Acids and Bases

• Acids and bases

• Acid-base properties of water (Kw)

• pH scale

• Strength of Acids and Bases

• Weak acid (Ka)

• Weak base (Kb)

• Relation between conjugate acid-base ionization constants (Ka and Kb)

• Chemical structure and acidity

• Acid-base properties of salt solutions

• Lewis acid-base

I. Acid and BaseA. Arrehenius acid - a compound that increases [H+] in water

B. Arrehenius Base : a compound that increases [OH-] in water

HCl(aq) +H2O(l) H3O+(aq) + Cl-(aq)

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

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

I. Acid and BaseC. Brønsted acid : a proton donor

D. Brønsted Base : a proton acceptor

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

acidbase acid base

E. Conjugate acid-base pair : acid-base pair that are related by a loss or gain of a proton H+

acid conjugate basebase conjugate acid

Example: Conjugate acid-base pair

• Classify each of the following species a Brønsted acid, base, or both. Give the conjugate acid or base.

CO32– base

species Acid/base/both Conjugate acid Conjugate base

HCO3–

H2O both OH–H3O+

H2SO4 acid HSO4–

• Amphoteric: a substance that can act either as an acid or a base

H2PO4– both HPO4

2–H3PO4

II. Acid-base properties of water

A. Autoionization of water: water is a very weak electrolyte

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

O

H

H + O

H

H O

H

H H OH-+[ ] +

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

acidconjugate base

base conjugate acid

B. Ion-product constant, Kw

H2O (l) H+ (aq) + OH- (aq) KC = Kw = [H+][OH-]

(1) Kw: the product of the molar concentrations of H+ and OH- ions at a particular temperature.

Kw = [H+][OH-] = 1.0 x 10-14 at 25oC

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Neutral solution

Acidic solution

Basic solution

(2) H+ and OH– ions are always present in aqueous solutions

Q: What is the concentration of H+ in a (a) neutral solution and (b) 2.0 x10-2 M NaOH solution at 25oC?

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

(b) [OH–] = 2.0 x10-2 M

[H+] =Kw

[OH–]

1.0 x 10-14

2.0 x10-2= = 5.0 x 10-13 M (basic)

(a) Neutral solution [H+] = [OH–]= x

= 1.0 x 10-7 M

Kw = [H+][OH-] = 1.0 x 10-14 = x2

€

[H+ ] = 1.0 ×10−14 M

III. pH scale – a measure of acidity

pH = -log [H+]

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution is

neutral

acidic

basic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = pOH = 7

pH < 7 pOH >7

pH > 7 pOH <7

At 250C

[H+] pH

pOH = -log [OH–]

As the solution is more acidic when

At 25oC [H+][OH-] = Kw = 1.0 x 10-14

-log [H+] – log [OH-] = 14.00 pH + pOH = 14.00

pOH

16.1

Q: If a solution has a pOH value of 10.5, what is the H+ ion concentration?

pH = 3.5 = -log [H+]

[H+] = 10-pH = 10-3.5 = 3.2 x 10-4 M

pH = 14.0 - pOH = 14.0-10.5 = 3.5

Q: The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

(acidic)

(basic)

IV. Strength of Acid and Base

A. Strong Electrolyte – 100% dissociation (ionization)- soluble ionic compound, strong acid, strong base

NaCl (s) Na+ (aq) + Cl- (aq)H2O

B. Weak Electrolyte – not completely

dissociated- weak acid, weak base

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

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

NaOH (s) Na+ (aq) + OH- (aq)H2O

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

HCl, strong acid HF, weak acid

C. Strong acid• Strong electrolyte, completely ionize (dissociate) in

water

• HCl, HBr, HI, HNO3, H2SO4, HClO4 HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

Q: What is the pH of the solution when 8.4 g of HCl is dissolved in 1.2 L of water? HCl is a strong acid – 100% dissociation.

