1. molecular ------------------- chapter 15 2. ionic (weak ... · 3. ionic (“insoluble” salts)...

Equilibrium

1. Molecular ------------------- Chapter 15

2. Ionic (Weak Acid / Base ) - Chapter 16

3. Ionic (“Insoluble” Salts) - Chapter 17

STRONG Acids & Bases

1. COMPLETELY IONIZED1. COMPLETELY IONIZED1. COMPLETELY IONIZED1. COMPLETELY IONIZED

(a) HCl(aq) �� H+ (aq) + Cl- (aq)

(b) NaOH(aq) �� Na+(aq) + OH-(aq)

2. ONLY IONS PRESENT2. ONLY IONS PRESENT2. ONLY IONS PRESENT2. ONLY IONS PRESENT

(a) H+ (aq) & Cl- (aq)

(b) Na+(aq) & OH-(aq)

3. NO EQUILIBRIUM3. NO EQUILIBRIUM3. NO EQUILIBRIUM3. NO EQUILIBRIUM

WEAKWEAKWEAK Acids & Bases

1.1.1.1. NOT COMPLETELY IONIZEDNOT COMPLETELY IONIZEDNOT COMPLETELY IONIZEDNOT COMPLETELY IONIZED

HF(aq) ↔↔↔↔ H+ (aq) + F- (aq)

2. Both IONSIONSIONSIONS and MOLECULESMOLECULESMOLECULESMOLECULES Present

HF(aq) ↔↔↔↔ H+ (aq) + F- (aq)

3. EQUILIBRIUM exists

for Weak Acids & Bases

The [H] of 0.50 M HCN(aq) is ? {Ka = 4.9 x 10–10

What do you do 1st ?

1ST Write & Balance “Reaction”

HCN(aq) ↔ H+ (aq) + CN- (aq)

2nd Write Equilibrium Expression

][

]][[H

HCN

CNKa

−+

=

Initial Change Equilibrium

Find [H] of 0.50 M HCN {Ka = 4.9 x 10–10

HCN(aq) ↔ H+ (aq) + CN- (aq)

I NITIALNITIALNITIALNITIAL ……. 0.5 0 0

C HANGEHANGEHANGEHANGE ……. - x + x + x

E QUILIBRIUMQUILIBRIUMQUILIBRIUMQUILIBRIUM 0.5 – x x + 0 x + 0

Ka = 4.9 x 10-10 ; [HCN] = 0.5 – x ; [H+]=[CN-] = x

x2≅ (4.9 x 10-10) (0.50) ≅ 2.45 x 10-10

[H+]=[CN-] = x ≅ 1.565 x 10-5 ≅ 1.6 x 10-5

][

]][[H

HCN

CNKa

−+

=

0.5

x

x -0.5

x

]5.0[

]][[x 109.4

2210

≈=

−

=×−

x

x

What approximation?

Yes!!

Why?

Since x ≅ 1.6 x 10-5 is so small

0.50 - x = 0.50

0.5

x

x -0.5

x

22

≈

pH – A measure of Acidity

The pH of a solution is defined as the

negative logarithm of the hydrogen ion

concentration (in mol/L).

pH = –log [H+]

pOH = –log [OH-]

pH + pOH = 14

Determine the pH & pOH of 0.50 M

HCN(aq)

pH = - log [H+]

x = [H+] = [CN-] ≅ 1.6 x 10-5

pH = - log [1.6 x 10-5]

pH = 4.8

pOH = 14.0 – 4.8 = 9.2

Definitions of Acids & Bases

1. Arrhenius

2. Bronsted-Lowry

3. Lewis

Acid / Base

ACIDACIDACIDACID: A substance which dissociates to form hydronium ions (H3O

+) in solution :

HA(aq) → H+(aq) + A–(aq)

BASEBASEBASEBASE: A substance that dissociates to form hydroxide ions (OH–) in solution :

MOH(aq) → M+(aq) + OH–(aq)

M = a Metal

-LOWRY Acid / Base

ACID:ACID:ACID:ACID: Substance that can donate H+

for example HF(aq), HCl(aq), HNO3(aq), etc

BASE:BASE:BASE:BASE: Substance that can accept H+

for example OH- , F- , Cl- , Br- , I- , etc

LEWIS LEWIS LEWIS Acid / Base

• Acid is an electron-pair acceptor. These are

generally cations and neutral molecules with

vacant valence orbitals. For example H+ & BF3.

