14 Equilibrium part 2.pdf

(1)

CHAPTER 14:

ACIDS AND BASES

Arrhenius Acids and Bases

There are a few definitions of acids and bases,

some are somewhat narrow and others are much broader.

Arrhenius Acids dissociate when dissolved in water

and produce hydrogen ions (H

+

).

Hydrogen chloride (HCl) is an Arrhenius acid.

Arrhenius Bases

dissociate when dissolved in water

and

produce hydroxide ions (OH

–

)

.

Sodium hydroxide (Na

OH

) is an Arrhenius base.

HCl(aq) H+_{(aq) + Cl}–_(aq)

NaOH(aq) Na+(aq) + OH–(aq)

Arrhenius Acids and Bases

The Arrhenius definition correctly predicts the

behavior of many acids and bases.

However, this definition is limited and

sometimes inaccurate.

For example, H

+

does not exist in water. Instead, it

reacts with water to form the hydronium ion, H

3

O

+

.

H

3

O

+

(

aq

)

H

+

(

aq

) + H

2

O(

l

)

hydrogen ion:

does not really exist

in solution

hydronium ion:

actually present in

aqueous solution

Bronsted-Lowry Acids and Bases

The Brønsted–Lowry definition is more widely used:

•

A Brønsted–Lowry acid is a proton (H

+

) donor.

•

A Brønsted–Lowry base is a proton (H

+

) acceptor.

H

3

O

+

(

aq

) + Cl

−

(

aq

)

HCl(

g

) + H

2

O(

l

)

This proton is donated.

•

HCl is a Brønsted–Lowry acid because it donates

a proton to the solvent water.

•

H

2

O is a Brønsted–Lowry base because it accepts

a proton from HCl.

A Brønsted–Lowry acid

must contain a hydrogen atom.

•

Common Brønsted–Lowry acids (HA):

HCl

hydrochloric acid

HBr

hydrobromic acid

C

H

C

O

H

acidic H

atom

acetic acid

H

2

SO

4

sulfuric acid

HNO

3

nitric acid

Acids Can Have Different # of H-Atoms

•

A monoprotic acid contains one acidic proton.

HCl

•

A diprotic acid contains two acidic protons.

H

2

SO

4

•

A triprotic acid contains three acidic protons.

H

3

PO

4

•

A Brønsted–Lowry acid may be neutral or it may

carry a net positive or negative charge.

(2)

Bronsted-Lowry Bases

•

A Brønsted–Lowry base is a proton acceptor,

so it must be able to form a bond to a proton.

•

A base must contain a lone pair of electrons that

can be used to form a new bond to the proton.

N

H

+ H

2

O(

l

)

N

H

+

+ OH

−

(

aq

)

Brønsted–Lowry

base

This e

−

pair forms a new

bond to a H from H

2

O.

Common Bronsted-Lowry Bases

NaOH

sodium hydroxide

KOH

potassium hydroxide

Mg(OH)

2

magnesium hydroxide

Ca(OH)

2

calcium hydroxide

H

2

O

water

NH

3

ammonia

Lone pairs make these

neutral compounds bases.

The OH

−

is the base in

each metal salt.

The Reaction of a Bronsted-Lowry Acid

With a Bronsted-Lowry Base

H

A

+

B

A

−

+

H

B

+

gain of H

+

acid

base

loss of H

+

This e

−

pair

stays on A.

This e

−

pair forms

a new bond to H

+

.

Bronsted-Lowry Conjugate Acid-Base Pairs

The product formed by

loss of a proton from

an

acid

is called its

conjugate base

.

The product formed by

gain of a proton by

a

base

is called its

conjugate acid

.

H

A

+

B

A

−

+

H

B

+

gain of H

+

acid

base

conjugate

base

conjugate

acid

loss of H

+

H Br +

+

gain of H

+

acid

base

conjugate

base

conjugate

acid

loss of H

+

H

2

O

Br

−

H

3

O

+

HBr and Br

−

are a conjugate acid–base pair.

H

2

O and H

3

O

+

are a conjugate acid–base pair.

The net charge must be the same on both sides of the equation.

(3)

There are

2

conjugate acid-base pairs in each reaction.

Conjugate acid-base pairs are related to each other

by

H

+

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq)

acid base Conjugate acid Conjugate base

conjugate pair

H

F(

aq

) + H

2

O(

l

)



F

–

(

aq

) + H

3

O

+

(

aq

)

•

In this acid-base reaction

•

An acid, HF, donates H

+

to form its

conjugate base, F

–

•

A base, H

2

O, accepts H

+

to form its

conjugate acid, H

3

O

+

•

There are two conjugate acid-base

pairs (HF / F

–

) and (H

2

O / H

3

O

+

),

and each pair is related by an H

+

Identify each Bronsted-Lowry Acid and Base

and the Conjugate Acid-Base Pairs

OCl

–

(aq) + H

2

O(

l

) HOCl(aq) + OH

–

(aq)

H

2

C

2

O

4

(aq) + H

2

O(

l

) H

3

O

+

(aq) + HC

2

O

4–

(aq)

HPO

42–

(aq) + H

2

O(

l

) H

3

O

+

(aq) + PO

43–

(aq)

H

3

AsO

4

(aq) + H

2

O(

l

) H

3

O

+

(aq) + H

2

AsO

4–

(aq)

How to Write the Formulas of Conjugate Acids and

Bases

When a species gains a proton (H

+

) to form a conjugate acid,

it gains a +1 charge.

H

2

O

base

zero charge

add H

+

H

3

O

+

+1 charge

When a species loses a proton (H

+

) to form a conjugate base,

it effectively gains a −1 charge.

HBr

acid

zero charge

lose H

+

Br

−

−1 charge

Write the formulas of the conjugate acids

for the following Brønsted–Lowry bases:

Remember: Bases are substances that

accept

a proton (H

+

).

a.

ClCH

2

CO

2–

b.

C

5

H

5

N

c.

SeO

42–

d.

(CH

3

)

3

N

Remember: Add H

+

to write the conjugate acid.

ClCH

2

CO

2

H

C5H5NH

+

HSeO4

–

(CH

3

)

3

NH

+

+ H

+

+ H

+

+ H

+

+ H

+

Write the formulas of the conjugate bases

for the following Brønsted–Lowry acids:

Remember: Acids are substances that donate a proton (H

+

).

a.

HCN

b.

(CH

3

)

2

NH

2+

c.

H

3

PO

4

d.

HSeO

3–

Remember: Remove H

+

to write the conjugate base.

CN

–

(CH3)2NH

H2PO4

–

SeO3

2–

– H

+

– H

+

– H

+

(4)

Identifying Conjugate Pairs in a Bronsted Acid-Base

Reaction

Strategy To find the conjugate base of a species, remove a proton from the formula. To find the conjugate acid of a species, add a proton to the formula. The word proton, in this context, refers to H+_{. Thus, the formula and the charge will}

both be affected by the addition or removal of H+_.

