Chapter 12Acids and Bases

12.1 The Nature of Acids and Bases 12.2 Acid Strength 12.3 The pH Scale 12.4 Calculating the pH of Strong Acid Solutions 12.5 Calculating the pH of Weak Acid Solutions 12.6 Bases 12.7 Polyprotic Acids 12.8 Acid-Base Properties of Salts 12.9 Acid Solutions in Which Water Contributes to the H+

Concentration (skip)12.10 Strong Acid Solutions in Which Water Contributes to

the H+ Concentration (skip)12.11 Strategy for Solving Acid-Base Problems: A Summary

Acids and Bases and Their Reactions

Definitions1. Arrhenius Acids and Bases Acids are

• H+ donors• Bases are OH- donors

2. Arrhenius Broadened Definition • Acids increase H+ concentration or [H+]

increases• Bases increase OH- concentration or [OH-]

increases

– Brønsted-Lowry Acids and Bases (1923)• Acids donate H+

• Bases accept H+

Arrhenius 1903

Nobel Prize

Acids-BasesBrønsted-Lowry Acids and Bases

A Brønsted-Lowry acid is a substance that can donate a hydrogen ion (aka H+, proton).

A Brønsted-Lowry base is a substance that can accept a hydrogen ion.

Acids and bases occur as conjugate acid - base pairs.

Conjugate Base - subtract an H+ from the acidConjugate Acid add H+ to the base

Examples

1.OH– is the conjugate base of

2.H2O is the conjugated base of

3.H2O is the conjugated acid of

4.H3O+ (or often shown as H+) is the conjugate acid of

Pairs 1 2 2 1

Acetic Acid

Point of View #1

acid base Conjugate acid of H2O

Conjugate base of CH3CO2H

CH3CO2H + H2O H3O+ + CH3CO2-

Point of View #2

acid baseConjugate base of H3O+

Conjugate acid

of CH3CO2-

CH3CO2H + H2O H3O+ + CH3CO2-

basebaseacid acid

Acetate Ion

CH3CO2H + H2O H3O+ + CH3CO2-

Nomenclature

When H+ is hydrated, it is H3O+ and is called a hydronium ion. A hydronium ion has the same molecular geometry as NH3.

111.7°

+

• There is Competition for the proton between two bases H2O and A–

• If H2O is a much stronger base than A– the equilibrium lies far to the right.

• If A– is a much stronger base than H2O the equilibrium lies far to the left.

HA + H2O H3O+ + A –

HA = generic acid

Acid Strength: graphical representation of the behavior of acids of different strengths in aqueous solution.

A strong acid: equilibrium lies far to the right

HA + H2O H3O+ + A –

A weak acid: equilibrium lies far to the left

HA + H2O H3O+ + A –

A weak acid yields a relatively strong conjugate base

Relationship of acid strength and conjugate base strength

HA + H2O H3O+ + A-

€

Ka =[H3O

+][A−]

[HA]

HA + H2O H3O+ + A–

HA H+ + A–

SAME AS

Ka is the Acidity constant

H2O + H2O H3O+ + OH-

Pairs 1 2 2 1basebaseacid acid

Autoionization of H2O

Point of View #2

acid baseConjugate base of H3O+

Conjugate acid

of OH-

H2O + H2O H3O+ + OH-

Point of View #1

acid base Conjugate acid of H2O #2

Conjugate base of H2O #1

H2O + H2O H3O+ + OH-# 1 # 2

Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions

Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions

H2O + NH3 NH4+ + OH-

1.) Water in NH3 serves as an acid

acid acid basebase

2.) Water in acetic acid serves as a base

H2O + CH3CO2H H3O+ + CH3CO2-

base acid acid base

Conjugate acid

of H2O

Conjugate base

of acetic acid

Water as an Acid and a Base

Kw = [H3O+][OH-] = [H+][OH-] [H3O+][OH-]

[H2O(l)]2

Autoionization of water:

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

Kw = 1.0 x 10-14 (at 25oC)

In pure water [H+] = [OH-]

Kw = [H3O+][OH-]

NH4+ H+ + NH3

€

Ka =[H+][NH3]

