18-1 Chapter 18 Acid-Base Equilibria. 18-2 Acid-Base Equilibria 18.1 Acids and bases in water 18.2...

18-1

Chapter 18

Acid-Base Equilibria

18-2

Acid-Base Equilibria

18.1 Acids and bases in water

18.2 Auto-ionization of water and the pH scale

18.3 Proton transfer and the Brønsted-Lowry acid-base definition

18.4 Solving problems involving weak acid equilibria

18.5 Weak bases and their relationships to weak acids

18.6 Molecular properties and acid strength

18.7 Acid-base properties of salt solutions

18.8 Generalizing the Brønsted-Lowry concept: The Leveling Effect

18.9 Electron-pair donations and the Lewis acid-base definition

18-3 Figure 18.1

Etching with acids

The inside surfaces of these light bulbs are etched with HF.

Acids are used to wash away oxides of silicon and metals during the production of computer chips.

18-4

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

18-5

For reaction between a strong acid and strong base:

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

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

For acid dissociation:

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

In aqueous solution, the H+ ions bind covalently towater to form solvated hydronium ions.

18-6

Figure 18.2

The nature of the hydrated proton

18-7

The Classical (Arrhenius) Definition of Acids and Bases

An acid is a substance that has H in its formula and dissociates in water

to yield H3O+.

A base is a substance that has OH in its formula and dissociates in water

to yield OH-.

Arrhenius acids contain covalently bonded H atoms that ionize in water.

Neutralization occurs when the H+ ion from the acid and the OH- ionfrom the base combine to form water.

H+(aq) + OH- (aq) H2O(l) ∆Horxn = -55.9 kJ

18-8

Defining the acid dissociation constant

Weak acids dissociate very slightly into ions in water.

Strong acids dissociate completely into ions in water.

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

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

Kc >> 1

Kc << 1

Kc = [H3O+][A-]

[H2O][HA]

Kc[H2O] = Ka =[H3O+][A-]

[HA]weaker acid, lower [H3O

+],

smaller Ka

stronger acid, higher [H3O+],

larger Ka

18-9

Figure 18.3

strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-(aq)

The extent of dissociation of strong acids

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

18-10

Figure 18.3

The extent of dissociation of weak acids

weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq)


18-11

Figure 18.4

Reaction of zinc with a strong and a weak acid

1 M HCl(aq) 1 M CH3COOH(aq)

Zn(s) + 2H3O+(aq) Zn+2(aq)

+ 2H2O(l) + H2(g)

18-12

Acid strength decreases down

the table (smallerKa values)

18-13

SAMPLE PROBLEM 18.1

SOLUTION:

Classifying acid and base strength from the chemical formula

PROBLEM: Classify each of the following compounds as a strong acid, weak acid, strong base or weak base.

(a) H2SeO4 (b) (CH3)2CHCOOH (c) KOH (d) (CH3)2CHNH2

PLAN: Pay attention to the text definitions of acids and bases. Look at O for acids and for the -COOH group; watch for amine groups and cations in bases.

(a) strong acid - H2SeO4 - the number of oxygen atoms

exceeds the number of ionizable protons by a factor of 2.(b) weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group

(a carboxylic acid).

(c) strong base - KOH is a Group 1A hydroxide.

(d) weak base - (CH3)2CHNH2 has a lone pair of electrons on the

nitrogen and is an amine.

18-14

The auto-ionization of water and the pH scale

H2O(l) H2O(l)

H3O+(aq) OH-(aq)

+

+

18-15

Kc = [H3O+][OH-]

[H2O]2

Kc[H2O]2 = [H3O+][OH-]

Defining the ion-product constant of water

Kw =

A change in [H3O+] causes an inverse change in [OH-].

= 1.0 x 10-14 at 25 oC

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

In an acidic solution, [H3O+] > [OH-]

In a basic solution, [H3O+] < [OH-]

In a neutral solution, [H3O+] = [OH-]

18-16

Figure 18.5

The relationship between [H3O+] and [OH-] and the relative acidity of solutions

[H3O+] [OH-]divide into Kw

acidic solution

basic solution

[H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-]

neutral solution


18-17

SAMPLE PROBLEM 18.2 Calculating [H3O+] and [OH-] in an aqueous

solution

PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 25 oC and obtains a solution with [H3O

+] = 3.0 x 10-4 M.

