Chap 16 Acid Base Equilibrium - Bakersfield College B1B Notes Fall...Acid–Base Titration Curves...

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Acid-Base Equilibria

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

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).

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.

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.

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.

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).

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–Hasselbalchequation can be applied

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.

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.


• 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.

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)

3

• Acetic acid is a weak acid. It reacts with

water as follows:

• HC2H3O2(aq) + H2O(l) H3O+(aq) +

C2H3O2-(aq)

• Here A = C2H3O2-(aq)

• 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.

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

++++

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

−−−−++++

====

– 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.

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:

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 colligativeproperty 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.

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.

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.

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.

5

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

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

−−−−++++++++++++

x

+x0 00.012Starting

x0.012-xEquilibrium

+x-xChange

– 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 −−−−====

– 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.

– 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%.

Do Exercise 161 Look at Problems 16.35 and 16.36

• 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

0.10 – x

–x

0.10

HC6H5O(aq) +

xxEquilibrium

+x+xChange

00Initial

C6H5O-(aq)H3O

+(aq) +H2O(l)

• 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.

• 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

6

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%

•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.

•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-

0.200 – x

–x

0.200

HA(aq) +

xxEquilibrium

+x+xChange

00Initial

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

[ ][ ][ ]

.HA

A OH3a

−+

=K

x

x

-0.20010 2.6

25

=×−

0.200 - 0.200 so 1000, 10 2.6

0.2005

≈>×

−x

)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 =

Do exercise 16.2 See problems 16.37 and 16.38

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.

– The general method for doing this was discussed in Chapter 15.

7


• 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).

Figure 16.3: Variation of

Percent Ionization of a

Weak Acid with Concentration


• 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 >>>>

– 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).

– 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.

– These data are summarized below.

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

−−−−++++++++++++

x

+x0 00.0036Starting

x0.0036-xEquilibrium

+x-xChange

– 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

8

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 ====××××−−−−××××++++

−−−−−−−−

– 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

−−−−++++××××========

– 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

Do Exercise 16.3 See problems 16.43-44

• 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.

Polyprotic Acids

• Some acids have two or more protons

(hydrogen ions) to donate in aqueous solution. These are referred to as polyproticacids.

– 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

−−−−++++−−−−++++++++

– 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

−−−−

−−−−++++

××××========

9

– 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.

– 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.

– 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.

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

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

−−−−++++ ++++++++

x

+x0 00.10Starting

x0.10-xEquilibrium

+x-xChange

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

−−−−++++ ++++++++


1a

2

3 K]AscH[

]HAsc][OH[====

−−−−++++

– 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 ====−−−−====

– 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

−−−−++++−−−− ++++++++

0.0028+y

+y

0.0028 00.0028Starting

y0.0028-yEquilibrium

+y-yChange

10


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( −−−−××××====

– Hence,

122106.1]Asc[y

−−−−−−−− ××××====≅≅≅≅

– The concentration

– of

– the ascorbate ion equals Ka2.

Do Exercise 16.4 See Problems 16.47-48

• 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 [C4H4O6

2-].

0.10 – x

–x

0.10

H2A(aq)

+

xxEquilibrium

+x+xChange

00Initial

HA-

(aq)

H3O+(aq) +H2O(l)

[ ][ ][ ]AH

HA OH

2

31a

−+

=Kx

x

-0.1010 9.2

24

=×−

.assumption gsimplifyin the use cannot We

100 10 9.2

0.104

<×

−

-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

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

11

0.0091 – x

–x

0.0091

HA-(aq) +

x0.0091 + xEquilibrium

+x+xChange

00.0091Initial

A2-

(aq)

H3O+(aq) +H2O(l)

[ ][ ][ ]

.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

≈≈+ xx

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 3

3

−+×=

M 10 9.0 ]OHC[H 46442

−×=

)10 (9.1 logpH 3−×−= 2.04pH =

Do Example 16.4 See Problems 16.47-48

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

−−−−++++

====

– 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.

• 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-

12

• 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-

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.


