8-1

Chapter 1 Acid-base Equilibrium and

Titrations

(酸碱平衡与酸碱滴定）

1.1 Strong Electrolyte(强电解质)

1.2 The Theory of Acids-Bases

1.3 Ionization Equilibrium of Weak

Electrolyte (弱电解质电离平衡)

1.4 Acid-base Titrations (酸碱滴定)

8-2

1.1 Strong Electrolyte(强电解质)

1.1.1 Ion-atmosphere(离子氛) and Ionic strength(离子强度)

Ion-atmosphere(离子氛)

I-Ionic strength; m-molality(质量摩尔浓度); z-electric charge of ion

In 1923, Debye and Huckel

Ionic strength(离子强度) I = ½ ∑mizi2

Example What is the I of 0.01mol•kg-1 BaCl2 solution?

A : I = ½ ∑mizi2 = ½(0.01×22+0.02×12)

= 0.03 mol•kg-1

8-3

1.1.2 Activity(活度) and Activity coefficient(活度系数)

Activity(活度): the effective concentration of electrolyte

a = f • c a - Activity(活度); c- concentration of ion

f- Activity coefficient(活度系数)

Usually the f <1, then a < c.

The higher of the electric charge of ion and Ionic strength (离子强度), the smaller the f, and the a is smaller than the c.

When f = 1, a = c. This is the status of lower electric charge of

ions and dilute(稀的) solution.

8-4

1.1.3 General Properties of Aqueous Solution

(水溶液的一般性质)

An electrolyte(电解质) may be either a strong electrolyte or a

weak electrolyte. A strong electrolyte exists in solution

completely (or almost completely) as ions, while a weak

electrolyte produces only a small concentration of ions when it

dissolves in solution.

Ionic compounds that are soluble in water dissociate(离解)

completely and exist in solution entirely as ions. Soluble ionic

compounds are strong electrolytes. Molecular compounds such

as sugar and alcohol are nonelectrolytes(非电解质). They have

no tendency(趋势) to come apart, and they exist in solution

entirely as aqueous molecules (水合分子).

8-5

The acids HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4 are

molecular compounds that ionize (电离) in aqueous solution

and exist completely as ions. For example, hydrogen chloride

(HCl) gas dissolves in water and ionizes in solution to give

aqueous hydrogen(氢) ion(= Hydronium-水合氢离子) and

aqueous chloride ion.

All of the seven acids listed above are strong electrolytes.

They are also called strong acids. In both instances the word

strong denotes complete ionization. (Strong bases are ionic

and are also strong electrolytes.)

HCl(g) H+(aq) + Cl- (aq) H2O

Some molecular compounds, most notably(著名的) acids and weak

bases, are electrolytes.

8-6

Weak acids and weak bases make up a group of molecular

compounds that are weak electrolytes. Acetic acid (CH3COOH ,

醋酸) is a weak acid. Although it ionizes in water, the reverse

(反) process occurs more readily. At any given time most of the

acetic acid exists as aqueous molecules in solution. Weak acids

are weak electrolytes. To represent the equilibrium between the

molecular acid and its ionized form, we use a double arrow in

the equation.

Weak bases are also weak electrolytes. Ammonia(氨) ionizes

in water to produce aqueous ammonium ions(水合氨离子)

and aqueous hydroxide(氢氧化物) ions(水合氢氧根离子).

CH3COOH(aq) H+(aq) + CH3COO- (aq)

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

8-7

1. Which of the following is a strong electrolyte?

H2O2, HF, LiF, C2H5OH

2. Carbonic acid (H2CO3) is a

weak electrolyte

nonelectrolyte

strong electrolyte

Answer

Answer

LiF

weak electrolyte

Question

8-8

1.2 Arrhenius concept of acids and bases

(阿累尼乌斯的酸碱理论)

Acids are substances that when dissolved in water, increase the

concentration of H+ ions.

Bases are substances that when dissolved in water, increase the

concentration of OH- ions.

The strength of acids or Bases are determined by the Kaθ or Kb

θ

在水中能电离出氢离子（H+ 或H3O+ ）的物质叫酸，电离出

氢氧根（OH- ）的物质叫做碱。酸碱可以发生中和反应，可使指示剂变色。有强酸（碱）、弱酸（碱）之分。酸碱的强弱用电离常数（Ka

θ 或 Kbθ ）表示。

阿累尼乌斯的酸碱理论是应用最广泛的酸碱理论。也称阿氏理论，或电离学说

8-9

1.3.1 The Autoionization of Water and The pH Scale

(水的电离与pH标度)

The Autoionization of Water

Pure water has a very small tendency to ionize, acting both as

an acid (donating a proton) and as a base (accepting a proton).

At 25 oC, the Kw for this process is 1.0×10–14, which means that

only about one molecule per billion(十亿) undergoes this

autoionization. The equilibrium expression for the

autoionization of water is

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

Kw = [H3O+] [OH-]

8-10

Because the autoionization of water is a very important equilibrium,

its equilibrium constant is given a special subscript, w. For any

aqueous solution at 25℃, the product of hydronium and hydroxide

ion concentrations is equal to Kw. In neutral(中性) water, where the

only source of either ion is the autoionization, the hydronium (H+)

and hydroxide ion (OH- ) concentrations are equal.

Kw = [H3O+] [OH-] = 1.0×10-14 (25 ℃)

Question

What is the hydroxide ion (OH-) concentration in an aqueous solution

in which the hydronium ion (H+) concentration is 0.13 M ?

