Chapter 17 Additional Aspects of Acid-Base Equilibria Dr. Peter Warburton [email protected]

Chapter 17Additional Aspects of Acid-Base Equilibria

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

2

The common-ion effect

Say we make two acid solutions:

0.100 M HCl (a strong acid) and

0.100 M CH3COOH (a weak acid).

A 0.100 M HCl solution by itself would have a pH 1.0 ([H3O+] = 0.100 M) since

the reaction goes to completion.

Cl OH OH HCl 32

3


A 0.100 M CH3COOH solution

(Ka = 1.8 x 10-5) by itself would have a pH 2.8 since an equilibrium is established

where the equilibrium concentration of acetic acid 0.100 M

x = [H3O+] = [CH3COO-] = 1.3 x 10-3 M

COOCH OH OH COOHCH 3323

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

Initial conc. 0.100 N/A 0.00 0.00 Conc. change -x N/A +x +x Equil. conc. 0.100 – x N/A +x + x

4


5


Say we put together a solution that is BOTH

0.100 M HCl (a strong acid) and

0.100 M CH3COOH (a weak acid).

The two reactions of

the acids with water take place

in the same container

at the same time!

6

Both reactions are a source of H3O+ and so we could expect that

[H3O+] = 0.100 M + 1.3 x 10-3 M

[H3O+] = 0.1013 M

and pH 0.99

which appears to be true


COOCH OH OH COOHCH

Cl OH OH HCl

3323

32

7


However, both reactions share the common ion H3O+, and so they cannot be treated as

independent reactions. Something that affects one reaction must also affect the other

reaction.

Le Chatalier’s Principle!

COOCH OH OH COOHCH

Cl OH OH HCl

3323

32

8


Imagine we start with the 0.100 M CH3COOH and then we add 0.100 M H3O+

by adding 0.100 M HCl.

Le Chatalier’ Principle tells us our reaction will shift back towards

reactants!


9


If we assume x is much smaller than 0.100 M we will quickly find that x = Ka

x = [CH3COO-] = 1.8 x 10-5 M

The value of x has decreased because of the added H3O+!



[HCl]initial 0.000 N/A 0.100 0.000 [CH3COOH]initial 0.100 N/A 0.000 0.000 Conc. changes +x N/A -x +x Equil. conc. 0.100 + x N/A 0.100 - x + x

10

H3O+ as a common-ion

11

OH- as a common-ion

12

Salts as source of a basic common-ion

13

Salts as source of a basic common-ion

On the left is a solution of 0.100 M CH3COOH while on the right is a solution

that is both 0.100 M CH3COOH and 0.100 M CH3COONa which is a

source of the basic common ion CH3COO-.

The reaction has shifted back towards reactants!

14

Salts as source of an acidic common-ion

15

Salts as source of an acidic common-ion

On the left is a solution of 0.100 M NH3 while on the right is a solution that is both 0.100 M NH3 and

0.100 M NH4Cl which is a source of the acidic common ion NH4

+.

The reaction has shifted back towards reactants!

16

Buffer solutions

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

buffer solutions.

These solutions are resistant to changes in pH.

17

Buffer solutions

The system has enough of the

original acid and conjugate base molecules

in the solution to react with

added acid or added base,

so the new equilibrium mixture will be

very close in composition to the original equilibrium mixture.

18

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[H1.8x10K

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

19

Buffer solutions

If we rearrange the Ka expression

and make the assumption that x is much less than 0.10, we see

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

and pH = pKa = 4.74

]COO[CH

COOH][CH1.8x10

]COO[CH

COOH][CHK]O[H

3

35

3

3a3

20

Buffer solutions

21

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 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

22

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


Initial conc. 0.09 N/A 0.0 0.11 Conc. change -x N/A +x +x Equil. conc. 0.09 – x N/A +x 0.11 + x

x)(0.09

x)(x)(0.111.8x10 so

COOH][CH

]COO][CHO[H1.8x10K 5

3

335a

M 1.5x100.11

0.091.8x10

]COO[CH

COOH][CHK]O[H 55

3

3a3

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

pH = - log 1.5 x 10-5

pH = 4.82

23

Buffer solutions

Adding 0.01 mol of OH- to 1.00 L of water would have given us a

pH of 12.0!

There is no significant amount of acid in water for the base to

react with.

24

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 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

25

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)(0.091.8x10 so

COOH][CH

]COO][CHO[H1.8x10K 5

3

335a

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

pH = - log 2.2 x 10-5

pH = 4.66



M 2.2x100.09

0.111.8x10

]COO[CH

COOH][CHK]O[H 55

3

3a3

26

Buffer solutions

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

of 2.0!

There is no significant amount of base in water for the acid to

react with.

