Chapter 18 – Acid-Base Equilibria 18.1 – Acids and Bases in Water 18.2 – Autoionization of...

Chapter 18 – Acid-Base Equilibria

• 18.1 – Acids and Bases in Water• 18.2 – Autoionization 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 Relation 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 Donation and the Lewis Acid-Base

Definition1

18.1 – Acids and Bases in Water

2


3

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

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

Zn with 1M HCl(aq)

vs. 1M CH3COOH(aq)


4

ACID

STR

ENG

TH


5

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 as well as the -COOH group; watch for amine groups and cations in bases.

(a) Strong acid - H2SeO4 - the number of O atoms exceeds the number of ionizable protons by 2.

(b) Weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group.

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

(d) Weak base - (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.

18.2 – Autoionization of Water and the pH Scale

6

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

+ +


7

Kc = [H3O+][OH-]

[H2O]2

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

The Ion-Product Constant for Water, Kw:

Kw =

A change in [H3O+] causes an inverse change in [OH-], and vice versa.

= 1.0 x 10-14 at 25 C

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-]


8

[H3O+] [OH-]Divide into Kw

ACIDIC SOLUTION

BASIC SOLUTION

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

NEUTRAL SOLUTION


9

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 °C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic?

SOLUTION:

PLAN: Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].

Kw = 1.0x10-14 = [H3O+] [OH-] so

[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =

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

3.3x10-11M


• pH = -log [H3O+]• # of sig figs• pH of a neutral soln = 7.00• pH of an acidic soln < 7.00• pH of a basic soln > 7.00• 1 pH unit = 10x change• [H3O+] = 10-pH

• p-scales– pOH = -log [OH-]– pK = -log K

10


11

Table 18.3 The Relationship Between Ka and pKa

Acid Name (Formula) Ka at 25 C pKa

Hydrogen sulfate ion (HSO4-) 1.02x10-2

Nitrous acid (HNO2)

Acetic acid (CH3COOH)

Hypobromous acid (HBrO)

Phenol (C6H5OH)

7.1x10-4

1.8x10-5

2.3x10-9

1.0x10-10

1.991

3.15

4.74

8.64

10.00


12


13

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

PROBLEM: In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 25 °C.

SOLUTION:

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

For 2.0M HNO3, [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30

[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48

For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH

[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80

For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH

18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition

14

(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+


15

Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions

Base Acid+Acid Base+

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

-+


16

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


17


18

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 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 with Kc > 1.

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

stronger baseweaker base stronger acid

weaker acid

Net direction is to the left with Kc < 1.

18.4 – Solving Problems Involving Weak-Acid Equilibria

19

SAMPLE PROBLEM 18.6: Finding the Ka of a Weak Acid from the pH of

Its Solution

PROBLEM: Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.13M HPAc is 2.62. What is the Ka of phenylacetic acid?

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

Ka =[H3O+][PAc-]

[HPAc]

Assumptions: With a pH of 2.62, the [H3O+]HPAc >> [H3O+]water.

[PAc-] ≈ [H3O+]; since HPAc is weak, [HPAc]initial ≈ [HPAc]initial - [HPAc]dissociation

SOLUTION: HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)


20

SAMPLE PROBLEM 18.6: Finding the Ka of a Weak Acid from the pH of

Its Solutioncontinued

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

Initial 0.12 - 1x10-7 0

Change --x +x +x

Equilibrium -0.12-x xx +(<1x10-7)

[H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water)

x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-] [HPAc]equilibrium = 0.12-x ≈ 0.12 M

So Ka =(2.4x10-3) (2.4x10-3)

0.12= 4.8 x 10-5

Be sure to check for % error. = 4x10-3 %

x100[HPAc]dissn;

2.4x10-3M0.12M

[H3O+]from water; 1x10-7M

2.4x10-3Mx100

= 2.0 %


21

SAMPLE PROBLEM 18.7: Determining Concentrations from Ka and

Initial [HA]

PROBLEM: Propanoic acid (CH3CH2COOH, which we simplify and HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10M HPr (Ka = 1.3x10-5)?

SOLUTION:

PLAN: Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; 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 will assume that x << 0.10


22

SAMPLE PROBLEM 18.7: Determining Concentrations from Ka and

Initial [HA]

continued

(x)(x)

0.101.3x10-5 =

[H3O+][Pr-]

[HPr]=

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

Check: [HPr]diss = 1.1x10-3M/0.10 M x 100 = 1.1%


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Polyprotic acids

acids with more than more 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+][HPO4

2-]

[H2PO4-]

Ka3 =[H3O+][PO4

3-]

[HPO42-]

Ka1 > Ka2 > Ka3

= 7.2x10-3

= 6.3x10-8

= 4.2x10-13


25

ACID

STR

ENG

TH


26

SAMPLE PROBLEM 18.8: Calculating Equilibrium Concentrations for a Polyprotic Acid

PROBLEM: Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the pH of 0.050M 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 so [H2Asc]initial ≈ [H2Asc]diss

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

Ka1 = [HAsc-][H3O+]

[H2Asc]= 1.0x10-5

Ka2 = [Asc2-][H3O+]