Calculate [OH-]. Is the solution neutral, acidic or basic?

SOLUTION:

PLAN: Use the Kw at 25 oC and the [H3O+] to find the corresponding [OH-].

Kw = 1.0 x 10-14 = [H3O+] [OH-]

[OH-] = Kw/ [H3O+] = 1.0 x 10-14/3.0 x 10-4 = 3.3 x 10-11 M

[H3O+] > [OH-]; the solution is acidic.

18-18

Figure 18.6

The pH values of some familiar

aqueous solutions

pH = -log [H3O+]

pOH = -log [OH-]

pK = -log K

Related Expressions

18-19

A low pK corresponds to a high K.

18-20



Relationshipsbetween [H3O+],

pH, [OH-] and pOH

Figure 18.7

18-21

SAMPLE PROBLEM 18.3 Calculating [H3O+], pH, [OH-], and pOH

PROBLEM: In a restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated nitric acid, HNO3, to 2.0 M, 0.30 M,

and 0.0063 M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the

three solutions at 25 oC.

SOLUTION:

PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-]

and then convert to pH and pOH.

For 2.0 M HNO3, [H3O+] = 2.0 M and -log [H3O

+] = -0.30 = pH

[OH-] = Kw/ [H3O+] = 1.0 x 10-14/2.0 = 5.0 x 10-15 M; pOH = 14.30

[OH-] = Kw/ [H3O+] = 1.0 x 10-14/0.30 = 3.3 x 10-14 M; pOH = 13.48


+] = 0.52 = pH

[OH-] = Kw/ [H3O+] = 1.0 x 10-14/6.3 x 10-3 = 1.6 x 10-12 M; pOH = 11.80


+] = 2.20 = pH

18-22

Figure 18.8

Methods for measuring the pH of an aqueous solution

pH meterpH (indicator) paper


18-23

The Brønsted-Lowry Definition of Acids and Bases

An acid is a proton donor, that is, any species that donates an H+ ion.All Arrhenius acids are Brønsted-Lowry acids.

A base is a proton acceptor, that is, any species

that accepts an H+ ion.

In the Brønsted-Lowry definition, an acid-base reaction is a protontransfer process.

18-24

Figure 18.9

Proton transfer is the essential feature of a Brønsted-Lowry acid-base reaction

(acid, H+ donor) (base, H+ acceptor)

HCl H2O

+

Cl- H3O+

+

lone pair

binds H+

(base, H+ acceptor) (acid, H+ donor)

NH3 H2O

+

NH4+ OH-

+

lone pair

binds H+


18-25

An acid reactant produces a base product and the two constitute an acid-base conjugate pair.

The Conjugate Acid-Base Pair

H2S + NH3 HS- + NH4+

H2S and HS- are a conjugate acid-base pair.

HS- is the conjugate base of the acid H2S.

NH3 and NH4+ are a conjugate acid-base pair.

NH4+ is the conjugate acid of the base NH3.

Every acid has a conjugate base, and every base has a conjugate acid.

18-26

The conjugate base of the pair has one fewer H and one more negative charge than the acid.

The conjugate acid of the pair has one more H and one less negative charge than the base.

A Bronsted-Lowry acid-base reaction occurs when an acid and a basereact to form their conjugate base and conjugate acid, respectively.

acid1 + base2 base1 + acid2

18-27

Table 18.4 Conjugate Pairs in Some Acid-Base Reactions

base1 acid2+acid1 base2+

conjugate pair

conjugate pair

reaction 4 H2PO4- OH-+

reaction 5 H2SO4 N2H5++

reaction 6 HPO42- SO3

2-+

reaction 1 HF H2O+ F- H3O++

reaction 3 NH4+ CO3

2-+

reaction 2 HCOOH CN-+ HCOO- HCN+

NH3 HCO3-+

HPO42- H2O+

HSO4- N2H6

2++

PO43- HSO3

-+


18-28

SAMPLE PROBLEM 18.4 Identifying conjugate acid-base pairs

PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs.