[NH4+]

NH3 + H2O OH- + NH4+

€

Kb =[OH−][NH4

+]

[NH3]

Ka/Kb/Kw

NH4+ H+ + NH3

€

Ka =[H+][NH3]

[NH4+]

NH3 + H2O OH- + NH4+

€

Kb =[OH−][NH4

+]

[NH3](1)

Ka/Kb/Kw

H2O OH- + H+

€

Kw = KaKb =[OH−][H+]

(1)

The pH ScalepH = -log10[H3O+]

pH < 7 acidic solution

[H3O+] > [OH-]

pH = 7 neutral solution

[H3O+] = [OH-]pH > 7 basic solution

[H3O+] < [OH-]

Sig figs: for logs: the number of decimal places in the log is equal to the number of sig figs in the original number

pH = -log10[H+]SAME AS

Calculate the pH of Strong Acid-Base Solutions

pH = -log10[H3O+]EXAMPLE

Calculate the pH (at 25oC) of an aqueous solution that has an

OH-(aq) concentration of 1.2 x 10-6 M (i.e., mol/liter)

Solution

The concentration of H+ (aq) is

[H+][OH-] = KW

[H+] = KW/[OH-] = 10-14/1.2x10-6 = 8.3 x 10-9

pH = -log[8.3 x 10-9] = 8.1

Strong Acids

A strong acid is one that dissociates completely in water to produce H+(aq).

E.g., Hydrochloric acid (HCl) is a strong acid:

HCl (aq) → H+ (aq) + Cl- (aq) (reaction essentially complete)

Dissolving 0.10 mol of HCl in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for H+(aq).

[H+][OH-] = Kw

[OH-] = Kw / [H+] = 10-14/10-1 = 10-13

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

Weak Acids

Write the major species in solution and Ka values. Which is dominant?

#1: HA H+ + A –

#2: H2O H+ + OH –

€

Kw =[H+ ][OH −]

Compare the value for Ka and Kw. Which is

larger? This is the dominate source of H+

€

Ka =[H+ ][A −][ ]HA

Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. The concentration of CH3CO2H is 1 M (initial, or total). The Ka for acetic acid is 1.8 x 10-5

Estimate major species in solution CH3CO2H (a weak acid) and H2O.

CH3CO2H H+ + CH3CO2─ Ka = 1.8 x 10-5

H2O H+ + OH – Kw = 1.0 x 10-14

Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.

CH3CO2H H+ + CH3CO2─

Initial 1.0M ~ 0 0

Change - y + y + y

Equilibrium 1.0 – y y y

Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.

CH3CO2H H+ + CH3CO2─

Initial 1.0M ~ 0 0

Change - y + y + y

Equilibrium 1.0 – y y y

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Ka =( )( )y y

(1.0 - )yassume y << 1.0

Ka =(y)(y)

(1.0)

Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.

CH3CO2H H+ + CH3CO2─

Initial 1.0M ~ 0 0

Change - y + y + y

Equilibrium 1.0 – y y y

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Ka =( )( )y y

(1.0 - )yassume y << 1.0

Ka =(y)(y)

(1.0)

€

Ka =1.8x10−5 =(y)(y)

(1.0)

y = 1.8x10−5

y = 4.3x10−3

check assumption : is y << 1.0?

Effect of dilution on the % dissociation and [H+] and pH

HA H+ + A –

The more dilute the weak acid solution, the greater the percent dissociation

Le Chatelier’s Priniciple:

If a chemical system at equilibrium experiences a change in concentration (or temperature, volume, or partial pressure) then the system shifts to counteract the imposed change.

Strong Acids and Bases

Strong BasesA strong base reacts completely with water to produce OH-(aq) ions.

Sodium hydroxide (NaOH) is a strong base:

Others are KOH–, NH2– (amide ion) and H– (Hydride ion)

NaOH (s) → Na+ (aq) + OH- (aq) (reaction essentially complete)

Dissolving 0.10 mol of NaOH in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for OH- (aq). From this you can calculate pH and pOH.

