1 Chapter 16. Acid –Base Equilibria... 2 Equilibria in Solutions of Weak Acids The dissociation of a weak acid is an equilibrium situation with an equilibrium

1

Chapter 16. Acid –Base Equilibria

. .

2

Equilibria in Solutions of Weak Acids

The dissociation of a weak acid is an equilibrium situation with an equilibrium

constant, called the acid dissociation constant, Ka based on the equation

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

HA

AOHK 3

a

3

Equilibria in Solutions of Weak Acids

The acid dissociation constant, Ka is always based on the reaction of one mole

of the weak acid with water.

If you see the symbol Ka, it always refers to a balanced equation of the form

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

4

Problem

The pH of 0.10 mol/L HOCl is 4.23. Calculate Ka for hypochlorous acid.

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

HOCl

ClOOHK 3

a

5

Calculating Equilibrium Concentrations in Solutions of Weak Acids

We can calculate equilibrium concentrations of reactants and

products in weak acid dissociation reactions with known values for Ka.

To do this, we will often use the ICE table technique we saw in the last

chapter on equilibrium.

6

Calculating Equilibrium Concentrations in Solutions of Weak Acids

We need to figure out what is an acid and what is a base in our system.

For example, if we start with 0.10 mol/L 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

7

Like our previous equilibrium problems, we then create a table of the initial concentrations of all chemicals, the change in their concentration, and their equilibrium concentrations in terms of known and unknown values.

(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)4.9x10so

[HCN]

]][CNO[H4.9x10K 10310

a

8

We can ALWAYS solve this equation using the quadratic formula and get the right

answer, but it might be possible to do it more simply.



x)(0.10

(x)(x)10 x 4.9 so

[HCN]

]][CNO[H10 x 4.9K 10310

a

9

Every time we do an weak acid equilibrium problem, divide the initial

concentration of the acid by Ka.

For this example

0.10 / 4.9 x 10-10 = 2 x 108

x)(0.10

(x)(x)10 x 4.9 so

[HCN]

]][CNO[H10 x 4.9K 10310

a

10

0.10 / 4.9 x 10-10 = 2 x 108

Since this value is greater than 100, we can assume that the initial concentration of

the acid and the equilibrium concentration of the acid are the same.

This assumption will lead to answers with less than 5% error since this pre-check is

greater than 100.

x)(0.10

(x)(x)10 x 4.9 so

[HCN]

]][CNO[H10 x 4.9K 10310

a

11

The assumption we will make is that

x << [HCN]i so [HCN]eqm [HCN]I

4.9 x 10-10 = x2 / 0.10

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

x = 4.9 x 10-11

x = 7.0 x 10-6 mol/L

x)(0.10

(x)(x)10 x 4.9 so

[HCN]

]][CNO[H10 x 4.9K 10310

a

12

Based on the assumption we’ve made, at equilibrium

x = [H3O+]eqm = [CN-]eqm

= 7.0 x 10-6 mol/L

(-ve value isn’t physically possible)

[HCN]eqm = 0.10 mol/L.

x)(0.10

(x)(x)10 x 4.9 so

[HCN]

]][CNO[H10 x 4.9K 10310

a

13

.

Any time we make an assumption,

we MUST check it.

We assumed x << [HCN]i

To check the assumption, we divide x by [HCN]i and express it as a

percentage

%007010 x 7.0M 0.10

M 10 x 7.0

HCN

x 5--6

i

.

14

As long as the assumption check is less than 5%, then

the assumption is valid!

If the assumption was not valid, we would have to go back and use the

quadratic formula!

15

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

is always taking place in water whether or not we have added

an acid or base.This reaction also contributes

H3O+ (aq) and OH- (aq)

to our system at equilibrium

Remember!

16

Since at 25 CKw = [H3O+] [OH-] = 1.0 x 10-14

it turns out that if our acid-base equilibrium we’re interested in gives a pH value between about 6.8 and 7.2 then the auto-dissociation of water contributes a significant amount of [H3O+] and [OH-] to our system and the real pH would not be what we calculated in the

problem.

