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Chapter 18 Acids and Bases. 1. Acids and Bases in Water. 2. Autoionization of Water and the pH Scale. 3. Proton Transfer and the Brønsted-Lowry Acid-Base Definition. 4. Solving Problems Involving Weak-Acid Equilibria. 5. Weak Bases and Their Relations to Weak Acids.

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slide1

Chapter 18 Acids and Bases

1. Acids and Bases in Water

2. Autoionization of Water and the pH Scale

3. Proton Transfer and the Brønsted-Lowry Acid-Base Definition

4. Solving Problems Involving Weak-Acid Equilibria

5. Weak Bases and Their Relations to Weak Acids

6. Molecular Properties and Acid Strength

7. Acid-Base Properties of Salt Solutions

8. Electron-Pair Donation and the Lewis Acid-Base Definition

slide2

Table Selected Acids and Bases

Acids

Bases

Strong

Strong

hydrochloric acid, HCl

sodium hydroxide, NaOH

hydrobromic acid, HBr

potassium hydroxide, KOH

hydroiodic acid, HI

calcium hydroxide, Ca(OH)2

nitric acid, HNO3

strontium hydroxide, Sr(OH)2

sulfuric acid, H2SO4

barium hydroxide, Ba(OH)2

perchloric acid, HClO4

Weak

Weak

hydrofluoric acid, HF

ammonia, NH3

amine, (CH3)2CHNH2

phosphoric acid, H3PO4

acetic acid, CH3COOH (or HC2H3O2)

slide3

Brønsted-Lowry Acid-Base Definition

Molecules as Lewis Acids and base

An acid is an electron-pair acceptor.

An acid is a proton donor, any species which donates a H+.

A base is an electron-pair donor.

A base is a proton acceptor, any species which accepts a H+.

An acid-base reaction can now be viewed from the standpoint of the reactants AND the products.

Acid and base reaction is a proton transfer process.

acid

base

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

adduct

slide4

Lone pair binds H+

+

+

HCl

H2O

Cl-

H3O+

Lone pair binds H+

+

+

NH4+

NH3

H2O

OH-

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

(acid, H+ donor)

(base, H+ acceptor)

(base, H+ acceptor)

(acid, H+ donor)

slide5

Acid

+

Base

Base

+

Acid

Reaction 1

HF

+

H2O

F-

+

H3O+

Reaction 2

HCOOH

+

CN-

Reaction 3

NH4+

+

CO32-

NH3

+

HCO3-

Reaction 4

H2PO4-

+

OH-

HPO42-

+

H2O

Reaction 5

H2SO4

+

N2H5+

HSO4-

+

N2H62+

Reaction 6

HPO42-

+

SO32-

PO43-

+

HSO3-

The Conjugate Pairs in Some Acid-Base Reactions

Conjugate Pair

Conjugate Pair

HCOO-

+

HCN

slide6

PROBLEM:

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

(a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)

(a)H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)

PLAN:

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

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

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

Identifying Conjugate Acid-Base Pairs

conjugate pair2

conjugate pair1

SOLUTION:

proton donor

proton acceptor

proton acceptor

proton donor

conjugate pair2

conjugate pair1

proton donor

proton acceptor

proton acceptor

proton donor

slide7

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):

(a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)

PLAN:

Identify the conjugate acid-base pairs to determine the relative strength of each. The stronger the species, the more preponderant its conjugate.

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

stronger acid

stronger base

weaker base

weaker acid

weaker acid

weaker base

stronger base

stronger acid

Predicting the Net Direction of an Acid-Base Reaction

SOLUTION:

(a)H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)

Net direction is to the right with Kc > 1.

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

Net direction is to the left with Kc < 1.

slide8

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

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

The extent of dissociation for strong acids.

Strong acids dissociate completely into ions in water.

Kc >> 1

The extent of dissociation for weak acids.

Weak acids dissociate slightly (partially) into ions in water.

Kc << 1

Weak aicd has a Ka value ranging from 10-1 to 10-12

slide9

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

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

[H3O+][A-]

Kc =

[H2O][HA]

[H3O+][A-]

Kc[H2O] = Ka =

[HA]

smaller Ka lower [H3O+]

weaker acid

Strong acids dissociate completely into ions in water.

Kc >> 1

Weak acids dissociate very slightly into ions in water.

Kc << 1

The Acid-Dissociation Constant

stronger acid higher [H3O+]

larger Ka

slide10

PROBLEM:

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

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.

Classifying Acid and Base Strength from the Chemical Formula

(a) H2SeO4

(b) (CH3)2CHCOOH

(c) KOH

(d) (CH3)2CHNH2

SOLUTION:

slide11

Autoionization of Water and the pH Scale

+

H2O(l)

H2O(l)

  • Ball-and-stick model:
  • The two polar O-H bonds.
  • The bent molecule shape give rise to the polar nature of H2O.