[H+][OH-] = Kw

[OH-] = 10-1 [H+] = 10-13 pOH = 1 pH = 13

[H+] = Kw/[OH-] =10-14/10-1 = 10-13

Weak Bases

B (aq) + H2O (l) BH+ (aq) + OH– (aq)B = Base

Kb is the basicity constant

Kb = [BH+][OH–]/[B]

Calculations for solutions of weak bases are similar to those for weak acids. Bases (B) compete with OH–, a very strong base, for H+ ions.

B (aq) + H2O (l) BH+ (aq) + OH– (aq)

Acid-Base Equilibria

Base Strength– strong acids have weak

conjugate bases– weak acids have strong

conjugate bases

pKa + pKb = pKwThis equation applies to an acid and its conjugate base.

The strength of a base is inversely related to the strength of its conjugate acid; the weaker the acid, the stronger its conjugate base, and vice versa

Weak Bases

NH3 is a base Kb = 1.8 x 10-5

NH3 + H20 NH4+ + OH-

acid1 base2 acid2 base1

NH4+

is an acid Ka = 5.6 x 10-10

NH4+ NH3 + H+

acid1 base1

Weak Bases

NH3 is a base Kb = 1.8 x 10-5

NH3 + H20 NH4+ + OH-

NH4+

is an acid Ka = 5.6 x 10-10

NH4+ NH3 + H+

Ka Kb = Kw = (5.6 x 10-10) (1.8 x 10-5) = 10-14

pKa + pKb = pKw = (-9.25) + (-4.75) = -14

Weak Bases with Weak Acids

NH4+

is an acid Ka(am) = 5.6 x 10-10

change the direction of the aa reaction and add them together

NH4+ + CH3COO- NH3 + CH3COOH

K = Ka(am)/Ka(aa) = 5.6 x 10-10/1.8 x 10-5

=3.1 x 10-5

CH3COOH is an acid Ka(aa) = 1.8 x 10-5

CH3COOH CH3COO- + H+

NH4+ NH3 + H+

Assuming 0.1M NH3 initial, calculate the pH of the resulting solution

H20 + NH3 NH4+ + OH-

Init. conc. 0.1M 0 ~0

Change - y + y + y

Equil. conc. 0.1– y y y

Assuming 0.1M NH3 initial, calculate the pH of the resulting solution

H20 + NH3 NH4+ + OH-

Init. conc. 0.1M 0 ~0

Change - y + y + y

Equil. conc. 0.1– y y y

Assuming 0.1M NH3 initial, calculate the pH of the resulting solution

H20 + NH3 NH4+ + OH-

- 24

b 3

5b 3

25

b

- -34

[NH ][OH ] yK

[NH ][1] 0.1 y

K for NH 1.8x10

assume y is small

yK 1.8x10

0.1

y [OH ] [NH ] 1.3x10

+

−

−

+

= =−

=

= =≈

= = =

[H+] = Kw/[OH-] =10-14/y

pH=-log[H+]

Polyprotic Acids

Note: Sulfuric acid is a strong acid in its first dissociation step and a weak acid in its second step.

H2SO4 (aq) → H+ (aq) + HSO4– (aq) Ka1 ≈ 1.0x102

HSO4– (aq) H+ (aq) + SO4

2– (aq) Ka2 ≈ 1.2x10-2

It is always easier to remove the first proton in a     polyprotic acid than the second. That is, Ka1 > Ka2 > Ka3

Polyprotic Acids1. Polyprotic acids have more than one ionizable proton.2. The protons are removed in steps, not all at once.3. The extent of the steps can be calculated sequentially.4. From the successive acidity constants, the equilibrium concentrations of all

species present can be calculated at any value of the pH.

     H2SO3 (aq) + H2O (l) → H3O+ (aq) + HSO3- (aq)

Ka1 = 103

     HSO3- (aq) + H2O (l) H3O+ (aq) + SO3

2- (aq)                                                                       

Ka2 = 1.2 x 10-2

Polyprotic Acids

• pH calculations for solutions of Polyprotic acids appear complicated, most common cases (with weak acids) are surprising straightforward.

• Setup equations, K values, set up ICE type table, assume dissociation is small (check assumption) and solve.

• For typical weak Polyprotic acids.