Remember!

17

Problem

Acetic acid CH3COOH (or HAc) 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 mol/L CH3COOH

b) 0.00100 mol/L CH3COOH

18

Problem a)

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


x)(1.00

(x)(x)10 x 1.8 so

COOH][CH

]COO][CHO[H10 x 1.8K 5

3

335a

Let’s check the initial acid concentration / Ka ratio.

1.00 / 1.8 x 10-5 55000 is larger than 100.

19

Problem a)

x)(1.00

(x)(x)10 x 1.8 so

COOH][CH


3

335a

We can probably assume that x << [HAc]i so [HAc]eqm [HAc]i

1.8 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 mol/L (but must be + value since x = [H3O+])

20

Problem a)

So at equilibrium,[H3O+] = [CH3COO-] = 4.2 x 10-3 mol/L

[CH3COOH] = 1.00 mol/L.

The assumption was valid and sopH = - log [H3O+]

pH = - log 4.2 x 10-3

pH = 2.38

%42010 x 24M 1.00

M 10 x 4.2

COOHCH

x 3--3

i3

..

21

Problem b)

x)(0.00100

(x)(x)10 x 1.8 so

COOH][CH


3

335a

Let’s check the initial acid concentration / Ka ratio.

0.00100 / 1.8 x 10-5 56is smaller than 100.



0.00100

0.00100

22

Problem b)

We can probably CAN NOT assume that x << [HAc]i so [HAc]eqm [HAc]i

852

25

10 x 1.8-x10 x 1.8x00xx)](0.0010010 x [1.8

x)(0.00100

(x)(x)10 x 1.8 so

COOH][CH


3

335a



0.00100

0.00100

23

Problem b)

2

10 x 82xor

2

10 x 52x

2

10 x 6210 x 1.8-xor

2

10 x 6210 x 1.8-x

2

10 x 7.210 x 3.210 x 1.8-xor

2

10 x 7.210 x 3.210 x 1.8-x

2(1)

)10 x 4(1)(-1.8)10 x (1.8)10 x (1.8xso

2a

4acbbx

47

41

49

54

95

8104

58104

5

82552

..

..

mol/L10 x 41- xor mol/L10 x 21x so 44

45

..

24

Problem b)

Since [H3O+] = x we must use the positive value, so

[H3O+] = [CH3COO-] = 1.3 x 10-4 mol/L[CH3COOH]

= 0.00100 mol/L – 1.3 x 10-4 mol/L = 0.00087 mol/L.

25

Problem b)

Let’s confirm that x << [HAc]i IS NOT TRUE

pH = - log [H3O+]pH = - log 4.2 x 10-4

pH = 3.38

% 13130M 0.00100

M 10 x 1.3

COOHCH

x -4

i3

.

26

Problem

A vitamin C tablet containing 250 mg of ascorbic acid (C6H8O6; Ka = 8.0 x 10-5 is dissolved in a 250 mL glass of water to give a solution where [C6H8O6] = 5.68 x 10-3 mol/L.

What is the pH of the solution?

27

Problem

Check the initial acid concentration / Ka ratio. 5.68 x 10-3/ 8.0 x 10-5 71

which is not larger than 100 so

(all in mol/L) C6H8O6 (aq) + H2O (l) H3O+ (aq) + C6H7O6

- (aq) Initial conc. 5.68 x 10-3 N/A 0.0 0.0 Conc. change -x N/A +x +x Equil. conc. 5.68 x 10-3 – x N/A +x +x

x)10 x (5.68

(x)(x)10 x 8.0 so

]OH[C

]OH][CO[H10 x 8.0K

3-5

686

67635a

74

52

235

x104.5-x10 x 8.0x00xx)10 x (5.6810 x 8.0

28

Problem

2

10x 1.2xor

2

10x 1.4x

2

10x 1.310 x 8.0-xor

2

10x 1.310 x 8.0-x

2

)x10(1.86.4x1010 x 8.0-xor

2

)10 x (1.810 x 6.410 x 8.0-x

2(1)