+

OH-(aq)

H3O+(aq)

slide12

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

[H3O+][OH-]

Kc =

[H2O]2

The Ion-Product Constant for Water

Kc[H2O]2 =

Kw =

[H3O+][OH-]

= 1.0 x 10-14 at 25 C

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

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

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

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

slide13

Divide into Kw

[H3O+] > [OH-]

[H3O+] = [OH-]

[H3O+] < [OH-]

The relationship between [H3O+] and [OH-] and the

relative acidity of solutions.

[H3O+]

[OH-]

ACIDIC SOLUTION

BASIC SOLUTION

NEUTRAL SOLUTION

slide14

Calculating [H3O+] and [OH-] in an Aqueous Solution

PROBLEM:

A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.010-4 M. Calculate [OH-]. Is the solution neutral, acidic, or basic?

PLAN:

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

SOLUTION:

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

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

3.3x10-11M

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

slide15

The pH values of some aqueous solutions

pH = - log [H3O+]

pOH = - log [OH-]

pH + pOH = 14

summary

Summary

Acid and base are essential substances in home, industry and environment.

In aqueous solution, water combines with the proton released from acid to form hydronium ions, H3O+.

Acid contains H and yields H3O+, while base contain OH and yield OH- in aqueous solution.

Strong acid dissociate completely and weak acid dissociate partially.

The extend of dissociation is expressed by the acid-dissociation constant, Ka.

Weak aicd has a Ka value ranging from 10-1 to 10-12.

Water undergoes the self-ionization.

Autoionization (Autoprotolysis) process is described by an equilibrium

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

The equilibrium constant is the ion-product for water, Kw=1.0  10-14

We use pH scale to express the values of H3O+ .

pH<7, acidic; pH=7, neutral; pH>7, basic.

slide18

PLAN:

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

Take-home message: Calculating [H3O+], pH, [OH-], and pOH

PROBLEM:

In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0 M, 0.30 M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 250C.

SOLUTION:

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

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

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

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

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

slide20

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.

SOLUTION:

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

[H3O+][PAc-]

Ka =

[HPAc]

Finding the Ka of a Weak Acid from the pH of

Its Solution

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

Assumptions:

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

slide21

Initial

0.12

-

1x10-7

0

Change

-x

-

+x

+x

Equilibrium

0.12-x

-

x +(<1x10-7)

x

(2.4x10-3) (2.4x10-3)

0.12

1x10-7M

[H3O+]from water;

X100%

2.4x10-3M

2.4x10-3M

[HPAc]dissn;

X100%

0.12M

Finding the Ka of a Weak Acid from the pH of

Its Solution

continued

Concentration(M)

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

[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 =

= 4.8 x 10-5

Be sure to check for % error.

= 4x10-3 %

= 2.0%

slide22

PLAN:

Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; substitute.

Assumptions:

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

Ka =

[H3O+][Pr-]

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

[HPr]

Initial

0.10

-

0

0

Change

-x

-

+x

+x

Equilibrium

0.10-x

-

x

x

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)?

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

SOLUTION:

Concentration(M)

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

slide23

[H3O+][Pr-]

(x)(x)

1.3x10-5 =

=

[HPr]

0.10

Determining Concentrations from Ka and

Initial [HA]

continued

= 1.1x10-3 M = [H3O+]

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

slide24

[H3O+][H2PO4-]

[H3O+][PO43-]

[H3O+][HPO42-]

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

Ka2 =

Ka1 =

Ka3 =

[H2PO4-]

[H3PO4]

[HPO42-]

H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq)

HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq)

Polyprotic acids

acids with more than more ionizable proton

= 7.2x10-3

= 6.3x10-8

= 4.2x10-13

Ka1 > Ka2 > Ka3

slide25

PLAN:

Write out expressions for both dissociations and make assumptions.

[HAsc-][H3O+]

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

Ka1 =

[H2Asc]

[Asc2-][H3O+]

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

Ka2 =

[HAsc-]

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.

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.

SOLUTION:

= 1.0x10-5

= 5x10-12

slide26

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

Initial

0.050

-

0

0

Change

- x

-

+ x

+ x

Equilibrium

0.050 - x

-

x

x

x

Concentration(M)

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

Initial

7.1x10-4M

-

0

0

Change

- x

-

+ x

+ x

Equilibrium

7.1x10-4 - x

-

x

x

x

Calculating Equilibrium Concentrations for a

Polyprotic Acid

continued

Concentration (M)

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

x = 7.1x10-4 M

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

= 6x10-8 M

summary for last lecture

Summary for last lecture

Acid and base are essential substances in home, industry and environment.

In aqueous solution, water combines with the proton released from acid to form hydronium ions, H3O+.

Acid contains H and yields H3O+, while base contain OH and yield OH- in aqueous solution. (Arrhenius)

Strong acid dissociate completely and weak acid dissociate partially.

The extend of dissociation is expressed by the acid-dissociation constant, Ka.

Weak acid has a Ka value ranging from 10-1 to 10-12.

Water undergoes the self-ionization.

