Chap 11B Acid Base Equilibrium.ppt - Bakersfield … Spring_10...3 • Acetic acid is a weak acid....

1

Acid-Base Equilibria

1

Solutions of a Weak Acid or Base

1. Acid-Ionization Equilibria

2. Polyprotic Acids

3. Base-Ionization Equilibria

4. Acid–Base Properties of Salt Solutions

Solutions of a Weak Acid or Base with Another Solute

5. Common-Ion Effect

6. Buffers

7. Acid–Base Titration Curves

Contents and Concepts

2

Learning Objectives


• Acid-Ionization Equilibria

– a. Write the chemical equation for a weak acid undergoing acid ionization in aqueous solution.

– b. Define acid-ionization constant and degree of ionization.

– c. Determine Ka from the solution pH.

– d. Calculate concentrations of species in a weak acid solution using Ka

(approximation method).3

1. Acid-Ionization Equilibria (cont)

a. State the assumption that allows for using approximations when solving problems.

b. Calculate concentrations of species in a weak acid solution using Ka (quadratic formula).

2. Polyprotic Acids

a. State the general trend in the ionization constants of a polyprotic acid.

b. Calculate concentrations of species in a solution of

a diprotic acid. 4

3. Base-Ionization Equilibria

– a. Write the chemical equation for a weak base undergoing ionization

in aqueous solution.

– b. Define base-ionization constant.

– c. Calculate concentrations of

species in a weak base solution using Kb.

5

4. Acid–Base Properties of Salt Solutions

– a. Write the hydrolysis reaction of an ion

to form an acidic solution.

– b. Write the hydrolysis reaction of an ion

to form a basic solution.

– c. Predict whether a salt solution is acidic, basic, or neutral.

– d. Obtain Ka from Kb or Kb from Ka.

– e. Calculating concentrations of species in a salt solution.

6

2

Solutions of a Weak Acid or Base with

Another Solute

5. Common-Ion Effect

– a. Explain the common-ion effect.

– b. Calculate the common-ion effect on acid ionization (effect of a strong acid).

– c. Calculate the common-ion effect on

acid ionization (effect of a conjugate base).

7

6. Buffers

– a. Define buffer and buffer capacity.

– b. Describe the pH change of a buffer

solution with the addition of acid or base.

– c. Calculate the pH of a buffer from given

volumes of solution.

– d. Calculate the pH of a buffer when a strong

acid or a strong base is added.

– e. Learn the Henderson–Hasselbalch equation.

– f. State when the Henderson–Hasselbalch equation can be applied

8

7. Acid–Base Titration Curves

– a. Define equivalence point.

– b. Describe the curve for the titration of a

strong acid by a strong base.

– c. Calculate the pH of a solution of a

strong acid and a strong base.

– d. Describe the curve for the titration of a

weak acid by a strong base.

– e. Calculate the pH at the equivalence

point in the titration of a weak acid by a strong base.

9

7. Acid–Base Titration Curves (cont)

– f. Describe the curve for the titration of a weak base by a strong acid.

– g. Calculate the pH of a solution at several points of a titration of a weak base by a strong acid.

10


• The simplest acid-base equilibria are those in which a single acid or base solute reacts with

water.

– In this chapter, we will first look at solutions of weak acids and bases.

– We must also consider solutions of salts, which can have acidic or basic properties

as a result of the reactions of their ions with water.

11

The simplest acid–base equilibria are those

in which a weak acid or a weak base reacts with water.

• We can write an acid equilibrium reaction

for the generic acid, HA.

• HA(aq) + H2O(l) H3O+(aq) + A-(aq)

12

3

• Acetic acid is a weak acid. It reacts with

water as follows:

• HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2

-(aq)

13

• The equilibrium constant for the reaction of

a weak acid with water is called the acid-ionization constant (or acid-dissociation

constant), Ka.

[ ][ ][ ]HA

A OH3a

−+

=K

• Liquid water is not included in the

equilibrium constant expression.

WHY? 14

Acid-Ionization Equilibria

• Acid ionization (or acid dissociation) is the reaction of an acid with water to produce

hydronium ion (hydrogen ion) and the conjugate base anion.

– Because acetic acid is a weak electrolyte, it ionizes to a small extent in water.

– When acetic acid is added to water it reacts as follows.