)10x 4(1)(-4.5)10 x (8.0)10 x (8.0xso

2a

4acbbx

37

33

35

535

5

62

9562

95

74

2552

mol/L106.3x xor mol/L107.1x x so 44

29

Problem

Since [H3O+] = x the answer must be the positive value

[H3O+] = [C6H7O6-] = 6.3 x 10-4 mol/L

[C6H8O6] = (5.68 x 10-3 - 6.3 x 10-4) mol/L = 5.05 x 10-3 mol/L

pH = - log [H3O+]pH = - log 6.3 x 10-4

pH = 3.20

30

Degree of ionization

The pH of a solution of a weak acid like acetic acid will depend on the initial concentration of

the weak acid and Ka. Therefore, we can define a second measure of the strength of a

weak acid by looking of the

degree (or percent) ionization of the acid.

%ionization = [HA]ionized / [HA]initial x 100%

31

Percent ionization

In part a) of an earlier problem an acetic acid solution with initial concentration of 1.00 mol/L at equilibrium had

[H3O+]eqm = [HA]ionized = 4.2 x 10-3 mol/L


%ionization = 4.2 x 10-3 mol/L / 1.00 mol/L x 100%

%ionized = 0.42%

32

Percent ionization

In part b) of an earlier problem an acetic acid solution with initial concentration of 0.00100 mol/L at equilibrium had

[H3O+] = [HA]ionized = 1.3 x 10-4 mol/L


%ionization = 1.3 x 10-4 mol/L / 0.00100 mol/L x 100%

%ionization = 13%

33

Figure

34

Equilibria in Solutions of Weak Bases

The dissociation of a weak base is an equilibrium situation with an equilibrium

constant, called the base dissociation constant, Kb based on the equation

B (aq) + H2O (l) BH+ (aq) + OH- (aq)

[B]

]][OH[BHKb

35


The base dissociation constant, Kb is always based on the reaction of one mole

of the weak base with water.

If you see the symbol Kb, it always refers to a balanced equation of the form

B (aq) + H2O (l) BH+ (aq) + OH- (aq)

36


Our approach to solving equilibria problems involving bases is exactly the same as for acids.

1. Set up the ICE table

2. Establish the equilibrium constant expression

3. Make a simplifying assumption when possible

4. Solve for x, and then for eq’m amounts

37

Problem

Strychnine (C21H22N2O2), a deadly poison used for killing rodents, is a weak base having Kb = 1.8 x 10-6. Calculate the pH if

[C21H22N2O2]initial = 4.8 x 10-4 mol/L

38

Problem(all in mol/L) C21H22N2O2 (aq) + H2O (l) C21H23N2O2

+ (aq) + OH- (aq) Initial conc. 4.8 x 10-4 N/A 0.0 0.0 Conc. change -x N/A +x +x Equil. conc. 4.8 x 10-4 – x N/A +x +x

x)10 x (4.8

(x)(x)10 x 1.8 so

]ONH[C

]][OHONH[C10 x 1.8K

4-6

222221

2223216b

Check the initial base concentration / Kb ratio

4.8 x 10-4 / 1.8 x 10-6 267

which is greater than 100

We are probably good to make a simplifying assumption that x << [C21H22N2O2]i

39

The assumption we will make is that x << [C21H22N2O2]i so

[C21H22N2O2]eqm [C21H22N2O2]I

1.8 x 10-6 = x2 / 4.8 x 10-4

x2 = (1.8 x 10-6)(4.8 x 10-4)

x = 8.64 x 10-10

x = 2.94 x 10-5 mol/L

x)10 x (4.8

(x)(x)10 x 1.8 so

]ONH[C

]][OHONH[C10 x 1.8K

4-6

222221

2223216b

40

Problem

Since x = [OH-], the answer must be the positive value,

x = [C21H23N2O2+] = [OH-] = 2.9 x 10-5 mol/L

[C21H22N2O2] = 4.8 x 10-4 mol/L – 2.9 x 10-5 mol/L

= 4.5 x 10-4 mol/L.

