CHAPTER 16 – ACIDS AND BASES

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CHAPTER 16 – ACIDS AND BASES. 1884 SVANTE ARRHENIUS Proposed the first definitions of acids and bases. ACID – A compound the produces hydrogen ions in a water solution HCl (g) → H + (aq) + Cl - (aq) BASE – A compound that produces hydroxide ions in a water solution - PowerPoint PPT Presentation

Transcript of CHAPTER 16 – ACIDS AND BASES

CHAPTER 16 – ACIDS AND BASES

ACID – A compound the produces hydrogen ions in a water solutionHCl (g) → H+(aq) + Cl-(aq)

BASE – A compound that produces hydroxide ions in a water solutionNaOH (s) → Na+(aq) + OH-(aq)

8A-1 (of 41)

1884 SVANTE ARRHENIUSProposed the first definitions of acids and bases

1923

Expanded the definitions of acids and bases

THOMAS LOWRY

8A-2

ACID – A hydrogen ion (or proton) donorBASE – A hydrogen ion (or proton) acceptor

1923

Expanded the definitions of acids and bases

JOHANNES BRØNSTED

8A-3

HCl + H2O →

acid base

HYDRONIUM ION – H3O+, formed when a hydrogen ion attaches to a water

+-

Cl- + H3O+

8A-4

NH3 + H2O →

base acid

AMPHOTERIC – A substance that can act as an acid or a base

+ -

NH4+ + OH-

8A-5

Acids turn into bases, and bases turn into acids

HCl + H2O → Cl- + H3O+

acid base

+-

acid base

8A-6

Acids turn into bases, and bases turn into acids

HCl + H2O → Cl- + H3O+

acid base conjugate base

of HCl

conjugate acid

of H2O

+-

acid basebase acid

8A-7

NH3 + H2O → NH4+ + OH-

base acid

+ -

base acid

8A-8

NH3 + H2O → NH4+ + OH-

base acid conjugate acid

of NH3

conjugate base of H2O

+ -

base acidacid base

8A-9

Conjugate base of HClO4

When HClO4 acts as an acid, it becomes: ClO4-

Conjugate base of H2CO3

When H2CO3 acts as an acid, it becomes: HCO3-

Conjugate acid of CF3NH2

When CF3NH2 acts as a base, it becomes: CF3NH3+

8A-10

POLYPROTIC ACID – An acid with more than one ionizable hydrogen ion

H2CO3

H3PO4

Hydrogen ions become successively more difficult to ionize

diprotictriprotic

H2SO4 (aq) → H+(aq) + HSO4-(aq)

Strong Acid

HSO4- (aq) → H+(aq) + SO4

2-(aq)

Weaker Acid

8A-11

WATER

Water ionizes to a small extent

8A-12

WATER

Water ionizes to a small extent

2H2O (l) → H3O+(aq) + OH-(aq)

-+

Makes solutions acidic Makes solutions basic

Equal amounts of H3O+ and OH- make a solution NEUTRAL pure water is neutral

8A-13

Ionization: 2H2O (l) → H3O+(aq) + OH-(aq)Association: H3O+(aq) + OH-(aq) → 2H2O (l) Equilibrium 2H2O (l) ↔ H3O+(aq) + OH-(aq)

Because of this equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant

K = [H3O+][OH-]

8A-14

Ionization: 2H2O (l) → H3O+(aq) + OH-(aq)Association: H3O+(aq) + OH-(aq) → 2H2O (l) Equilibrium 2H2O (l) ↔ H3O+(aq) + OH-(aq)

Because of this equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant

Kw = [H3O+][OH-]

ION-PRODUCT CONSTANT FOR WATER (Kw) – The product of the hydronium ion and hydroxide ion molarities in any water solution

8A-15

The Kw depends on temperature

10

2540

0.29 x 10-14 M2

1.00 x 10-14 M2

2.92 x 10-14 M2

Temp (ºC) Kw

8A-16

Calculate the hydronium ion and hydroxide ion molarities in pure water at 25ºC.

(x)(x)

1.00 x 10-14 M2 =

2H2O (l) → H3O+(aq) + OH-(aq)Molarities before Ionization:Molarities at Equilibrium:

Kw = [H3O+][OH-]

= x2 x

1.00 x 10-7 M =

In pure water, [H+] = [OH-] = 1.00 x 10-7 M

0 0

x x

8A-17

Calculate the hydroxide ion molarity in a water solution in at 25ºC in which the [H3O+] = 1.00 x 10-5.

(1.00 x 10-5 M)(x)

1.00 x 10-14 M2 =

Kw = [H3O+][OH-]

1.00 x 10-9 M = x

1.00 x 10-14 M2 _________________

1.00 x 10-5 M

= x

= [OH-]

8A-18

Calculate the hydronium ion molarity in a water solution in at 25ºC in which the [OH-] = 8.00 x 10-3.

8A-19

[H3O+] = [OH-] :

[H3O+] > [OH-] :

[H3O+] < [OH-] :

the solution is NEUTRALthe solution is ACIDICthe solution is BASIC

All water solutions contain both hydronium and hydroxide ions

For all water solutions: Kw = [H3O+][OH-]

8A-20

THE pH SCALE

pH – The negative logarithm of the hydronium ion molarity of a solution

The common logarithm of a number is:the exponent to which 10 must be raised to equal the number

100 1000 0.001

100 = 102 1000 = 103 0.001 = 10-3

200

200 = 102.3

log 100 = 2 log 1000 = 3 log 0.001 = -3 log 200 = 2.3

8A-21

[H3O+] log [H3O+] pH

0.1 M0.01 M0.001 M0.02 M

-1.0-2.0-3.0-1.7

1.02.03.01.7

For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC)

8A-22

Calculate the pH of pure water at 25ºC.