(a) H2PO4-(aq) + CO3

2-(aq) HPO42-(aq) + HCO3

-(aq)

(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3

-(aq)

SOLUTION:

PLAN: Identify proton donors (acids) and proton acceptors (bases).

(a) H2PO4-(aq) + CO3

2-(aq) HPO42-(aq) + HCO3

-(aq)proton donor

proton acceptor

proton acceptor

proton donor

conjugate pair1conjugate pair2

(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3

-(aq)

conjugate pair2conjugate pair1

proton donor

proton acceptor

proton acceptor

proton donor

18-29

Relative Acid-Base Strength and Reaction Direction

General Rule: An acid-base reaction proceeds to the greater extent in the direction in which a stronger acid and stronger base form aweaker acid and a weaker base.

H2S + NH3 HS- + NH4+ Kc > 1

A competition for the proton between the two bases!

HF + H2O F- + H3O+ Kc < 1

a b b a

a b b a

18-30

Figure 18.10

Strengths of conjugate acid-base

pairs

An acid-base reaction proceeds to the right if the acid reacts with a

base that is lower on the list because this combination produces a weaker

conjugate base and a weaker conjugate acid.

A weaker acid has a strongerconjugate base.

18-31

SAMPLE PROBLEM 18.5 Predicting the net direction of an acid-base reaction

PROBLEM: Predict the net direction and whether Ka is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species):

(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)

(a) H2PO4-(aq) + NH3(aq) HPO4

2-(aq) + NH4+(aq)

SOLUTION:

PLAN: Identify the conjugate acid-base pairs and then consult Figure 18.10 to determine the relative strength of each. The stronger the species, the more preponderant will be its conjugate.

(a) H2PO4-(aq) + NH3(aq) HPO4

2-(aq) + NH4+(aq)

stronger acid weaker acidstronger base weaker base

Net direction is to the right; Kc > 1.

(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)stronger baseweaker base stronger acidweaker acid

Net direction is to the left; Kc < 1.

18-32

Weak Acid Equilibrium Problems

Follow the general procedures described in Chapter 17

Two Key Assumptions

[H3O+] from the auto-ionization of water is negligible.

[HA]eq = [HA]init - [HA]dissoc ≈ [HA]init

(because a weak acid has a small Ka)

[HA]dissoc is very small

18-33

SAMPLE PROBLEM 18.6Find the Ka of a weak acid from the pH of its

solution

PROBLEM: Phenylacetic acid (C6H5CH2COOH, denoted as HPAc) builds up in

the blood of people afflicted with phenylketonuria, an inherited genetic disorder that, if left untreated, causes mental retardation and death. A study of the acid shows that the pH of a 0.12 M solution of HPAc is 2.60. What is the Ka of phenylacetic acid?

PLAN: Write the dissociation equation. Use pH and solution concentration to find Ka.

Ka = [H3O+][PAc-]

[HPAc]

Assumptions: At a pH of 2.60, [H3O+]HPAc >> [H3O

+]water

[PAc-]eq ≈ [H3O+]eq, and since HPAc is weak, [HPAc]eq ≈

[HPAc]initial = [HPAc]initial - [HPAc]dissociation SOLUTION: HPAc(aq) + H2O(l) H3O

+(aq) + PAc-(aq)

18-34

SAMPLE PROBLEM 18.6 (continued)

concentration (M) HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)

initial 0.12 - 1 x 10-7 0

change --x +x +x

equilibrium -0.12 - x xx +(<1 x 10-7)

[H3O+] = 10-pH = 2.5 x 10-3 M, which is >> 10-7 ([H3O

+] from water)

x ≈ 2.5 x 10-3 M ≈ [H3O+] ≈ [PAc-] [HPAc]eq = 0.12 - x ≈ 0.12 M

Ka =(2.5 x 10-3) (2.5 x 10-3)

0.12= 5.2 x 10-5

Testing the assumptions:

= 4 x 10-3 %

x 100[HPAc]dissn: 2.5 x 10-3 M0.12 M

[H3O+]water:

1 x 10-7 M

2.5 x 10-3 Mx100

= 2.1 %

(the 5% rule is not violated in either case)

18-35

SAMPLE PROBLEM 18.7 Determining concentrations from Ka and initial [HA]

PROBLEM: Propanoic acid (CH3CH2COOH, simplified as HPr) is an organic

acid whose salts are used to retard mold growth in foods. What is

the [H3O+] of a 0.10 M aqueous solution of HPr (Ka = 1.3 x 10-5)?