We should check the assumption!

%060600M 10 x 4.8

M 10 x 2.9

ONHC

x4-

-5

i222221

..

41

Problem

In this case, the error is more than 5%.

I will leave it to you to go back and use the quadratic formula.

Compare the two answers

%060600M 10 x 4.8

M 10 x 2.9

ONHC

x4-

-5

i222221

..

42

Problem

To continue towards the answer of the problem AS IF the assumption WERE

VALID

pOH = - log [OH-]pOH = - log 2.9 x 10-5

pOH = 4.54

pH + pOH = 14.00pH = 14.00 - pOHpH = 14.00 - (4.54)

pH = 9.46

43

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

Since Brønsted-Lowry acid-base reactions involve conjugate acid-base pairs there

should be a connection between the

Ka value and the Kb value of a conjugate acid-base pair.

44

Relation Between Ka and Kb

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

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

HA

AOHK 3

a

A

HAOHKb

45

Since these reactions take place in the same beaker at the same time let’s

add them together

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

2 H2O (l) H3O

+ (aq) + OH- (aq)

46

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)

C25at1.0x10KOHOHA

HAOH

HA

AOHKxK 14

w33

ba

47

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.

48

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

49

Problem

a) – Piperidine (C5H11N) is an amine found in black pepper. Find Kb for piperidine in Appendix C, and then calculate Ka for the C5H11NH+ cation.

Kb = 1.3 x 10-3

b) Find Ka for HOCl in Appendix C, and then calculate Kb for OCl-.

Ka = 3.5 x 10-8

50

Acid-Base Properties of Salts

When acids and bases react with each other,

they form ionic compounds called salts.

Salts, when dissolved in water, can lead to acidic, basic, or neutral solutions, depending on the relative strengths of the acid and base we derive them from.

Strong acid + Strong base Neutral salt solution

Strong acid + Weak base Acidic salt solution

Weak acid + Strong base Basic salt solution

51

Salts that Yield Neutral Solutions

Strong acids and strong bases react to form neutral salt

solutions. When the salt dissociates in water, the cation and anion do not appreciably react with water to form H3O+

or OH-.

52


Strong base cations like the alkali metal cations (Li+, Na+, K+) or

alkaline earth cations (Ca2+, Sr2+, Ba2+, but NOT Be2+) and strong acid anions such as Cl-, Br-, I-, NO3

-, and ClO4- will combine

together to give neutral salt solutions with pH = 7.

53


Sodium chloride (NaCl) will dissociate into Na+ and Cl- in water.

Cl- has no acidic or basic tendencies.

Cl- (aq) + H2O (l) no reaction⇌

Chloride ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong acid,

which makes it very, very weak.

54


Na+ has no acidic or basic tendencies.

Na+ (aq) + H2O (l) no reaction⇌

Sodium ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong base,


55

Salts that Yield Acidic Solutions

The reaction of a strong acid with anions like

Cl-, Br-, I-, NO3-, and ClO4

-

with a weak base will lead to an

acidic salt solution.

The solution is acidic because the anion shows no acidic or basic tendencies, but

the cation does, as it is the conjugate acid of a weak base.

56


Ammonium chloride (NH4Cl) will dissociate into NH4

+ and Cl- in water.

Cl- has no acidic or basic tendencies.

Cl- (aq) + H2O (l) no reaction⇌

Chloride ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong acid,


57


NH4+ has acidic tendencies.

That is:

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

Ammonium ions hydrolyze in water because it is the conjugate acid of the weak base NH3, which

means ammonium is a weak acid.