(1) 0.13 M (2) 7.7×10–14 M (3) 0.36 M (4) 1.3×10–15 M

Kw – ion-product constant(离子积常数)

8-11

The pH Scale(pH标度)

The pH scale is often used to express the hydronium ion

(H+) concentration in an aqueous solution.

Using this formula, we can calculate the pH of pure water.

Therefore, at 25 ℃, An acidic solution has a pH less than 7.00,

a basic solution has a pH greater than 7.00. However, at other

temperature, An acidic solution is one in which the hydronium ion

concentration is more than the hydroxideion (OH-) concentration, a

basic solution is one in which the hydronium ion concentration is

less than the hydroxideion concentration.

Attention: 25 ℃

8-12

We can also express the acidity of a solution in terms of pOH.

By taking the log of both sides of the under equation

we get

Questions

What is the pH of a solution that is 5.4×10–11M NaOH?

10.27 3.73 7.00 11.5

8-13

1.3.2 Indicator of acid-base(酸碱指示剂)

Some organic acids or bases (weak) have different color with the

change of pH in solution. It is helpful in indicating the pH values.

It is the Indicator of acid-base(酸碱指示剂)

H++ In- HIn e.x: methyl orange(甲基橙)

Red Yellow

Kiθ= [H+][In-]/[HIn] [In-]/[HIn] = Ki

θ/[H+]

When [In-]=[HIn], Kiθ= [H+], pH=pKi

θ –指示剂的理论变色点

When [In-]/[HIn]≥10, it is pure [In-] color.

When [In-]/[HIn] ≤1/10, it is pure [HIn] color.

transition interval of indication

(指示剂的理论变色范围) pH=pKi

θ ±1

8-14

1.3.3 Ionization Equilibrium of Weak Mono-

Electrolyte (一元弱酸弱碱的电离平衡)

Weak Acid (弱酸)

A weak acid is one that ionizes partially in water to produce

hydronium ion. Acetic acid(醋酸) is a weak acid.

The ionization equilibrium of a weak acid has an associated

equilibrium constant called the acid-dissociation contant(酸离解常数) or ionization constant(电离常数). The equilibrium constant

for a weak acid ionization is subscripted with an a for acid. The

equilibrium expression for the above equation is

Kaθ= [H+][Ac-]/[HAc]

HAc + H2O Ac- + H3O +

8-15

Ionization constant Kaθ and ionization degree α

(电离常数,电离度)

Kaθ= [H+][Ac-]/[HAc] = [H+]2/(c0- [H

+])

Solution of the quadratic(二次) equation gives [H+]

When c0 / Kaθ＞400

c0- [H+]≈ c0

0

2θ ]H[

CKa

0

θ]H[ CKa

0

θ

αC

K aα= [H+]/c0

ionization degree α is the percent ionization of a weak acid

C0 /Ka＞380，南京大学版本

C0/Ka＞400，北京大学版本

C0 /Ka＞102.81 ≈646 ,中国农大版本

8-16

As with all equilibria, the larger the value of KaӨ, the further the

equilibrium lies to the right. This means that the larger the value of

Kaθ, the stronger the acid is. Table below lists some weak acids,

their conjugate(共轭) bases, and their acid-dissociation constants.

8-17

What is the pH of a 1.75×10–3 M nitrous acid(亚硝酸) solution?

and its equilibrium expression is

The ionization equilibrium for nitrous acid is

Initial

Change

Equilibrium

1.75 ×10-3M

-x M

1.75 ×10-3 – x M

0

+x M

x M

0

+x M

x M

HNO2(aq) H+(aq) + NO2-(aq)

HNO2(aq) H+(aq) + NO2-(aq)

4104.5][

]][[

2

2

HNO

NOHK

a

Example

8-18

Solution of the quadratic(二次) equation gives x = 6.90×10–4.

According to our equilibrium table, [H+]= x. Therefore,

Note that the initial concentration of H+ is entered as zero. In

fact, it is 1.0×10–7 M, but that amount is generally insignificant

compared to the amount produced by the ionization.

4

3

2

105.41075.1

x

xKa

pH = - log(6.90×10–4) = 3.16

The percent ionization α

For a given weak acid, percent ionization α

increases with decreasing concentration.

( )39.4%=100%×

)(initiallyHNO M10×1.75

mequilibriuat H M10×6.90

2

3-

+4

α=

8-19

Question

1. What is the percent ionization of phenol(苯酚) at a

concentration of 0.50 M ? (Ka(phenol)=1.3×10-10)

(1) 50 % (2) 8.1 % √(3) 0.0016 % (4) 0.13 %

2. What is the pH of a 0.50 M solution of phenol (苯酚)?

√(1) 5.09 (2) 1.40× 10–5 (3) 0.176 (4) 9.71

3. The ionization degree of 0.01 M HAc solution is? Ka(HAc)

=1.76×10–5)

A. 1% √B. 4% C.15% D. 10%

4. What is the pH of a 0.100 M solution of acetic acid ? (Ka =

1.76 ×10–5)?