27

Adding acid or base to a buffer

28

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) Initial conc. 0.25 N/A 0.0 0.50 Conc. change -x N/A +x +x Equil. conc. 0.25 – x N/A +x 0.50 + x

x)(0.25

x)(x)(0.503.5x10 so

[HF]

]][FO[H3.5x10K 434

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

29

Problem

M x101.70.50

0.253.5x10

][F

[HF]K]O[H 4

54a

3

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

pH = 3.76

30

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)(0.483.5x10 so

[HF]

]][FO[H3.5x10K 434

a

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


31

Problem

M x101.90.48

0.273.5x10

][F

[HF]K]O[H 4

74a

3

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

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

pH = 3.71

32

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 0.10 0.004 N/A 0.10 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

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


33

Problem

M x101.30.54

0.213.5x10

][F

[HF]K]O[H 4

64a

3

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

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

pH = 3.87

34

Predicting whether a solution is a buffer

Any solution that becomes a mixture of a conjugate acid-base pair will be a buffer.

1) Weak acid-base conjugate pairs like CH3COOH and CH3COO- or NH4

+ and NH3.

2) Weak acid reacting with small amounts of strong base like CH3COOH and NaOH.

3) Weak base reacting with small amounts of strong acid like NH3 and HCl.

35

Problem

Describe how a mixture of a strong acid such as HCl and a salt of a weak acid such as CH3COONa can be a buffer solution.

36

Problem

What is the pH of a buffer solution prepared by dissolving 23.1 g of NaCHO2

(molar mass is 68.01 gmol-1) in a sufficient volume of 0.432 M HCHO2 to make 500.0 mL of the buffer?

Ka of formic acid is 1.8 x 10-4

Answer: pH = 3.94

37

The Henderson-Hasselbalch equation

We’ve seen that, for solutions with both 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]loglogK

[base]

[acid]Klog]Olog[H aa3

38

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.

39



ALWAYS remember when you use the H-H eqn that there is the assumption

that the equilibrium concentrations of acid and base are relatively unchanged from the initial

concentrations.

That is we have assumed x is very small compared to the initial

concentrations!

40



Generally this assumption is valid as long as we know

0.10 < [base] / [acid] < 10

AND

[base] / Ka > 100 AND [acid] / Ka > 100

41

Alternate Henderson-Hasselbalch equation

We can always look at a buffer solution as a base combined with its conjugate acid

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

pOH = pKb + log [acid] / [base]

[acid]

[base]K

][BH

[B]K][OH so

[B]

]][OH[BHK bbb

[acid]

[base]logK log

[acid]

[base]Klog]log[OH bb

42

Alternate Henderson-Hasselbalch equation

If we have a buffer solution of a conjugate acid-base pair, then the pOH of the solution will be close to the pKb of the base.

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

43

Problem

Use 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.

Ka of HCO3- = 4.7 x 10-11

(see Table 16.4) We should also check the validity of

using H-H at the end to be sure!

44

Problem answer

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 concentrations will be half the given values.

pH = pKa + log [base] / [acid]pH = (-log 4.7 x 10-11) + log (0.05) / (0.10)

pH = 10.33 – 0.30pH = 10.03

45

Preparing buffer solutions

If we want to create a buffer solution of a specific pH,

the H-H equation tells us we need to pick a conjugate

acid-base pair with a pKa for the acid close to the pH we

want, and then we adjust the amounts of the

conjugate acid and base.

46

How do we adjust the concentrations?

47

Problem

How many grams of (NH4)2SO4 (molar mass is 132.141gmol-1) must be dissolved in 0.500 L of 0.35 M NH3 to produce a solution with pH = 9.00? Assume that the solution volume remains at 0.500 L.

Kb for ammonia is 1.8 x 10-5.Answer: 21 g

48

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 higher buffer capacity, and buffer solutions of higher initial concentrations of the conjugate acid-base pair have a larger buffer capacity.

49

Buffer range

We’ve seen that as long as 0.10 < [base] / [acid] < 10

then the assumption that the Henderson-Hasselbalch equation

is based upon(x << [base] and x << [acid])

is likely to be valid.

50

Buffer range

pH = pKa + log [base] / [acid]pH = pKa + log 10

pH = pKa + 1.0

OR

pH = pKa + log [base] / [acid]pH = pKa + log 0.10

pH = pKa - 1.0

51

Buffer range

In general buffer solutions have a useful range of pH that is pKa 1.0

For instance an acetic acid - acetate buffer has a useful pH range of about

3.7 to 5.7since pKa is 4.7 (Ka = 1.8 x 10-5)

52

Application of buffers

Many biological processes can only occur at very specific pH values

(usually between pH = 6 and pH = 8). The reactions often take place in

buffered environments (e.g. human blood is buffered to pH = 7.4 – see

text pg. 734).

