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Chapter 16 Acid-Base Equilibria

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Learning goals and key skills:

Understand the nature of the hydrated proton, represented as either H+(aq) or H3O+(aq) Define and identify Arrhenuis acids and bases.

Define and identify Bronsted-Lowry acids and bases, and identify conjugate acid-base pairs.

Relate the strength of an acid to the strength of its conjugate base.

Understand how the equilibrium position of a proton transfer reaction relates the strengths of acids and bases involved.

Describe the autoionization of water and understand how [H3O+] and [OH-] are related Calculate the pH of a solution given [H3O+] or [OH-]

Calculate the pH of a strong acid or strong base given its concentration

Calculate Kaor Kbfor a weak acid or weak base given its concentration and the pH of the solution

Calculate pH of a weak acid or weak base or its percent ionization given its concentration and Kaor Kb.

Calculate Kbfor a weak base given Kaof its conjugate acid, and similarly calculate Kafrom Kb.

Predict whether and aqueous solution of a salt will be acidic, basic, or neutral Predict the relative strength of a series of acids from their molecular structures Define and identify Lewis acids and bases.

Chapter 16 Acid-Base Equilibria

Acids and Bases

Arrhenius

-An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions.

-A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.

Brønsted-Lowry -An acid is a proton donor.

-A base is a proton acceptor.

Acids and bases may be inorganic (7 strong acids, 8 strong bases) or organic (acids have –COOH group):

Amphiprotic – a substance that is capable of acting as

an acid or a base e.g. HCO

-

, HSO

-

,H O

(2)

What happens when an acid dissolves in water?

HCl

(aq)

+ H

2

O

(l)

→ Cl

(aq)

+ H

3

O

+(aq)

• Water acts as a Brønsted-Lowry base and abstracts a proton (H

+

) from the acid.

• As a result, the conjugate base of the acid and a hydronium ion are formed.

Conjugate acids and bases

HCN

(aq)

+ H

2

O

(l)

CN

-(aq)

+ H

3

O

+(aq)

HClO

(aq)

+ H

2

O

(l)

ClO

-(aq)

+ H

3

O

+(aq)

NH

3 (aq)

+ H

2

O

(l)

NH

4+

(aq)

+ OH

-(aq)

Acid and base strength

•Strong acids are completely dissociated in water.

–Their conjugate bases are weak.

•Weak acids only dissociate partially in water.

–Their conjugate bases

are strong.

(3)

Acid and Base Strength

• In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.

HCl

(aq)

+ H

2

O

(l)

→ H

3

O

+(aq)

+ Cl

-(aq)

• H

2

O is a much stronger base than Cl

-

, so the equilibrium lies so far to the right that K is not measured (K>>1).

• Acetate is a stronger base than H

2

O, so the equilibrium favors the left side (K<1).

CH

3

CO

2

H

(aq)

+ H

2

O

(l)

⇌ H

3

O

+(aq)

+ CH

3

CO

2-(aq)

(4)

Auto-ionization of water

K

w

= [H

3

O

+

] [OH

]

• This special equilibrium constant is referred to as the ion-product constant for water, K

w

.

• At 25°C, K

w

= 1.0 × 10

−14

• As we have seen, water is amphoteric.

• In pure water, a few molecules act as bases and a few act as acids.

• This is referred to as autoionization.

H

2

O

(l)

+ H

2

O

(l)

⇌ H

3

O

+(aq)

+ OH

-(aq)

The ion-product constant for water Temperature K

w

= [H

3

O

+

] [OH

]

0 °C 0.114 × 10

-14

25 °C 1.008 × 10

−14

50 °C 5.476 × 10

-14

75 °C 15.85 × 10

-14

100 °C 51.3 × 10

-14

pH and pOH scale pH = -log [H 3 O + ] pOH = -log [OH - ]

• At 25 °C in pure water,

K

w

= [H

3

O

+

] [OH

] = 1.0 × 10

−14

[H

3

O

+

] = 1.0 × 10

-14

= 1.0 × 10

-7

Since in pure water [H

3

O

+

] = [OH

-

],

(5)

pH = -log [H 3 O + ]

These are the pH values for several common substances.