√A. 2.87 B. 1.76×10–6 C. 5.75 D. 9.71

8-20

Weak Base (弱碱)

A weak base(弱碱) is one that ionizes partially in water to produce

hydroxide ion and a conjugate acid. Ammonia is an example of a

group of weakly basic compounds called amines (胺). An amine

contains a nitrogen atom with a lone pair of electrons. The site of

the lone pair is where a proton can be accepted. In general,

Being an equilibrium, the ionization of a weak base has an

equilibrium constant called the base-dissociation constant(碱离解常数) associated with it. The equilibrium constant for a

weak base ionization has the subscript b for base. The

equilibrium expression for the above equation is

8-21

As with all equilibria, the larger the value of K, the further the

equilibrium lies to the right. This means that the larger the value

of Kb, the stronger the base is. Table below lists some weak bases,

their conjugate acids, and their base-dissociation constants.

8-22

Determining the pH of a weak base solution is very similar to

solving the equilibrium problems encountered previously.

Sample: Determine the pH of a 0.018 M solution of

methylamine NH2CH3(甲胺)。

As before, we set up an equilibrium table.

NH2CH3 + H2O NH3CH3+ + OH

-

Initial

Change

Equilibrium

0.018 M

-x M

0.018– x M

0

+x M

x M

0

+x M

x M

4

2

104.4018.0

x

xKb

8-23

Questions

1. What is the percent ionization of 0.018 M methylamine(甲胺)?

(1) 14.4 % (2) 85.6 % (3) 0.158 % (4) 16.9 %

2. In which solution would trimethylamine (CH3)3N (三甲胺) exhibit

the higher percent ionization?

(1) 0.001 M (2) 0.01 M

(3) Both would have the same percent ionization

Solving for x gives 0.0026. According to our table, x is the

equilibrium concentration of OH– . Taking minus the log of x

gives pOH = 2.58. pH = 14 – pOH = 11.42.

8-24

Relationship Between Ka and Kb (Ka与Kb的关系)

a weak acid has a strong conjugate base. It is important to

distinguish between the meaning of the word strong in the

context of acids and bases and its meaning in the context

of conjugate acids and bases. A strong acid is one that

ionizes completely in water. A strong conjugate acid is

one that is sufficiently acidic compared with water to

protonate the water to some extent. Note again the

difference in equations used to represent each. The

ionization of HCl is written with an arrow in one direction.

The ionization of the ammonium ion (the conjugate acid

of the weak base ammonia) shows a double arrow,

denoting an equilibrium.

8-25

The strength of a conjugate base or acid depends on the strength

of the original acid or base, respectively.

For any conjugate pair

8-26

Question

What is the Kb for the conjugate base of a weak acid with

Ka = 2.30 ×10–9 ?

√(1) 4.35×10–6 (2) 2.30×10–23 (3) 4.35×10–11 (4) 2.30×10–9

some conjugate pairs and their Ka and Kb values

8-27

1.3.4 The Common-Ion Effect (同离子效应）

Le Châtelier's principle says that if a product is added to a

system at equilibrium, the reaction equilibrium will shift toward

the reactants. Consider the ionization of acetic acid in water.

The addition of either product, hydrogen ion or acetate ion, would

shift this equilibrium to the left, resulting in less ionization of

acetic acid. One way to add a product to this equilibrium is to add

a salt of acetic acid, such as sodium acetate (CH3COONa). Being

a soluble ionic compound, sodium acetate dissociates completely

in aqueous solution to give aqueous sodium ions and aqueous

acetate ions. The addition of acetate ions causes the equilibrium

reaction to shift to the left. In this example acetic acid and sodium

acetate have the acetate ion in common. This sort of shift in

equilibrium is called the common-ion effect(同离子效应）.

CH3COOH(aq) H+ (aq) + CH3COO-(aq)

8-28

In general, the ionization of a weak electrolyte is

diminished (减少) by the addition of a strong electrolyte

that has a common ion with the weak electrolyte.

Question

1. Which of the following can be added to an aqueous solution of

ammonia(氨) to diminish the ionization of ammonia?

(1) acetic acid (2) sodium chloride (3) ammonium chloride(氯化铵)

2. What is the Kb for the conjugate base of a weak acid with

Ka =1.76×10–5?

A. 1.76×10–5 B. 4.35×10–6 C. 2.30×10–9 D. 5.68×10–10

8-29

Questions

3. What is the Ka for the conjugate acid of a weak base with

Kb= 1.77 × 10–5?

(A) 5.65×10–10 (B) 1.38×10–6 (C) 6.28×10–8 (D) 3.14×10–12

4. How would the addition of sodium hydroxide (NaOH) affect the

ionization equilibrium of ammonia?

(1) It would have no effect.

(2) It would shift the equilibrium to the left.

(3) It would shift the equilibrium to the right

8-30

1.3.5 Ionization of Polyprotic acids(多元酸的电离)

Polyprotic acids(多元酸) are those that have more than one hydrogen

that can be donated as a proton. Each proton on a polyatomic acid has

a Ka value associated with it—except for the first proton on sulfuric

acid, because it is strongly acidic. Sulfuric acid(H2SO4), carbonic acid

(H2CO3), and phosphoric acid (H3PO4) are all polyprotic acids.

Solving an equilibrium problem involving a polyprotic acid requires

the construction of multiple equilibrium tables.

8-31

Calculate the pH, and the concentration of all species in solution at

equilibrium, for a 0.100 M solution of H3PO4.