What is (a) the conjugate base on HNO3, (b) the conjugate acid of O2-, (c) the

conjugate base of HSO4-, and (d) the conjugate acid of HCO3-.

Solution (a) NO3-

(b) OH-

(c) SO42-

(d) H2CO3

Think About It A species does not need to be what we think of as an acid in order for it to have a conjugate base. For example, we would not refer to the hydroxide ion (OH-_{) as an acid – but it does have a conjugate base, the oxide ion}

(O2-_{). Furthermore, a species that can either lose or gain a proton, such as}

HCO3-, has both a conjugate base (CO32-) and a conjugate acid (H2CO3).

Identifying Conjugate Pairs in a Bronsted Acid-Base

Reaction

Strategy In each equation, the reactant that loses a proton is the acid and the reactant that gains the proton is the base. Each product is the conjugate of one of the reactants. Two species that differ only by a proton constitute a conjugate pair. Label each of the species in the following equations as an acid, base, conjugate base, or conjugate acid:

(a) HF(aq) + NH3(aq) ⇌ F-(aq) + NH4+(aq)

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

Solution (a) HF loses a proton a becomes F-_{; NH}

3 gains a proton and becomes

NH4+.

HF(aq) + NH3(aq) ⇌ F-(aq) + NH4+(aq)

(b) CH3COO- gains a proton to become CH3COOH; H2O loses a proton to

become OH-_.

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

acid base conjugate base conjugate acid

acid

base conjugate acid conjugate base

Think About It In a Brønsted acid-base reaction, there is always an acid and a base, and whether a substance behaves as an acid or a base depends on what it is combined with. Water, for example, behaves as a base when combined with HCl but behaves as an acid when combined with NH3.

Amphoteric Compounds Can Behave as Either

an Acid or a Base

An Amphoteric compound contains both a H-atom and a lone

pair of e

−

and can behave either as an acid or a base.

H O H

H

2

O as a base

add H

+

H O H

H

+

conjugate acid

H O H

H

2

O as an acid

remove H

+

H O

−

conjugate base

Two Sets of Rules for Naming Acids

Set #1:

Binary acids

Hydrogen

bonded to a

nonmetal

Examples: HCl, HBr, HI, H

2

S

Set #2:

Oxyacids

Hydrogen

bonded to a polyatomic ion

Examples: H

2

SO

4

, HNO

3

, HClO

4

, HCH

3

CO

2

“

Is the compound binary or not binary?

”

Naming Binary Acids

“

hydro - nonmetal root – ic

” acid

1. The word

hydrogen

in the anhydrous compound name is

dropped.

2. The prefix

hydro-

is attached to the stem of the name of

the nonmetal that is combined with hydrogen.

3. The suffix

-ide

on the stem of the nonmetal that is

combined with hydrogen is replaced with the suffix

-ic

.

4. The word

acid

is added the the end of the name as a

separate word.

5. If the nonmetal is sulfur or phosphorus, the stems

sulf-and

phosph-

are expanded to

sulfur-

and

phosphor-

for

pronunciation reasons before the

-ic

suffix is added.

Naming Binary Acids

HCl

hydrogen chloride

hydro + chlor + ic acid hydrochloric acid

HBr

hydrogen bromide

hydro + brom + ic acid hydrobromic acid

H

2

S

hydrogen sulfide

hydro + sulfur + ic acid

(5)

A compound must contain

at least one

ionizable hydrogen atom

to be an acid upon dissolving.

Naming Oxyacids

1. All hydrogens that are written as the first part of the

formula of the acid are dropped.

2. The polyatomic ion that remains is named.

3. If the polyatomic ion ends with the suffix

-ate

, the suffix is

replaced by the suffix

-ic

, and the word

acid

is added.

4. If the polyatomic ion ends with the suffix

-ite

, the suffix is

replaced by the suffix

-ous

, and the word

acid

is added.

5. If the polyatomic ion contains sulfur or phosphorus, the

stems

sulf-

and

phos-

are expanded to

sulfur-

and

phosphor-

for pronunciation reasons before the

-ic

or

-ous

suffixes are added.

Naming Oxyacids

H

2

SO

4

SO

42–

= sulfate ion

sulfuric acid

HNO

3

NO

3–

= nitrate ion

nitric acid

HNO

2

NO

2–

= nitrite ion

nitrous acid

Names and Formulas of Common Acids

Naming Bases

Bases with

OH

–

ions

are named as

the

hydroxide of the metal

in the formula.

Na

OH

sodium

hydroxide

K

OH

potassium

hydroxide

Ba(

OH

)2

barium

hydroxide

Al(

OH

)3

aluminum

hydroxide

Fe(

OH

)3

iron(III)

hydroxide

Name Each of the Following

Acids and Bases

•Al(

OH

)

3

•K

OH

•

H

Br

•

H

NO

2

•

H

2

SO

4

•

H

BrO

2

(6)

Write Formulas for Each of the Following

Acids and Bases

•Barium hydroxide

•Nitric acid

•Sodium hydroxide

•Hydroiodic acid

•Strontium hydroxide

•

Ba(OH)

2

•

HNO

3

•

NaOH

•

HI

•

Sr(OH)

2

Comparing Acids and Bases

Characteristic

Acids

Bases

Arrhenius

Produces H

+

in

water

Produces OH

–

in

water

Bronsted-Lowry

Donates H

+

Accepts H

+

Electrolytes

Yes

Taste

Sour

Bitter

Feel

May sting

Slippery

Litmus

Red

Blue

Phenolphthalein

Colorless

Pink

Strength of Acids

The strength of an acid is a measure of how completely

it

dissociates

(ionizes) when dissolved.

HCl(aq) + H

2

O(

l

)



H

3

O

+

(aq) + Cl

–

(aq)

When a

strong acid

dissolves in water,

100% of the acid dissociates

into ions.

A

single reaction arrow

is used, because the

product is greatly favored at equilibrium.

Common strong acids are

HI

,

HBr

,

HCl

,

H

2

SO

4

, and

HNO

3

.

Strong acid dissociations are not treated as equilibria,

rather as processes that go to completion.

Hydrochloric acid HCl(aq) + H2O(l) H3O+(aq) + Cl–(aq)

HBr(aq) + H2O(l) H3O+(aq) + Br–(aq)

Hydrobromic acid

HI(aq) + H2O(l) H3O+(aq) + I–(aq)

Hydroiodic acid

HNO3(aq) + H2O(l) H3O+(aq) + NO3–(aq)

Nitric acid

HClO3(aq) + H2O(l) H3O+(aq) + ClO3–(aq)

Chloric acid

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

Perchloric acid

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

Sulfuric acid

Strength of Acids

The strength of an acid is a measure of how completely

it

dissociates

(ionizes) when dissolved.

