1

Chapter 17a

Ionic Equilibria: Part II Buffers and Titration Curves

2

Chapter Goals

1. The Common Ion Effect and Buffer Solutions

2. Buffering Action

3. Preparation of Buffer Solutions

4. Acid-Base Indicators

Titration Curves

5. Strong Acid/Strong Base Titration Curves

6. Weak Acid/Strong Base Titration Curves

7. Weak Acid/Weak Base Titration Curves

8. Summary of Acid-Base Calculations

3

The Common Ion Effect and Buffer Solutions If a solution is made in which the same ion is produced by

two different compounds the common ion effect is exhibited.

Buffer solutions are solutions that resist changes in pH when acids or bases are added to them. Buffering is due to the common ion effect.

4

The Common Ion Effect and Buffer Solutions There are two common kinds of buffer solutions:1 Solutions made from a weak acid plus a soluble ionic

salt of the weak acid.2 Solutions made from a weak base plus a soluble ionic

salt of the weak base

5

The Common Ion Effect and Buffer Solutions1. Solutions made of weak acids plus a soluble ionic

salt of the weak acid One example of this type of buffer system is:

The weak acid - acetic acid CH3COOH

The soluble ionic salt - sodium acetate NaCH3COO

acids. with reacts base) (aanion salt The

COOCH Na COOCHNa

H COOCH COOHCH

bases. with reacts acid weak The

-3

%1003

-33

6

The Common Ion Effect and Buffer Solutions Example 19-1: Calculate the concentration of H+and the pH of a

solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate. This is another equilibrium problem with a starting concentration for both

the acid and anion.

MMM

MxMxMx

15.0 15.0 15.0

COOCHNaCOONaCH

)15.0(

COOCHHCOOHCH

3100%

3

-33

7

The Common Ion Effect and Buffer Solutions Substitute the quantities determined in the previous

relationship into the ionization expression.

x

xx

15.0

15.0108.1

COOHCH

COOCH HK 5

3

-3

+

a

8

The Common Ion Effect and Buffer Solutions Apply the simplifying assumption to both the

numerator and denominator.

74.4pH

H108.1

108.10.15

0.15

gives sassumption theseMaking

15.015.0 and 15.015.0

5

5

Mx

x

xx

9

The Common Ion Effect and Buffer Solutions This is a comparison of the acidity of a pure acetic acid

solution and the buffer described in Example 19-1.

10

The Common Ion Effect and Buffer Solutions Compare the acidity of a pure acetic acid solution and

the buffer described in Example 19-1.

Solution [H+] pH

0.15 M CH3COOH 1.6 x 10-3 2.80

0.15 M CH3COOH &

0.15 M NaCH3COO buffer

1.8 x 10-5 4.74

[H+] is 89 times greater in pure acetic acid than in buffer solution.

11

The Common Ion Effect and Buffer Solutions The general expression for the ionization of a weak

monoprotic acid is:

The generalized ionization constant expression for a weak acid is:

HA H A+

KH A

HAa

12

The Common Ion Effect and Buffer Solutions If we solve the expression for [H+], this relationship

results:

By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to:

H KHA

A

acid

salta

H Kacidsalta

13

The Common Ion Effect and Buffer Solutions The relationship developed in the previous slide is

valid for buffers containing a weak monoprotic acid and a soluble, ionic salt.

If the salt’s cation is not univalent the relationship changes to:

H Kacid

n salt

where n = charge on cation

a

14

The Common Ion Effect and Buffer Solutions Simple rearrangement of this equation and

application of algebra yields the

Henderson-Hasselbach equation.

acid

saltlogpKpH

acid

saltlogKlogHlog

1-by multiply

salt

acidlogKlogHlog

a

a

a

The Henderson-Hasselbach equation is one method to calculate the pHof a buffer given the concentrations of the salt and acid.

15

Weak Bases plus Salts of Weak Bases2. Buffers that contain a weak base plus the salt of a

weak base One example of this buffer system is ammonia plus

ammonium nitrate.

5

3

+4

3+4

%10034

-+423

108.1NH

NH

NONHNONH

OHNHOH NH

OHKb

16

Weak Bases plus Salts of Weak Bases Example 19-2: Calculate the concentration of OH-

and the pH of the solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in ammonium nitrate, NH4NO3.