[HAsc-]= 5x10-12

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

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


27

- x - + x + x

continued

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

Initial 0.050 - 0 0

Equilibrium 0.050 - x - x x

Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0x10-5 = (x)(x)/0.050 M

pH = -log(7.1x10-4) = 3.15

7.1x10-4M - 0 0

Change - x - + x + x

7.1x10-4 - x - x xEquilibrium

Change

Initial

(0.050)(1.0x10 5)x x = 7.1x10-4 M

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

(7.4x10 5)(5x10 12)x = 6x10-8 M

18.5 – Weak Bases and Their Relation to Weak Acids

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BASE

STR

ENG

TH

Kb =[BH+][OH-]

[B]


29

+

CH3NH3+ OH-

methylammonium ion

+

CH3NH2 H2O

methylamine

Lone pair binds H+


30

SAMPLE PROBLEM 18.9: Determining pH from Kb and Initial [B]

PROBLEM: Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of 5.9x10-4. What is the pH of 1.5M (CH3)2NH?

SOLUTION:

PLAN: Perform this calculation as you did those 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)2NH2

+] - x ≈ [(CH3)2NH]initial

Kb >> Kw so [OH-]from water is neglible

Initial 1.50M 0 0-

Change - x - + x + x

Equilibrium 1.50 - x - x x

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


31

Kb = 5.9x10-4 = [(CH3)2NH2

+][OH-]

[(CH3)2NH]

5.9x10-4 = (x) (x)

1.5Mx = 3.0x10-2M = [OH-]

Check assumption: 3.0x10-2M/1.5M x 100 = 2%

[H3O+] = Kw/[OH-] = 1.0x10-14/3.0x10-2 = 3.3x10-13M

pH = -log 3.3x10-13 = 12.48


32

SAMPLE PROBLEM 18.10: Determining the pH of a Solution of A-

PROBLEM: Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is 1.8x10-5.

SOLUTION:

PLAN: Sodium salts are soluble in water so [Ac-] = 0.25M.

Write the association equation for acetic acid; use the Ka to find the Kb.

Initial 0.25M - 0 0

Change -x +x +x-

Equilibrium -0.25M-x x x

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

Kb = [HAc][OH-]

[Ac-]=

Kw

Ka

= 5.6x10-10MKb = 1.0x10-14

1.8x10-5


33

Kb = [HAc][OH-]

[Ac-]

[Ac-] = 0.25M-x ≈ 0.25M

5.6x10-10 = x2/0.25M

x = 1.2x10-5M = [OH-]

Check assumption: 1.2x10-5M/0.25M x 100 = 4.8x10-3 %

[H3O+] = Kw/[OH-] = 1.0x10-14/1.2x10-5 = 8.3x10-10M

pH = -log 8.3x10-10M = 9.08

18.6 – Molecular Properties and Acid Strength

34

6A(16)

H2O

H2S

H2Se

H2Te

7A(17)

HF

HCl

HBr

HI

Electronegativity increases, acidity increases

Bond

str

engt

h de

crea

ses,

ac

idity

incr

ease

s

Figure 18.12 The effect of atomic and molecular properties on nonmetal hydride acidity.


35

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


36

Table 18.7 Ka Values of Some Hydrated Metal Ions at 250C

Free Ion Hydrated Ion Ka

Fe3+ Fe(H2O)63+(aq) 6 x 10-3

Sn2+ Sn(H2O)62+(aq) 4 x 10-4

Cr3+ Cr(H2O)63+(aq) 1 x 10-4

Al3+ Al(H2O)63+(aq) 1 x 10-5

Cu2+ Cu(H2O)62+(aq) 3 x 10-8

Pb2+ Pb(H2O)62+(aq) 3 x 10-8

Zn2+ Zn(H2O)62+(aq) 1 x 10-9

Co2+ Co(H2O)62+(aq) 2 x 10-10

Ni2+ Ni(H2O)62+(aq) 1 x 10-10

ACID

STR

ENG

TH


37

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

Figure 18.13 The acidic behavior of the hydrated Al3+ ion.

H2O H3O+

Electron density drawn toward Al3+

Nearby H2O acts as base

18.7 – Acid-Base Properties of Salt Solutions

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39

SAMPLE PROBLEM 18.11: Predicting Relative Acidity of Salt Solutions

PROBLEM: Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction of any ion with water:

(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- , both of which come from a strong

base(KOH) and a strong acid(HClO4). Therefore the 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 it’s hydrated form will react with water to produce H3O+. Cl- comes from the strong acid HCl. Acidic solution.

(d) Na+ comes from a strong base. HSO4- can react with water to form H3O+.

So the salt solution will be acidic.


40

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 out the dissociation reactions and use the information in Tables 18.2 and 18.7.

Ka Zn(H2O)62+ = 1x10-9

Ka HCOO- = 1.8x10-4 ; Kb = Kw/Ka = 1.0x10-14/1.8x10-4 = 5.6x10-11

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

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

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

18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition

41

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


42

Figure 18.15

The Mg2+ ion as a Lewis acid in the chlorophyll molecule.


43

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- H2Oacceptor

donor

(b) Cl- + BCl3 BCl4-

donor

acceptor

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

acceptor

donor

Chapter 18 – Acid-Base Equilibria 18.1 – Acids and Bases in Water 18.2 – Autoionization of...

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Transcript of Chapter 18 – Acid-Base Equilibria 18.1 – Acids and Bases in Water 18.2 – Autoionization of...