Example

Step 1: The first ionization of phosphoric acid is governed by Ka1

which is 7.5×10–3. Our first equilibrium table is

H3PO4 H+ + H2PO4

-

Initial

Change

Equilibrium

0.100 M

-x M

0.100 – x M

0

+x M

x M

0

+x M

x M

Using the quadratic equation, we get x = 0.0239

32

43

-

421 105.7

100.0

)(

POH

POH H

x

xKa

8-32

Step 2:

H2PO4 - H+ + HPO4

2 -

Initial

Change

Equilibrium

0.0239 M

-y M

0.0239 – y M

0.0239 M

+y M

0.0239+y M

0

+y M

y M

y

yyKa

0239.0

)0239.0(

POH

HPO H-

42

-2

42

0.0239+y ≈ 0.0239-y

The y is so small in contrast with the concentration of H+ and

H2PO4 - that it can be insignificant in this system.

[HPO4-] = y = Ka2 = 6.2×10–8

8-33

Step 3: HPO4 2

- H+ + PO4

3 -

Initial

Change

Equilibrium

6.2×10–8 M

-z M

(6.2×10–8- z) M

0.0239 M

+z M

(0.0239+z) M

0

+z M

z M

[PO43– ] =z = Ka2 • Ka3 / 0.0239 = 1.09×10–18

So at equilibrium, [H+] = x = 0.024 M, pH = 1.62;

[H3PO4] = 0.100 – x = 0.076 M;

[H2PO4– ] = x = 0.024 M;

[HPO42– ] = y = 6.2×10–8 M;

[PO43– ] = z = 1.09×10–18 M.

88-2

4

-3

43

102.6

0239.0

102.6

)0239.0(

HPO

PO H

z

z

zzKa

Ka2

8-34

1.3.6 Buffered Solutions（缓冲溶液）

A solution that contains a weak acid and its conjugate base or

one that contains a weak base and its conjugate acid is called a

buffer, or a buffered solution（缓冲溶液）. Buffers resist(抵抗) drastic(激烈的) changes in pH. Such as HAc-Ac- or NH4

+-

NH3 etc.

Human blood is an important example of a complex

aqueous medium with a pH buffered at about 7.4. There

are many buffer-pairs in it, such as H2CO3-NaHCO3,

NaH2PO4 -Na2HPO4 and K protein-H etc..

Soil is an another important example of a complex buffer

system which can supply the best pH for plant growing.

8-35

Composition and Action of Buffered Solution (缓冲溶液的组成与作用)

Buffers are often prepared by mixing a weak acid or a weak

base with a salt of that acid or base. It resist changes in pH

because they contain both an acidic species to neutralize(中和)

OH- ions or a basic one to neutralize H+ ions.

For example, hydrofluoric acid(HF) and sodium fluoride(NaF)

can be used to prepare a buffer. HF is a weak acid, and F– is its

conjugate base.

HF(aq) H+ (aq) + F -(aq)

8-36

When acid is added to a solution containing both HF and F–,

the acid reacts with the F– ion to produce HF.

Ka = [H+][F–] / [HF] [H+] = Ka [HF ] / [F–]

To understand how a buffer resists drastic pH change,

consider the equilibrium between HF and its conjugate base.

The concentration of fluoride ion F– decreases, and the concen-

tration of hydrofluoric acid HF increases. The pH decreases

slightly, but as long as the amount of acid added is small com-

pared to the amount of fluoride ion available to react with it, the

ratio(比值) [HF ] / [F–] doesn’t change much, the pH change is

very small.

HF(aq) H+ (aq) + F -(aq)

H+ (aq) + F -(aq) HF(aq)

8-37

A base added to the solution reacts with the HF to produce

more F– ion. (This can also be thought of in terms of the added

base reacting with the H+ to produce water, drawing the

equilibrium to the right via consumption消耗 of a product.)

In this case the concentration of fluoride ion increases and the

concentration of hydrofluoric acid decreases. The pH increases

slightly, but, again, as long as the amount of base added is

small compared to the amount of hydrofluoric acid in solution,

the pH change is slight.

OH- (aq) + HF (aq) H2O(l) + F -(aq)

[H+] = Ka [HX ] / [X–] For any weak acid

When [HX ] equals [X–], [H+] equals Ka. For this reason, We

usually try to select a buffer whose acid form has a pKa close to

the desired pH.

8-38

Questions

1. What is the pH of a solution in which the concentrations of

weak acid and conjugate base are equal?

(1) 1 (2) 0 (3) pH = pKa + 1 (4) pH = pKa

2. What is the pH of a solution containing 50 g of hypochlorous

acid(次氯酸) and 25 g of sodium hypochlorite(次氯酸钠) in 2 L

of water? (HClO的 Ka = 3.0×10-8)

(1) 7 (2) 7.5 (3) 5 (4) 8

8-39

pH and Buffer Capacity (缓冲溶液的pH值和缓冲容量)

Two important characteristics(性质) of a buffer are its pH and

its capacity (容量) .

[H+] = Ka [HX ] / [X–]

– log [H+] = – log Ka – log ([HX ] / [X–])

Because – log [H+] = pH, – log Ka = pKa, we have

pH = pKa – log [HX ] / [X–] = pKa + log [X–] / [HX ]

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

This equation is known as the Henderson-Hasselbalch equation.

The pH of the buffer depends on the Ka for the acid and on the

concentrations of the acid and base that comprise(组成) the buffer.