H

3

O

+

(

aq

) + CH

3

COO

−

(

aq

)

CH

3

COOH(

l

) + H

2

O(

l

)

When a

weak acid

dissolves in water, only a

small fraction

of the acid dissociates into ions.

Unequal reaction arrows

may be used, because the

reactants are usually favored at equilibrium.

Common weak acids are

H

3

PO

4

,

HF

,

H

2

CO

3

, and

HCN

.

(7)

Comparing Strong and Weak Acids

A strong acid, HCl, is

completely dissociated

into H

3

O

+

(

aq

) and Cl

−

(

aq

).

A weak acid, CH

3

COOH,

contains mostly

undissociated acid.

6 Strong Acids (Memorize)

HCl hydrochloric acid

HBr hydrobromic acid

HI

hydroiodic acid

HClO

4

perchloric acid

HNO

3

nitric acid

H

2

SO

4

sulfuric acid

Strong acids have weak conjugate bases.

Strength of Bases

When a

strong base

dissolves in water,

100% of the base dissociates

into ions.

Na

+

(

aq

) +

−

OH(

aq

)

NaOH(

s

) + H

2

O(

l

)

NH

4+

(

aq

) +

−

OH(

aq

)

NH

3

(

g

) + H

2

O(

l

)

When a

weak base

dissolves in water, only a

small fraction

of the base dissociates into ions.

Comparing Strong and Weak

Bases

A strong base, NaOH, is

completely dissociated

into Na

+

(

aq

) and

−

OH(

aq

).

A weak base contains

mostly undissociated

base, NH

3

.

Strong Bases are

Hydroxides

of

Groups 1A and 2A Metals

LiOH lithium hydroxide

NaOH sodium hydroxide

KOH potassium hydroxide

RbOH

rubidium hydroxide

CsOH cesium hydroxide

Ca(OH)

2

calcium hydroxide

Sr(OH)

2

strontium hydroxide

Ba(OH)

2

barium hydroxide

Relating Acid and Base Strength

•

A strong acid readily donates a proton, forming

a weak conjugate base.

HCl

strong acid

Cl

−

weak conjugate base

•

A strong base readily accepts a proton, forming

a weak conjugate acid.

OH

−

(8)

Strong acids

have

weak conjugate bases

.

Weak acids

have

strong conjugate bases

.

The

position of the equilibrium

depends on the

strengths of the acids and bases

.

(Competition for the Proton)

The

stronger acid

reacts with the

stronger base

to form the

weaker acid

and the

weaker base

.

H

A

+

B

A

−

+

H

B

+

acid

base

conjugate

base

conjugate

acid

Stronger acid + Stronger base Weaker acid + Weaker base

The

depends on the

.

When the stronger acid and base are the reactants

on the left side, the reaction readily occurs and

the reaction proceeds to the right.

H

A

+

B

A

−

+

H

B

+

stronger

acid

stronger

base

weaker

base

weaker

acid

A larger forward arrow means that products are favored.

If you mix equal concentrations of reactants and

products, will the reaction proceed to the left or the

right (where does the equilibrium lie)?

H2SO4(aq) + NH3(aq) NH4+(aq) + HSO4–(aq)

acid base acid base

H

2

SO

4

is a stronger acid than NH

4+

,

and NH

3

is a stronger base than HSO

4–

.

Therefore, NH

3

gets the proton

and the reaction proceeds to the right.

H2SO4(aq) + NH3(aq) NH4+(aq) + HSO4–(aq)

The

depends on the

.

Since an acid-base reaction forms the stronger acid and

base, equilibrium favors the reactants and only a small

amount of product forms.

H

A

+

B

A

−

+

H

B

+

weaker

acid

weaker

base

stronger

base

stronger

acid

A larger reverse arrow means that reactants are favored.

How to Predict Whether Reactants or Products are

Favored in an Acid-Base Reaction

Step [1]

Identify the acid in the reactants and the

conjugate acid in the products.

acid

base

conjugate

base

conjugate

acid

+

HCN(

g

)

−

OH(

aq

)

−

CN(

aq

)

H

2

O(

l

)

gain of H

+

(9)

How to Predict Whether Reactants or Products are

Favored in an Acid-Base Reaction

Step [2]

Determine the relative strength of the acid

and the conjugate acid.

From Table, HCN is a stronger acid than H

2

O.

Step [3]

Equilibrium favors the formation of the

weaker acid.

+

HCN(

g

)

−

OH(

aq

)

−

CN(

aq

)

H

2

O(

l

)

stronger

acid

weaker

acid

Products are favored.

The Relative Strength of Weak Acids is Determined

by the Value of the Acid Dissociation Constant,

K

a

For the reaction where an acid (HA) dissolves in water,

Ka is the acid dissociation constant. The concentration of water is not included in the equation because its concentration is essentially constant.

K

a

=

H

₃

O

+

[

]

[ ]

A

–

HA

[ ]

K

eq

=

H

₃

O

+

[

]

[ ]

A

–

HA

[ ]

[

H

2

O

]

the following equilibrium constant

can be written:

H

3

O

+

(

aq

) + (

aq

)

HA(

g

) + H

2

O(

l

)

A

−

Stronger Acids Have Larger

K

a

Values;

Weaker acids Have Smaller

K

a

Values

The Relative Strength of Weak Bases is Determined

by the Value of the Base Dissociation Constant,

K

b

For the reaction where an base (B) dissolves in water,

Kb is the base dissociation constant. The concentration of water is not included in the equation because its concentration is essentially constant.

K

b

=

BH

+

éë

ùû

éë

OH

–

ùû

B

[ ]

K

eq

=

BH

+

éë

ùû

éë

OH

–

ùû

B

[ ]

[

H

2

O

]

the following equilibrium constant

can be written:

BH

+

(

aq

) + (

aq

)

B(

aq

) + H

2

O(

l

)

OH

−

Strength of Acids

Because K

a

values are often small and it is hard to work with

small numbers, we often use the pK

a

value to measure the

strength of an acid.

pKa = –log Ka

Like with pH, the smaller the pK

a

value, the stronger the acid.

Example: pK

a

of acetic acid is –log(1.8 x 10

–5

) = 4.74

AutoIonization of Water

One water molecule acts as an acid and donates H

+

to the other

water molecule which acts as a base and accepts the H

+

.

= concentration of H3O+ in units of molarity (mol/L)

= concentration of OH–_{in units of molarity (mol/L)}

H O H

base

H O H

H

+

conjugate

acid

H O H

acid

conjugate

base

H

O

−

+

loss of H

+

(10)

The Ion Product Constant, K

w

, for Water

Experimentally it can be shown that in

pure water at 25

°

C,

[H3O

+

] = [OH

–

] = 1.00 x 10

–7

M

.

=

(

1

.