58

Salts that Yield Basic Solutions

The reaction of a strong base with cations like Li+, Na+, K+, Ca2+, Sr2+, and Ba2+

with a weak acid will lead to an

basic salt solution.

The solution is acidic because the cation shows no acidic or basic tendencies, but

the anion does, as it is the conjugate base of a weak acid.

59


Sodium fluoride (NaF) will dissociate into Na+ and F- in water.

Na+ (aq) + H2O (l) no reaction⇌

Sodium ions DO NOT HAVE hydrolysis reactions with water since it is the “conjugate” of a strong base,


60


F- has basic tendencies.That is:

F- (aq) + H2O (l)⇌ HF (aq) + OH- (aq)

Fluoride ions hydrolyze in water because it is the conjugate base

of the weak acid HF, which means fluoride is a weak base.

61

Problem

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

0.25 mol/L NH4Br – NH3 has a Kb value of 1.8 x 10-5

(all in mol/L) NH4+ (aq) + H2O (l) H3O

+ (aq) + NH3 (aq) Initial conc. 0.25 N/A 0.0 0.0 Conc. change -x N/A +x +x Equil. conc. 0.25 – x N/A +x +x

x)(0.25

(x)(x)x105.5so

][NH

]][NHO[Hx105.5K 10

6

4

33106a

62

Problem

Initial acid [HA] / Ka ratio is

0.25 / 5.56 x 10-10 4.5 x 108

we can probably assume 0.25 >> x

5.56 x 10-10 = x2 / 0.25

x2 = (5.56 x 10-10)(0.25)

x2 = 1.39 x 10-10

x = 1.39 x 10-10

x = 1.18 x 10-5 mol/L

63

Problem

Negative answer not physically possible so

therefore, [H3O+] = 1.18 x 10-5 mol/L

Since we’ve shown the assumption is valid

pH = -log [H3O+] = - log 1.18 x 10-5 = 4.93.

% 10 x 8410 x 84

M 0.25

M 10 x 1.2

NH

x 3-5--5

i4

..

64

Salts that Contain Acidic Cations and Basic Anions

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.

65


If the acid cation is “stronger” than the base anion, it “wins” and the salt solution is acidic.

If the base anion is “stronger” than the acid cation, it “wins” and the salt solution is basic.

66


Ka > Kb

the acid cation is “stronger” and the salt solution is acidic.

Ka < Kbthe base anion is “stronger” and the salt solution is

basic.

Ka Kbthe salt solution is close to neutral.

67

Problem

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

a) KBr

b) NaNO2

c) NH4Br

d) NH4FKa for HF = 6.6 x 10-4

Kb for NH3 = 1.8 x 10-5

68

The Common-Ion Effect

Solutions consisting of both an acid and its conjugate base are very important because they are very resistant to changes in pH. Such buffer solutions regulate pH in a variety of biological systems.

69

The Common-Ion Effect

Let’s consider a solution made of 0.10 moles of acetic acid and 0.10 moles of sodium acetate with a total volume of 1.00 L, making the initial [CH3COOH] = [CH3COO-] = 0.10 mol/L.

First we must identify all potential acids and bases in the system.

CH3COOH CH3COO- Na+ H2O acid base neutral acid or base

70

Our reaction will beCH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)

Ka = 1.8 x 10-5


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

Note that the initial concentration of our product CH3COO- is NOT ZERO!