SOLUTION:

PLAN: Write the dissociation equation and Ka expression; make assumptions about concentration that are valid; substitute.

x = [HPr]diss = [H3O+]from HPr = [Pr-]

Assumptions: For: HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)

Ka = [H3O+][Pr-]

[HPr]

HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)concentration (M)

initial 0.10 - 0 0

change --x +x +x

equilibrium -0.10 - x xx

Since Ka is small, we assume that x << 0.10

18-36


(x)2

0.101.3 x 10-5 =

[H3O+][Pr-]

[HPr]=

€

x= (0.10)(1.3x10−5) = 1.1 x 10-3 M = [H3O+]

Checking assumptions:

[HPr]diss: 1.1 x 10-3 M / 0.10 M x 100 = 1.1%

Ka =

18-37

percent HA dissociation = [HA]dissociated

[HA]initial

x 100

As the initial concentration of a weak acid decreases, the percent dissociation of the acid increases!

In the prior problem: for 0.10 M HPr, 1.1% dissociation for 0.010 M HPr, 3.6% dissociation

Rationale: larger solution volume accommodates more ions(analogous to pressure effects on gas equilibria when ∆ngas is non-zero)

18-38

Polyprotic acids

Acids that contain more than one ionizable proton

H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O

+(aq)

H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O+(aq)

HPO42-(aq) + H2O(l) PO4

3-(aq) + H3O+(aq)

Ka1 =[H3O

+] [H2PO4-]

[H3PO4]

Ka2 =[H3O

+] [HPO42-]

[H2PO4-]

Ka3 =[H3O

+] [PO43-]

[HPO42-]

Ka1 > Ka2 > Ka3

= 7.2 x 10-3

= 6.3 x 10-8

= 4.2 x 10-13

phosphoric acid, H3PO4

18-39


18-40

Since successive acid dissociation constants typically differby several orders of magnitude, pH calculations for aqueous

solutions of the free acids can be simplified

by neglecting H3O+ generated by subsequent dissociations.

Why are the Ka values successively smaller as more H+ ions

dissociate from a polyprotic acid like phosphoric acid?

18-41

SAMPLE PROBLEM 18.8 Calculating equilibrium concentrations for a polyprotic acid

PROBLEM: Ascorbic acid (H2C6H6O6; abbreviated H2Asc), known as vitamin

C, is a diprotic acid (Ka1 = 1.0 x 10-5 and Ka2 = 5 x 10-12) found in

citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the pH of a

0.050 M aqueous solution of H2Asc.

SOLUTION:

PLAN: Write out expressions for both dissociations and make assumptions.

Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+.

Ka1 is small, thus [H2Asc]initial ≈ [H2Asc]eq

After finding concentrations of the various species for the first dissociation, use them as initial concentrations for the second dissociation.

H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)

Ka1 = [HAsc-][H3O

+]

[H2Asc]= 1.0 x 10-5

HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq) Ka2 =

[Asc2-][H3O+]

[HAsc-]= 5 x 10-12

18-42


H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)concentration (M)

initial 0.050 - 0 0

change - x - + x + x

equilibrium 0.050 - x - x x

Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0 x 10-5 = (x)(x)/0.050 M

pH = -log(7.1 x 10-4) = 3.15

HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq)

7.1 x 10-4 - 0 0


equilibrium 7.1 x 10-4 - x - x x

initial

€

= (0.050)(1.0x10−5)x x = 7.1 x 10-4 M = [HAsc-]

concentration (M)

x = [(7.1 x 10-4)(5 x 10-12)]0.5 = 6 x 10-8 M = [H3O+]

18-43

We can ignore the hydronium ion generated by second ionization.