53

Acid-base indicators

We often measure the pH of a solution with a chemical acid-base indicator. Such indicators are weak acids in their own right (symbolized HIn) and indicate pH because the acid form has a different colour than the conjugate base form (In-)

HIn (aq) + H2O (l) H3O+ (aq) + In- (aq) colour A colour B

54

Diprotic indicator!

55


HIn (aq) + H2O (l) H3O+ (aq) + In- (aq) colour A colour B

If we increase the [H3O+] we shift this reaction towards reactants. The colour will be that of HIn.

If we decrease the [H3O+] we shift this reaction towards products. The colour will be that of In-.

A [H3O+] in between these two extremes will give a colour that is a mixture of the two colours

because both HIn and In- are present in significant amounts!

56


HIn (aq) + H2O (l) H3O+ (aq) + In- (aq)

colour A colour B

More specifically, the HIn and In- form a buffer so the indicator works in a pH range

of about 1 around the pKHIn of HIn.

57

58

Since indicators work in a range of about 2 pH units we often put several indicators that have different ranges in a single solution to give a universal indicator for pH range of about 1 - 12

59

Universal indicator

60

Applications of indicators

Indicators are useful when we want a general idea of the value of the pH without using a pH meter. Pool chlorination (with Cl2 or NaOCl) is done to avoid algae growth. This works best at pH = 7.4, so we might need to add some acid or base…

61

Neutralization reactions

A reaction of an acid and a base often produces water and an

aqueous salt as products. Such reactions are called neutralization

reactions, and can be categorized by the strengths of the acid and base

involved.

62

Strong acid – strong base neutralization

The reaction of a strong acid such as HCl and a strong base such as NaOH becomes a reaction of H3O+ and OH-,

Therefore while the overall reaction is

HCl (aq) + NaOH (aq) H2O (l) + NaCl (aq)

the actual net ionic equation is

H3O+ (aq) + OH- (aq) 2 H2O (l)

where the Na+ and Cl- ions are not involved (they are neutral spectator ions which don’t react with water!)

See slides 102, 103 and 106 of last chapter!

63


If we mix equal numbers of moles of HCl and NaOH, we will create twice as much water, while leaving an excess of NEITHER ion.

In this case the [H3O+] = [OH-] = 1.0 x 10-7 M at 25 °C (reaction goes to completion).

H3O+ (aq) + OH- (aq) 2 H2O (l)

is the reverse of the autoionization of water reaction so

C 25at 10 x 1.01.0x10

1

K

1

OHOH

1K 14

14w3

n

64


Since the equilibrium constant is very large, we see the reaction goes to completion, and at the end, the equilibrium mixture will consist of water and an aqueous salt of ions that are neutral in character because they don’t react with water. The pH after the reaction will be 7 (a neutral solution).

65

Strong acid titration by strong base

Say we add the strong base to the strong acid solution drop by drop. Each drop of base (the titrant) will react with some of the acid to completion until the added base (the limiting reagent!) is all gone. If we measure the pH after we add each drop of base, we can plot a pH versus total volume of added base graph.

This is called a titration curve.

66

Titration curve – strong acid titrated by strong base

67

End point versus equivalence point

How do we know when we’ve reached the end of our titration? We often will use an indicator to tell us when the titrated solution reaches a specific pH. This is the end point of the titration.

The end point DEPENDS on the indicator we use!

68


The equivalence point is that point in the titration where we have added equal numbers of MOLES of acid and base. For strong acid – strong base neutralizations

the equivalence point occurs when the pH is 7

(a neutral solution)

69

Titration curve – strong acid titrated by strong base

Different end points depending

on indicator

70


Bromothymol blue is a good indicator for strong acid-strong

base titrations because the end point is very close to the equivalence

point.Methyl red and phenolphtalein are pretty good choices too because the titration curve is very steep in

their effective pH ranges.

71

Titration curve – strong base titrated by strong acid

Equivalence point

The curve looks exactly the same, just

flipped vertically!

72

Millimoles (mmol)

Since we often set up titrations where our titrant concentration is less than 1 M and total titrant volume used is less than 50 mL this means we are adding about 5.00 x 10-3 moles of acid or base in our titration.

Because of this, we often do calculations in millimoles

(1 mmol = 1 x 10-3 mol)

73

Millimoles (mmol)

M = mol / L

M = mmol / mL

74

Problem

For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate

a) the initial pH;

b) the pH when the neutralization is 50.0% complete;

c) the pH when the neutralization is 100% complete; and

d) the pH when 1.00 mL of NaOH is added beyond the equivalence point.