• What is the pH of a 0.20 M solution of

pyridine, C5H5N, in aqueous solution? The

Kb for pyridine is 1.4 x 10-9.

1. Write the equation and make a table of concentrations.

2. Set up the equilibrium constant expression.

3. Solve for x = [OH-].

– As before, we will follow the three steps in solving an equilibrium.

– Pyridine ionizes by picking up a proton from water (as ammonia does).

)aq(OH)aq(NHHC)l(OH)aq(NHC 55255

−−−−++++++++++++

x+x0 00.20Starting

x0.20-xEquilibrium

+x-xChange

– Note that

89

b

b 104.1104.1

20.0K

C ××××====××××

==== −−−−

which is much greater than 100, so we may

use the simplifying assumption that

(0.20-x) ≅≅≅≅ (0.20).

– The equilibrium expression is

b

55

55 K]NHC[

]OH][NHHC[====

−−−−++++

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

92

104.1)x20.0(

x −−−−××××====−−−−

– Using our simplifying assumption that the x in the denominator is negligible, we get

92

104.1)20.0(

x −−−−××××≅≅≅≅

– Solving for x we get

)104.1()20.0(x92 −−−−××××××××≅≅≅≅

59107.1)104.1()20.0(]OH[x

−−−−−−−−−−−− ××××====××××××××≅≅≅≅====

13

8.4)107.1log(]OHlog[pOH5 ====××××−−−−====−−−−−−−−====

−−−−

2.98.400.14pOH00.14pH ====−−−−====−−−−====

– Since pH + pOH = 14.00

– Solving for pOH

Work exercise 17.6 See problems 17.47&48

• Salt Solutions

• We will look at the cation and the anion separately, and then combine the result to

determine whether the solution is acidic, basic, or neutral.

Acid-Base Properties of a Salt Solution

• One of the successes of the Brønsted-Lowry

concept of acids and bases was in pointing out that

some ions can act as acids or bases.

• Hydrolysis: the reaction of an ion with water to

produce the conjugate acid and hydroxide or the

conjugate base and hydronium ion.

– A 0.1 M solution has a pH of 11.1 and is therefore fairly basic.

– Consider a solution of sodium cyanide, NaCN.

)aq(CN)aq(Na)s(NaCNOH2

−−−−++++++++→→→→

– Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH-.

)aq(OH)aq(HCN)l(OH)aq(CN 2

−−−−−−−− ++++++++

– From the Brønsted-Lowry point of view, the CN- ion acts as a base, because it accepts a proton from H2O.


−−−−−−−−++++++++


−−−−−−−− ++++++++

– You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic.

– The reaction of the CN- ion with water is referred to as the hydrolysis of CN-.

– The CN- ion hydrolyzes to give the conjugate acid and hydroxide.


−−−−−−−− ++++++++

– The hydrolysis reaction for CN- has the form of a base ionization so you writ the Kb expression for it.


−−−−−−−−++++++++

– The NH4+ ion hydrolyzes to the conjugate

base (NH3) and hydronium ion.


++++++++++++++++

– This equation has the form of an acid ionization so you write the Ka expression for it.


++++++++++++++++

14

Predicting Whether a Salt is Acidic, Basic,

or Neutral

• How can you predict whether a particular

salt will be acidic, basic, or neutral?

– The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs.

– Consequently, the anions of weak acids (poor proton donors) are good proton acceptors.

– Anions of weak acids therefore, are basic.

– One the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze.

– For example, the Cl- ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water.

reactionno)l(OH)aq(Cl 2→→→→++++

−−−−

– Conversely, the cations of weak bases are acidic.

– One the other hand, the cations of strong bases have virtually no acidic character, that is, they do not hydrolyze. For example,

reactionno)l(OH)aq(Na 2→→→→++++

++++


or Neutral

• To predict the acidity or basicity of a salt, you

must examine the acidity or basicity of the ions

composing the salt.

– Consider potassium acetate, KC2H3O2.

The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

reactionno)l(OH)aq(K 2→→→→++++

++++

– Consider potassium acetate, KC2H3O2.

The acetate ion, however, is the anion of a

weak acid (HC2H3O2) and is basic.