Also: 7.1 x 10-4/0.050 x 100 = 1.4% This value is < 5%, thus the assumption made in the analysis of the first

dissociation reaction is justified.

[Asc-2] = (Ka2 x HAsc-)/[H3O+] = 5 x 10-12 M

18-44

Weak Bases: Their Relationship to Weak Acids

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

[BH+][OH-]

[B][H2O]

Base-dissociation constant, Kb

[BH+][OH-][B]

pKb decreases with increasing Kb (i.e., increasing base strength)

Kc =

Kb =

18-45

Main Classes of Weak Bases: nitrogen-containing molecules (ammoniaand amines) and anions of weak acids.

Ammonia:

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb = 1.76 x 10-5 (25 oC)

Amines:

RNH2, R2NH and R3N: all have a lone pair of electrons thatcan bind a proton donated by an acid

18-46

+

Figure 18.11

Abstraction of a proton from water by methylamine

+

lone pair

binds H+


CH3NH2

methylamineH2O

CH3NH3+

methylammonium ion

OH-

18-47

Base strengthdecreases

downthe table(smaller

Kb values)

18-48

SAMPLE PROBLEM 18.9 Determining pH from Kb and initial [B]

PROBLEM: Dimethylamine, (CH3)2NH, a key intermediate in detergent

manufacture, has a Kb = 5.9 x 10-4. What is the pH of a 1.5 M

aqueous solution of (CH3)2NH?

SOLUTION:

PLAN: Perform this calculation as done for acids. Keep in mind that you are working with Kb and a base.

(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)

Assumptions:

[(CH3)2NH2+] = [OH-] = x [(CH3)2NH]eq≈ [(CH3)2NH]initial

Kb >> Kw so [OH-]water is negligible

initial 1.50 0 0-


equilibrium 1.50 - x - x x

(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)concentration

18-49


Kb = 5.9 x 10-4 = [(CH3)2NH2

+][OH-]

[(CH3)2NH]

5.9 x 10-4 = (x) (x)

1.5 Mx = 3.0 x 10-2 M = [OH-]

Check assumption: [3.0 x 10-2 M / 1.5 M] x 100 = 2% (error is < 5%; thus, assumption is justified)

[H3O+] = Kw/[OH-] = 1.0 x 10-14/3.0 x 10-2 = 3.3 x 10-13 M

pH = -log (3.3 x 10-13) = 12.48

18-50

Anions of Weak Acids as Weak Bases

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

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

Kb = ([HF][OH-]) / [F-]

Rationale for the basicity of A-(aq): e.g., 1 M NaF

Acidity is determined by [OH-] generated from the F- reaction and

[H3O+] generated from the auto-ionization of water; since

[OH-] >> [H3O+], the solution is basic.

Ka x Kb = Kw

For a conjugate acid-base pair:

18-51

SAMPLE PROBLEM 18.10 Determining the pH of a solution of A-

PROBLEM: Sodium acetate (CH3COONa, abbreviated NaAc) has

applications in photographic development and textile dyeing. What is the pH of a 0.25 M aqueous solution of NaAc? Ka of

acetic acid (HAc) is 1.8 x 10-5.

SOLUTION:

PLAN: Sodium salts are soluble in water so [Ac-] = 0.25 M.Use Ka to find Kb.

initial 0.25 - 0 0

change -x +x +x-

equilibrium -0.25 - x x x

Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)concentration

Kb = [HAc][OH-]

[Ac-]=

Kw

Ka

= 5.6 x 10-10Kb = 1.0 x 10-14

1.8 x 10-5

18-52


Kb = [HAc][OH-]

[Ac-]

[Ac-] = 0.25 M - x ≈ 0.25 M (since Kb is small)

5.6 x 10-10 ≈ x2/0.25 M

x ≈ 1.2 x 10-5 M = [OH-]