Point of view of the acid

71

Let’s check the

initial acid concentration / Ka ratio. 0.10 / 1.8 x 10-5 5500

It’s probably safe to assume that

x << [HAc]i so [HAc]eqm [HAc]I

and x << [Ac-]i so [Ac-]eqm [Ac-]i

1.8 x 10-5 = x (0.10 + x) / (0.10 –x)1.8 x 10-5 = x (0.10) / (0.10)

x = 1.8 x 10-5 mol/L

x)(0.10

x)(x)(0.10 10 x 1.8

COOH][CH

]COO][CHO[HK 5

3

33a

72

At equilibrium,

[H3O+] = 1.8 x 10-5 mol/L

[CH3COO-] = 0.10 + 1.8 x 10-5 = 0.10 mol/L

[CH3COOH] = 0.10 - 1.8 x 10-5 = 0.10 mol/L

Assumption was valid! Check for yourself!

pH = - log [H3O+]

pH = - log 1.8 x 10-5

pH = 4.74

x)(0.10

x)(x)(0.10 10 x 1.8

COOH][CH

]COO][CHO[HK 5

3

33a

73

If we had started out with only 0.10 mol/L acetic acid, the pH would be found from (all in mol/L) CH3COOH (aq) + H2O (l) H3O

+ (aq) + CH3COO- (aq) Initial conc. 0.10 N/A 0.0 0.00 Conc. change -x N/A +x +x Equil. conc. 0.10 – x N/A +x +x

x)(0.10

(x)(x)10 x 1.8

COOH][CH

]COO][CHO[HK 5

3

33a

74

The initial acid concentration / Ka will still be the same, so we can assume

x << [HAc]i so [HAc]eqm [HAc]i

1.8 x 10-5 = x2 / (0.10 –x)1.8 x 10-5 = x2/ (0.10)x = 1.8 x 10-6 mol/L

x = 1.3 x 10-3 mol/L (can’t be –ve)

pH = - log [H3O+]pH = - log 1.3 x 10-3

pH = 2.89

75

Without the acetate ion the pH of

0.10 M acetic acid is 2.89.

With an equal concentration of acetate ion present, the pH of

0.10 M acetic acid – 0.10 M acetate is 4.74

The acetate ion makes a large difference on the equilibrium pH!

76

CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)

Adding the conjugate base (a stress!) to the equilibrium system of an acid dissociation shows the common-ion effect, where the addition of a common

ion causes the equilibrium to shift. This is an example of

Le Chatalier’s Principle.

Addition of the weak base to the acid

dissociation

77

Problem

Calculate the concentrations of all species present, and the pH in a solution that is 0.025 mol/L HCN and 0.010 mol/L NaCN.

(Ka of HCN = 4.9 x 10-10)



78

Problem

x)(0.025

x)(x)(0.01010 x 4.9

[HCN]

]][CNO[HK 103

a

The initial base concentration / Ka ratio is 0.010 / 4.9 x 10-10 2 x 107

It’s probably safe to assume that x << [HCN]i so [HCN]eqm [HCN]I

and x << [CN-]i so [CN-]eqm [CN-]i

79

Problem

4.9 x 10-10 = x (0.010 + x) / (0.025 –x)4.9 x 10-10 = x (0.010) / (0.025)

x = 1.2 x 10-9 mol/L

So at equilibrium, [H3O+] = 1.2 x 10-9 mol/L

[CN-] = 0.010 + 1.2 x 10-9 = 0.010 mol/L [HCN] = 0.025 - 1.2 x 10-9 = 0.025 mol/L.Assumption was valid! Check this for

yourself!

80

Problem

pH = - log [H3O+]pH = - log 1.2 x 10-9

pH = 8.91

81

Buffer Solutions

Solutions that contain both a weak acid and its conjugate base are buffer solutions.

These solutions are resistant to changes in pH.

82

Buffer Solutions

If more acid (H3O+) or base (OH-) is added to the system, the system has enough of

the original acid and conjugate base molecules in the solution to react with the

added acid or base, and so the new equilibrium mixture will be

very close in composition to the original equilibrium mixture.

83

Buffer solutions

A 0.10 molL-1 acetic acid – 0.10 molL-1 acetate mixture has a pH of 4.74 and is a buffer solution!

CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)

COOH][CH

]COO][CHO[H10 x 1.8K

3

335a


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

84

Buffer solutions

If we rearrange the Ka expression to solve for

[H3O+]

]COO[CH

COOH][CH10 x 1.8

]COO[CH

COOH][CHK]O[H

3

35

3

3a3

85

Buffer solutions

Assume x << [HAc]i so [HAc]eqm [HAc]I

and x << [Ac-]i so [Ac-]eqm [Ac-]i, and we should see

If [CH3COOH]i = [CH3COO-]i, then [H3O+] = 1.8 x 10-5 M = Ka

and pH = pKa = 4.74

]COO[CH

COOH][CH10 x 1.8

]COO[CH

COOH][CHK]O[H

3

35

3

3a3

86

Buffer solutions

What happens if we add 0.01 mol of NaOH (strong base) to 1.00 L of the acetic acid – acetate buffer solution?

CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq)

This reaction goes to completion and keeps occurring until we run out of the limiting reagent OH-

(all in moles) CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq) Initial 0.10 0.01 N/A 0.10 Change -x -x N/A +x Final (where x = 0.01 due to limiting OH-)

0.10 – x = 0.09

0.01 – x = 0.00

N/A 0.10 + x = 0.11

New [CH3COOH] = 0.09 M and new [CH3COO-] = 0.11 M

87

Buffer solutions

With the assumption that x is much smaller than 0.09 mol (an assumption we always need to

check after calculations are done!), we find



x)(0.09

x)(x)(0.1110 x 1.8

COOH][CH

]COO][CHO[HK 5

3

33a

M 10 x 1.50.11

0.0910 x 1.8

]COO[CH

COOH][CHK]O[H 55

3

3a3

Note we’ve made the assumption that x << 0.09 M!pH = - log [H3O+]

pH = - log 1.5 x 10-5

pH = 4.82

88

Buffer solutions

Adding 0.01 mol of OH- to 1.00 L of water would have given us a pH of 12.0 because there is no significant amount of acid in

water for the base to react with. Our buffer solution resisted this change

in pH because there is a significant amount of acid (acetic acid) for the

added base to react with.

89

Buffer solutions

What happens if we add 0.01 mol of HCl (strong acid) to 1.00 L of the acetic acid – acetate buffer solution?

CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)

This reaction goes to completion and keeps occurring until we run out of the limiting reagent H3O+

New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M

(all in moles) CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)

Initial 0.10 0.01 N/A 0.10 Change -x -x N/A +x Final (where x = 0.01 due limiting H3O

+) 0.10 – x = 0.09

0.01 – x = 0.00

N/A 0.10 + x = 0.11

90

Buffer solutions

With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check after

calculations are done!), we find

x)(0.11

x)(x)(0.0910 x 1.8

COOH][CH

]COO][CHO[HK 5

3

33a

Note we’ve made the assumption that x << 0.09 M!pH = - log [H3O+]

pH = - log 2.2 x 10-5

pH = 4.66



M 10 x 2.20.09

0.1110 x 1.8

]COO[CH

COOH][CHK]O[H 55

3

3a3

91

Buffer solutions

Adding 0.01 mol of H3O+ to 1.00 L of water would have given us a pH of 2.0

because there is no significant amount of acid in water for the base

to react with. Our buffer solution resisted this change in pH because there is a

significant amount of base (acetate) for the added acid to react with.

92

.

93

Buffer capacity

Buffer capacity is the measure of the ability of a buffer to absorb acid or base without

significant change in pH.

Larger volumes of buffer solutions have a larger buffer capacity than smaller

volumes with the same concentration.Buffer solutions of higher concentrations have a larger buffer capacity than a buffer solution of the same volume with smaller

concentrations.

94

Problem

Calculate the pH of a 0.100 L buffer solution that is 0.25 mol/L in HF and 0.50 mol/L in NaF.