• Base is an electron-pair donor. These are

generally anions and neutral molecules with

available pairs of electrons, such as H2O, NH3, O2–

• The bond formed is called a coordinate bond.

STRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASES

HCl(aq) is a strong acid but Cl- is a “weak” base

Why?

Because Cl- will not react with H+

Why Not?

If it did, HCl(aq) {a strong acid} would form

which does not exist


HF(aq) is a weak acid and F- is a “strong” base

Why?

Because F- will react with H+

Why ?

HF(aq) {a weak acid} would form

which does exist


HClO4

HI

HBr

HCl

HNO3

HF

HNO2

HCOOH

NH4+

HCN

H2O

ClO4–

I–

Br –

Cl –

NO3 –

F –

NO2 –

HCOO –

NH3

CN –

OH –

ACID BASE ACID BASE

Incre

asin

g A

cid

Str

ength

Str

on

g A

cid

s

–BASE PAIRS

An Acid and a Base that

differ only in the presence

{or absence} of a proton.

Conjugate acid–base pairs

HF + H2O = H3O+ + F-


–BASE PAIRS

HClO4

HI

HBr

HCl

H2SO4

HNO3

H3O+

HSO4–

HSO4–

HF

HNO2

HCOOH

NH4+

HCN

H2O

NH3

ClO4–

I–

Br –

Cl –

HSO4 –

NO3 –

H2O

SO42–

SO42–

F –

NO2 –

HCOO –

NH3

CN –

OH –

NH2 –

ACID CONJ. BASE ACID CONJ. BASE

What are the conjugate bases of the following

Brønsted–Lowry acids

H2SO4

HClO4

H2SO4

HSO4–

H2O

+ H2O � H3O+ + HSO4

-

+ H2O �

+ H2O �

+ H2O �

+ H2O �

Which one of the conjugate bases of the

following Brønsted-Lowry acids is incorrect?

(a) ClO- for HClO

(b) HS- for H2S

(c) NH3 for NH4+

(d) SO42- for HSO4

-

(e) H2SO4 for HSO4-

Aqueous Equilibria:

WEAK ACIDS

Dissociation (or ionization) of Water

H2O(l) + H2O(l) = H3O+(aq) + OH–(aq)

or simply

H2O(l) = H+(aq) + OH–(aq)

Keq = [H+] [OH–]

This a product (concentration x concentration)

and it is water,

therefore called the Ion Product Constant of water


The equilibrium gives us the ion product of water.

Kw = Kc = [H+] [OH–] = 1.0 x10–14

If either [H+] or [OH–] is known the other can be

determined

When does the constant Kw change ?

If the concentration of OH– ions in a certain

cleaning solution is 0.0001 M, what is the

concentration of H+ ions?

[H+] [OH–] = 1.0 x10–14

[OH-] = 1.0 x10–4

[H+] = 1.0 x10–14 / 1.0 x10–4

[H+] = 1.0 x10–10

Calculate the concentration of OH– ions in a HCl

solution whose hydrogen ion concentration is 0.1M

[H+] [OH–] = 1.0 x10–14

[H+] = 1.0 x10–1

[OH-] = 1.0 x10–14 / 1.0 x10–1

[OH-] = 1.0 x10–13

pH – A measure of Acidity

The pH of a solution is defined as the

negative logarithm of the hydrogen ion

concentration (in mol/L).

pH = –log [H+]

pOH = –log [OH-]

pH + pOH = 14

PERCENT DISSOCIATIONPERCENT DISSOCIATIONPERCENT DISSOCIATION

• A measure of the strength of an acid.

• Stronger acids have higher percent

dissociation.

• Percent dissociation of a weak acid

decreases as its concentration increases.