8-40

(1) add 0.010 mol H+ to the solution,then

H+ + F - HF

Initial amounts

Final amounts 0 mol

0.010 mol

0.45mol

0.46mol 0.20 mol

0.21 mol

[HF] = 0.21mol/2.0L = 0.105M, [F–] = 0.45mol/2.0L = 0.225M

pH = pKa + log [F -] / [HF] = 3.17 + log (0.225 /0.105) = 3.50

2.0 L solution, [HF] = 0.10M, [NaF] = 0.23M, pKa = 3.17

What would happened when addition of little strong

acids, little strong bases or slightly diluted to buffers?

Example

pH = pKa + log [F -] / [HF] = 3.53

ΔpH = 3.50 - 3.53 = -0.03

8-41

OH- + HF H2O + F-

Initial amounts

Final amounts 0 mol

0.010 mol

0.19mol

0.20mol 0.46 mol

0.47 mol

[HF] = 0.19mol/2.0L = 0.095M,

[F–] = 0.47mol/2.0L = 0.235M

pH = pKa + log [F -] / [HF]

= 3.17 + log (0.235 /0.095)

= 3.56

(2) add 0.010 mol NaOH to the solution,then

ΔpH = 3.56 - 3.53 = +0.03

8-42

pH = pKa + log [F -] / [HF]

= 3.17 + log (0.115 /0.05)

= 3.53

HF ~ F-

Initial concentration/mol·L-1 0.10 0.23

Final concentration/mol·L-1 0.05 0.115

ΔpH = 3.53 - 3.53 = 0

(3) the solution was diluted twice, then

8-43

The buffer capacity (缓冲容量) is the amount of strong acid

(or strong base) and the multiple of dilution that a buffer can

neutralize. It can be explained as follows:

(1) When the component of a buffer necessary to neutralize

added strong acid has been consumed mostly, further addition

of acid will cause a significant (显著的) change in pH.

(2) When the component necessary to neutralize added strong

base has been consumed mostly, further addition of base will

increase the pH significantly.

(3) Likewise(同样地), excessive dilution of a buffer will also

change its pH significantly.

pH = pKa + log [X–] / [HX ]

The buffer capacity(缓冲容量) is biggest when [X–]=[HX].

The buffer range(缓冲范围) is : [X–]:[HX ] = 0.1~10,

i.e. (即) pH = pKa±1

8-44

Questions

1. What is the pH of a buffer prepared with 0.15 mol acetic acid and

0.25 mol sodium acetate in 1.0 L after the addition of 0.030 mol of

strong acid?

(1) 4.97 (2) 4.83 (3) 4.65 (4) 4.52

2. What would be the pH of a liter of pure water after the addition of

0.030 mol of strong acid?

(1) 1.52 (2) 4.74 (3) 3.00 (4) 5.48

3. What is the pH of a solution in which the concentrations of weak

acid and conjugate base are 10:1?

(A) 1 (B) pH = pKa－1 (C) pH = pKa (D) pH = pKa + 1

8-45

The Theory of Acid and Base

(酸碱理论)

Arrhenius acids and bases

(阿累尼乌斯的酸碱理论—复习)

Acids are substances that when dissolved in water, increase

the concentration of H+ ions.

Bases are substances that when dissolved in water, increase

the concentration of OH- ions.

The strength of acids or Bases are determined by the Kaθ or

Kbθ

8-46

BrØnsted-Lowry Acids and Bases

(布朗斯特-劳莱酸碱理论)

The Arrhenius definitions of acids and bases are limited to

aqueous reactions. A broader definition is the Brønsted-Lowry

approach(方法). According to Brønsted-Lowry, an acid is a

substance that donates an H+ ion to another substance; a base is a

substance that accepts an H+ ion.

Chemists use the term proton to refer to the aqueous hydrogen

ion, H+. They also refer to the hydronium ion,H3O+ in the context

of acids. All four of these terms, H+, hydrogen ion, H3O+, and

hydronium ion, are used interchangeably. In water a proton inter-

acts with the lone pairs on oxygen atoms of water molecules and

becomes hydrated(水合). So it is slightly more realistic(实际) to

represent an aqueous proton as H3O+, which depicts(描述) the

proton attached to a water molecule.

8-47

The Brønsted-Lowry approach can be applied to proton-transfer

reactions that do not happen in an aqueous medium, such as the

reaction of ammonia gas with hydrogen chloride gas.

In this reaction the HCl molecule donates a proton to the ammo-

nia molecule, making HCl a Brønsted-Lowry acid.The ammonia

molecule accepts the proton, making it a Brønsted-Lowry base. In

order for one substance to behave as an acid, another substance

must behave as a base (and vice versa). A Brønsted-Lowry acid

must have a proton to donate, and a Brønsted-Lowry base must

have a lone pair of electrons in order to accept the proton. Some

substances are capable of acting as an acid in one reaction and as

a base in another. Such substances are called amphoteric(两性).

8-48

When an acid molecule ionizes in water, it donates its proton to

a water molecule, producing a hydronium ion and an anion(阴离子). The anion produced by the ionization is the conjugate base

(共轭碱)of that acid. (HX is used to denote a generic(一类的)

acid.)

HX(aq) + H2O(l) H3O+(aq) + X-(aq)

acid conjugate base

HX (the acid) and X– (the conjugate base) differ (不一致) only

by a proton. They are a conjugate acid-base pair(共轭酸碱对).

Every acid has a conjugate base. By the same token(出于同样的原因), every base has a conjugate acid .