= -log(1.00 x 10-7 M)

pH = -log[H3O+]

For pure water:[H3O+] = 1.00 x 10-7 M

= 7.000

8A-23

pH 7 = Neutral

< 7Acidic

> 7Basic

10. M1. M0. 1 M

0.01 M0.001 M0.0001 M0.00001 M0.000001 M0.0000001 M0.00000001 M0.000000001 M0.0000000001 M0.00000000001 M0.000000000001 M0.0000000000001 M0.00000000000001 M

[H3O+] [OH-]

0.000000000000001 M0.00000000000001 M0.0000000000001 M

0.000000000001 M0.00000000001 M0.0000000001 M0.000000001 M0.00000001 M0.0000001 M0.000001 M0.00001 M0.0001 M0.001 M0.01 M0.1 M1 M

8A-24

pH 7 = Neutral

< 7Acidic

> 7Basic

Battery Acid

Gastric Fluid

SodaOrange Juice

Tap Water

“Pure” WaterSea WaterBaking Soda

AmmoniaBleachDrain Cleaner

10. M1. M0. 1 M

0.01 M0.001 M0.0001 M0.00001 M0.000001 M0.0000001 M0.00000001 M0.000000001 M0.0000000001 M0.00000000001 M0.000000000001 M0.0000000000001 M0.00000000000001 M

[H3O+] Common Substances

8A-25

Calculate the pH of orange juice if it has a hydronium ion concentration of 1.6 x 10-3 M.

= -log(1.6 x 10-3 M)

pH = -log[H3O+]

[H3O+] = 1.6 x 10-3 M

= 2.80

8A-26

Calculate the pH of toothpaste that has a hydroxide ion concentration of 5.6 x 10-5 M.

= -log(1.79 x 10-10 M)

pH = -log[H3O+]

= 9.74

(x)(5.6 x 10-5 M)

Kw = [H3O+][OH-]

= 1.79 x 10-10 M1.00 x 10-14 M2 = x _________________

5.6 x 10-5 M

1.00 x 10-14 M2 =

[OH-] = 5.6 x 10-5 M

= [H3O+]

8A-27

Calculate the pH of milk if it has a hydroxide ion concentration of 3.2 x 10-9 M.

8A-28

Calculate the hydrogen ion concentration in blood, which has a pH of 7.4.

= 0.0000000398 M

pH = -log[H3O+]

-pH = log[H3O+]

antilog (-pH) = [H3O+]

antilog (-7.4) = [H3O+]

For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC)

= 3.98 x 10-8 M = 4 x 10-8 M

8A-29

Calculate the hydroxide ion concentration in egg yolks, which have a pH of 5.65.

8A-30

pH OF STRONG ACID SOLUTIONS

Strong acids completely ionize

HCl (g)

HCl (aq)

H+(aq) + Cl-(aq)

8A-31

Calculate the pH of a 0.015 M hydrochloric acid solution.

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

1 M

0.015

1 M

= -log(0.015 M)

pH = -log[H3O+]

= 1.82

x 1 M H3O+

____________

1 M HCl

= 0.015 M H3O+ 0.015 M HCl

8A-32

Calculate the pH of a 0.0400 M sulfuric acid solution.

H2SO4 (aq) + 2H2O (l) → 2H3O+(aq) + SO4 2-(aq)

x

2 M

0.0400

1 M

= -log(0.0800 M)

pH = -log[H3O+]

= 1.097

x 2 M H3O+

______________

1 M H2SO4

= 0.0800 M H3O+ 0.0400 M H2SO4

8A-33

pH OF STRONG BASE SOLUTIONS

Strong bases completely dissociate

NaOH (s)

NaOH (aq)

Na+(aq) + OH-(aq)

8A-34

Calculate the pH of a 0.40 M sodium hydroxide solution.

NaOH (aq) → Na+(aq) + OH-(aq) x

1 M0.40

1 M

= -log(2.50 x 10-14 M)

pH = -log[H3O+] = 13.60

(x)(0.40 M)

Kw = [H3O+][OH-]

= 2.50 x 10-14 M1.00 x 10-14 M2 = x _________________

0.40 M

1.00 x 10-14 M2 =

= [H3O+]

x 1 M OH-

______________

1 M NaOH

= 0.40 M OH-0.40 M NaOH not [H3O+]

8A-35

BUFFERS

BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it

7.00

pH

1.00 L H2O0.10 moles HCl

8A-36

BUFFERS

BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it

1.00

pH

1.00 L H2O0.10 moles HCl

3.45

pH

1.00 L 1.0 M HF, 1.0 M F-

0.10 moles HCl

8A-37

BUFFERS

BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it

1.00

pH

1.00 L H2O0.10 moles HCl

3.37

pH

1.00 L 1.0 M HF, 1.0 M F-

0.10 moles HCl

8A-38

A solution is a buffer if is contains a weak acid and its conjugate base

Weak acid:Conjugate base:

HFF-

1.00 L 1.0 M HF, 1.0 M F-

0.10 moles HCl

If a strong acid is added to the buffer:

8A-39

H+ + F- → HF

The strong acid (H+) is reacted away by the conjugate base (F-)

A solution is a buffer if is contains a weak acid and its conjugate base

Weak acid:Conjugate base:

HFF-

1.00 L 1.0 M HF, 1.0 M F-

0.10 moles HCl

no more strong acid in the solution

If a strong acid is added to the buffer:

8A-40

A solution is a buffer if it contains a weak acid and its conjugate base

Weak acid:Conjugate base:

HFF-

1.00 L 1.0 M HF, 1.0 M F-

If a strong base is added to the buffer:

OH- + HF → H2O + F-

The strong base (OH-) is reacted away by the weak acid (HF)

no more strong base in the solution

0.10 moles OH-

8A-41