Percent Dissociation

For Acids

100%[HA]

][HonDissociati % ×=

+

Calculate the % dissociation of 0.5 M HCN (aq)

From the equilibrium calculation

x ≅ 1.6 x 10-5 = [H+]

100%[HA]


+

3-5

3.2x10100%[0.50]

][1.6x10onDissociati % =×=

−

Ka for 0.5 M HCN (aq)

If % DISSOCIATION = 3.2 x 10-3

[H+] = [CN-] = (% dissociation) ([HA]) = 1.6 x 10-5

][

]][[H

HCN

CNKa

−+

= 100%[HA]


+

1055

101.5]50.0[

]106.1][[1.6x10 −

−−

== xx

Ka

0.1 M AMMONIA SOLUTION, if pH = 11.1

NH3 (aq) = NH4+ (aq) + OH - (aq)

[NH4+ ] = [OH - ]

100%][NH

][OHonDissociati %

3

×=

−

1.34%100%[0.10]

][1.34x10onDissociati %

-3

=×=

Aqueous Equilibria:

WEAK BASES

WEAKWEAKWEAK Bases

NH3(gas) = Ammonia

NH4OH = Ammonium Hydroxide

NH3 (aq) = Aqueous Ammonia ≡ NH4OH

NH4OH = NH4+ (aq) + OH - (aq)

or

NH3 (aq) = NH4+ (aq) + OH - (aq)

Calculate pOH & pH for a 0.10 M aqueous

ammonia solution {Kb = 1.8 x 10-5

NH3 (aq) = NH4+ (aq) + OH - (aq)

x = [OH-] = [NH4+] = 1.34 x 10-3

][NH

]][OH[NH

3

4

−+

=bK

5

3

4 108.1x]-[0.10

[x][x]

][NH

]][OH[NH−

−+

=== xKb

Calculate pOH & pH for a 0.10 M NH3(aq)

[OH-] = 1.34 x 10-3

pOH = -log [OH-] = 2.9

pH = 14 – 2.9 = 11.1

pH of some common fluids

• Gastric juice in stomach 1.0 – 2.0

• Lemon juice 2.4

• Vinegar 3.0

• Grapefruit juice 3.2

• Orange juice 3.5

• Urine 4.8 – 7.5

• Saliva 6.4 – 6.9

• Milk 6.5

• Blood 7.35 – 7.45

• Tears 7.4

• Milk of Magnesia 10.6

• Household Ammonia 11.5

pH + pOH = 14

Neutral solutions: [H+] = [OH-] = 1.0 x 10 –7 M

pH = pOH = 7

Acidic solutions: [H+] > 1.0 x 10 –7 M,

pH < 7.00

Basic solutions: [H+] < 1.0 x 10 –7 M,

pH > 7.00

The OH– ion concentration of a sample of

blood is 2.5x10–7 M. What is its pH

[H+] [OH–] = 1.0 x10–14

[H+] (2.5x10–7) = 1.0 x10–14

[H+] = 4.0 x10–8

pH = –log [H+] = –log (4.0 x10–8)

pH = 7.39794

Really ?

Calculate the pH of a 0.01M HNO3 (aq) solution

pH = –log [H+]

Nitric is a STRONG acid !

100% Ionized: 1 HNO3 → 1 H+ + 1 NO3-

[H+] = [NO3-] = [HNO3]

[HNO3] = 1.0 x 10 –2 M

pH = –log (1.0 x 10 –2) = 2

Acid Ionization Constants

7.1 x 10 –4

4.5 x 10 –4

3.0 x 10 –4

1.7 x 10 –4

8.0 x 10 –5

6.5 x 10 –5

1.8 x 10 –5

4.9 x 10 –10

1.3 x 10 –10

HF

HNO2

C9H8O4 (aspirin)

HCO2H (formic)

C6H8O6 (ascorbic)

C6H5CO2H (benzoic)

CH3CO2H (acetic)

HCN

C6H5OH (Phenol)

F–

NO2 –

C9H7O4 –

HCO2 –

C6H7O6 –

C6H5CO2 –

CH3CO2 –

CN –

C6H5O –

ACID Ka CONJ. BASE Kb

1.4 x 10 –11

2.2 x 10 –11

3.3 x 10 –11

5.9 x 10 –11

1.3 x 10 –10

1.5 x 10 –10

5.6 x 10 –10

2.0 x 10 –5

7.7 x 10 –5

Relationship between Ka and Kb:

Ka × Kb = Kw

Ka × Kb = Kw

For HF(aq)

Ka x Kb = (7.1 x 10-4)(1.4 x 10-11) = 9.94 x 10-15

For HCN(aq)

Ka x Kb = (4.9 x 10-10)(2.0 x 10-5) = 9.8 x 10-15

For C6H5OH (aq) (Phenol)

Ka x Kb = (1.3 x 10–10)(7.7 x 10–5) = 10.01 x 10-15

Determine the concentration of a HNO3 (aq)

solution with a pH of 2

pH = –log [H+]

pH = 2 therefore

[H+] = 1.0 x 10 –2 M

Nitric is a STRONG acid therefore

[H+] = [NO3-] = [HNO3]

[HNO3] = 1.0 x 10 –2 M

Acids , Bases

&

SALTS

Why does NaOH(aq) react with HCl(aq) ?