8-49

The conjugate base of hydrochloric acid(HCl), chloride ion, has

almost no tendency to recombine with a proton to produce the

original acid. (This would constitute the reverse reaction.) By

contrast, the conjugate base of acetic acid, acetate ion, has a

considerable tendency to do so. In fact, the stronger the acid,

the less likely the reverse reaction is to occur—the weaker its

conjugate base. The weaker the acid, the more apt the

reverse reaction is to occur—the stronger its conjugate base.

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

In this example, B (the base) and HB+ (the conjugate acid) are

the conjugate acid-base pair or simply the conjugate pair.

conjugate acid base

8-50

Acidic and basic are merely relative (相对的) terms. Whether

or not a compound behaves as an acid depends on the relative

ability of the compound with which it is combined to behave as

an acid. In the proton-transfer reaction,

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

conjugate base acid base conjugate acid

acetic acid behaves as an acid and water behaves as a base. In

another proton transfer reaction, the ionization of ammonia in

water,

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

conjugate base base acid conjugate acid

8-51

ammonia behaves as a base and water behaves as an acid.

Whether water behaves as an acid (proton donor) or as a base

(proton acceptor) depends on the substance with which it is

combined.

Questions

1. Which pair of species is a conjugate pair?

HCl and Cl2

OH– and O2–

CO2 and H2CO3

C2H4 and C2H6

2. Which of the following acids has a strong

conjugate base?

HCl HNO3 HF HClO4

8-52

3. Which pair of species is not a conjugate pair?

(A) NH3, NH2－ (B) H2PO4

－, PO43－

(C) HS－, S2 (D) H3O, H2O

4. What is the conjugate acid of HPO42-? 。

(A) H3PO4 (B) H2PO4- (C) H+ (D) PO4

3-

5. Which ion is the weakest base in follow ions?

(A) ClO4- (B) ClO3

- (C) ClO2- (D) ClO-

6. Which of the following can act as amphoteric (两性的) according the Bronsted-Lowry theory?

A. CO32- B. HCl C. HPO4

2- D. Ac-

Question

8-53

Lewis Acids and Bases(路易斯酸碱理论)

The Brønsted-Lowry acid-base theory broadens the definitions of

acids and bases to include reactions that do not occur in aqueous

media. Lewis acid-base theory further broadens the definition to

include reactions other than proton-transfer reactions.

A Lewis acid is defined as an electron-pair acceptor, and a Lewis

base as an electron-pair donor. In the examples we have seen of

Arrhenius and Brønsted-Lowry acid-base behavior, the bases are

all acting as Lewis bases. In the ionization of an acid in water, the

water molecule donates electrons (a lone pair on the oxygen atom)

to the hydrogen atom of the acid. As a bond forms between the

oxygen atom on water and the acidic proton, the bond between

the acidic proton and its original molecule is broken.

8-54

A similar description can be given for the gas-phase reaction

between ammonia and hydrogen chloride. The ammonia

molecule donates the lone pair of electrons on nitrogen to the

hydrogen atom on hydrogen chloride.

The Lewis definition of an acid does not require that an acid

have a proton to donate, only that it be able to accept a pair of

electrons from a Lewis base.

8-55

Metal ions in aqueous solution behave as Lewis acids. They

accept electrons from the oxygen atoms in water. When they do,

the positive charge on the metal ion draws electron density(电子密度) from the O–H bond in water, making it more polar, and

easier to break. The breakage of an O–H bond produces an

aqueous proton, resulting in an acidic solution.

Which of the following can act as a Lewis base?

(1) CH4 (2) H2O (3) H2 (4) H+

Question

Answer H2O

8-56

1.4 (Acid-base titrations)酸碱容量分析

1.4.1 Quantitative relation in titration容量分析基本定量关系

1.4.2 Titration of Strong Acids and Strong Bases

强碱(酸)滴定强酸(碱)

1.4.3 Titration of strong bases (acids) and weak acids (bases)

强碱(酸)滴定弱酸(碱)

1.4.4 Titrations in polyprotic systems多元弱酸(碱)的滴定

1.4.5 Preparation of standard solution酸碱标准溶液的配制与标定

8-57

1.4.1 Quantitative relation in titration容量分析基本定量关系

物质的量浓度c (mol·L-1) ×体积V(L) = 溶质物质的量n(mol)

物质的量浓度c (mol·L-1) ×体积V(mL) = 溶质物质的量n(mmol)

稀释与浓缩：c1· V1 = c2· V2

（稀释或浓缩前后的溶质物质的量相等）

容量分析计算：

待测物质量(g) = [摩尔质量(g·mol-1) / 1000] × c ×V(mL)

c —已知浓度的标准溶液, V—完全反应后消耗的体积

定量的溶液反应： c1· V1 = c2· V2

（互相反应的两种物质的物质的量相等）

8-58

酸碱滴定基本原理

(Acid-base titrations)

滴定曲线：溶液pH 随中和百分数(T%)变化的曲线。

选择指示剂的原则:指示剂的变色点尽可能与化学计

量点接近，以减小滴定误差。在滴定突跃范围变色

的指示剂可使滴定(终点)误差小于0.1%。

直接滴定：指示剂在化学计量点前变色, 误差为 - ；

指示剂在化学计量点后变色, 误差为 ＋。

8-59

1.4.2 强碱(酸)滴定强酸(碱)

(Strong base-strong acid titrations)

A. 滴定前(Before titration)

强碱滴定强酸滴定曲线的计算过程

加入滴定剂(NaOH)体积为 0.00 ml时：

—— 0.1000 mol/L 盐酸溶液的pH=1.00

let’s consider the titration of 20.00 mL of 0.1000

M HCl with 0.1000 M NaOH.