00

´

10

–7

M

)

(

1

.

00

´

10

–7

M

)

K

w

= 1.00 x 10

–14

=

1.00

´

10

–14

M

2

K

w

=

[

H

3

O

+

]

OH

[ ]

–

H

2

O(

l

) + H

2

O(

l

) H

3

O

+

(aq) + OH

–

(aq)

Pure Water is Neutral: [H

3

O

+

] = [OH

–

]

•

In pure water, the ionization of

water molecules produces

small but equal quantities of

H

3

O

+

and OH

–

ions

•

Molar concentrations are

indicated in brackets

[H

3

O

+

] = 1.0 x 10

–7

M

[OH

–

] = 1.0 x 10

–7

M

Acidic Solutions:

[H

3

O

+

] > [OH

–

]

Adding acid to pure water

•

Increases the [H

3

O

+

]

•

Causes the [H3O

+

] > 1.0 x 10

–7

M

•

Decreases the [OH

–

]

Basic Solutions:

[OH

–

] > [H

3

O

+

]

Adding a base to water

•

Increases the [OH

–

]

•

Causes the [OH

–

] > 1.0 x 10

–7

M

•

Decreases the [H

3

O

+

]

Acidic, Neutral, and Basic Solutions

Comparison of [H

3

O

+

] and [OH

–

]

Using the

K

w

to Calculate

[H

3

O

+

] and [OH

–

] in Solution

To calculate [OH

−

] when

[H

3

O

+

] is known:

To calculate [H

3

O

+

] when

[OH

−

] is known:

(11)

Using the

K

w

to Calculate

[H

3

O

+

] and [OH

–

] in Solution

If the [H

3

O

+

] in a cup of coffee is 1.0 x 10

−5

M,

Then the [OH

−

] can be calculated as follows:

[OH

−

]

=

K

w

[H

3

O

+

]

=

1.0 x 10

−14

1.0 x 10

−5

=

1.0 x 10

−9

M

In this cup of coffee, therefore,

[H

3

O

+

] > [OH

–

]

And the solution is acidic overall.

Using

K

w

and [H

3

O

+

]

to

Calculate

[OH

–

]

in Solution

1. [H

3

O

+

] = 1.2 x 10

–5

2. [H

3

O

+

] = 0.27

OH–

[ ]

=1.00´10–14 H3O+

[

]

=1.00´10

–14

1.2´10–5 =8.3´10–10 M

OH–

[ ]

=1.00´10–14 H3O+

[

]

=1.00´10

–14

0.27 =3.7´10

–14_M

Using

K

w

and [OH

–

]

to

Calculate

[H

3

O

+

]

in Solution

1. [OH

–

] = 0.0071

2. [OH

–

] = 4.2 x 10

–4

H3O+

[

]

=1.00´10–14 OH–

[ ]

=1.00´10

–14

0.0071 =1.4´10

–12_M

H3O+

[

]

=1.00´10–14 OH–

[ ]

=1.00´10

–14

4.2´10–4 =2.4´10–11 M

Using K

w

to Calculate [H

3

O

+

] and [OH

–

] in Solution

Strategy Use the value of Kw to determine [OH-] when [H3O+] = 0.10 M.

The concentration of hydronium ions in stomach acid is 0.10 M. Calculate the concentration of hydroxide ions in stomach acid at 25°C.

SolutionKw = [H3O+][OH-] = 1.0×10-14 at 25°C. Rearranging to solve for [OH

-],

[OH-_{] =}1.0×10-14

[H3O+]

[OH-_{] =}1.0×10-14

0.10 [OH-_{] = 1.0×10}-13_M

Think About It Remember that the equilibrium constants are temperature dependent. The value of Kw = 1.0×10-14 only at 25°C.

pH

It

’

s hard to work with numbers like 1.00 x 10

–7

M, so chemists

have adopted a shortcut notation known as

pH

.

pH is the negative logarithm of the hydronium ion

concentration.

To simplify the equations, we substitute H

+

for H

3

O

+

.

[H

+

] = 10

–pH

= antilog(–pH)

pH = –log[H

+

]

The

Lower

the

pH

,

the

Higher

the Concentration of

H

3

O

+

Acidic solution

:

pH < 7



[H

3

O

+

] > 1 x 10

−7

Basic solution: pH > 7



[H

3

O

+

] < 1 x 10

−7

(12)

The pH Scale Has Values That Range From 0-14

7 neutral

<7 acidic >7 basic

Lemon juice (acidic), pH = 2 Household ammonia (basic), pH = 11

Blood (basic), pH = 7.5 Oven cleaners (basic), pH = 14

Milk (acidic), pH = 6.6

Beer & wine (acidic), pH = 4 Cola (acidic), pH = 3 Milk of Magnesia (basic), pH = 10 Detergents (basic), pH = 9

Identify Each Solution as

Acidic

,

Basic

,

Neutral

HCl with a pH = 1.5

Pancreatic fluid, [H3O

+

] = 1 x 10

–8

M

Sprite soft drink, pH = 3.0

pH = 7.0

[OH

–

] = 3 x 10

–10

M

[H

3

O

+

] = 5 x 10

–12

M

Variation in pH

Values in the

Human Body

Hydrangea

They produce

blue flowers

in

acidic

soil. Just add

aluminum sulfate

[Al

2

(SO

4

)

3

] or peat moss.

Hydrangeas produce

pink flowers

in

alkaline (

basic

) soil.

Just add lime (CaO).

Significant Figures in pH

What is the pH of a solution with

[H

3

O

+

] = 1.2 x 10

–5

M

?

pH = –log [H

3

O

+

]

pH = –(–4.92)

The number of figures to the

right of the decimal

in a pH is the same as the

number of significant figures

in the [H

+

] value.

pH = –log (1.2 x 10

–5

)

2 digits

pH = 4.92

2 decimal places

a. [H+_{] = 3.0 x 10}–8_M

pH = –log [H+] = –log (3.0 x 10–8) =

b. [OH–_{] = 7.0 x 10}–3_M _{(HINT: Use}_K

w to obtain [H3O+].)

pH = –log [H+_{] = –log (1.4 x 10}–12_{) = 11.85, basic} 7.52, basic

Calculate the pH of the following solutions. Classify each

solution as acidic or basic.

H3O+

[

]

=1.00´10–14 OH–

[ ]

=1.00´10

–14

7.0´10–3 =1.4´10– 12_M 1. Enter the [H3O+] value using the EE key. 2. Press the log key and change the sign.

(13)

Calculating pH from [H

3

O

+

]

Strategy Given [H3O+], use pH = –log[H3O+] to solve for pH.

Determine the pH of a solution at 25°C in which the hydronium ion concentration is (a) 3.5×10-4_M_{, (b) 1.7}×10-7_M_{, and (c) 8.8}×10-11_M_.