HCl(aq) + NaOH(aq) → H2O + NaCl(aq)

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → H2O +

Na+(aq) + Cl-(aq)

Na+(aq) & Cl-(aq) are spectator ions

Does NaOH(aq) react with HCl(aq) ?

Yes & the reaction is:

H+(aq) + OH–(aq) = H2O(l)

How Strongly does this reaction take place?

The strong acid strong base reaction


Is the reverse of the ionization of water


With Kw = [H+] [OH–] = 1.0 x10–14

The reverse reaction H+(aq) + OH– (aq) = H2O(l)

Would have a K = 1.0 x 10+14

Acid–Base Properties of Salts

SALTS IN WATER EITHER FORM

neutral solutions

basic solutions

or acidic solutions

Acids REACT with Bases

Will NaOH (aq) react with HCl( aq) ?

Yes , 100%

HCl(aq) + NaOH(aq) → H2O + NaCl(aq)

Why does NaOH(aq) react with HCl(aq) ?

Because water is formed


Will NaCl(aq) react with water?

No

Why Not ?

Because if it did a strong acid HCl(aq) and a

strong base NaOH(aq) would be formed and

neither exist in water

H2O + NaCl(aq) � HCl(aq) + NaOH(aq)

Will sodium ion react with water?

No because a strong base would be formed

What base ?

NaOH

Will potassium ion react with water?

No because a strong base would be formed

What base ?

KOH

Will chloride ion react with water?

No because a strong acid would be formed

What acid ?

HCl(aq)

Will nitrate ion react with water?

No because a strong acid would be formed

What acid ?

HNO3(aq)

What ions will react with water?

The ones that will not form

strong acids or bases

Can you name them?

Salts that produce neutral solutions

are those formed from strong acids

and strong bases.

For example NaCl(aq)

Can you think of others ?

Salts that produce basic solutions are

those formed from weak acids and

strong bases.

For example NaF

Can you think of others?

Salts that produce acidic solutions

are those formed from strong acids

and weak bases.

For Example NH4Cl

Can you think of others?

Acid–Base Properties of Salts

• Salts that produce neutral solutions are those

formed from strong acids and strong bases.

• Salts that produce basic solutions are those

formed from weak acids and strong bases.

• Salts that produce acidic solutions are those

formed from strong acids and weak bases.

Strength of Acids

HClO4

HI

HBr

HCl

H2SO4

HNO3

H3O+

HSO4–

HF

HNO2

HCOOH

NH4+

HCN

ClO4–

I–

Br –

Cl –

HSO4 –

NO3 –

H2O

SO42–

F –

NO2 –

HCOO –

NH3

CN –

ACID CONJ. BASE ACID CONJ. BASE

Predict whether the following solutions will

be acidic, basic or nearly neutral:

(a) NH4I

(b)SrCl2

(c) KCN

(d) Fe(NO3)3

(e) LiClO4

(f) Na3PO4

(a) ______

(b) _______

(c) _______

(d) ______

(e) ______

(f) ______

Last Topic in Chapter 16

The strength of an acid depends

on its tendency to ionize.

Molecular Structure

and Acid Strength

The strength of an acid depends

on its tendency to ionize.

The stronger the bond, the weaker the acid.

Or

The more polar the bond, the stronger the acid.

For Acids in Same Group

Bond strength plays the key roll

Oxoacids of The Halogens

Cl–O–H (aq) Strongest

Br–O–H (aq)

I–O–H (aq) weakest

as electronegativity increases,

acid strength increases

Oxoacids of The Halogens

as electronegativity increases, acid strength

increases

Oxoacids of Chlorine

HClO (aq)

HClO2 (aq)

HClO3 (aq)

HClO4 (aq)

acid strength increases with

increasing oxidation number

Oxoacids of Chlorine:

Predict the relative strengths of the

following groups of oxoacids:

• a) HClO, HBrO, and HIO

• b) HNO3, and HNO2

• c) H3PO3, and H3PO4.

End

Chapter 16

1. molecular ------------------- chapter 15 2. ionic (weak ... · 3. ionic (“insoluble” salts)...

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