Check assumption: [1.2 x 10-5 M / 0.25 M] x 100 = 4.8 x 10-3 %

[H3O+] = Kw/[OH-] = 1.0 x 10-14/1.2 x 10-5 = 8.3 x 10-10 M

pH = -log (8.3 x 10-10 M) = 9.08

18-53

Molecular Properties and Acid Strength

Non-metal hydrides: two factors determine acid strength, namely,the electronegativity of the central non-metal atom E and the strength

of the E-H bond

Non-metal hydride acid strength increases across a period (E electronegativity effect)

Non-metal hydride acid strength increases down a group(E-H bond strength effect)

18-54

Figure 18.12

The effect of atomic and molecular properties on non-metal hydride acidity

6A(16)

H2O

H2S

H2Se

H2Te

7A(17)

HF

HCl

HBr

HI

electronegativity increases, acidity

increases

bon

d s

tre

ngth

de

cre

ase

s,

aci

dity

incr

ea

ses


18-55

H O I H O Br H O Cl> >

H O Cl

O

O

O<<

Figure 18.13

The relative strengths of oxoacids

H O Cl


Electronegativity increases, acidity increases

number of O atoms increases,acidity increases

18-56

Acidity of Hydrated Metal Ions

Aqueous solutions of certain metal ions are acidic because the hydrated

metal ion transfers an H+ ion to water.

Generalized Reactions

M(NO3)n(s) + xH2O(l) M(H2O)xn+(aq) + nNO3

-(aq)

M(H2O)xn+(aq) + H2O(l) M(H2O)x-1OH(n-1)+(aq) + H3O

+(aq)

18-57

Al(H2O)5OH2+Al(H2O)63+

Figure 18.14

The acidic behavior of the hydrated Al3+ ion

H2O H3O+

electron density

drawn toward Al3+solvent H2O acts

as base


18-58



high chargeand small size

enhance acidity

18-59

Acid-Base Properties of Salt Solutions

A. Salts That Yield Neutral Solutions: the anion of a strong acid and the cation of a strong base (the ions do not react with water)

The anion of a strong acid is a much weaker base than water (HNO3).

The cation of a strong base only becomes hydrated (NaOH).

HNO3(l) + H2O(l) NO3-(aq) + H3O

+(aq)

NaOH(s) Na+(aq) + OH-(aq)

18-60

B. Salts That Yield Acidic Solutions: (1) the anion of a strong acidand the cation of a weak base (the cation acts as a weak acid); (2) small, highly charged metal ions; (3) cations of strong bases and anions of polyprotic acids with another ionizable proton.

NH4Cl(s) NH4+ (aq) + Cl-(aq)

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

+ (aq)Case 1

Fe(NO3)3(s) + 6H2O(l) Fe(H2O)63+(aq) + 3NO3

-(aq)

Fe(H2O)63+(aq) + H2O(l) Fe(H2O)5OH2+(aq) + H3O

+(aq)

Case 2

NaH2PO4(s) Na+(aq) + H2PO4- (aq)

H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O+(aq)

Case 3

18-61

C. Salts That Yield Basic Solutions: the anion of a weak acid and thecation of a strong base (the anion acts as a weak base)

CH3COONa(s) Na+(aq) + CH3COO-(aq)

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

18-63

SAMPLE PROBLEM 18.11 Predicting the relative acidity of salt solutions

PROBLEM: Predict whether aqueous solutions of the following compounds are acidic, basic, or neutral (write an equation for the reaction of the appropriate ion with water to explain pH effect).

(a) potassium perchlorate, KClO4 (b) sodium benzoate, C6H5COONa

(c) chromium trichloride, CrCl3 (d) sodium hydrogen sulfate, NaHSO4

SOLUTION:

PLAN: Consider the acid-base nature of the anions and cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic.

(a) The ions are K+ and ClO4-, which come from a strong base

(KOH) and a strong acid (HClO4). The salt solution will be neutral.(b) Na+ comes from the strong base NaOH while C6H5COO- is the anion of a

weak organic acid. The salt solution will be basic.(c) Cr3+ is a small cation with a large + charge, so its hydrated form will react

with water to produce H3O+. Cl- comes from the strong acid HCl. The

salt solution will be acidic.(d) Na+ comes from a strong base. HSO4

- can react with water to form H3O+.