Solution

(a) pH = –log(3.5×10-4_{) = 3.46}

(b) pH = –log(1.7×10-7_{) = 6.77}

(c) pH = –log(8.8×10-11_{) = 10.06}

Think About It When a hydronium ion concentration falls between two “benchmark” concentrations in Table 16.4, the pH falls between the two corresponding pH values. In part (c), for example, the hydronium ion concentration (8.8×10-11_M_{) is greater than 1.0}×10-11_M_{but less than}

1.0×10-10_M_{. Therefore, we expect the pH to be between 11.00 and 10.00.}

Calculating [H

3

O

+

] from pH

If the pH of a solution is 8.50, what is the [H

3

O

+

]?

pH = −log [H

3

O

+

]

8.50 = −log [H

3

O

+

]

−8.50 = log [H

3

O

+

]

antilog (−8.50 ) = [H

3

O

+

]

[H

3

O

+

] = 3.2 x 10

−9

M

The solution is basic because [H

3

O

+

] > 1 x 10

–7

M.

2 decimal places

2 digits

Summary of Steps:

Calculating [H

3

O

+

] From pH:

[H

3

O

+

] = 1 x 10

–pH

Calculate the [H

3

O

+

] for a pH value of 8.0

1.

Enter the pH value, then change the sign: –8.0

2.

Convert –pH to concentration: Use 2

nd

function key

and then 10

x

key or inverse key and then log key.

3.

Adjust the sig figs (1 digit following the decimal point

equals 1 digit in the [H

3

O

+

]).

The pH values of specific samples of food items are

listed below. Convert each value to [H

+

].

a. Soft drink, pH = 2.91

[H

+

] = 10

–pH

= antilog(–2.91) =

b. Tomato juice, pH = 4.11

[H

+

] = 10

–pH

= antilog(–4.11) =

c. Lemon juice, pH = 2.32

[H

+

] = 10

–pH

= antilog(–2.32) =

4.8 x 10–3_M

7.8 x 10–5_M

1.2 x 10–3_M

Calculating

[H

3

O

+

] From pH

Calculating [H

3

O

+

] from pH

Strategy Given pH, use [H3O+] = 10-pH to calculate [H3O+].

Calculate the hydronium ion concentration in a solution at 25°C in which the pH is (a) 4.76, (b) 11.95, and (c) 8.01.

Solution

(a) [H3O+] = 10-4.76 = 1.7×10-5 M

(b) [H3O+] = 10-11.95 = 1.1×10-12 M

(c) [H3O+] = 10-8.01 = 9.8×10-9 M

Think About It Think About It If you use the calculated hydronium ion concentrations to recalculate pH, you will get numbers slightly different from those given in the problem. In part (a), for example, −log(1.7×10-5_{) = 4.77. The small difference between this and}

4.76 (the pH given in the problem) is due to a rounding error. Remember that a concentration derived from a pH with two digits to the right of the decimal point can have only two significant figures. Note also that the benchmarks can be used equally well in this circumstance. A pH between 4 and 5 corresponds to a hydronium ion concentration between 1.7×10-4_M_{and 1.0}×10-5_M.

Fun with Logarithms

H

3

O

+

éë

ùû

OH

éë ùû

–

=

1.0

´

10

–14

Take the negative logarithm of both sides…

-

log

(

éë

H

3

O

+

ùû

OH

éë ùû

–

)

=

-

log

(

1.0

´

10

–14

)

-

log

éë

H

3

O

+

ùû-

log

éë ùû

OH

–

=

14.00

(14)

Calculating pOH from [OH

–

]

Strategy Given [OH-_{], use pOH = –log[OH}-_{] to calculate pOH.}

Determine the pOH of a solution at 25°C in which the hydroxide ion concentration is (a) 3.7×10-5_M_{, (b) 4.1}×10-7_M_{, and (c) 8.3}×10-2_M_.

Solution

(a) pOH = –log(3.7×10-5_{) = 4.43}

(b) pOH = –log(4.1×10-7_{) = 6.39}

(c) pOH = –log(8.3×10-2_{) = 1.08}

Think About It Remember that the pOH scale is, in essence, the reverse of the pH scale. On the pOH scale, numbers below 7 indicate a basic solution, whereas number above 7 indicate an acidic solution. The pOH benchmarks (abbreviated in Table 16.6) work the same way the pH benchmarks do. In part (a), for example, a hydroxide ion concentration between 1×10-4_M_{and 1}×10-5_M_corresponds

to a pOH between 4 and 5.

Calculating [OH

-

] from pOH

Strategy Given pOH, use [OH-_{] = 10}-pOH_{to calculate [OH}-_].

Calculate the hydroxide ion concentration in a solution at 25°C in which the pOH is (a) 4.91, (b) 9.03, and (c) 10.55.

Solution

(a) [OH-_{] = 10}-4.91_{= 1.2}×10-5_M

(b) [OH-_{] = 10}-9.03_{= 9.3}×10-10_M

(c) [OH-_{] = 10}-10.55_{= 2.8}×10-11_M

Think About It Use the benchmark pOH values to determine whether these solutions are reasonable. In part (a), for example, the pOH between 4 and 5 corresponds to [OH-_{] between 1}×10-4_M_{and 1}×10-5_M_.

Testing the pH of a Solution

The pH of solutions can be measured using

•

A pH meter

•

pH paper

•

Indicators that have specific colors at different pH values

pH in Solutions of

Strong

Acids

Calculate the pH of a

0.050 M HClO

4

solution.

HClO

4

(aq) + H

2

O(

l

) H

100% 3

O

+

(aq) + ClO

4–

(aq)

Source of hydronium ions:

1. from HClO

4

(0.050 M)

2. from autoionization of H

2

O

small and can be ignored

[H

3

O

+

] = [HClO

4

] = 0.050 M

pH = –log(0.050) = 1.30

Relating the Concentration of a Strong Acid to the pH of

an Aqueous Solution

Strategy HI, HNO3, and HClO4 are all strong acids, so the concentration of

hydronium ions in each solution is the same as the stated concentration of the acid. Use pH = –log[H3O+] to calculate pH.

Calculate the pH of an aqueous solution at 25°C that is (a) 0.035 M in HI, (b) 1.2×10-4_M_{in HNO}

3, and (c) 6.7×10-5M in HClO4.

Solution (a) [H3O+] = 0.035 M

pH = –log(0.035) = 1.46

(b) [H3O+] = 1.2×10-4M

pH = –log(1.2×10-4_{) = 3.92}

(c) [H3O+] = 6.7×10-5M

pH = –log(6.7×10-5_{) = 4.17}

Think About It Again, note that when a hydronium ion concentration falls between two of the benchmark concentrations in Table 16.4, the pH falls between the two corresponding pH values. In part (b), for example, the hydronium ion concentration of 1.2×10-4_M_{is greater than 1}×10-4_M_{and less than 1}×10-3_M_.