H2O (l) + HCl (aq) H3O+ (aq) + Cl- (aq)

pH = -log [H+] = -log [H3O+] = -log(0.19) = 0.72

Initial

Equilibrium

0.19 M

0.19 M 0.19 M0.00 M

0.00 M 0.00 M

€

MHCl = 8.4 g HCl × 1 mol HCl36.46 g HCl × 1

1.2 L = 0.19 M

Change -0.19 M 0.19 M 0.19 M

D. Strong Base• Hydroxide of alkali (IA) and alkaline earth (IIA) metal

• Ionic metal oxide (Na2O, CaO), metal hydride, and nitrides

Na2O (s) + H2O (l) 2Na+ (aq) + 2OH- (aq)

Ex: What is the pH of a 0.048 M Ca(OH)2 solution?

Ca(OH)2 is a strong base – 100% dissociation.

Ca(OH)2 (aq) Ca2+ (aq) + 2OH- (aq)

pOH = -log [OH-] = -log(0.096) = 1.02

Initial

Equilibrium

0.048 M

0.048 M 0.096 M0.000 M

0.000 M 0.000 M

Change -0.048 M 0.048 M 2 x 0.048 M

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O

pH = 14.00 - pOH = 12.98

E. Conjugate acid-base pair

• The conjugate base of a strong acid is extremely weak

• The conjugate acid of a strong base is extremely weak.

• The stronger the acid, the weaker the conjugate base

• The stronger the base, the weaker the conjugate acid

Ex: Predict the direction of the following reaction

HCN (aq) + F– (aq) HF (aq) + CN– (aq)

Which is a stronger acid? HCN or HF

Equilibrium will proceed from right to left since HF is a stronger acid and CN– is a stronger base.

Which is a stronger base? CN– or F–

KC < 1

* Reaction goes to the side that produces gas, solid, or weak/nonelectrolyte

16.2

V. Weak Acid

A. Ka: acid ionization constant

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

HA (aq) H+ (aq) + A- (aq)

Ka =[H+][A-]

[HA]

Ka

acidstrength

Ex: What is the pH of a 0.80 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-]

[HF]= 7.1 x 10-4

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

Initial (M)Change (M)

Equilibrium (M)

0.80 0.00-x +x

0.80 - x

0.00+x

x x

Ka =x2

0.80 - x= 7.1 x 10-4

Ka x2

0.80= 7.1 x 10-4

0.80 – x 0.80Ka << 1

x2 = 5.68 x 10-4 x = 0.024 M

[H+] = [F-] = 0.024 MpH = -log [H+] = 1.62

[HF] = 0.80 – x = 0.78 M

When will this approximation work?

0.80 – x 0.80Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.0240.024 M0.80 M

x 100% = 3.0%Less than 5%

Approximation ok.

Ex. What is the pH of a 0.010 M HF solution (at 250C)?

Ka x2

0.010= 7.1 x 10-4 x = 0.0027 M

0.0027 M0.010 M

x 100% = 27%More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation or other method.

Ka =x2

0.010 - x= 7.1 x 10-4 x2 + 7.1x10-4 x -7.1x10-6 = 0

[H+] = x = 0.0023 M pH = -log [H+] = 2.64

Steps to solve weak acid ionization problems:

1. Identify the major species that can affect the pH.

• In most cases, you can ignore the autoionization of water.

• Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of single unknown x.

3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.

4. Calculate concentrations of all species and/or pH of the solution.

V. Weak Acid

B. Percent ionization

% ionization = ionized acid concentration at equilibrium

initial concentration of acidx 100%

For a monoprotic acid HA

% ionization = [H+]

[HA]0

x 100%

[HA]0 = initial concentration

*Percent ionization asinitial concentration increases

decreases

V. Weak Acid

C. Polyprotic acid

Acids that have more than one ionizable H

H2S (aq) H+ (aq) + HS- (aq) Ka1= 9.5 x10-8

HS– (aq) H+ (aq) + S2- (aq) Ka2= 1 x10-19

H3PO4(aq) H+ (aq) + H2PO4- (aq) Ka1= 7.5 x10-3

H2PO4–(aq) H+ (aq) + HPO4

2– (aq) Ka2= 6.2 x10-8

HPO42–(aq) H+ (aq) + PO4

3– (aq) Ka3= 4.8 x10-13

Ka3 Ka2 Ka1<<

diprotic

triprotic

Ex: What is the pH of a 0.024 M H2SO3 solution at 25oC?