−−−−−−−−++++++++ OHOHCH)l(OH)aq(OHC 2322232

A solution of potassium acetate is predicted to be basic.

Do Exercise 16.6 See Problems 16.53-54


or Neutral

• These rules apply to normal salts (those in

which the anion has no acidic hydrogen)

1. A salt of a strong base and a strong acid.

The salt has no hydrolyzable ions and so gives a neutral aqueous solution.

An example is NaCl.

2. A salt of a strong base and a weak acid.

The anion of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic solution.

An example is NaCN.

3. A salt of a weak base and a strong acid.

The cation of the salt is the conjugate of the weak base. It hydrolyzes to give an

acidic solution.

An example is NH4Cl.

15

4. A salt of a weak base and a weak acid.

Both ions hydrolyze. You must compare the Ka

of the cation with the Kb of the anion.

If the Ka of the cation is larger the solution is

acidic.

If the Kb of the anion is larger, the solution is basic.

How about?

The pH of a Salt Solution

• To calculate the pH of a salt solution would

require the Ka of the acidic cation or the Kb of the

basic anion. (see Figure 17.8)

– The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species.

– Thus the Kb for CN- is related to the Ka for HCN.

• The conjugate acid of a strong base is

very weak and does not react with water. It

is therefore considered to be neutral.

• Na+ + H2O � No Reaction (NR)

• The conjugate base of a strong acid is very weak and does not react with water. It

is therefore considered to be neutral.

• Cl- + H2O � NR

• The conjugate acid of a weak base reacts

with water to form an acidic solution:

• NH4+ + H2O <==> NH3 + H3O

+

• The conjugate base of a weak acid reacts with water to form a basic solution:

• F- + H2O <==> HF + OH-

•We classify each salt by examining its cation and its anion, and then combining the result.

•NaBr (Where did it come from?

Na+ is the conjugate acid of NaOH, a strong base. It does not react with water, so it is neutral.

Br- is the conjugate base of HBr, a strong acid. It does not react with water, so is neutral.

The cation is neutral; the anion is neutral. Thus NaBr is neutral.

• NaC2H3O2

Na+ is the conjugate acid of NaOH, a strong base. It does not react with water,

so it is neutral.

C2H3O2- is the conjugate base of

HC2H3O2-, a weak acid. It reacts with

water to give a basic solution.

The cation is neutral; the anion is basic.

• NaC2H3O2 is basic.

16

• NH4Cl

NH4+ is the conjugate acid of NH3, a

weak base. It reacts with water to give an acidic solution.

Cl- is the conjugate base of HCl, a

strong acid. It does not react with water, so it is neutral.

The cation is acidic; the anion is neutral. NH4Cl is acidic.

• NH4F

• NH4+ is the conjugate acid of NH3, a weak

base. It reacts with water to give an acidic

solution.

• F- is the conjugate base of HF, a weak

acid. It does not react with water, so it is basic.

• The cation is acidic; the anion is basic. We

need more information in this case. We

compare Ka for HF to Kb for NH3.

• If Ka > Kb, the solution is acidic. If Ka < Kb,

the solution is basic. If Ka = Kb, the solution is neutral.

10

5

14

b

wa4 10 5.6

10 1.8

10 1.0 :NH For −

−

−+

×=×

×==

K

KK

acidic. is FNH4

11

4

14

a

wb 10 1.5

10 6.8

10 1.0 :F For −

−

−

−×=

×

×==

K

KK

−+> F for KNH for K b4a

• Ammonium nitrate, NH4NO3, is

administered as an intravenous solution to

patients whose blood pH has deviated from

the normal value of 7.40.

• Would this substance be used for acidosis

(blood pH < 7.40) or alkalosis (blood pH >

7.40)?

• NH4+ is the conjugate acid of a NH3, a

weak base. NH4+ is acidic.

• NO3- is the conjugate base of HNO3, a

strong acid. NO3- is neutral.

• NH4NO3 is acidic, so it could be used for alkalosis.