Therefore, we expect the pH to be between 4.00 and 3.00.

Relating the Concentration of a Strong Acid to the pH of

an Aqueous Solution

Strategy Use [H3O+] = 10-pH to convert from pH to [H3O+]. In a strong acid

solution, [H3O+] is equal to the acid concentration.

Calculate the concentration on HCl in a solution at 25°C that has pH (a) 4.95, (b) 3.45, and (c) 2.78.

Solution

(a) [HCl] = [H3O+] = 10-4.95 = 1.1×10-5 M

(b) [HCl] = [H3O+] = 10-3.45 = 3.5×10-4 M

(c) [HCl] = [H3O+] = 10-2.78 = 1.7×10-3 M

(15)

pH in Solutions of

Weak

Acids

The ionization of a weak monoprotic acid HA in water is represented by:

K

a

is called the

acid ionization constant

.

The larger the value of

K

a

, the stronger the acid.

HA(

aq

) + H

2

O(

l

)

⇌

H

3

O

+

(

aq

) + A

–

(

aq

)

 

+ 3 a

H O

A

HA

K





 





 





Solution (at 25

°C) Ka pH

0.10 M HF 7.1 x 10–4 _2.09

0.10 M CH3COOH 1.8 x 10–5 2.87

pH in Solutions of

Weak

Acids

K

a

for CH

3

COOH is 1.8 billion times larger than K

w

.

Thus, H

3

O

+

from H

2

O is so small, it can be ignored.

0.10 M CH

3

COOH

solution.

CH

3

COOH(aq) + H

2

O(

l

) H

3

O

+

(aq) + CH

3

COO

–

(aq)

Source of hydronium ions:

1. from CH

3

COOH (0.10 M)

2. from autoionization of H

2

O

K

a

= 1.8 x 10

–5

K

w

= 1.0 x 10

–14

[H

3

O

+

] ≠ [CH

3

COOH] = 0.10 M

pH in Solutions of

Weak

Acids

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.10 ~0 0

x x

–x

x

0.10 – x x

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO–(aq)

K

a

=

H

3

O

+

éë

ùû

éë

CH

3

COO

-

ùû

CH

3

COOH

[

]

=

x

( )

x

0.10

-

x

(

)

=

1.8

´

10

-5

pH in Solutions of

Weak

Acids

x

2

0.10

-

x

(

)

»

x

2

0.10

(

)

=

1.8

´

10

-5

x

2

=

1.8

´

10

-6

x

= [H

3

O

+

] = 1.3 x 10

–3

M

pH = –log(1.3 x 10

–3

) = 2.89

K

a

=

H

3

O

+

éë

ùû

éë

CH

3

COO

-

ùû

CH

3

COOH

[

]

=

x

( )

x

0.10

-

x

(

)

=

1.8

´

10

-5

pH in Solutions of

Weak

Acids

Calculate the pH of a 0.50

M

HF solution at 25

°

C.

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

 

+

3 4

a

H O

F

7.1 10

HF

K

 



  



  





Initial concentration (M)

Change in concentration (M)

Equilibrium concentration (M)

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

–x

0.50 – x

+x +x

x x

0.50 ₀ ₀

pH in Solutions of

Weak

Acids

 

+ 3 a

H O

F

HF

K





  



  



Initial concentration (M) 0.50 0 0

Change in concentration (M) –x +x +x

Equilibrium concentration (M) 0.50 – x x x

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

  

4

a

0.50

x x

7.1 10

K

x









Use quadratic formula to solve –or –

(16)

pH in Solutions of

Weak

Acids

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

  

4 a _0.50x x 7.1 10 K _ _x_ _ 

 Ka

  

x x_0.50 7.1 104 

  

x2_{= (0.50)(7.1 x 10}–4_{) = 3.55 x 10}–4

x = 1.9 x 10–2

simplifies

pH in Solutions of

Weak

Acids

Change in concentration (M) –1.9 x 10–2 _{+1.9 x 10}–2_{+1.9 x 10}–2

Equilibrium concentration (M) 0.48 1.9 x 10–2 _{1.9 x 10}–2

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

a

0.019

0.50

M

100% 3.8%

K

M







pH = –log(0.019) = 1.72

The shortcut is acceptable to use if

the calculate value of

x

is less than

5% of the initial acid concentration

Using K

a

to Determine the pH of a weak Acid Solution

Strategy Construct an equilibrium table, and express the equilibrium and concentration of each species in terms of x. Solve for x using the approximation shortcut, and evaluate whether or not the approximation is valid. Use pH = –log[H3O+] to determine pH.

The Ka of hypochlorous acid (HClO) is 3.5×10-8_{. Calculate the pH of a solution}

at 25°C that is 0.0075 M in HClO.

HClO(aq) + H2O(l) ⇌ H3O+(aq) + ClO–(aq)

Using K

a

to Determine the pH of a weak Acid Solution

Solution These equilibrium concentrations are then substituted into the equilibrium expression to give

Assuming that 0.0075 – x ≈ 0.0075,

Solving for x, we get

x = = 1.62×10-5_M

According to the equilibrium table, x = [H3O+]. Therefore,

pH = –log(1.62×10-5_{) = 4.79}

Ka = _{0.0075 –}(x)(x)_x= 3.5×10-8

x2

0.0075 = 3.5×10-8 x2= (3.5×10-8)(0.0075)

10

10 625 . 2 _ 

Think About It We learned in Section 16.3 that the concentration of hydronium ion in pure water at 25°C is 1.0×10-7_M_{, yet we use 0}_M_as

the starting concentration to solve for the pH of a weak acid. The reason for this is that the actual concentration of hydronium ion in pure water is insignificant compared to the amount produced by the ionization of the weak acid. We could use the actual concentration of hydronium as the initial concentration, but doing so would not change the result because (x + 1.0×10-7₎_M_≈_{x M}_{. In solving problems of this type, we neglect the}

small concentration of H+_{due to the autoionization of water.}

pH in Solutions of

Strong

Bases

Calculate the pH of a 0.0050 M Ca(OH)

2

solution.

Ca(OH)

2

(aq) Ca

100%

2+

(aq) + 2OH

–

(aq)

[OH

–

] =

2

[Ca(OH)

2

] = 0.010 M

pOH = –log(0.010) = 2.00

pH = 14.00 – pOH = 12.00

Calculations Involving [OH

–

], pOH, and pH

Strategy LiOH, Ba(OH)2, and KOH are all strong bases. Use reaction

stoichiometry to determine the hydroxide ion concentration and pOH = –log[OH-_]

to determine pOH.

Calculate the pOH of the following aqueous solutions at 25°C: (a) 0.013 M LiOH, (b) 0.013 M Ba(OH)2, and (c) 9.2×10-5M KOH.