(aq)OHC(aq)OH 2323

−−−−++++++++)l(OH)aq(OHHC 2232

++++

15

Acid-Ionization Equilibria

• For a weak acid, the equilibrium concentrations of ions in solution are

determined by the acid-ionization constant(also called the acid-dissociation constant).

– Consider the generic monoprotic acid, HA.

)aq(A)aq(OH)l(OH)aq(HA 32

−−−−++++ ++++++++

– The corresponding equilibrium expression is:

]OH][HA[

]A][OH[K

2

3c

−−−−++++

====

16

– Since the concentration of water remains relatively constant, we rearrange the equation to

get:

]HA[

]A][OH[K]OH[K 3

c2a

−−−−++++

========

– Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.

]HA[

]A][OH[K 3

a

−−−−++++

====

– Table 17.1 lists acid-ionization constants for various weak acids.

17

18

4

HCN

Highest pH

Lowest pH

HF

HNO2

HC2H3O2

HCN

HF

HNO2

HC2H3O2

1.7 × 10-5

4.5 × 10-4

Ka

6.8 × 10-4

4.9 × 10-10

In order of decreasing Ka:

19

B-8

Experimental Determination of Ka

• The degree of ionization of a weak electrolyte is the fraction of molecules that

react with water to give ions.

– Electrical conductivity or some other colligative property can be measured to determine the degree of ionization.

– With weak acids, the pH can be used to determine the equilibrium composition of ions in the solution.

20

Calculations with Ka

• Given the value of Ka and the initial concentration of HA, you can calculate the equilibrium concentration of all species.

• Given the value of Ka and the initial

concentration of HA, you can calculate the degree of ionization and the percent ionization.

• Given the pH of the final solution and the initial

concentration of HA, you can find the value

of Ka and the percent ionization.21

We can be given pH, percent or degree ionization, initial concentration, and Ka.

From pH, we can find [H3O+].

From percent or degree ionization, we can find Ka.

Using what is given, we can find the other quantities.

22

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C and the percent ionization

– It is important to realize that the solution was made 0.012 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.012 M.

– We will abbreviate the formula for nicotinic acid as HNic.

23

– Let x be the moles per liter of product formed.

)aq(Nic)aq(OH)l(OH)aq(HNic 32

−−−−++++++++++++

Starting 0.012 0 0Change -x +x +x

Equilibrium 0.012-x x x

– The equilibrium-constant expression is:

]HNic[

]Nic][OH[K 3

a

−−−−++++

====

– Substituting the expressions for the equilibrium concentrations, we get

)x012.0(

xK

2

a −−−−====

24

5

– We can obtain the value of x from the given pH.

)pHlog(anti]OH[x 3−−−−========

++++

)39.3log(antix −−−−====

00041.0101.4x4 ====××××====

−−−−

– Substitute this value of x in our equilibrium expression.

– Note first, however, that

012.001159.0)00041.0012.0()x012.0( ≅≅≅≅====−−−−====−−−−

the concentration of unionized acid remains virtually unchanged.

25

– Substitute this value of x in our equilibrium expression.

522

a 104.1)012.0(

)00041.0(

)x012.0(

xK

−−−−××××====≅≅≅≅−−−−

====

– To obtain the degree of dissociation:

034.00.012

0.00041ondissociatiofDegree ========

– The percent ionization is obtained by multiplying by 100, which gives 3.4%.

26

• Sore-throat medications sometimes contain the weak acid phenol, HC6H5O. A 0.10 M solution

of phenol has a pH of 5.43 at 25°C.

• a. What is the acid-ionization constant, Ka, for phenol at 25°C?

• b. What is the degree of ionization?

Another Problem

27

HC6H5O(aq) + H2O(l) H3O+(aq) + C6H5O

-(aq)

Initial 0.10 0 0

Change –x +x +x

Equilibrium 0.10 – x x x

• We were told that pH = 5.43. That allows us to find [H3O

+] = 3.7 × 10-6 M = x = [C6H5O-].

• Now we find [HC6H5O] = 0.10 – x = 0.10 M.28

• Finally, we write the expression for Ka and

substitute the concentrations we now know.