Before the equivalence point, HCl is present in excess and

the pH is determined by the concentration of excess HCl.

8-60

B.滴定中(Before the stoichiometric point—— 指到达化学计量点之前)

(相对误差RE = -0.10%, T=99.9%)

[H+] = c VHCl/V

= 0.1000 (20.00 - 19.98) / (20.00+19.98)

= 5.0 10-5 mol/L

溶液pH=4.30

[H+] = 0.1000 (20.00-18.00) / (20.00+18.00)

= 5.3 10-3 mol/L

溶液 pH=2.28

After adding 18.00 mL of NaOH, therefore, the concentration

of excess HCl is

After adding 19.98 mL of NaOH, therefore, the concentration

of excess HCl is

8-61

Kw = 1.00 × 10–14 = c(H3O+)× c(OH–) = (H3O

+)2

c（H3O+） = 1.00 ×10–7 M

pH = –logc（H3O+）= 7.00 T=100%

At the equivalence point the moles of HCl and the moles of

NaOH are equal. Since neither the acid nor the base is in excess,

the pH is determined by the dissociation of water:

after adding 20.02 mL of titrant the concentration of OH– is:

10

( ) ( )( ) 0.1000

( ) ( )

20.02 20.000.1000 5.0 10

20.02 20.00

V NaOH V HClc OH

V NaOH V HCl

pH = 9.70 T=100.1%

8-62

Data for Titration of 20.00 mL of 0.1000 M HCl with

0.1000 M NaOH

Volume (mL) of NaOH pH

0.00 1.00

10.00 1.48

18.00 2.28

19.98 4.30

20.00 7.00

20.02 9.70

22.00 11.70

40.00 12.50

突跃范围

8-63

强酸碱滴定曲线的讨论

pH

12

10

8

6

4

2

0 0 100 200%

滴定百分数,T%

9.7 sp+0.1%

4.3 sp-0.1%

sp 7.0 突跃

9.0

5.0

4.0

9.8

8.0

3.1

*6.2

*4.4

PP

MR

MO

0.10M HCl

滴定0.10M

NaOH

0.10M

NaOH 滴定0.10M HCl

PP-酚酞-phenolphthalein

MR-甲基红-methyl red

MO-甲基橙-methyl orange

滴定突跃是指示

剂选择的依据。

指示剂变色点(

滴定终点)与化

学计量点并不一

定相同，但突跃

范围内变色时相

差不超过

±0.02mL，相

对误差不超过

±0.1%， 符合

滴定分析要求。

8-64

酸(碱)浓度对滴定曲线的影响

0 100 200

pH

12

10

8

6

4

2

10.7

9.7

8.7

7.0

5.3

4.3

3.3

PP

MR

MO

5.0

4.4

9.0

6.2

4.4

3.1

0.01M

0.1M

1.0M

浓度增大9倍, 突跃增加

2个pH单位。

因此, 在酸碱滴定中, 标

准溶液和被测溶液的浓

度均不宜过低, 通常控

制在0.1mol·L-1左右。

Concentration Titration jump ΔpH

0.01M: 5.3 - 8.7 3.4

0.1M: 4.3 - 9.7 5.4

1.0M: 3.3 - 10.7 7.4

8-65

1.4.3 Titration of Weak Acid with Strong Base

Before adding any NaOH the pH is that for a solution of 0.1000

M acetic acid.

let’s consider the titration of 20.00 mL of 0.1000 M acetic acid,

CH3COOH, with 0.1000 M NaOH.

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

pH=2.87 T=0%

35

0 1034.11000.01075.1)HAc(]H[ cKa

与强酸相比，滴定开始点的pH抬高。

8-66

Adding NaOH converts a portion of the acetic acid to its

conjugate base. Any solution containing comparable amounts

of a weak acid, HA, and its conjugate weak base, A–, is a

buffer. the pH of a buffer using the Henderson–Hasselbalch

equation.

开始滴定后,溶液即变为HAc(ca)-NaAc(cb) 缓冲溶液; 按缓冲溶液的pH进行计算。

8-67

after adding 19.98 mL of NaOH，

5

3

2

3

20.00 19.98( ) 0.1000 5.00 10

20.00 19.98

19.98( ) 0.1000 5.00 10

20.00 19.98

c CH COOH

c CH COO

pH=7.76 T=99.9%

8-68

( ) 0.1000/ 2 0.05000c Ac

5

6

[ ] 0.05000 1.75 10

5.3 10

bOH K c

pH=8.72 T=100%

At the equivalence point, the moles of acetic acid initially

present and the moles of NaOH added are identical. To

calculate the pH we first determine the concentration of

CH3COO–：

化学计量点(stoichiometric point)

8-69

计量点之后

After the equivalence point NaOH is present in excess, and the

pH is determined in the same manner as in the titration of a

strong acid with a strong base. For example, after adding 20.02

mL of NaOH, the concentration of OH– is：

10

( ) ( )[ ] 0.1000

( ) ( )