Solution (a) The hydroxide ion concentration is simply equal to the concentration of the base. Therefore, [OH-_{] = [LiOH] = 0.013}_M_.

pOH = –log(0.013) = 1.89 (b) The hydroxide ion concentration is twice that of the base:

Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq)

Therefore, [OH-_{] = 2[Ba(OH)}₂_{] = 2(0.013}_M_{) = 0.026}_M_.

pOH = –log(0.026) = 1.59

(c) The hydroxide ion concentration is equal to the concentration of the base. Therefore, [OH-_{] = [KOH] = 9.2}×10-5_M_.

pOH = –log(9.2×10-5_{) = 4.04}

(17)

Calculations Involving [OH

–

], pOH, and pH

Strategy Use pH + pOH = 14.00 to convert from pH to pOH and [OH-_{] =}

10-pOH_{to determine the hydroxide ion concentration. Consider the stoichiometry}

of dissociation in each case to determine the concentration of the base itself. An aqueous solution of a strong base has pH 8.15 at 25°C. Calculate the original concentration of base in the solution (a) if the base is NaOH and (b) if the base is Ba(OH)2.

Solution

pOH = 14.00 – 8.15 = 5.85 [OH-_{] = 10}-5.85_{= 1.41}_×₁₀-6_M

(a) The dissociation of 1 mole of NaOH produces 1 mole of OH-. Therefore, the concentration of base is equal to the concentration of hydroxide ion.

[NaOH] = [OH-_{] = 1.41}×10-6_M

(b) The dissociation of 2 mole of Ba(OH)2 produces 2 moles of OH-. Therefore,

the concentration of base is only one-half the concentration of hydroxide ion. [Ba(OH)2] = [OH1 2 -] = 7.1×10-7 M

Think About It Alternatively, we could determine the hydroxide ion concentration using [H3O+] = 10-8.15 = 7.1×10-7M and

Once [OH-] is known, the solution is the same as shown previously. [OH-_{] =}1.0×10-14

7.1×10-9 = 1.4×10-6M

pH in Solutions of

Weak

Bases

K

b

for strychnine is 180 million times larger than K

w

.

Thus, OH

–

from H

2

O is so small, it can be ignored.

0.012 M strychnine (C

21

H

22

N

2

O

2

)

solution.

C

21

H

22

N

2

O

2

(aq) + H

2

O(

l

) C

21

H

23

N

2

O

2+

(aq) + OH

–

(aq)

Source of hydroxide ions:

1.from strychnine (0.012 M)

2.from autoionization of H

2

O

K

b

= 1.8 x 10

–6

K

w

= 1.0 x 10

–14

[OH

–

] ≠ [strychnine] = 0.012 M

pH in Solutions of

Weak

Bases

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.012 0 ~0

x x

–x

x

0.012 – x x

K

b

=

C

21

H

23

N

2

O

2

+

éë

ùû

éë

OH

-

ùû

C

21

H

22

N

2

O

2

[

]

=

x

( )

x

0.012

-

x

(

)

=

1.8

´

10

-6

C21H22N2O2(aq) + H2O(l) C21H23N2O2+(aq) + OH–(aq)

pH in Solutions of

Weak

Bases

x

2

0.012

-

x

(

)

»

x

2

0.012

(

)

=

1.8

´

10

-6

x

2

=

2.2

´

10

-8

x

= [OH

–

] = 1.5 x 10

–4

M

pOH = –log(1.5 x 10

–4

) = 3.83

pH = 14.00 – pOH = 10.17

K

b

=

C

21

H

23

N

2

O

2

+

éë

ùû

éë

OH

-

ùû

C

21

H

22

N

2

O

2

[

]

=

x

( )

x

0.012

-

x

(

)

=

1.8

´

10

-6

Percent Ionization in Solutions of Weak Acids

Using the example of a weak acid a few slides back…

=

1.3

´

10

-3

(

)

0.10

(

)

´

100%

=

1.3%

[CH

3

COO

–

] = 1.3 x 10

–3

M

[CH

3

COOH] = 0.10 M

Percent ionization=amount dissociated (mol/L)

initial concentration (mol/L)´100%

Change in concentration (M) –1.9 x 10–2 _{+1.9 x 10}–2_{+1.9 x 10}–2

Equilibrium concentration (M) 0.48 1.9 x 10–2 _{1.9 x 10}–2

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

A quantitative measure of the degree of ionization is

percent

ionization.

 

3 0eq H O percent ionization _HA 100%



 

 

 

0.019

percent ionization

100% 3.8%

0.50

M

(18)

In general, the percent dissociation depends on the

acid and INCREASES

with increasing value of Ka

Also, for a given weak acid, the percent dissociation

INCREASES

with decreasing concentration

.

0.00% 0.50% 1.00% 1.50% 2.00% 2.50% 3.00% 3.50% 4.00% 4.50%

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000

Per

ce

nt

d

iss

oci

at

io

n

Concentration of CH3COOH (M)

Percent Ionization in Solutions of Weak Acids

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

0.027

percent ionization 100% 2.7% 1.0

M M

  

Solution (at 25

°C) pH % ionization

0.5 M HF 1.72 3.8

1.0 M HF 1.57 2.7

Calculate the percent ionization of a 1.0

M

HF solution at 25

°

C.

–2.7 x 10–2 ₊_{2.7 x 10}–2₊_{2.7 x 10}–2

2.7 x 10–2 _{2.7 x 10}–2

0.97

HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)

Solution (at 25

°C) pH % ionization

0.5 M HF 1.72 3.8

1.0 M HF 1.57 2.7

Calculating Percent Ionization of Weak Acid Solutions

Strategy Using the procedure described in Worked Example 16.13, we construct an equilibrium table and for each concentration of acetic acid, we solve for the equilibrium concentration of H+_{. We use pH = –log[H}

3O+] to find pH, and

the equation below to find percent ionization. Ka for acetic acid is 1.8×10-5.

Determine the pH and percent ionization for acetic acid solutions at 25°C with concentrations (a) 0.15 M, (b) 0.015 M, and (c) 0.0015 M.

 

3 0eq H O percent ionization 100%

HA



 

 

 

Calculating Percent Ionization of Weak Acid Solutions

Solution (a)

Solving for x gives [H3O+] = 0.0016 M and pH = –log(0.0016) = 2.78.

(b) Solving the same way as part (a) gives [H3O+] = 5.2×10-4M and pH = 3.28.

CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq)

0.0016 M 0.15 M

percent ionization= × 100% = 1.1%

5.2×10-4_M

0.015 M

percent ionization= × 100% = 3.5%

Calculating Percent Ionization of Weak Acid Solutions

Solution (c) Solving the quadratic equation, or using successive approximation [Appendix 1] gives [H3O+] = 1.6×10-4M and pH = 3.78.