• HC6H5O + H2O H3O+ + C6H5O

-

• [H3O+] = [C6H5O

-] = 3.7 × 10-6 M

• [HC6H5O] = 0.10 M

[[[[ ]]]][[[[ ]]]][[[[ ]]]]OHHC

OHC OH

56

563a

−−−−++++

====K

0.10

)10 (3.7 26

a

−×

=K

-10

a 10 1.4 ×=K 29

5

6

10 3.7ionization of Degree

0.10

10 3.7

0.100 ionization of Degree

−

−

×=

×==

x

• The degree of ionization is the ratio of

ionized concentration to original concentration:

• Percent ionization is the degree of ionization

× 100%:

• Percent ionization = 3.7 x 10-3% or 0.0037%

30

6

•Simplifying Assumption for Acid and Base Ionizations

•The equilibrium concentration of the acid is most often ([HA]0 – x).

•If x is much, much less than [HA]0, we can assume that subtracting x makes no difference to [HA]:

•([HA]0 – x) = [HA]0

•This is a valid assumption when the ratio of [HA]0to Ka is > 103. If it is not valid, you must use the quadratic equation to solve the problem.

31

•Para-hydroxybenzoic acid is used to make certain dyes. What are the concentrations of

this acid, of hydronium ion, and of para-hydroxybenzoate ion in a 0.200 M aqueous solution at 25°C? What is the pH of the

solution and the degree of ionization of the acid? The Ka of this acid is 2.6 × 10-5.

We will use the generic formula HA for para-hydroxybenzoic acid and the following equilibrium:

HA + H2O H3O+ + A-

32

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

Initial 0.200 0 0

Change –x +x +x


[ ][ ][ ]

.HA

A OH3a

−+

=K

x

x

-0.20010 2.6

25

=×−

0.200 - 0.200 so 1000, 10 2.6

0.2005

≈>×

−x

33

)10 (2.3 pH ]O[H log pH -33 -- ×==

+

2610 5.2 x=×−

0.200

2

a

xK =

]O[H 10 2.3 3

3 +−=×= Mx

2.64 pH =

34

Calculations With Ka

• Once you know the value of Ka, you can

calculate the equilibrium concentrations of species HA, A-,

and H3O+ for solutions of different

molarities.

35


• Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid.

• It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression.

• The degree of ionization of a weak acid depends on both the Ka and the concentration of the acid solution (see Figure 16.3).

36

7

Figure 16.3: Variation of

Percent Ionization of a Weak Acid with Concentration

37


• How do you know when you can use this simplifying assumption?

– then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%.

– It can be shown that if the acid concentration, Ca, divided by the Ka exceeds 100, that is,

100K

Cif

a

a >>>>

38

– If the simplifying assumption is not valid, you can solve the equilibrium equation

exactly by using the quadratic equation.

– The next example illustrates this with a solution of aspirin (acetylsalicylic acid), HC9H7O4, a common headache remedy.

– The molar mass of HC9H7O4 is 180.2 g.

From this we find that the sample contained 0.00180 mol of the acid.

Hence, the concentration of the acetylsalicylic acid is 0.00180 mol/0.500 L = 0.0036 M (Retain two significant figures, the same number of significant figures in Ka). 39

– Note that

11103.3

0036.0K

C4

a

a ====××××

==== −−−−

which is less than 100, so we must solve the equilibrium equation exactly.

Abbreviate the formula for acetylsalicylic acid asHAcs and let x be the amount of H3O

+ formed per liter.

– The amount of acetylsalicylate ion is also x mol; the amount of nonionized acetylsalicylic acid is (0.0036-x) mol.

40

– These data are summarized below.

)aq(Acs)aq(OH)l(OH)aq(HAcs 32

−−−−++++++++++++



– The equilibrium constant expression is

a3 K

]HAcs[

]Acs][OH[====

−−−−++++

– If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get 41

a3 K

]HAcs[

]Acs][OH[====

−−−−++++

42

103.3)x0036.0(

x −−−−××××====−−−−

– You can solve this equation exactly by using the quadratic formula.

Rearranging the preceding equation to put it in the form ax2 + bx + c = 0, we get

0)102.1(x)103.3(x642 ====××××−−−−××××++++

−−−−−−−−

42

8

– Now substitute into the quadratic formula.

a2

ac4bbx

2 −−−−±±±±−−−−====

– Now substitute into the quadratic formula.

2

)102.1(4)103.3()103.3(x

6244 −−−−−−−−−−−− ××××−−−−××××±±±±××××−−−−====

– The lower sign in ± gives a negative root which we can ignore

– Taking the upper sign, we get

43 104.9]OH[x

−−−−++++××××========

43

– Now we can calculate the pH.