MMM

MxMxMx

30.0 30.0 30.0

NONHNONH

)15.0(

OHNH OH NH

3+4

%10034

+423

17

Weak Bases plus Salts of Weak Bases Substitute the quantities determined in the previous

relationship into the ionization expression for ammonia.

8.95pH and 5.05pOH

OH100.9

108.115.0

30.0K

applied. becan assumption gsimplifyin The

108.115.0

30.0K

108.1NH

OH NHK

6

5b

5b

5

3

4b

Mx

x

x

xx

18

Weak Bases plus Salts of Weak Bases A comparison of the aqueous ammonia concentration

to that of the buffer described above shows the buffering effect.

Solution [OH-] pH

0.15 M NH3 1.6 x 10-3 M 11.20

0.15 M NH3 &

0.15 M NH4NO3 buffer

9.0 x 10-6 M 8.95

The [OH-] in aqueous ammonia is 180 times greater than in the buffer.

19

Weak Bases plus Salts of Weak Bases We can derive a general relationship for buffer

solutions that contain a weak base plus a salt of a weak base similar to the acid buffer relationship. The general ionization equation for weak bases is:

:B H O BH OH

where B represents a weak base2

20

Weak Bases plus Salts of Weak Bases The general form of the ionization expression is:

Solve for the [OH-]

KBH OH

Bb

OH K

B

BHbase

salt

b

21

Weak Bases plus Salts of Weak Bases For salts that have univalent ions:

For salts that have divalent or trivalent ions:

OH Kbasesaltb

OH Kbase

n salt

where n = charge on anion

b

22

Weak Bases plus Salts of Weak Bases Simple rearrangement of this equation and

application of algebra yields the

Henderson-Hasselbach equation.

base

saltlogpKpOH

base

saltlogKlogOHlog

1-by multiply

salt

baselogKlogOHlog

b

b

b

23

Buffering Action

These movies show that buffer solutions resist changes in pH.

24

Buffering Action

Example 19-3: If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl.

1 Calculate the pH of the original buffer solution.

25

Buffering Action

8.95pH 5.05pOH

100.9OH

20.0

10.0108.1OH

ClNH

NHKOH

6-

5-

4

3b

-

M

M

M

26

Buffering Action

2 Next, calculate the concentration of all species after the addition of the gaseous HCl. The HCl will react with some of the ammonia and

change the concentrations of the species. This is another limiting reactant problem.

mol 0.220 mol 0.080 mol 0 rxn.After

mol 0.020+ mol 0.020-mol 0.020- Change

mol 0.200 mol 0.100 mol 0.020 Initial

ClNH NH HCl 43

27

Buffering Action

HCl NH NH Cl

Initial 0.020 mol 0.100 mol 0.200 mol

Change - 0.020 mol - 0.020 mol + 0.020 mol

After rxn. 0 mol 0.080 mol 0.220 mol

mol1.0 L

mol1.0 L

3 4

NH

NH Cl

3

4

M M

M M

0 0800 080

0 2200 220

..

..

28

Buffering Action

3 Using the concentrations of the salt and base and the Henderson-Hassselbach equation, the pH can be calculated.

M

M

220.0

080.0108.1OH

ClNH

NHKOH

5

4

3b

29

Buffering Action

OH KNH

NH Cl

OH

OH

pOH 5.19 pH 8.81

b3

4

18 100 0800 220

6 5 10

5

6

...

.

MM

M

30

Buffering Action

4 Finally, calculate the change in pH.

-0.14=8.95-8.81=pH

pHpH pH originalnew

31

Buffering Action

Example 19-4: If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH.

You do it!You do it!

32

Buffering Action

pH of the original buffer solution is 8.95, from above.

