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Acids and Bases. Acids and bases Acid-base properties of water (K w ) pH scale Strength of Acids and Bases Weak acid (K a ) Weak base (K b ) Relation between conjugate acid-base ionization constants (K a and K b ) Chemical structure and acidity Acid-base properties of salt solutions

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Acids and bases

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 base

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 Base

A. Arrehenius acid - a compound that increases [H+] in water

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


I acid and base1

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

I. Acid and Base

C. Brønsted acid : a proton donor

D. Brønsted Base : a proton acceptor

base

acid

acid

base

conjugateacid

conjugate base

base

acid

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


Example conjugate acid base pair

species

Acid/base/both

Conjugate acid

Conjugate base

Example: Conjugate acid-base pair

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

HCO3–

CO32–

base

HSO4–

H2SO4

acid

H2O

both

OH–

H3O+

H2PO4–

both

H3PO4

HPO42–

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


Ii acid base properties of water

H

H

O

O

-

H

H

H

O

+

[

]

+

H

H

+

O

H

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

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

II. Acid-base properties of water

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

base

conjugateacid

conjugatebase

acid


B ion product constant k w

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

B. Ion-product constant, Kw

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

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

[H+] = [OH-]

Neutral solution

[H+] > [OH-]

Acidic solution

[H+] < [OH-]

Basic solution


Acids and bases

=

1.0 x 10-14

Kw

2.0 x10-2

[OH–]

[H+] =

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

Q:

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

(a) Neutral solution [H+] = [OH–]= x

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

= 1.0 x 10-7M

(b) [OH–] = 2.0 x10-2M

= 5.0 x 10-13 M (basic)


Iii ph scale a measure of acidity

[H+]

pH

pOH

III. pH scale – a measure of acidity

pH = -log [H+]

pOH = -log [OH–]

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

pH + pOH = 14.00

-log [H+] – log [OH-] = 14.00

Solution is

At 250C

pH = pOH = 7

neutral

[H+] = [OH-]

[H+] = 1 x 10-7

pH < 7 pOH >7

[H+] > 1 x 10-7

acidic

[H+] > [OH-]

pH > 7 pOH <7

basic

[H+] < [OH-]

[H+] < 1 x 10-7

As the solution is more acidic when


Acids and bases

16.1


Acids and bases

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

Q:

pH = 14.0 - pOH = 14.0-10.5 = 3.5

pH = 3.5 = -log [H+]

= 10-3.5

= 3.2 x 10-4M

[H+] = 10-pH

(acidic)

  • 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

(basic)

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40


Iv strength of acid and base

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

H2O

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

H2O

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

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

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

IV. Strength of Acid and Base

Strong Electrolyte – 100% dissociation (ionization)

- soluble ionic compound, strong acid, strong base

Weak Electrolyte – not completely dissociated

- weak acid, weak base


Acids and bases

HCl, strong acid

HF, weak acid


C strong acid

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

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

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

  • Strong electrolyte, completely ionize (dissociate) in water

  • HCl, HBr, HI, HNO3, H2SO4, HClO4

C. Strong acid

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.

= 0.19 M

0.19 M

0.00 M

0.00 M

Initial

-0.19 M

0.19 M

0.19 M

Change

0.19 M

0.19 M

Equilibrium

0.00 M

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


D strong base

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

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

H2O

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

  • Hydroxide of alkali (IA) and alkaline earth (IIA) metal

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

D. Strong Base

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

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

0.048 M

0.000 M

0.000 M

Initial

0.048 M

-0.048 M

Change

2 x 0.048 M

0.048 M

0.096 M

Equilibrium

0.000 M

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

pH = 14.00 - pOH = 12.98


E conjugate acid base pair

HCN (aq) + F– (aq) HF (aq) + CN– (aq)

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

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

Which is a stronger acid? HCN or HF

Which is a stronger base? CN– or F–

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

KC < 1


Acids and bases

16.2


V weak acid

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

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

[H+][A-]

Ka =

[HA]

acid

strength

Ka

V. Weak Acid

Ka: acid ionization constant


Acids and bases

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

= 7.1 x 10-4

= 7.1 x 10-4

= 7.1 x 10-4

[H+][F-]

x2

x2

Ka

Ka =

Ka =

0.80 - x

[HF]

0.80

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

What is the pH of a 0.80M HFsolution (at 250C)?