03.3)104.9log(pH4 ====××××−−−−====

−−−−

Check your answer by substituting the hydrogenIon concentration into the equilibrium expression andCalculate Ka

44

• Polyprotic Acids

• A polyprotic acid has more than one acidic

proton—for example, H2SO4, H2SO3,

H2CO3, H3PO4.

• These acids have successive ionization

reactions with Ka1, Ka2, . . .

• The next example illustrates how to do

calculations for a polyprotic acid.

45

Polyprotic Acids

• Some acids have two or more protons (hydrogen ions) to donate in aqueous

solution. These are referred to as polyprotic acids.

– The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO4

-.

– Sulfuric acid, for example, can lose two protons in aqueous solution.

)aq(HSO)aq(OH)l(OH)aq(SOH 43242

−−−−++++ ++++→→→→++++

)aq(SO)aq(OH)l(OH)aq(HSO2

4324

−−−−++++−−−−++++++++

46

– For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered.

)aq(HCO)aq(OH)l(OH)aq(COH 33232

−−−−++++ ++++++++

)aq(CO)aq(OH)l(OH)aq(HCO2

3323

−−−−++++−−−−++++++++

– Each equilibrium has an associated acid-ionization constant.

– For the loss of the first proton

7

32

331a 103.4

]COH[

]HCO][OH[K

−−−−

−−−−++++

××××========

47

– Each equilibrium has an associated acid-ionization constant.

– For the loss of the second proton

11

3

2

332a 108.4

]HCO[

]CO][OH[K

−−−−

−−−−

−−−−++++

××××========

– In general, the second ionization constant, Ka2, for a polyprotic acid is smaller than the first ionization constant, Ka1.

– In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2.

48

9

– When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.

– However, reasonable assumptions can be made that simplify these calculations as we show in the next example.

49

– For diprotic acids, Ka2 is so much smaller than Ka1

that the smaller amount of hydronium ion produced in the second reaction can be neglected.


• Ascorbic acid (vitamin C) is a diprotic acid,

H2C6H6O6. What is the pH of a 0.10 M solution?

What is the concentration of the ascorbate ion,

C6H6O62- ?

The acid ionization constants are Ka1 = 7.9 x 10-5

and Ka2 = 1.6 x 10-12.

– The pH can be determined by simply solving the equilibrium problem posed by the first ionization.

50

– If we abbreviate the formula for ascorbic acid as H2Asc, then the first ionization is:

(aq)AscH(aq)OH)l(OH)aq(AscH 322

−−−−++++ ++++++++



(aq)AscH(aq)OH)l(OH)aq(AscH 322

−−−−++++ ++++++++


1a

2

3 K]AscH[

]HAsc][OH[====

−−−−++++

51

– Substituting into the equilibrium expression

52

109.7)x10.0(

x −−−−××××====−−−−

– Assuming that x is much smaller than 0.10, you get

0028.0108.2x

)10.0()109.7(x

3

52

====××××≅≅≅≅

××××××××≅≅≅≅

−−−−

−−−−

– The hydronium ion concentration is 0.0028 M, so

55.2)0028.0log(pH ====−−−−====

52

– The ascorbate ion, Asc2-, which we will call y, is produced only in the second ionization of H2Asc.

(aq)Asc(aq)OH)l(OH)aq(HAsc2

32

−−−−++++−−−− ++++++++

– Assume the starting concentrations for HAsc- and H3O

+ to be those from the first equilibrium.


32

−−−−++++−−−− ++++++++


32

−−−−++++−−−− ++++++++

Starting 0.0028 0.0028 0Change -y +y +y

Equilibrium 0.0028-y 0.0028+y y

53


2a

23 K

]HAsc[

]Asc][OH[====

−−−−

−−−−

++++

– Substituting into the equilibrium expression

12106.1

)y0028.0(

)y)(y0028.0( −−−−××××====−−−−

++++

– Assuming y is much smaller than 0.0028, the equation simplifies to

12106.1

)0028.0(

)y)(0028.0( −−−−××××====

54

10

– Hence,

122106.1]Asc[y

−−−−−−−− ××××====≅≅≅≅

– The concentration

– of

– the ascorbate ion equals Ka2.

55

• Tartaric acid, H2C4H4O6, is a diprotic acid used in food products. What is the pH of a

0.10 M solution? What is the concentration of the C4H4O6

2− ion in the same solution?