1. First, calculate the concentration of all species after the addition of NaoH. NaOH will react with some of the ammonium chloride. The limiting reactant is the NaOH.

mol 0.120 mol 0 mol 0.180 rxn.After

mol 0.020+ mol 0.020- mol 0.020- Change

mol 0.100 mol 0.020 mol 0.200 Initial

NaCl OH NH NaOH ClNH 234

33

Buffering Action

MM

MM

180.0L 1.0

mol 180.0

120.0L 1.0

mol 120.0

mol 0.120 mol 0 mol 0.180 rxn.After

mol 0.020+ mol 0.020- mol 0.020- Change

mol 0.100 mol 0.020 mol 0.200 Initial

NaCl OH NH NaOH ClNH

ClNH

NH

234

4

3

34

Buffering Action

2 Calculate the pH using the concentrations of the salt and base and the Henderson-Hasselbach equation.

9.08pH 4.92pOH

102.1OH

180.0

120.0108.1OH

ClNH

NHKOH

5

5

4

3b

M

M

M

35

Buffering Action

3 Calculate the change in pH.

0.13=8.95-9.08=pH

pHpH =pH originalnew

36

Buffering Action

This table is a summary of examples 19-3 and 19-4.

Notice that the pH changes only slightly in each case.

Original SolutionOriginal

pH

Acid or base

added

New pH

pH

1.00 L of solution containing

0.100 M NH3 and 0.200 M NH4Cl

8.95

0.020 mol NaOH

9.08 +0.13

0.020 mol HCl

8.81 -0.14

37

Preparation of Buffer Solutions

This move shows how to prepare a buffer.

38

Preparation of Buffer Solutions

Example 19-5: Calculate the concentration of H+ and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions.

Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction.

NaOH mmol 0.10mL

mmol 0.100mL 100=NaOH mmol ?

COOHCH mmol 0.30mL

mmol 0.15mL 200=COOHCH mmol ? 33

39

Preparation of Buffer Solutions

Sodium hydroxide and acetic acid react in a 1:1 mole ratio.

NaOH + CH COOH Na CH COO + H O

Initial 10.0 mmol 30.0 mmol

Change -10.0 mmol -10.0 mmol +10.0 mmol

After rxn. 0 20.0 mmol 10.0 mmol

3 3 2

40

Preparation of Buffer Solutions

After the two solutions are mixed, the total volume of the solution is 300 mL (100 mL of NaOH + 200 mL of acetic acid). The concentrations of the acid and base are:

COONaCH 0333.0mL300

mmol 0.10

COOHCH 0667.0mL300

mmol 0.20

3COONaCH

3COOHCH

3

3

MM

MM

41

Preparation of Buffer Solutions

Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.

KH CH COO

CH COOH

H

pH

a3

3

18 10

18 10 0 0667

0 03333 6 10

4 44

5

55

.

. .

..

.

M

42

Preparation of Buffer Solutions

For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH.

Example 19-6:Calculate the number of moles of solid ammonium chloride, NH4Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15. Because pH = 9.15, the pOH can be determined.

pOH = 14.00 - 9.15 = 4.85

OH- 10 14 104 85 5. . M

43

Preparation of Buffer Solutions

The appropriate equilibria representations are:

MxMxMx

MMM

Cl NH ClNH

104.1 104.1 104.110.0

OH NH OH + NH

44

555

-423

44

Preparation of Buffer Solutions

Substitute into the ionization constant expression (or Henderson-Hasselbach equation) for aqueous ammonia

assumption gsimplifyin apply the

104.110.0

104.1 104.1K

108.1NH

OH NHK

5

55

b

5

3

4b

x

45

Preparation of Buffer Solutions

g/L 9.6mol

g 53

L

mol 13.0

L

ClNH g ?

ClNH=ClNH 13.0

108.110.0

104.1 K

applied. becan assumption gsimplifyin The

104.110.0

104.1 104.1K

108.1NH

OH NHK

4

original44

55

b

5

55

b

5

3

4b

Mx

x

x

46

Acid-Base Indicators

The point in a titration at which chemically equivalent amounts of acid and base have reacted is called the equivalence point.

The point in a titration at which a chemical indicator changes color is called the end point.

A symbolic representation of the indicator’s color change at the end point is:

HIn H In

Color 1 Color 2

47

Acid-Base Indicators

The equilibrium constant expression for an indicator would be expressed as:

HIn

In HKa

48

Acid-Base Indicators

If the preceding expression is rearranged the range over which the indicator changes color can be discerned.