20.02 20.000.1000 5.0 10

20.02 20.00

V NaOH V HAcOH

V NaOH V HAc

pH = 9.70 T=100.1%

8-70

Data for Titration of 20.00 mL of 0.1000 M HAc with

0.1000 M NaOH

Volume (mL) of NaOH pH

0.00 2.87

18.00 5.71

19.98 7.76

20.00 8.72

20.02 9.70

22.00 11.70

40.00 12.50

突跃范围

8-71

强碱滴定弱酸和强酸滴定强碱的滴定曲线特点

0.10 mol·L-1

NaOH

↓

HAc

0.10 mol·L-1

0 100 200 T%

pH

12

10

8

6

4

2

0

HAc Ac- Ac-+OH-

突跃

6.2

4.4

3.1

9.7

8.7

7.7

4.3 5.0

4.0

PP

MR

MO HAc

HCl

曲线起点提高，

突跃处于弱碱

性,只能选酚酞

作指示剂。

8-72

pH

NaOH滴定HAc(浓度不同)

浓度增大9倍,

突跃增加1个pH单位

浓 度 滴定突跃 ΔpH

0.01M: 7.74-8.7 0.96

0.1M: 7.74- 9.7 1.96

1.0M: 7.74-10.7 2.96

8-73

Ka 对滴定曲线的影响

The magnitude of

titration jump is

smaller with the

decrease of Ka. When

Ka is about 10-9,

titration jump

disppear.对于0.1M 的

HB, Ka≥10-7才能准确

滴定. 即cKa≥10-8

8-74

弱酸滴定曲线的讨论

1）滴定前，弱酸在溶液中部分电离，与强酸相比，曲线开始点提高；

2）滴定开始时，溶液pH升高较快，这是由于中和生成的Ac-产生同离子效应，使HAc更难离解，[H+]降低较快；

3）继续滴加NaOH，溶液形成缓冲体系，曲线变化平缓；

4）接近化学计量点时，溶液中剩余的HAc已很少，pH变化加快。

5）化学计量点前后产生pH突跃，与强酸相比，突跃变小；

6）甲基橙指示剂不能用于弱酸滴定；

7）随着弱酸Ka变小，突跃变小，Ka 在10-9左右突跃消失;

8）直接滴定条件：c·Ka≥10-8

8-75

0.1mol·L-1

HCl

Strong Acid-Weak Base Titration

0 50 100 150 200%

6.25

4.30

5.28

6.2

4.4

3.1

pH NaOH

NH3

突跃处于

弱酸性,

选甲基红

作指示剂.

8.0

NH3

0.1mol·L-1

pKb=4.75

8-76

1.4.4 多元弱酸(碱)的滴定

1. 分步滴定可行性判断

如前所述，多元弱酸在溶液中分步电离，因此要进行分步滴定：

(1) 哪一级的c·Ka≥10-8, 则哪一级的H+就可能被准确滴定。

(2) 两个相邻的Ka都符合时，则两个Ka比值要满足≥104, 则可以形成独立的突跃，两个H+可以分步被准确滴定。不满足条件，则只能同时被滴定。

(3)多元弱碱判断同上类似。

8-77

2. 典型的多元弱酸分析

H3PO4 H+ + H2PO4

- Ka1= 10-2.12

H2PO4- H+ + HPO4

2- Ka2= 10-7.21

HPO42- H+ + PO4

3- Ka3= 10-12.7

H2C2O4 = H+ + HC2O4- Ka1=10-1.25

HC2O4- = H+ + C2O4

2- Ka2 =10-4.29

第一、二级电离的H+均可分步滴定，第三步电离的H+不能被直接滴定。

不能分步滴定，只能一步滴定完全。

8-78

The titration of 0.1000 mol/LH3PO4 with NaOH

（1）When the first proton is completely titrated, the

species in the solution is NaH2PO4, which concentration is

0.050 mol ·L-1.

（2）When the second proton is completely titrated, the

species in the solution is Na2HPO4, which concentration is

0.033 mol ·L-1.

66.42

)(21

aapKpK

pH

94.92

)(32

aapKpK

pH

甲基红

酚酞

8-79

pKa △lgKa

2.16

7.21

12.32

5.05

5.11

pHsp1= 4.66

pHsp2= 9.94

8-80

NaOH滴定H3PO4 时指示剂的选择

pKa1(2.16) pKa2(7.21) pKa3(12.32)

H3PO4 H2PO4

- HPO4

2- PO43-

pHsp1= 4.66 pHsp2= 9.94

PP MR

∵Ka3＜＜10-7

∴ H3PO4不能被直接滴定至第三终点。

8-81

1.4.5 酸碱标准溶液的配制与标定

酸碱滴定中最常用的标准溶液是0.1M的NaOH和HCl溶液。但是它们都不能直接配制，需要配制成近似浓度，然后用基准物进行标定。

盐酸标准溶液的配制与标定

• 配制:用市售HCl(12 mol·L-1)稀释.

• 标定:

1. 无水Na2CO3：吸湿性强，使用前在270-300℃烘1h, 保存在干

燥器中备用。缺点是摩尔质量较小(106g·mol-1)，称量误差大。

2. 硼砂(Na2B4O7·10H2O)：摩尔质量大(381.4 g·mol-1),称量误差

小，稳定。缺点是在空气中容易风化失水，要保存在相对湿

度60%的容器中。

8-82

NaOH标准溶液的配制与标定

• 配制: 以饱和的NaOH(约19 mol·L-1), 用除去CO2 的

去离子水稀释.

• 标定:

• 1.邻苯二甲酸氢钾(KHC8H4O4), M=204.2g·mol-1

• 2.草酸(H2C2O4·2H2O), M=126.07g·mol-1