• Ka1 = 9.2 × 10−4; Ka2 = 4.3 × 10−5.

First, we will use the first acid-ionization equilibrium to find [H+] and [HC4H4O6

-]. In these calculations,

we will use the generic formula H2A for the acid.

Next, we will use the second acid-ionization

equilibrium to find [C4H4O62-].

56

H2A(aq) +

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

(aq)

Initial 0.10 0 0

Change –x +x +x


[ ][ ][ ]AH

HA OH

2

31a

−+

=Kx

x

-0.1010 9.2

24

=×−

.assumption gsimplifyin the use cannot We

100 10 9.2

0.104

<×

−

57

-5-4 10 9.2- 10 9.21 ×=×== cba

a

acbbx

2

42−±−

=

10 9.210 9.2 245 xx =×−×−−

24 )(0.1010 9.2 xx =−×−

0 10 9.2 - 10 9.2 -5-42=××+ xx

2(1)

)10 4(1)(-9.2)10 (9.2)10 (9.2 -5244×−×±×−

=

−−

x

3424

10 9.6 10 4.62

)10 (1.92)10 (9.2 −−

−−

×±×−=×±×−

=x

58

2-3 10 1.0and10 9.1 −×−=×= xx

ion.concentrat negative a have to impossible physically

is it because value negative the eliminate We

M 10 9.1]O[H 3

3

−+×=

M 10 9.1][HA 3−−×=

m.equilibriu ionization

-acid second the for these use Now we

M 10 9.0A][H 4

2

−×=

are ionsconcentrat

the m,equilibriu ionization acid first the of end the At

59

HA-(aq) + H2O(l) H3O+(aq) + A2-

(aq)

Initial 0.0091 0.0091 0

Change –x +x +x

Equilibrium

0.0091 – x 0.0091 + x x

[ ][ ][ ]

.HA

A OH-

23

2a

−+

=K)- (0.0091

) (0.009110 4.3 5

x

xx+=×

−

0.0091 - 0.0091 and 0.0091 0.0091

that assume can We

≈≈+ xx60

11

0.0091

0.009110 4.3 5 x

=×−

valid. asw assumption urO

Mx 10 4.3 5−×=

M 10 4.3 ]OH[C 5-2

644−

×=

M 10 9.1 ]OH[HC 3-

644−

×=

M 10 9.1 ]O[H 33

−+×=

M 10 9.0 ]OHC[H 46442

−×=

)10 (9.1 logpH 3−×−= 2.04pH =

61

Base-Ionization Equilibria

• Equilibria involving weak bases are treated

similarly to those for weak acids.

– Ammonia, for example, ionizes in water as follows.

)aq(OH)aq(NH)l(OH)aq(NH 423

−−−−++++++++++++

– The corresponding equilibrium constant is:

]OH][NH[

]OH][NH[K

23

4c

−−−−++++

====

62

– The concentration of water is nearly constant.

]NH[

]OH][NH[K]OH[K

3

4c2b

−−−−++++

========

– In general, a weak base B with the base ionization

)aq(OH)aq(HB)l(OH)aq(B 2

−−−−++++ ++++++++

has a base ionization constant equal to

]B[

]OH][HB[Kb

−−−−++++

====Table 17.2 lists

ionization constants for

some weak bases.63

64

• Writing Kb Reactions

• The bases in Table 16.2 are nitrogen

bases; that is, the proton they accept adds

to a nitrogen atom. Next we’ll practice writing the Kb reactions.

• Ammonium becomes ammonium ion:

NH3+ H2O � NH4+ + OH-

• Ethylamine becomes ethyl ammonium ion:

C2H5NH2 + H2O � C2H5NH3+ + OH-

65

• Dimethylamine becomes

dimethylammonium ion:

(CH3)2NH2 + H2O � (CH3)2NH3+ + OH-

• Pyridine becomes pyridinium ion:

C5H5N+ H2O � C5H5NH+ + OH-

• Hydrazine becomes hydrazinium ion:

N2H4+ H2O � N2H5+ + OH-

66

12

We can be given pH, initial concentration,

and Kb.

From the pH, we can find first [H3O+] and

then [OH-]. Using what is given, we can

find the other quantities.

We can also use a simplifying assumption: When [B]0 / Kb > 103, the expression

([B]0 – x) = [B]0.

67

68

Go To 11C Slides

Figure 16.12: Curve for the titration of a

strong acid by a strong base.