• The hydrolysis equilibrium constant can be used in problems to determine the pH of a

salt solution. To use the hydrolysis equilibrium, we need to compute the K

value for it.

17

•We can determine K for the reaction of the ion with water (hydrolysis reaction).

•NH4+ + H2O <==> NH3 + H3O

+

•is the sum of

•NH4+ + OH- <==> NH3 + H2O; K1 = 1/Kb

•2H2O <==> H3O+ + OH-; K2 = Kw

•Overall reaction

•NH4+ + H2O <==> NH3 + H3O

+

b

w21a

K

KK KK ==

•What is Kb for the F- ion, the ion added to

the public water supply to protect teeth?

•For HF, Ka = 6.8 × 10-4.

• F- + H2O <==> HF + OH-

• Is the sum of

• H3O+ + F- <==> HF + H2O; K1 = 1/Ka

• 2H2O <==> H3O+ + OH-; K2 = Kw

• Overall reaction:

• F- + H2O <==> HF + OH-

11

b

- 10 1.5 :F For −×=K

-4

14

a

w21b

106.8

101.0

×

×===

−

K

KKKK

• Household bleach is a 5% solution of

sodium hypochlorite, NaClO. This

corresponds to a molar concentration of

about 0.70 M NaClO.

• What is the [OH-] and the pH of the

solution?

• For HClO, Ka = 3.5 × 10-8.

• We are told that Ka = 3.5 × 10-8. That

means that Kb= 2.9 × 10-7.

• This allows us to substitute into the Kb

expression to solve for x.

0.70 – x

–x

0.70

ClO-(aq) +

xxEquilibrium

+x+xChange

00Initial

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

• ClO- + H2O HClO + OH-

• [H3O+] = [A-] = x and [HA] = 0.200 – x

[ ][ ][ ]NHC

HO NHHC

56

56b

−+

=K

)-(0.7010 2.9

2

x

x=×

−7

0.70. - 070 that assume can weso 1000, 10 2.9

0.707-

=>×

x

-72 10 2.0 ×=x

0.70 10 2.9

27- x

=×

Mx 10 4.5 -4×=

18

• The question asks for [OH-] and the pH:

)4-10 (4.5 log - 14.00 pH

][OH log14.00pOH14.00 pH

×=

−===−

35.300.41 pH −=

65.10pH =

Mx 10 4.5 ][OH -4×==

−


• To see the relationship between Ka and Kb for

conjugate acid-base pairs, consider the acid

ionization of HCN and the base ionization of CN-.

)aq(CN)aq(OH)l(OH)aq(HCN 32

−−−−++++++++++++


−−−−−−−−++++++++

Ka

Kb

– When these two reactions are added you get the ionization of water.

)aq(OH)aq(OH)l(OH2 32

−−−−++++++++ Kw


−−−−++++++++++++


−−−−−−−−++++++++

Ka

Kb

– When two reactions are added, their equilibrium constants are multiplied.


−−−−++++++++ Kw


−−−−++++++++++++


−−−−−−−−++++++++

Ka

Kb

– Therefore,


−−−−++++++++ Kw

wba KKK ====××××


• For a solution of a salt in which only one ion

hydrolyzes, the calculation of equilibrium

composition follows that of weak acids and bases.

– The only difference is first obtaining the Ka

or Kb for the ion that hydrolyzes.

– The next example illustrates the reasoning

and calculations involved.

• What is the pH of a 0.10 M NaCN

solution at 25 °C? The Ka for HCN is 4.9

x 10-10.

– Only the CN- ion hydrolyzes.


−−−−−−−−++++++++


– The CN- ion is acting as a base, so first, we must calculate the Kb for CN-.

5

10

14

a

wb 100.2

109.4

100.1

K

KK

−−−−

−−−−

−−−−

××××====××××

××××========

19

– Now we can proceed with the equilibrium calculation.

– Let x = [OH-] = [HCN], then substitute into the equilibrium expression.

bK]CN[

]OH][HCN[====

−−−−

−−−−

52

100.2)x10.0(

x −−−−××××====

−−−−

– Solving the equation, you find that

3104.1]OH[x

−−−−−−−−××××========

2.11)104.1log(00.14pOH00.14pH3 ====××××++++====−−−−====

−−−−

– Hence,

– As expected, the solution has a pH greater than 7.0.