The salt solution will be acidic.

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Salts of Weakly Acidic Cations and Weakly Basic Anions

Both components react with water!

Acidity is determined by the relative acid strength and base strength of the separated ions.

example: NH4HS

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

+(aq)

HS-(aq) + H2O(l) H2S(aq) + OH-(aq)

Ka(NH4+) = 5.7 x 10-10 Kb(HS-) = 1 x 10-7

Since Kb > Ka, the solution is basic.

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SAMPLE PROBLEM 18.12 Predicting the relative acidity of salt solutions from Ka and Kb of the ions

PROBLEM: Determine whether an aqueous solution of zinc formate, Zn(HCOO)2, is acidic, basic, or neutral.

SOLUTION:

PLAN: Both Zn2+ and HCOO- come from weak conjugates. In order to find the relatively acidity, write the dissociation reactions and use the information in Tables 18.2 and 18.7.

Ka Zn(H2O)62+ = 1 x 10-9

Ka HCOOH = 1.8 x 10-4 ; Kb = Kw/Ka = 1.0 x 10-14/1.8 x 10-4 = 5.6 x 10-11

Ka for Zn(H2O)62+ > Kb HCOO-; the solution is acidic.

Zn(H2O)62+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O

+(aq)

HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq)

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The Leveling Effect

In water, the strongest acid possible is H3O+ and the

strongest base possible is OH-.

Any acid stronger than H3O+ donates its proton to H2O, and

any base stronger than OH- accepts a proton from H2O; thus,water exerts a leveling effect (levels the strengths of all strong acids and bases).

To rank strong acids: must dissolve in a solvent that is a weakerbase than water (i.e., one that accepts their protons less readily).

HCl(g) + CH3COOH(l) Cl-(acet) + CH3COOH2+(acet)

HBr(g) + CH3COOH(l) Br-(acet) + CH3COOH2+(acet)

HI(g) + CH3COOH(l) I-(acet) + CH3COOH2+(acet)

KHI > KHBr > KHCl

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Three Definitions of Acids and Bases

The Arrhenius Definition

The Brønsted-Lowry Definition

The Lewis Definition

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F

B

F F

H

N

H H

+

F

B

F F

H

N

H H

acid base adduct

An acid is an electron-pair acceptor. A base is an electron-pair donor.

M2+

H2O(l)

M(H2O)42+(aq)

adduct

The Lewis Acid-Base Definition

The adductcontains a

new covalentbond.

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Lewis Acids with Electron-Deficient Atoms

Lewis Acids with Polar Multiple Bonds

Metal Cations as Lewis Acids

ROR’ + AlCl3 R-O-R’

AlCl3base acid adduct

SO2 + H2O H2SO3

M2+ + 4H2O M(H2O)42+

baseacid adduct

adductbaseacid

Metal ions actas Lewis acidswhen dissolved

in water.

Lewis acidscontain (or can

generate) a vacant orbital.

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Figure 18.15

Mg2+ ion as a Lewis acid in the chlorophyll molecule


The ion accepts electron pairs fromfour nitrogen atoms comprising part

of the chlorophyll molecule

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SAMPLE PROBLEM 18.13 Identifying Lewis acids and bases

PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions:

(a) H+ + OH- H2O

(b) Cl- + BCl3 BCl4-

(c) K+ + 6H2O K(H2O)6+

SOLUTION:

PLAN: Look for electron pair acceptors (acids) and donors (bases).

(a) H+ + OH- H2O

acceptor

donor

(b) Cl- + BCl3 BCl4-

donor

acceptor

(c) K+ + 6H2O K(H2O)6+

acceptor

donor

18-1 Chapter 18 Acid-Base Equilibria. 18-2 Acid-Base Equilibria 18.1 Acids and bases in water 18.2...

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Transcript of 18-1 Chapter 18 Acid-Base Equilibria. 18-2 Acid-Base Equilibria 18.1 Acids and bases in water 18.2...