1.6×10-4_M

0.0015 M

percent ionization= × 100% = 11%

Think About It Check your work by using the calculated value of Ka to solve

(19)

Determine the

K

a

of a

Weak Acid

Determine the

K

a

of a weak acid that has a concentration of 0.25

M

and a pH of 3.47 at 25

°

C.

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)

 

+ 3 a

H O A ? HA K           

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)

0.2497 3.39 x 10–4 _{3.39 x 10}–4

+3.39 x 10–4_{+3.39 x 10}–4

–3.39 x 10–4

H3O+ = 10–3.47 = 3.39 x 10–4 M

Determine the

K

a

of a

Weak Acid

Determine the Ka of a weak acid that has a concentration of 0.25 M and a

pH of 3.47 at 25°C.

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)

0.2497 3.39 x 10–4 _{3.39 x 10}–4

+3.39 x 10–4_{+3.39 x 10}–4

–3.39 x 10–4

 

+ 3 a

H O

A

HA

K





 





 









2 4 7 a

3.39 10

4.6 10

0.2497

K

 







Using pH to Determine K

a

Strategy Determine the hydronium ion concentration from the pH. Use the hydronium ion concentration to determine the equilibrium concentrations of the other species, and plug the equilibrium concentrations into the equilibrium expressions to evaluate Ka.

Aspirin (acetylsalicylic acid, HC9H7O4) is a weak acid. It ionizes in water

according to the equation

HC9H7O4(aq) + H2O(l) ⇌ H3O+(aq) + C9H7O4-(aq)

A 0.10-M aqueous solution of aspirin has a pH of 2.27 at 25°C. Determine the Ka of aspirin.

Using pH to Determine K

a

Solution [H3O+] = 10–2.27 = 5.37×10-3M

To calculate Ka, though, we also need the equilibrium concentrations of C9H7O4

-and HC9H7O4. The stoichiometry of the reaction tells us that [C9H7O4-] = [H3O+].

Furthermore, the amount of aspirin that has ionized is equal to the amount of hydronium ion in solution. Therefore, the equilibrium concentration of aspirin is (0.10 – 5.37×10-3₎_M_{= 0.095}_M_.

The Ka of aspirin is 3.0×10-4.

Change in concentration (M) –0.005 +5.37×10-3₊_5.37×10-3

Equilibrium concentration (M) 0.095 5.37×10-3 _5.37×10-3

HC9H7O4(aq) + H2O(l) ⇌ H3O+(aq) + C9H7O4-(aq)

Ka = [H3O +_][C

9H7O4-]

[HC9H7O4] = 3.0×10 -4

=(5.37_0.095×10-3)2

Think About It Check your work by using the calculated value of Ka to solve for the pH of a 0.10-M solution of aspirin.

Using K

b

to Calculate the pH of a Weak Base Solution

Strategy Construct an equilibrium table, and express equilibrium concentrations in terms of the unknown x. Plug these equilibrium concentrations into the equilibrium expression, and solve for x. From the value of x, determine the pH. What is the pH of a 0.040 M ammonia solution at 25°C.

NH

3

(

aq

) + H

2

O(

l

)

⇌

NH

4+

(

aq

) +

OH

-

(

aq

)

Using K

b

to Calculate the pH of a Weak Base Solution

Solution The equilibrium concentrations are substituted into the equilibrium expression to give

Assuming that 0.040 – x ≈ 0.040 and solving for x gives

x2_{= (1.8}_×₁₀-5_{)(0.040) = 7.2}_×₁₀-7

x = = 8.5×10-4_M

According to the equilibrium table, x = [OH-_{]. Therefore, pOH = – log(}_x_):

–log(8.5×10-4_{) = 3.07}

and pH = 14.00 – pOH = 14.00 – 3.07 – 10.93. The pH of a 0.040-M solution of NH3 at 25°C is 10.93.

Kb = [NH4 +_][OH-_]

[NH3] = 1.8×10 -5

=_{0.040 –}(x)(x)_x

= 1.8×10-5

(x)(x) 0.040 – x ≈(0.040 x)(x)

7 10 2 . 7 _ 

Think About It It is a common error in Kb problems to forget that

(20)

Using pH to Determine K

b

Strategy Use pH to determine pOH, and pOH to determine the hydroxide ion concentration. From the hydroxide ion concentration, use reaction stoichiometry to determine the other equilibrium concentrations and plus those concentrations into the equilibrium expression to evaluate Kb.

Caffeine, the stimulant in coffee and tea, is a weak base that ionizes in water according to the equation

C8H10N4O2(aq) + H2O(l) ⇌ HC8H10N4O2+(aq) + OH-(aq)

A 0.15-M solution of caffeine at 25°C has a pH of 8.45. Determine the Kb of

caffeine.

Using pH to Determine K

b

Solution pOH = 14.00 – 8.45 – 5.55; [OH-_{] = 10}-5.55_{= 2.82}×10-6_M

Based on the reaction stoichiometry, [HC8H10N4O2+] = [OH-], and the amount of

hydroxide ion in solution at equilibrium is equal to the amount of caffeine that has ionized. At equilibrium, therefore,

[C8H10N4O2] = (0.15 – 2.82×10-6) M ≈ 0.15 M

Kb = [HC8H10N4O2 +_][OH-_]

[C8H10N4O2] = 5.3×10 -11

= (2.82×10_0.15-6)2

Change in concentration (M) –2.82×10-6 ₊_2.82×10-6 ₊_2.82×10-6

Equilibrium concentration (M) 0.15 2.82×10-6 _2.82×10-6

C8H10N4O2(aq) + H2O(l) ⇌ HC8H10N4O2+(aq) + OH-(aq)

Think About It Check your answer using the calculated Kb to

determine the pH ofa 0.15-M solution.

The pH of a Polyprotic Acid

Polyprotic acids can donate more than one proton. There is a K

a

value for each successive dissociation ( , , etc.).

For a typical weak polyprotic acid,

Ka1 Ka2

K

a1

>

K

a2

>

K

a3

Sulfuric acid (H

2

SO

4

) is unique in that it is a

strong

acid in its first

dissociation step and a

weak

acid in its second step.

The pH of a Polyprotic Acid

Diprotic and polyprotic acids undergo successive ionizations, losing

one proton at a time, and each has a

K

a

associate with it.

K

a1

>

K

a2

For a given acid, the first ionization constant is much larger than the

second, and so on.

H

2

CO

3

(

aq

)

⇌

H

+

(

aq

) + HCO

3–

(

aq

)

HCO

3–

(

aq

)

⇌

H

+

(

aq

) + CO

32–

(

aq

)

+ 2 3 a2

3

H CO

HCO K

 

        

 

 





+ 3 a1

2 3

H HCO

H CO K



        

Calculating Equilibrium Concentrations of All Species for

a Polyprotic Acid

Strategy Follow the same procedure for each ionization as for the determination of equilibrium co