(Ka1= 1.3 x 10-2 and Ka2=6.3 x 10-8)

Initial (M)

Change (M)

Equilibrium (M)

0.024 0.00

-x +x

0.024 - x

0.00

+x

x x

Ka1 =x2

0.024 - x= 1.3 x 10-2

x2 + 1.3 x 10-2 - 3.12 x 10-4 = 0

[H+] = [HSO3-] = 0.0123 M

[H2SO3] = 0.024 – x = 0.0117 M

H2SO3(aq) H+ (aq) + HSO3– (aq)

x = 0.0123 or -0.025

Not done yet!

Next step, let us consider the second ionization of H

Treat this type problem with step-by-step ionization

Initial (M)

Change (M)

Equilibrium (M)

0.0123 0.0123

-y +y

0.0123 - y

0.00

+y

0.0123+y y

Ka2 =(0.0123+y)y0.0123 - y

= 6.3 x 10-8

[H+] = 0.0123 + y = 0.012 M

[HSO3–] = 0.0123 – y = 0.012 M

HSO3–(aq) H+ (aq) + SO3

– (aq)

y= 6.3 x10-8

0.0123 – y 0.0123Ka2 << 1 0.0123 + y 0.0123

[SO32–] = y = 6.3 x10-8 M

[H2SO3] = 0.012 M

pH = -log[H+] =1.92

6.3 x10-8 M0.0123 M

x 100% << 5 %

VI. Weak Base

A. Kb: base ionization constant

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

Kb =[HB+][OH-]

[B]

Kb

basestrength

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

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

Ex: What is the pH of a 0.50 M C2H5NH2 solution at 250C(Kb=5.6 x10-4)?

C2H5NH2(aq) +H2O(l) C2H5NH3+ (aq) + OH- (aq)

Initial (M)

Change(M)

Equilibrium(M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Kb =x2

0.50 - x= 5.6 x 10-4

Kb x2

0.50= 5.6 x 10-4

0.50 – x 0.50Kb << 1

x2 = 2.8 x 10-4 x = 0.017 M

[OH–] = [C2H5NH3+] = 0.017 M

pH = 14.00 - pOH = 14.00 + log [OH–] = 12.23

[C2H5NH2] = 0.50 – x = 0.48 M

0.017 M0.50 M

x 100% = 3.4 % assumption okay

16.5

VII. Ionization Constants of Conjugate Acid-Base Pairs

HA (aq) H+ (aq) + A- (aq)

C6H5COO– (aq) + H2O (l) OH- (aq) + C6H5COOH (aq)

Ka

Kb

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

KaKb = Kw

Weak acid (base) and its conjugate base (acid)

Ka = Kw

Kb

Kb = Kw

Ka

Ex. What is the reaction and the equilibrium constant of C6H5COO– in water?

C6H5COO– is a base and is the conjugate base of C6H5COOH

Kb = = Kw

Ka

= 1.5 x 10-101.0 x 10-14

6.5 x 10-5

A– (aq) + H2O (l) OH- (aq) + HA (aq)

VIII. Molecular Structure and Acid Strength

A. Factors affecting acidity- bond strength : weaker bond, stronger acid

- bond polarity: more polar bond, stronger acid

B. Binary acid : bond strength is more important

- hydrohalic acid HF HCl HBr HI

Ex: H2O H2S H2Se< <

<<<<

C. Oxoacids

Z O H Z O- + H+- +

The O-H bond will be more polar and easier to break if:

• Z is very electronegative or

• Z is in a high oxidation state

C. Oxoacids• Oxoacids having different central atoms (Z) that are from the

same group and that have the same oxidation number (O atom)- Acid strength increases with increasing EN of Z

• Oxoacids having the same central atom (Z) but different numbers of attached groups.- Acid strength increases as the

oxidation no of Z increases

H O Cl O

O••

••••••

••••••

••••

H O Br O

O••

••••••

••

••••

••••

Cl is more electronegative than Br

> HBrO HIO>

HClO4 HClO3 HClO2 HClO> > >

D. Carboxylic acids

• Acids with functional group -COOH

Ex. Predict the strength of the following acids.

