 12.5%

89

90

Polyprotic acids

Acids that contain more than one dissociable protons are polyprotic acids.

Each dissociable proton has its own Ka value.

Carbonic acid (H2CO3) regulates blood pH.

H2CO3 (aq) + H2O (l) H3O+ (aq) + HCO3- (aq)

HCO3- (aq) + H2O (l) H3O+ (aq) + CO3

2- (aq)

7

32

33a 10 x 4.4

COH

HCOOHK

11

3

233

a 10 x 7.4HCO

COOHK

91

Polyprotic acids

Ka1 > Ka2 > Ka3 is always true!

92

Problem

Calculate the pH and the concentration of all species, including OH- present in 0.10 molL-1 H2SO3. Values of Ka are given in the table on the previous slide.

Ka1 = 1.3 x 10-2 and Ka2 = 6.2 x 10-8

93

Problem

Potential reactions that can occur in our system are

H2SO3 (aq) + H2O (l) H3O+ (aq) + HSO3- (aq)

HSO3- (aq) + H2O (l) H3O+ (aq) + SO3

2- (aq)

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)We’ll first have to assume

[H3O+] [H3O+] (H2SO3 dissociation)

(all in mol/L) H2SO3 (aq) + H2O (l) H3O+ (aq) + HSO3

- (aq) Initial conc. 0.10 N/A 0.0 0.0 Conc. change -x N/A +x +x Equil. conc. 0.10 – x N/A +x +x

x)(0.10

(x)(x)10 x 1.3 so

]SO[H

]][HSOO[H10 x 1.3K 2

32

332a1

94

Problem

We can’t assume x << 0.10 so

mol/L10 x 3.0xormol/L10 x 4.3xso2

10 x 6.0xor

2

10x 8.6x

2

10x 7.310 x 1.3xor

2

x107.310 x 1.3x

2

)10 x (5.210x 1.610 x 1.3xor

2

)10 x (5.210x 1.610 x 1.3x

1)2(

)1)(1.3x104()1.3x10()10 x 1.3(xso

2a

4acbbx

222

32

3

23

223

2

349

2349

2

32222

0x x)10 x 1.3()10 x (1.3

0xx)(0.1010 x 1.3223

22

95

Problem

x = [H3O+]principal = [HSO3-] = 3.0 x 10-2 molL-1

[H2SO3] = (0.10 – 3.0 x 10-2) molL-1 = 0.07 molL-1

The second dissociation reaction is

HSO3- (aq) + H2O (l) H3O+ (aq) + SO3

2- (aq)

With our assumption that almost all H3O+ comes from the first dissociation, if we substitute our equilibrium concentrations from our first reaction into the equilibrium constant expression for this reaction we should see little change if the assumption is true…

96

Problem

x = [SO32-] = [H3O+]sub = 6.3 x 10-8 mol/L

Our assumption is valid!

Since the second proton dissociation contributes a insignificant amount of H3O+

then the autoionization of water won’t either, because it has an even smaller K value.

)(3.2x10

)(x)(3.2x1010 x 6.3so

][HSO

]][SOO[H10 x 6.3K

2

28

3

2338

a2

97

Problem

pH = - log [H3O+]

pH = - log 3.0 x 10-2

pH = 1.52

[OH-] = Kw / [H3O+]

[OH-] = 1.0 x 10-14 / 3.0 x 10-2

[OH-] = 3.3 x 10-13 molL-1

98

Relation between Ka and Kb

The strength of an acid in water is expressed through Ka. while the strength of a base can be

expressed through Kb

But a Brønsted-Lowry acid-base reaction involves conjugate acid-base pairs so there should be a

connection between the Ka value and the Kb value.

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

A- (aq) + H2O (l) OH- (aq) + HA (aq)

HA

AOHK 3

a

A

HAOHKb

99

Let’s add the reactions together

The sum of the reactions is the dissociation of water reaction, which has the ion-product constant for water

Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25 °C

Closer inspection shows us that

HA (aq) + H2O (l) + A- (aq) + H2O (l) H3O+ (aq) + A- (aq) + OH- (aq) + HA (aq)

2 H2O (l) H3O

+ (aq) + OH- (aq)

C 25at1.0x10KOHOHA

HAOH

HA

AOHKxK 14

w33

ba

100

As the strength of an acid increases (larger Ka) the strength of the conjugate base must

decrease (smaller Kb) because their product must always be the dissociation constant for

water Kw.

Strong acids always have very weak conjugate bases. Strong bases always have very weak

conjugate acids.

Since Ka x Kb = Kw then Ka = Kw / Kb

and Kb = Kw / Ka

101

Problem

a) Piperidine (C5H11N) is an amine found in black pepper. If Kb = 1.6 x 10-3 for piperidine then calculate Ka for the C5H11NH+ cation.

b) If Ka = 2.9 x 10-8 for HOCl then calculate Kb for OCl-.Answers: a) Ka = 6.3 x 10-12 b) Kb = 3.4 x 10-7

102

Ions as acids and bases

Consider the strong acid anions such as Cl-, Br-, I-, NO3

-, and ClO4-. They must

be very weak bases because their conjugate acids are so strong. They will not react with water to form the strong acid…

Cl- (aq) + H2O (l) no reaction!

103


Consider the strong base cations such as Na+, K+, Ca2+, Mg2+, etc. They can not be acids because they have no protons. They will not react with water to form the strong base…

H2O (l) + Na+ (aq) no reaction!