Work Exercise 16.7 and 16.8 See Problems 16. 63-68

• Common-Ion Effect

• The common-ion effect is the shift in an

ionic equilibrium caused by the addition of a solute that takes part in the equilibrium.

• Buffers

• A buffer solution is characterized by the ability to resist changes in pH when limited

amounts of acid or base are added to it.

• A buffer is made by combining a weak acid with its conjugate base or a weak

base with its conjugate acid.

The Common Ion Effect

• The common-ion effect is the shift in an ionic

equilibrium caused by the addition of a solute that

provides an ion common to the equilibrium.

– Consider a solution of acetic acid (HC2H3O2), in which you have the following equilibrium.

(aq)OHC(aq)OH 2323

−−−−++++++++)l(OH)aq(OHHC 2232

++++

– If we were to add NaC2H3O2 to this solution, it would provide C2H3O2

- ions which are present on the right side of the equilibrium.

(aq)OHC(aq)OH 2323

−−−−++++++++)l(OH)aq(OHHC 2232

++++

– The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased.

– This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.

20


• An aqueous solution is 0.025 M in formic acid,

HCH2O and 0.018 M in sodium formate,

NaCH2O. What is the pH of the solution. The Ka

for formic acid is 1.7 x 10-4.

– Consider the equilibrium below.

(aq)OCH(aq)OH 23

−−−−++++ ++++)l(OH)aq(OHCH 22++++

x

+x

0 0.0180.025Starting

0.018+x0.025-xEquilibrium

+x-xChange

a

2

23 K]OHCH[

]OCH][OH[====

−−−−++++

– The equilibrium constant expression is:

– Substituting into this equation gives:

4107.1

)x025.0(

)x018.0(x −−−−××××====−−−−

++++

– Assume that x is small compared with 0.018 and 0.025. Then

025.0)x025.0(

018.0)x018.0(

≅≅≅≅−−−−

≅≅≅≅++++

– The equilibrium equation becomes

4107.1

)025.0(

)018.0(x −−−−××××≅≅≅≅

– Hence,

44104.2

018.0

025.0)107.1(x

−−−−−−−−××××====××××××××≅≅≅≅

– Note that x was much smaller than 0.018 or 0.025.

63.3)104.2log(pH4 ====××××−−−−====

−−−−

– For comparison, the pH of 0.025 M formic acid is 2.69.

Do Exercise 17.9,10,11 See problems 17.63,64,65,66

Buffers

• A buffer is a solution characterized by the ability

to resist changes in pH when limited amounts of

acid or base are added to it.

– Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid.

– Thus, a buffer contains both an acid species and a base species in equilibrium.

Buffers• A buffer is a solution characterized by the ability



– Consider a buffer with equal molar amounts of HA and its conjugate base A-.

– When H3O+ is added to the buffer it reacts with the

base A-.

)l(OH)aq(HA)aq(A)aq(OH 23++++→→→→++++

−−−−++++

Buffers

• A buffer is a solution characterized by the ability



– Consider a buffer with equal molar amounts of HA and its conjugate base A-.

When OH- is added to the buffer it reacts with the acid HA.

)aq(A)l(OH)aq(HA)aq(OH 2

−−−−−−−−++++→→→→++++

– Two important characteristics of a buffer are its buffer capacity and its pH.

21

– Buffer capacity depends on the amount of

acid and conjugate base present in the

solution.

– The next example illustrates how to

calculate the pH of a buffer.

The Henderson-Hasselbach Equation

• How do you prepare a buffer of given pH?

– A buffer must be prepared from a

conjugate acid-base pair in which the Ka of the acid is approximately equal to the

desired H3O+ concentration.

– To illustrate, consider a buffer of a weak

acid HA and its conjugate base A-.

The acid ionization equilibrium is:

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

−−−−++++ ++++++++

The Henderson-Hasselbalch Equation


– The acid ionization constant is:

– By rearranging, you get an equation for the H3O+

concentration.