CF3COOH CH3COOH

H C C O H

O••••

••

••H

H

withdraw electron

can be ionized

>

IX. Salt Solutions• Salts are strong electrolytes, completely ionizes in water. The

acidity/basicity of salt solutions depends on the ions.

• Hydrolysis: reaction of ions with H2O to produce OH– or H+

• NO2– + H2O <=> HNO2 + OH–

• NH4+ + H2O <=> NH3 + H3O+

A. Neutral Solutions:

Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl–, Br–, I–, and NO3

–).NaBr (s) Na+ (aq) + Br- (aq)

H2O

KNO3 (s) K+ (aq) + NO3– (aq)

H2O

B. Basic Solutions:

Salts derived from a strong base and a weak acid.

CH3COONa (s) Na+ (aq) + CH3COO- (aq)H2O

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

C. Acidic Solutions:

a. Salts derived from a strong acid and a weak base.

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

H2O

NH4+ (aq) NH3 (aq) + H+ (aq)

b. Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid. For example: AlCl3

Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)3+ 2+

Al(H2O)6 (aq) +H2O (l) Al(OH)(H2O)5 (aq) + H3O+ (aq)3+ 2+

Acid hydrolysis of Al3+

Acidity of metal ion increases with smaller size and higher charge.

D. Solutions in which both the cation and the anion hydrolyze:

• Kb for the anion > Ka for the cation, solution will be basic

• Kb for the anion < Ka for the cation, solution will be acidic

• Kb for the anion Ka for the cation, solution will be neutralExample: NH4NO2

NO2– : Kb = = = 2.2 x 10-11Kw

Ka

1.0 x 10-14

4.5 x 10-4

NO2– and NH4

+

NH4+ : Ka = = = 5.6 x 10-10Kw

Kb

1.0 x 10-14

1.8 x 10-5

Ka > Kb

Weak acid

Ka,HNO2 =4.5 x10-4 Kb,NH3 =1.8 x10-5

Ex: What is the pH of a 0.10 M solution of a NaF?

F– (aq) +H2O (l) HF (aq) + OH- (aq)

Initial (M)Change (M)Equilibrium (M)

0.10 0.00-x +x

0.10 - x

0.00+x

x x

Kb x2

0.10= 1.4 x 10-110.10 – x 0.10Kb << 1

x2 = 1.4 x 10-12 [OH–] = x = 1.2 x10-6 M

pH = 14.00 -pOH = 14.00 + log [OH–] = 8.08

Kb =[HF][OH-]

[F–]

Acid or basic?

H2O (l) + F– (aq) HF (aq) + OH- (aq)

Kb = = = 1.4 x 10-11=Kw

Ka

1.0 x 10-14

7.1 x 10-4

x2

0.10 - x

• Basic oxide: group 1A and 2A oxides (except BeO). BaO + H2O -> Ba(OH)2

• Amphoteric oxide: BeO and group 3A and 4A oxides.

• Acid oxide: nonmetal oxide with high oxidation number, such as N2O5, SO2 etc.

N2O5+ H2O --> 2 HNO3

• Nonmetal oxide with low oxidation state such as CO or NO are neutral)

X. Acidic, basic, and amphoteric oxides

XI. Lewis acid and base

A. Arrehenius acid/base: substance that increases [H+]/[OH–] in water

B. B. Brønsted acid/base : proton donor/acceptorC. A Lewis acid is a substance that can accept a pair of electrons (electron

pair acceptor)D. A Lewis base is a substance that can donate a pair of electrons (electron

pair donor)

H+ H O H••••+ OH-••

••••

acid base

N H••

H

H

H+ +

acid base

N H

H

H

H+

N H••

H

H

acid base

F B

F

F

+

F B

F

F

N H

H

H