104


Consider some weak acid anions such as OCl-, CH3COO-, HCO3

-, or NO2-. They

will be weak bases because their conjugate acids are weak. They can react with water to form the weak acid and OH-…

OCl- (aq) + H2O (l) OH- (aq) + HOCl (aq)

105


Consider some weak base cations such as NH4

+, C6H5NH+, CH3NH3+, etc. They

are weak acids because their conjugate bases are weak. They can react with water to form the weak base and H3O+ …

H2O (l) + NH4+ (aq) H3O+ (aq) + NH3 (aq)

106

Salts that yield neutral solutions

Any ionic salt that contains neither

an acidic cation OR a basic anion

will give a neutral solution because neither ion can react with water.

NaCl, KNO3, Ca(ClO4)2, MgI2, etc.

107

Salts that yield acidic or basic solutions

Any ionic salt that contains either

an acidic cation OR a basic anion

while the other ion can not react with water

will give an acidic or basic solution because of the ability of the ion to react with water.

NH4Cl, CH3COONa, C6H5NHNO3, KOCl

etc.

108

If a salt is composed of an acidic cation

and a basic anion,

the acidity or basicity of the salt solution

depends on the relative strengths of the acid and base.


109


If the acid cation is “stronger” than the base anion

the solution will be acidic.

If the base anion is “stronger” than the acid cation

the solution will be basic.

110


Ka > Kb the acid is better at

reacting with water than the base and the solution is acidic.

Ka < Kbthe base is better at

reacting with water than the acid and the solution is basic.

Ka Kbthe solution is close to neutral.

111

Problem

Predict whether the following salt solution is basic, neutral or acidic, and calculate the pH.

0.25 molL-1 NH4Br

Kb of NH3 is 1.8 x 10-5

Answer: The solution is acidic and pH is 4.93

112

Problem

Classify each of the following salts as acidic, basic, or neutral:

a) KBr

b) NaNO2

c) NH4Br

d) NH4F

neutral

basic

acidicKa for HF = 6.6 x 10-4

Kb for NH3 = 1.8 x 10-5 acidic

113

Factors that affect acid strength

There are several factors that can affect acid strength, and the importance of the factors can be variable. However, two trends are notable.

1) Bond strength – The strength of the bond between the acidic proton and the rest of the molecule will have an effect on acidity. The weaker the bond, the stronger the acid will generally be.

114

Factors that affect acid strength

2) Bond polarity – The polarity of a bond is the distribution of the electrons between the two bonded atoms.

A highly polar bond between an acidic hydrogen and another atom tends to make it more easy for the proton to leave the molecule than would happen for a non-polar bond.

115

Polarity versus strength

In the binary acids the bond strength is the more important factor. Bond strength tends to decrease down a column in the periodic table. HF is the weakest binary acid even though it has the most polar bond because it has the strongest bond.

116

Polarity versus strength

For acids of elements in the same row, the bond strengths tend to be similar, and so the polarity of the bond plays the greater role in determining acid

strength.

117

Polarity versus strengthCombining the decrease of bond strength down a column and increase of bond polarity across a row

we find the strongest acids tend to be those of the elements in the bottom right of the periodic table.

118

Oxoacids and acid strength

For oxoacids, acid strength tends to increase with the electronegativity of

the central atom, and with an increase in the central atom

oxidation number (which generally increases with the number of other atoms bonded to the central atom).

119


Here we see three oxoacids with different central atoms.

Oxoacid strength increases with electronegativity of X

120


Oxoacids with the same central atom X will be strongest when many other atoms are bonded

to X (the oxidation number of X increases.)

121

Organic compounds and acid strength

The acid strength of organic compounds can be rationalized by the bond strength between the proton and the atoms it’s bonded to (weaker bond - stronger acid) but more correctly it is determined by the stability of the conjugate base. Any factor that tends to stabilize the conjugate base increases organic acid strength.

122


123


fluoroacetic acid Ka = 2.7 x 10-3chlororoacetic acid Ka = 1.4 x 10-3

124

Amines and base strength

We require a lone pair of electrons to have a BL base, so any factor that tends to reduce the availability of

the lone pair will weaken the base while any factor that tends to increase the

availability of the lone pair will strengthen the base.

125


Alkyl groups are slightly electron donating, so secondary and tertiary amines tend to

be slightly stronger bases than ammonia and primary amines.

126


Factors that stablize the structure of amines will decrease the base strength

due to reduced electron availability.

127


Factors that stablize the structure of amines will decrease the base strength

due to reduced electron availability.

128

Lewis acids and bases

A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor.

These definitions are more general than the BL definitions because protons aren’t involved which means there exist

Lewis acids that are not Brønsted-Lowry acids.

129

Lewis acids and bases

We’ve seen that all Brønsted-Lowry bases must all have at least one lone

pair of electrons, so any Brønsted-Lowry base

must also be a Lewis base, and any Lewis base

must also be a Brønsted-Lowry base!

130

In general LA + :LB LA-LB

131

In general LA + :LB LA-LB

132

Coordination compounds

We’ve already seen that coordination compounds have

complex ions that are formed from a central

metal ion (a Lewis acid) surrounded by

ligands (Lewis bases)

133

Acidic solutions of metal ions

Small and/or highly charged metal ions, like Al3+, Be2+, and Li+

form acidic solutions because they form complex ions with water. The protons of the water ligands see less electron density than in free water, and so the O-H bond is

weaker, leading to increased acidity!

134

Acidic solutions of metal ions

There is a second benefit as well because the complex ion has a smaller charge in a larger volume….

Ka = 1.74 x 10-5 – about the same as acetic acid!

Documents

Chapter 16 Acids and Bases Dr. Peter Warburton [email protected]