Three ways to measure pH

• Litmus paper red-to-blue:

basic, pH > 8

blue-to-red:

acidic, pH < 5

• An indicator

• A pH meter

Seven strong acids

Strong acids completely ionize.

hydrochloric acid HCl

(aq)

H

+(aq)

+ Cl

-(aq)

hydrobromic acid hydroiodic acid

chloric acid

perchloric acid HClO

4 (aq)

H

+(aq)

+ ClO

4-(aq)

nitric acid

sulfuric acid H

2

SO

4 (aq)

“100%”

“100%”

“100%”

(6)

Anion name Acid name

S

2-

, sulfide ion H

2

S, hydrosulfuric acid CN

-

, cyanide ion HCN, hydrocyanic acid

CO

32-

, carbonate ion H

2

CO

3

, carbonic acid CrO

42-

, chromate ion H

2

CrO

4

, chromic acid

BrO

-

, hypobromite ion HBrO, hypobromous acid NO

2-

, nitrite ion HNO

2

, nitrous acid

Acids: nomenclature

Anion name Acid name

_______ ide ion

hydro

_____ ic acid Cl

-

, chloride ion HCl, hydrochloric acid

_______ ate ion _______ ic acid ClO

3-

, chlorate ion HClO

3

, chloric acid

_______ ite ion _______ ous acid ClO

2-

, chlorite ion HClO

2

, chlorous acid

Acids: nomenclature

Strong bases

Strong bases completely dissociate.

Hydroxides of the

• alkali metals (Li, Na, K, Rb, Cs)

NaOH

(aq)

→ Na

+(aq)

+ OH

-(aq)

• heavy alkaline earth metals (Ca, Sr, Ba) (although these have limited solubility)

“100%”

(7)

K a and K b

HA

(aq)

+ H

2

O

(l)

H

3

O

+(aq)

+ A

-(aq)

[H

3

O

+

] [A

-

] K

c

= [HA]

This equilibrium constant is called the acid-dissociation constant, K

a

.

[H

3

O

+

] [A

-

] K

a

= [HA]

Dissociation Constants

The greater the value of K

a

, the stronger is the acid.

Calculating K a from the pH

The pH of a 0.100 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K

a

for formic acid at this temperature.

We know that [H

3

O

+

] [HCOO

-

] [HCOOH]

K

a

=

To calculate K

a

, we need the equilibrium concentrations of all three things.

We can find [H

3

O

+

], which is the same as [HCOO

-

], from the pH.

HCOOH

(aq)

+ H

2

O

(l)

H

3

O

+(aq)

+ HCOO

-(aq)

(8)

Calculating K a from pH

Now we can set up an ICE table…

[HCOOH], M [H3O+], M [HCOO-], M

Initially 0.100 0 0

Change - 4.17 ×10-3 + 4.17 ×10-3 + 4.17 ×10-3 Equilibrium 0.10 - 4.17 ×10-3

= 0.0958

4.17 ×10-3 4.17 ×10-3

Calculating Percent Ionization

concentration ionized original concentration

In this example,

Percent Ionization = × 100%

[H

3

O

+

]

eq

= 4.2 × 10

-3

M [HCOOH]

initial

= 0.10 M

[H

3

O

+

]

eq

[HA]

initial

Percent Ionization = × 4.2 × 10

-3

100%

0.10

= 4.2%

× 100%

Percent Ionization =

Calculating pH from K a

Calculate the pH of a 0.30 M solution of acetic acid, HC

2

H

3

O

2

, at 25°C.

HC

2

H

3

O

2(aq)

+ H

2

O

(l)

H

3

O

+(aq)

+ C

2

H

3

O

2-(aq)

K

a

for acetic acid at 25°C is 1.8 × 10

-5

.

[H

3

O

+

] [C

2

H

3

O

2-

]

[HC

2

H

3

O

2

]

K

a

=

(9)

Calculating pH from K a

We next set up an ICE table…

[HC2H3O2], M [H3O+], M [C2H3O2-], M

Initially 0.30 ≈0 0

Change -x +x +x

Equilibrium 0.30 - x ≈0.30 x x

We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

• 100*K

a

= 100*(1.8 × 10

-5

) = 0.0018

• 0.30-x = 0.30-0.0018 ≈ 0.30 (using sign. figures) In general, the approximation that [HA]

eq

is effectively equal to [HA]

0

is valid whenever

[HA]

0

is greater than 100*K

a

.