75

Problem answers

a) pH = 0.824 (Vtotal = 25.00 mL)

b) pH = 1.238 (Vtotal = 32.50 mL)

c) pH = 7.00 (Vtotal = 40.00 mL)

d) pH = 11.79 (Vtotal = 41.00 mL)

76

Weak acid – strong base

Because a weak acid HA is largely undissociated in water, while the strong base is completely dissociated in water (becoming a source of OH-), the neutralization reaction of a weak acid and strong base becomes that of

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

Acetic acid is a weak acid, so when it reacts with NaOH, the net ionic equation is

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

Now remember thatKnet = K1 x K2 x K3 x … x Kn

77

Weak Acid – Strong Base

The equilibrium constant shows us the reaction goes to completion, so for equal numbers of moles of weak acid and strong base, we expect only water and the aqueous salt in the equilibrium mixture.

9

wan323

14

w23

5a3323

10 x 1.8

K

1 x KK(aq)COOCH(l)OH(aq)OH(aq)COOHCH

_____________________________________________________________

10 x 1.0K

1(l)OH2(aq)OH(aq)OH

10 x 1.8K(aq)COOCH(aq)OH(l)OH(aq)COOHCH

78

Weak acid – strong base

However, the salt comprises of a neutral cation (Na+ in this case) and a weakly basic anion (CH3COO-), meaning the equilibrium mixture will be BASIC and have a pH greater than 7.

See slide 107 of last chapter!In a weak acid titration by a strong base this means the equivalence point is NOT at pH = 7 but rather at pH > 7!

9

wan323

10 x 1.8

K

1 x KK(aq)COOCH(l)OH(aq)OH(aq)COOHCH

79

Titration curve – weak acid titrated by strong base

Titration of CH3COOH with

NaOH. 1. higher initial pH for a weak acid

2. creation of a buffer

3. pH = pKa when moles added OH- = ½ initial

moles weak acid

4. buffer breaks!5. pH > 7

because of water hydrolysis by

conjugate base

6. Added strong base dominates weak base and determines pH

80

Problem

For the titration of 20.00 mL of 0.150 M HF with 0.250 M NaOH, calculate

a) the initial pH; Ka = 6.6 x 10-4

b) the pH when the neutralization is 25.0% complete;

c) the pH when the neutralization is 50.0% complete; and

d) the pH when the neutralization is 100% complete.

81

Problem answers

a) pH = 2.00 (Vtotal = 20.00 mL)

b) pH = 2.70 (Vtotal = 23.00 mL)

c) pH = 3.18 (Vtotal = 26.00 mL)

d) pH = 8.08 (Vtotal = 32.00 mL)

82

Weak base – strong acid

Because a weak base B is largely undissociated in water, while the strong acid is completely dissociated in water (becoming a source of H3O+), the neutralization reaction of a strong acid and weak base becomes that of

H3O+ (aq) + B (aq) H2O (l) + BH+ (aq)

Ammonia is a weak base, so when it reacts with HCl, the net ionic equation is

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

Again

Knet = K1 x K2 x K3 x … x Kn

83


Again the equilibrium constant shows us the reaction goes nearly to completion, so for equal number of reactant moles, we expect only water and the aqueous salt in the equilibrium mixture.

9

wbn2433

14

w23

5b423

10 x 1.8

K

1 x KK (l)OH(aq)NH (aq)NH(aq)OH

10 x 1.0K

1 (l)OH2 (aq)OH (aq)OH

10 x 1.8K(aq)OH(aq)NH (l)OH(aq)NH

84


However, the salt comprises of a neutral anion (Cl- in this case) and a weakly acidic cation (NH4

+), meaning the equilibrium mixture will be ACIDIC and have a pH less than 7.

In a weak base titration by a strong acid this means the equivalence point is NOT at pH = 7 but rather at pH < 7!

9

wbn2433

10 x 1.8

K

1 x KK(l)OH(aq)NH(aq)NH(aq)OH

85

Titration curve – weak base titrated by strong acid

pH = 14.00 - pKb when moles added H3O+ = ½ initial moles weak base

86

Problem

For the titration of 50.00 mL of 0.106 M NH3 with 0.225 M HCl, calculate

a) the initial pH; Kb = 1.8 x 10-5

b) the pH when the neutralization is 25.0% complete;c) the pH when the neutralization is 50.0% complete; andd) the pH when the neutralization is 100% complete.

87

Problem answers

a) pH = 11.14 (Vtotal = 50.00 mL)

b) pH = 9.74 (Vtotal = 55.89 mL)

c) pH = 9.26 (Vtotal = 61.78 mL)

d) pH = 5.20 (Vtotal = 73.55 mL)

Documents

Chapter 17 Additional Aspects of Acid-Base Equilibria Dr. Peter Warburton [email protected]