(all in mol/L) HF (aq) + H2O (l) H3O+ (aq) + F- (aq)


x)(0.25

x)(x)(0.5010 x 3.5

[HF]

]][FO[HK 43

a

With the assumption that x is much smaller than 0.25 mol (an assumption we always need to check after calculations are done!), we find

Assume x << [HCN]i so [HCN]eqm [HCN]i

and x << [CN-]i so [CN-]eqm [CN-]i

95

Problem

M 10 x 1.70.50

0.2510 x 3.5

][F

[HF]10 x 3.5

][F

[HF]K]O[H

45

4

4a3

pH = - log [H3O+]pH = - log 1.75 x 10-4

pH = 3.76

96

Problem

a) What is the change in pH on addition of 0.002 mol of HNO3?(all in moles) F- (aq) + H3O

+ (aq) → H2O (l) + HF (aq) Initial 0.050 0.002 N/A 0.025 Change -x -x N/A +x Final (where x = 0.002 due to limiting H3O

+) 0.050 – x = 0.048

0.002 – x = 0.00

N/A 0.025 + x = 0.027

New [HF] = 0.27 M and new [F-] = 0.48 M

x)(0.27

x)(x)(0.4810 x 3.5

[HF]

]][FO[HK 43

a



97

Problem

M 10 x 1.90.48

0.2710 x 3.5

][F

[HF]K]O[H 4

74

a3

Notice we’ve made the assumption that x << 0.27 M. We should check this!

pH = - log [H3O+]pH = - log 1.97 x 10-4

pH = 3.71

98

Problem

b) What is the change in pH on addition of 0.004 mol of KOH? (all in moles) HF (aq) + OH- (aq) → H2O (l) + F- (aq) Initial 0.025 0.004 N/A 0.050 Change -x -x N/A +x Final (where x = 0.004 due to limiting OH-)

0.025 – x = 0.021

0.004 – x = 0.00

N/A 0.050 + x = 0.054

New [HF] = 0.21 M and new [F-] = 0.54 M

x)(0.21

x)(x)(0.543.5x10 so

[HF]

]][FO[H3.5x10K 434

a



99

Problem

M 10x 1.30.54

0.2110 x 3.5

][F

[HF]K]O[H 4

64

a3

Notice we’ve made the assumption that x << 0.21 M. We should check this!

pH = - log [H3O+]pH = - log 1.36 x 10-4

pH = 3.87

100

The Henderson-Hasselbalch Equation

We’ve seen that, for buffer solutions containing members of a conjugate acid-base pair, that

pH = pKa + log [base] / [acid]

This is called the Henderson-Hasselbalch Equation.

[base]

[acid]K]O[H a3

[base]

[acid]logK log

[base]

[acid]Klog]O[H log aa3

101

The Henderson-Hasselbalch Equation

If we have a buffer solution of a conjugate acid-base pair, then the pH of the solution

will be close to the pKa of the acid.

This pKa value is modified by the logarithm of ratio of the concentrations of the base and acid in the solution to give the actual

pH.

102

ProblemUse the Henderson-Hasselbalch Equation to calculate the pH of a buffer solution prepared by mixing equal volumes of 0.20 mol/L NaHCO3 and 0.10 mol/L Na2CO3.

We need the Ka and the concentrations of the acid (HCO3

-) and the base (CO32-).

Ka = 5.6 x 10-11 (we use the Ka for the second proton of H2CO3!).

103

Problem

NOTE: The concentrations we are given for the acid and the base are the

concentrations

before the mixing of equal volumes!

104

Problem

If we mix equal volumes, the total volume is TWICE the volume for the original acid or base solutions.

Since the number of moles of acid or base DON’T CHANGE on mixing,

the initial concentrations we use will be half the given values.

pH = pKa + log [base] / [acid]pH = (-log 5.6 x 10-11) + log (0.05) / (0.10)

pH = 10.25 – 0.30pH = 9.95

Documents

1 Chapter 16. Acid –Base Equilibria... 2 Equilibria in Solutions of Weak Acids The dissociation of a weak acid is an equilibrium situation with an equilibrium