Ex:

Initial (M)

0.80

0.00

0.00

Change (M)

-x

+x

+x

Equilibrium (M)

0.80 - x

x

x

0.80 – x 0.80

Ka << 1

x2 = 5.68 x 10-4

x = 0.024 M

[H+] = [F-] = 0.024 M

pH = -log [H+] = 1.62

[HF] = 0.80 – x = 0.78 M


When will this approximation work

= 7.1 x 10-4

= 7.1 x 10-4

0.0027 M

0.024 M

x2

x2

x 100% = 3.0%

x 100% = 27%

0.80 M

0.010 M

Ka

Ka =

0.010 - x

0.010

When will this approximation work?

0.80 – x 0.80

Ka << 1

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

Less than 5%

Approximation ok.

x = 0.024

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

x = 0.0027 M

More than 5%

Approximation not ok.

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

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

pH = -log [H+] = 2.64

[H+] = x = 0.0023 M


Steps to solve weak acid ionization problems

Steps to solve weak acid ionization problems:

  • 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+].

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

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

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


V weak acid1

ionized acid concentration at equilibrium

x 100%

x 100%

% ionization =

initial concentration of acid

[H+]

[HA]0

V. Weak Acid

B. Percent ionization

% ionization =

For a monoprotic acid HA

[HA]0 = initial concentration

decreases

*Percent ionization as

initial concentration increases


V weak acid2

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) + HPO42–(aq) Ka2= 6.2 x10-8

HPO42–(aq) H+(aq) + PO43–(aq) Ka3= 4.8 x10-13

<

<

V. Weak Acid

Polyprotic acid

Acids that have more than one ionizable H

diprotic

triprotic

Ka3 Ka2 Ka1


Acids and bases

= 1.3 x 10-2

x2

Ka1 =

0.024 - x

H2SO3(aq) H+(aq) + HSO3–(aq)

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

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

Ex:

Treat this type problem with step-by-step ionization

Initial (M)

0.024

0.00

0.00

Change (M)

-x

+x

+x

Equilibrium (M)

0.024 - x

x

x

x = 0.0123 or -0.025

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

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

[H2SO3] = 0.024 – x = 0.0117 M

Not done yet!

Next step, let us consider the second ionization of H


Acids and bases

= 6.3 x 10-8

(0.0123+y)y

Ka2 =

0.0123 - y

6.3 x10-8M

x 100% << 5 %

0.0123 M

HSO3–(aq) H+(aq) + SO3–(aq)

Initial (M)

0.0123

0.0123

0.00

Change (M)

-y

+y

+y

Equilibrium (M)

0.0123 - y

0.0123+y

y

Ka2<< 1

0.0123 – y 0.0123

0.0123 + y 0.0123

y= 6.3 x10-8

[H+] = 0.0123 + y = 0.012 M

pH = -log[H+] =1.92

[HSO3–] = 0.0123 – y = 0.012 M

[SO32–] = y = 6.3 x10-8M

[H2SO3] = 0.012 M


Vi weak base

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

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

[HB+][OH-]

Kb =

[B]

base

strength

Kb

VI. Weak Base

Kb: base ionization constant

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


Acids and bases

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

Initial (M)

Change(M)

= 5.6 x 10-4

= 5.6 x 10-4

Equilibrium(M)

x2

x2

Kb

Kb =

0.50 - x

0.50

0.017 M

x 100% = 3.4 % assumption okay

0.50 M

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

Ex:

0.50

0.00

0.00

-x

+x

+x

x

x

0.50 - x

Kb << 1

0.50 – x 0.50

x2 = 2.8 x 10-4

x = 0.017 M

[OH–] = [C2H5NH3+] = 0.017 M

[C2H5NH2] = 0.50 – x = 0.48 M

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


Acids and bases

16.5


Vii ionization constants of conjugate acid base pairs

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

A–(aq) + H2O (l) OH- (aq) + HA (aq)

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

Kw

Kw

Kw

Ka=

Kb=

Kb= =

Kb

Ka

Ka

1.0 x 10-14

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

= 1.5 x 10-10

6.5 x 10-5

VII. Ionization Constants of Conjugate Acid-Base Pairs

Ka

Kb

Kw

KaKb = Kw

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

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

C6H5COO– is a base

and is the conjugate base of C6H5COOH


Viii molecular structure and acid strength

<

<

<

<

<

<

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 acidHF HCl HBr HI

Ex: H2O H2S H2Se


C oxoacids

d-

d+

O-

Z

+ H+

O

Z

H

C. Oxoacids

  • 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 oxoacids1

>

O

Cl is more electronegative than Br

H O Cl O

O

H O Br O

>

>

>

  • 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

C. Oxoacids

>

HBrO HIO

  • Oxoacids having the same central atom (Z) but different numbers of attached groups.

  • - Acid strength increases as the

  • oxidation no of Z increases

HClO4 HClO3 HClO2 HClO


D carboxylic acids

withdraw electron

O

H

H C C O H

H

can be ionized

D. Carboxylic acids

  • Acids with functional group -COOH

Ex. Predict the strength of the following acids.

CF3COOH CH3COOH

>


Ix salt solutions

H2O

H2O

KNO3 (s) K+ (aq) + NO3– (aq)

NaBr (s) Na+ (aq) + Br- (aq)

  • 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+

IX. Salt Solutions

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–).


Acids and bases

H2O

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

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

H2O

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

3+

2+

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

2+

3+

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

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

B. Basic Solutions:

Salts derived from a strong base and a weak acid.

C. Acidic Solutions:

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

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

For example: AlCl3


Acid hydrolysis of al 3

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

1.0 x 10-14

1.0 x 10-14

4.5 x 10-4

1.8 x 10-5

Kw

Kw

Kb= = = 2.2 x 10-11

Ka= = = 5.6 x 10-10

Kb

Ka

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 neutral

Example: NH4NO2

NO2– and NH4+

Ka,HNO2 =4.5 x10-4

Kb,NH3 =1.8 x10-5

NO2– :

Ka >Kb

Weak acid

NH4+ :


Acids and bases

H2O (l) + F–(aq) HF (aq) + OH-(aq)

F–(aq) +H2O (l) HF (aq) + OH-(aq)

= 1.4 x 10-11

[HF][OH-]

x2

Kb

Kb =

0.10

[F–]

x2

0.10 - x

1.0 x 10-14

7.1 x 10-4

Kw

Kb= = = 1.4 x 10-11=

Ka

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

Acid or basic?

Ex:

Initial (M)

0.10

0.00

0.00

Change (M)

-x

+x

+x

Equilibrium (M)

0.10 - x

x

x

Kb << 1

0.10 – x 0.10

[OH–] = x = 1.2 x10-6M

x2 = 1.4 x 10-12

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


X acidic basic and amphoteric oxides

X. Acidic, basic, and amphoteric oxides

  • 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)


Xi lewis acid and base

H

F

F

F B

F B

N H

F

F

H

+ OH-

H

N H

H

H

H

+

H O H

+

H+

H

N H

N H

H

H

XI. Lewis acid and base

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

  • B.Brønstedacid/base : proton donor/acceptor

    C.A Lewis acid is a substance that can accept a pair of electrons (electron pair acceptor)

  • A Lewis base is a substance that can donate a pair of electrons (electron pair donor)

H+

+

acid

base

acid

base

acid

base


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