Autoionization (Autoprotolysis) process is described by an equilibrium

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

The equilibrium constant is the ion-product for water, Kw=1.0  10-14

We use pH scale to express the values of H3O+ .

pH<7, acidic; pH=7, neutral; pH>7, basic.

slide28

Summary

  • The Bronsted-Lowry acid and base definition : acid does not require to contain H or base contains OH.
  • Acid base reaction does not require to occur in aqueous solution.
  • An acid is a proton donor, any species which donates a H+.
  • A base is a proton acceptor, any species which accepts a H+.
  • Acid donates its proton and becomes its conjugate base.
  • Base accepts proton and becomes its conjugate acid.
  • A stronger acid gives a weaker conjugate base, or vise versa.
  • The reaction proceeds to the net reaction in which a stronger acid and base to form a weaker acid and base.
  • The common type of weak acid equilibrium problems involve finding Ka from known concentration or finding concentration from known Ka.
  • We can make assumption to simplify the calculation by neglecting x.
  • Monoprotic acid has one ionizable proton.
  • Polyprotic acid contains more than one ionizable proton.
slide29

Bronsted-Lowry base: any species accept a proton.

Abstraction of a proton from water by methylamine.

Lone pair binds H+

B + H2O BH+ + OH-

+

[HB+][OH-]

Kb =

[B][H2O]

H2O

CH3NH2

methylamine

Base dissociation constant

Base hydrolysis constant

+

CH3NH3+

OH-

methylammonium ion

slide30

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)

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

Concentration

Change

- x

-

+ x

+ x

Determining pH from Kb and Initial [B]

PROBLEM:

Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of 5.910-4. What is the pH of 1.5M (CH3)2NH?

Assumptions:

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

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

SOLUTION:

10-7 ≈ 0

Initial

1.50M

-

0

Equilibrium

1.50 - x ≈ 1.5

-

x

x

slide31

[(CH3)2NH2+][OH-]

Kb = 5.910-4 =

[(CH3)2NH]

(x) (x)

5.9x10-4 =

1.5M

continued

Determining pH from Kb and Initial [B]

x = 3.010-2M = [OH-]

Check assumption:

3.010-2M/1.5M  100 % = 2%

pH calculation:

Method 01: pH calculation:

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

pH = -log 3.3 10-13 = 12.48

Method 02: pH calculation:

pOH = -log 3.010-2= 1.52

pH = 14- pH = 14 - 1.52 = 12.48

slide32

The relation between Ka and kb of a conjugate acid-base pair

There exists an important relationship between the Ka of HA and the Kb of A-.

There is the reaction of dissociation of acid and hydrolysis of base.

HA + H2O  H3O+ + A- (1) Ka

A- + H2O  HA + OH- (2) Kb

The sum of the reaction

H2O + H2O  OH- + H+ Kw=?

Kw = KaKb =  = [H3O+][OH-]

[A-][H3O+]

[HA][OH-]

[A-][H2O]

[HA][H2O]

K w = Ka Kb

slide33

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.

PLAN:

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

Concentration

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

Initial

0.25M

-

0

0

Change

-x

-

+x

+x

Equilibrium

0.25M-x

-

x

x

[HAc][OH-]

K w

1.0x10-14

Kb =

=

[Ac-]

K a

1.8x10-5

Determining the pH of a Solution of A-

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

SOLUTION:

K b =

= 5.610-10 M

slide34

Kb =

Check assumption:

1.2 10-5M/0.25M  100% = 4.810-3 %

[H3O+] = Kw/[OH-]

= 1.0 10-14/1.2 10-5

= 8.3 10-10M

[HAc][OH-]

[Ac-]

Determining the pH of a Solution of A-

continued

[Ac-] = 0.25M-x ≈ 0.25M

5.610-10 = x2/0.25M

x= 1.210-5M = [OH-]

pH = -log 8.3 10-10 = 9.08

pH is unitless.

slide36

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:

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.

Predicting Relative Acidity of Salt Solutions

(a) Potassium perchlorate, KClO4

(b) Sodium benzoate, C6H5COONa

(c) Chromium trichloride, CrCl3

(d) Sodium hydrogen sulfate, NaHSO4

SOLUTION:

(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.

slide37

PROBLEM:

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

PLAN:

Both Zn2+ and HCOO- come from weak conjugates. In order to find the relatively acidity, write out the dissociation reactions.

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

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

Predicting the Relative Acidity of Salt

Solutions from Ka and Kb of the Ions

SOLUTION:

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.

slide38

Summary

  • We can construct the reaction table to calculate pH, concentration of a chemical species, or Ka/Kb.
  • We can make assumption to simplify the calculation by neglecting x.
  • Monoprotic acid has one ionizable proton.
  • Polyprotic acid contains more than one ionizable protons.
  • The first dissociation of polyprotic acid provides virtually all the H3O+.
  • The extent to which a weak base abstracts a proton from H2O to form OH- is expressed by a base hydrolysis constant Kb.
  • By multiplying the expression Ka of HA and Kb of A-, we obtain Kw.
  • The relationship between Ka and Kb allows us to calculate Ka and Kb or pH and pOH.