]A[

]HA[K]O[H a3 −−−−

++++ ××××====

]HA[

]A][OH[K 3

a

−−−−++++

====


– Taking the negative logarithm of both sides of the equation we obtain:

]A[

]HA[log)Klog(]Olog[H- a3 −−−−

++++ −−−−−−−−====

– The previous equation can be rewritten

]HA[

]A[logKppH a

−−−−

++++====

Henderson-Hasselbach Equation



– More generally, you can write

– This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalchequation.

]acid[

]base[logKppH a

++++====

This problem involves dilution first. Once we know

the concentrations of benzoic acid and benzoate

ion, we can use the acid equilibrium to solve for x.

We will use HBz to represent benzoic acid and Bz-

to represent benzoate ion.

• What is the pH of a buffer made by mixing

1.00 L of 0.020 M benzoic acid, HC7H5O2,

with 3.00 L of 0.060 M sodium benzoate,

NaC7H5O2? Ka for benzoic acid is 6.3 × 10-

5.

22

MM

0.0050.L 3.0) (1.0

L) )(1.0 (0.020]HBz[ =

+=

MM

0.045.L 3.0) (1.0

L) )(3.0 (0.060]Bz[ =

+=

−

0.0050 –

x

–x

0.0050

HBz(aq)

+

0.045 + xxEquilibrium

+x+xChange

0.0450Initial

Bz-(aq)H3O+(aq)

+

H2O(l)

[ ][ ][ ]HBz

B OH3a

−+

=z

K

)-(0.0050

) (0.04510 6.3 5

x

xx +=×

−

0.0050 - 0.0050 and 0.045 0.045

that assume can We

≈≈+ xx

Mx 10 7.0 6−×=

0.0050

0.04510 6.3 5 x

=×−

valid. wasassumption urO

Mx 10 7.0 ]O[H 63

−+×==

5.15 pH =

)10 (7.0 log ]O[H log pH 6

3

−+×−=−=



– So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid-base pair with a pKa close to the desired pH.

– The Ka for acetic acid is 1.7 x 10-5, and its pKa is 4.77.

– You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid].

Do Exercise 16.12 See problems 16. 75-76

Instructions for making a buffer say to mix 60 mL of 0.100 M NH3

With 40 mL of 0.100 M NH4Cl. What is the pH of the solution?

Example 16.11

NH3(aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq)

1. How many moles of NH3 and NH4+ are are added?

2. Fill in the concentration table.


Start

Change

Equilibrium

3. Substitute in to equilibrium constant equation.


4. Solve the equation (assume x is small compared with

the molar concentrations.

pH = 9.43

Do Exercise 16.12 See problems 16.75-76

23

Check concept check 16.4 page 682

Acid-Ionization Titration Curves

• An acid-base titration curve is a plot of the pH

of a solution of acid (or base) against the volume

of added base (or acid).

– Such curves are used to gain insight into the titration process.

– You can use titration curves to choose an appropriate indicator that will show when the titration is complete.

Titration of a Strong Acid by a Strong Base

• Figure 16.12 shows a curve for the titration of HCl

with NaOH.

– Note that the pH changes slowly until the

titration approaches the equivalence point.

– The equivalence point is the point in a

titration when a stoichiometric amount of

reactant has been added.

Figure 16.12: Curve for the titration of a

strong acid by a strong base.

• 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)


• 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.

24

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

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

• 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 ====−−−−====

• 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

Figure 17.13: Curve for the titration of a weak

acid by a strong base.


• 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.

– 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

25

• 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

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?

Look at examples 16.13 and 14

Do exercises 16.13, 14, 15, and 16

Look at problems 16.85-92

Figure 16.14: Curve for the titration of a weak

base by a strong acid.

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

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

Chap 16 Acid Base Equilibrium - Bakersfield College B1B Notes Fall...Acid–Base Titration Curves...

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Transcript of Chap 16 Acid Base Equilibrium - Bakersfield College B1B Notes Fall...Acid–Base Titration Curves...