69

• Figure 16.12 shows a curve for the titration of HCl

with NaOH.

– At the equivalence point, the pH of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze.

– However, the pH changes rapidly from a pH of about 3 to a pH of about 11.

Titration of a Strong Acid by a Strong Base

– To detect the equivalence point, you need an acid-base indicator that changes color within the pH range 3-11.

– Phenolphthalein can be used because it changes color in the pH range 8.2-10. (see Figure 15.10)

70


• Calculate the pH of a solution in which 10.0 mL of

0.100 M NaOH is added to 25.0 mL of 0.100 M

HCl.

– Because the reactants are a strong acid and a strong base, the reaction is essentially complete.

)l(OH)l(OH)aq(OH)aq(OH 223++++→→→→++++

−−−−++++

– We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities.

71

mol00250.0100.0L0250.0OHMol Lmol

3====××××====

++++

mol00100.0100.0L0100.0OHMol Lmol ====××××====

−−−−

– All of the OH- reacts, leaving an excess of H3O+

l0.00100)mo00250.0(OHExcess 3−−−−====

++++

++++==== OHmol0.00150 3

– You obtain the H3O+ concentration by dividing the

mol H3O+ by the total volume of solution (=0.0250 L

+ 0.0100 L=0.0350 L)

M0429.0L0.0350

mol00150.0]OH[ 3

========++++

72

13

• Calculate the pH of a solution in which 10.0

mL of 0.100 M NaOH is added to 25.0 mL

of 0.100 M HCl.

– Hence,

368.1)0429.0log(pH ====−−−−====

73

• The titration of a weak acid by a strong base gives

a somewhat different curve.

– The pH range of these titrations is shorter.

– The equivalence point will be on the basic sidesince the salt produced contains the anion of a weak acid.

– Figure 17.13 shows the curve for the titration of nicotinic acid with NaOH.

Titration of a Weak Acid by a Strong Base

74

Figure 17.13: Curve for the titration of a weak

acid by a strong base.

75


• Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M

acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5.

– At the equivalence point, equal molar amounts of acetic acid and sodium

hydroxide react to give sodium acetate.

– First, calculate the concentration of the acetate ion.

– In this case, 25.0 mL of 0.10 M NaOH is needed to react with 25.0 mL of 0.10 M acetic acid.

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– The molar amount of acetate ion formed equals the initial molar amount of acetic acid.

solnL1

e ionmol acetat0.10solnL1025

3 ××××××××−−−− e ionmol acetat102.5

3-××××====

– The total volume of the solution is 50.0 mL. Hence,

M050.0L1050

mol102.5ionconcentratmolar

3

3-

====××××

××××====

−−−−

– The hydrolysis of the acetate ion follows the method given in an earlier section of this chapter.

– You find the Kb for the acetate ion to be 5.9 x 10-10 and that the concentration of the hydroxide ion is 5.4 x 10-6. The pH is 8.73 77

• The titration of a weak base with a strong

acid is a reflection of our previous example.

– Figure 17.14 shows the titration of NH3 with HCl.

– In this case, the pH declines slowly at first, then falls abruptly from about pH 7 to pH 3.

– Methyl red, which changes color from yellow at pH 6 to red at pH 4.8, is a possible indicator.

Titration of a Strong Acid by a Weak Base

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14

What would the titration of a weak acid with a

weak base appear?

What would the titration of a weak base with a

weak acid appear?

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Figure 16.14: Curve for the titration of a weak

base by a strong acid.

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Operational Skills

• Determining Ka (or Kb) from the solution pH

• Calculating the concentration of a species in a weak acid solution using Ka

• Calculating the concentration of a species in a weak base solution using Kb

• Predicting whether a salt solution is acidic, basic, or neutral

• Obtaining Ka from Kb or Kb from Ka

• Calculating concentrations of species in a salt solution

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Operational Skills

• Calculating the common-ion effect on acid

ionization

• Calculating the pH of a buffer from given volumes

of solution

• Calculating the pH of a solution of a strong acid

and a strong base

• Calculating the pH at the equivalence point in the

titration of a weak cid with a strong base

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Chap 11B Acid Base Equilibrium.ppt - Bakersfield … Spring_10...3 • Acetic acid is a weak acid....

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Transcript of Chap 11B Acid Base Equilibrium.ppt - Bakersfield … Spring_10...3 • Acetic acid is a weak acid....