Ka1 >> Ka2 >> Ka3

• The first dissociation step dominates the H+ concentration and thus the pH

A plot of the fractions of H2CO3, HCO-3 and CO3

2-

At pH = 9.00 H2CO3 ≈ 0%, HCO-3 = 95% and CO3

2- = 5%

At pH = 10.00 H2CO3 ≈ 0%, HCO-3 = 68% and CO3

2- = 32%

Acid-Base Properties of Salts

“Hydrolysis”

Hydrolysis is a Brønsted-Lowry Acid and Base Reaction

Anion: A─ + H2O ↔ HA + OH─

Cation: M+2 + H2O ↔ H3O+ + M(OH)+

E.g., Cations Ni+2 and Fe+2

• Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7

• When NaCl is placed in water, the resulting solution is observed to be neutral (pH = 7)

• However when sodium acetate (NaC2H3O2) is dissolved in water the resulting solution is basic

• Other salts behave similarly, NH4Cl and AlCl3 give acid solutions.• These interactions between salts and water are called hydrolysis

Example problem:

Suppose a 0.1 mole solution sodium acetate is dissolved in 1 liter of water. What is the pH of the solution?

Init. conc. 0.1M 0 ~0

∆ conc. - y + y + y

Equil. conc. 0.1– y y y

1. Find Kb

2. Find [OH-]

3. Find [H+]

4. Find pH

CH3CO2– + H2O ↔ CH3CO2H + OH–

base baseacid acid

Example problem:

What is the pH of the solution?

CH3CO2– + H2O ↔ CH3CO2H + OH–

Init. conc. 0.1M 0 ~0

∆ conc. - y + y + y

Equil. conc. 0.1– y y y

1. Find Kb

2. Find [OH-]

3. Find [H+]

4. Find pH

Ka x Kb = Kw

• Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7

• When NaCl is placed in water, the resulting solution is observed to be neutral (pH = 7)

• However when sodium acetate (NaC2H3O2) is dissolved in water the resulting solution is basic (Problem: found at 0.1M NaAc, pH =8.89)

• a non-hydrolyzed cation (Na+)• a hydrolyzed anion (acetate ion)

• Other salts behave similarly, NH4Cl and AlCl3 give acid solutions.

• a non-hydrolyzed anion (Cl-)

• a hydrolyzed cation (NH4+ or Al+3)

• These interactions between salts and water are called hydrolysis

Hydrolysis of Result

Anions

Cations

Raise pH

Lower pH

Non-Hyrolyzed Ions (a few)

7 Anions, not hydrolyzed

Cl –, Br –, I –, HSO4–, NO3

–, ClO3–, ClO4

–

10 Cations, not hydrolyzed

Li+, Na+, K+, Rb+, Sc+, Mg++, Ca++, Sr++, Ba++, Ag+

Hydrolysis of Result

Anions

Cations

Raise pH

Lower pH

Predict pH of salts in water (relative pH)

Na3PO4 is basic (raised pH)

(a non hydrolyzed cation and a hydrolyzed anion)

FeCl3 is acidic (lowers pH)

(a hydrolyzed cation and a non hydrolyzed anion)

Na3PO4?

FeCl3?

Summary

Acid-Base Properties of Salts

“Hydrolysis”

Summary

• Acid-Base Equilibrium Problems– Which major species are present– Does a reaction occur that can be assumed to go to

completion?– Which equilibrium dominates the solution?– Set up ICE type table– Solve for equilibrium concentrations using know K

values– Check any simplifying assumptions– Typically solve for pH or % species in solution

Chapter 7Acids and Bases

7.1 The Nature of Acids and Bases 7.2 Acid Strength 7.3 The pH Scale 7.4 Calculating the pH of Strong Acid Solutions 7.5 Calculating the pH of Weak Acid Solutions 7.6 Bases 7.7 Polyprotic Acids 7.8 Acid-Base Properties of Salts 7.9 Acid Solutions in Which Water Contributes to the H+

Concentration (skip)7.10 Strong Acid Solutions in Which Water Contributes to the

H+ Concentration (skip)7.11 Strategy for Solving Acid-Base Problems: A Summary