Polyprotic Acids…

…have more than one acidic proton

If the difference between the K

a

for the first dissociation and subsequent K

a

values is 10

4

or more, the pH generally depends only on the first dissociation.

Example (polyprotic acids)

H

3

PO

4 (aq)

+ H

2

O

(l)

H

3

O

+(aq)

+ H

2

PO

4- (aq)

K

a1

= 7.5×10

-3

H

2

PO

4-

(aq)

+ H

2

O

(l)

H

3

O

+(aq)

+ HPO

42- (aq)

K

a2

= 6.2×10

-8

HPO

42-

(aq)

+ H

2

O

(l)

H

3

O

+(aq)

+ PO

43- (aq)

K

a3

= 3.6×10

-13

H

2

C

2

O

4

(aq) ⇌ H

3

O

+

(aq) + HC

2

O

4-

(aq)

K

a1

= 6.5x10

-2

HC

2

O

4-

(aq) ⇌ H

3

O

+

(aq) + C

2

O

42-

(aq)

K = 6.1x10

-5

(10)

Weak Bases

Bases react with water to produce hydroxide ion.

Weak Bases

The equilibrium constant expression for this reaction is:

[HB

+

] [OH

-

] K

b

= [B]

where K

b

is the base-dissociation constant.

B (aq) + H

2

O(l) HB

+

(aq) + OH

-

(aq)

Weak Bases

K

b

can be used to find [OH

-

] and, through it, pH.

(11)

pH of Basic Solutions

What is the pH of a 0.15 M solution of NH

3

at 25 °C?

[NH

4+

] [OH

-

] [NH

3

]

K

b

= = 1.8 × 10

-5

NH

3(aq)

+ H

2

O

(l)

⇌ NH

4+ (aq)

+ OH

-(aq)

pH of Basic Solutions

Tabulate the data.

[NH

3

], M [NH

4+

], M [OH

-

], M

Initially 0.15 0 0

At Equilibrium 0.15 - x ≈ 0.15 x x

K a and K b

K

a

and K

b

are related in this way:

K

a

× K

b

= K

w

Therefore, if you know one of them, you can calculate the other.

at 25 °C

(12)

Reactions of Anions with Water

• Anions are bases (conjugate base of an acid).

• As such, they can react with water in a hydrolysis reaction to form OH

-

and the conjugate acid:

X

-(aq)

+ H

2

O

(l)

⇌ HX

(aq)

+ OH

-(aq)

Reactions of Cations with Water

• Cations with acidic protons (like NH

4+

) will lower the pH of a solution.

• Most metal cations that are hydrated in solution also lower the pH of the solution.

Reactions of Cations with Water

• Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water.

• This makes the O-H bond more polar and the water more acidic.

• Greater charge and smaller

size make a cation more

acidic.

(13)

Effect of Cations and Anions

1. An anion that is the conjugate base of a strong acid will not affect the pH.

2. An anion that is the conjugate base of a weak acid will increase the pH.

3. A cation that is the conjugate acid of a weak base will decrease the pH.

Effect of Cations and Anions

4. Cations of the strong Arrhenius bases will not affect the pH.

5. Other metal ions will cause a decrease in pH.

6. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the K

a

and K

b

values.

Factors Affecting Acid Strength

• The more polar the H-X bond and/or the weaker the H-X bond strength, the more acidic the compound.

• So acidity increases from left to right across a row

and from top to bottom down a group.

(14)

Factors Affecting Acid Strength

In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

Factors Affecting Acid Strength

For a series of oxyacids, acidity increases with the number of oxygens.

• Strength of an acid increases as additional electronegative atoms are added

Arrange the following oxoacids in order of decreasing acid strength:

HClO, HClO

2

, HClO

3

, HBrO

(rank strongest to weakest)

(15)

Factors Affecting Acid Strength

Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

Lewis Acids

• Lewis acids are defined as electron-pair acceptors.

• Atoms with an empty valence orbital can be Lewis acids.

adduct

Lewis Bases

• Lewis bases are defined as electron-pair donors.

• Anything that could be a Brønsted-Lowry base is a Lewis base.

• Lewis bases can interact with things other than

protons, however.

(16)

References

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