In

HInK

H

-a

49

Acid-Base Indicators

Color change ranges of some acid-base indicators

IndicatorColor in

acidic range pH rangeColor in

basic range

Methyl violet Yellow 0 - 2 Purple

Methyl orange Pink 3.1 – 4.4 Yellow

Litmus Red 4.7 – 8.2 Blue

Phenolphthalein Colorless 8.3 – 10.0 Red

50

Titration CurvesStrong Acid/Strong Base Titration Curves

These graphs are a plot of pH vs. volume of acid or base added in a titration.

As an example, consider the titration of 100.0 mL of 0.100 M perchloric acid with 0.100 M potassium hydroxide. In this case, we plot pH of the mixture vs. mL of KOH added. Note that the reaction is a 1:1 mole ratio.

OHKClOKOHHClO 244

51

Strong Acid/Strong Base Titration Curves Before any KOH is added the pH of the HClO4

solution is 1.00. Remember perchloric acid is a strong acid that ionizes

essentially 100%.

00.1log(0.100)pH

100.0H

10000.100 100.0

ClOHHClO 4%100

4

M

M.M M

52

Strong Acid/Strong Base Titration Curves After a total of 20.0 mL 0.100 M KOH has been

added the pH of the reaction mixture is ___?

1.17pH 067.0H

067.0mL 120

HClO mmol 8.0

rxn.

mmol 2.0 mmol 0.0 mmol 8.0 After

mmol 2.0 mmol 2.0- mmol 2.0- :Change

mmol 2.0 mmol 10.0 :Start

OHKClO KOH HClO

4HClO

244

4

M

MM

53

Strong Acid/Strong Base Titration Curves After a total of 50.0 mL of 0.100 M KOH has been

added the pH of the reaction mixture is ___?

1.48pH 033.0H

033.0mL 150

HClO mmol 5.0

rxn.

mmol 5.0 mmol 0.0 mmol 5.0 After

mmol 5.0 mmol 5.0- mmol 5.0- :Change

mmol 5.0 mmol 10.0 :Start

OHKClO KOH HClO

4HClO

244

4

M

MM

54

Strong Acid/Strong Base Titration Curves After a total of 90.0 mL of 0.100 M KOH has been

added the pH of the reaction mixture is ____?

2.28pH 0053.0H

0053.0mL 190

HClO mmol 1.0

rxn.

mmol 9.0 mmol 0.0 mmol 1.0 After

mmol 9.0 mmol 9.0- mmol 9.0- :Change

mmol 9.0 mmol 10.0 :Start

OHKClO KOH HClO

4HClO

244

4

M

MM

55

Strong Acid/Strong Base Titration Curves After a total of 100.0 mL of 0.100 M KOH has been

added the pH of the reaction mixture is ___?

7.00pH

neutral baseor acid No

rxn.

mmol 10.0 mmol 0.0 mmol 0.0 After

mmol 10.0 mmol 10.0- mmol 10.0- :Change

mmol 10.0 mmol 10.0 :Start

OHKClO KOH HClO 244

56

Strong Acid/Strong Base Titration Curves We have calculated only a few points on the titration

curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

57

Weak Acid/Strong Base Titration Curves As an example, consider the titration of 100.0 mL of

0.100 M acetic acid, CH3 COOH, (a weak acid) with 0.100 M KOH (a strong base). The acid and base react in a 1:1 mole ratio.

1 mol 1mol 1mol

CH COOH + KOH K CH COO + H O

1mmol 1mmol 1mmol3

+3

-2

58

Weak Acid/Strong Base Titration Curves Before the equivalence point is reached, both CH3COOH and KCH3COO are

present in solution forming a buffer. The KOH reacts with CH3COOH to form KCH3COO.

A weak acid plus the salt of a weak acid form a buffer. Hypothesize how the buffer production will effect the titration curve.

59

Weak Acid/Strong Base Titration Curves1. Determine the pH of the acetic acid solution before

the titration is begun. Same technique as used in Chapter 18.

5

a

5

3

-3

+

a

-33

108.110.0

K

108.1COOHCH

COOCH HK

10.0

HCOOCHCOOHCH

x

xx

xMxMMx

60

Weak Acid/Strong Base Titration Curves

89.2pH 101.3H

101.3= 108.1

applied. becan assumption gsimplifyin The

108.110.0

K

108.1COOHCH

COOCH HK

10.0

HCOOCHCOOHCH

3-

3-62

5a

5

3

-3

+

a

-33

xx

x

xx

MxMxMx

61

Weak Acid/Strong Base Titration Curves After a total of 20.0 mL of KOH solution has been

added, the pH is:

KOH + CH COOH K CH COO H O

Initial: 2.00 mmol 10.0 mmol

Chg. due to rxn:-2.00 mmol - 2.00 mmol + 2.00 mmol

After rxn: 0.00 mmol 8.00 mmol 2.00 mmol

8.0 mmol120 mL

2.0 mmol120 mL

3+

3-

2

CH COOH

CH COO

3

3-

M M

M M

0 067

0 017

.

.

62

Weak Acid/Strong Base Titration Curves

KH CH COO

CH COOH

HCH COOH

CH COO

H

pH

a3

3

3

3

18 10

18 10

18 100 0670 017

7 1 10

4 15

5

5

5 5

.

.

...

.

.

M

Similarly for all other cases before the equivalence point is reached.

63

Weak Acid/Strong Base Titration Curves At the equivalence point, the solution is 0.500

M in KCH3COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution. This is a solvolysis process as discussed in

Chapter 18. Both processes make the solution basic.

The solution cannot have a pH=7.00 at equivalence point.

Let us calculate the pH at the equivalence point.

64

Weak Acid/Strong Base Titration Curves1. Set up the equilibrium reaction:

KOH + CH COOH K CH COO H O

Initial: 10.0 mmol 10.0 mmol

Chg. due to rxn:-10.0 mmol -10.0 mmol +10.0 mmol

After rxn: 0.0 mmol 0.0 mmol 10.0 mmol

3+

3-

2

65

Weak Acid/Strong Base Titration Curves2. Determine the concentration of the salt in solution.

COOCH 0500.00500.0

0500.0mL 200

mmol 10.0=

3COOKCH

COOKCH

3

3

MMM

MM

66

Weak Acid/Strong Base Titration Curves3. Perform a hydrolysis calculation for the potassium

acetate in solution.

8.72pH5.28pOH

OH1027.5108.2

106.50500.00500.0

=K

106.5COOCH

OHCOOHCH=K

0500.0

OH COOHCHOHCOOCH

6112

102

b

11

3

-3

b

-323

xx

x

x

xx

xMxMMx

67

Weak Acid/Strong Base Titration Curves4. After the equivalence point is reached, the pH is

determined by the excess KOH just as in the strong acid/strong base example.

11.68=pH and 2.32pOH

108.4OH

108.4mL 210

mmol 0.1

mmol 10.00 mmol 0.00 mmol 1.00 :rxnAfter

mmol 10.00+ mmol 10.0- mmol -10.0:rxn todue Chg.

mmol 10.0 mmol 11.0 :Initial

OHCOOCHKCOOHCH + KOH

3

KOH3

KOH

2-

3+

3

M

MM

68

Weak Acid/Strong Base Titration Curves We have calculated only a few points on the titration

curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

69

Strong Acid/Weak BaseTitration Curves Titration curves for Strong Acid/Weak Base

Titration Curves look similar to Strong Base/Weak Acid Titration Curves but they are inverted.

70

Weak Acid/Weak BaseTitration Curves Weak Acid/Weak Base Titration curves have

very short vertical sections. The solution is buffered both before and after

the equivalence point. Visual indicators cannot be used.

71

Synthesis Question

Bufferin is a commercially prepared medicine that is literally a buffered aspirin. How could you buffer aspirin? Hint - what is aspirin?

72

Synthesis Question

Aspirin is acetyl salicylic acid. So to buffer it all that would have to be added is the salt of acetyl salicylic acid.

73

Group Question

Blood is slightly basic, having a pH of 7.35 to 7.45. What chemical species causes our blood to be basic? How does our body regulate the pH of blood?

74

Group Question

75

End of Chapter 19

We have examined :1 Gas phase equilibria in Chapter 172 Hydrolysis equilibria in Chapter 183 Acid/base equilibria in Chapter 19

Chapter 20 is the last equilibrium chapter. It involves solid/solution equilibria.