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Chemistry 102 Chapter 15

1

ACID-BASE CONCEPTS

General Properties:

ACIDS BASES

Taste sour Bitter

Change color of

indicators

Blue Litmus turns red no change

Red Litmus no change turns blue

Phenolphtalein Colorless turns pink

Neutralization React with bases to

form salt and water

React with acids to

form salt and water

Arrhenius Concept of Acids and Bases:

Arrhenius definition of acids and bases in based on their behavior in water (aqueous solutions)

ACIDS BASES

Substances that dissolved in water

to increase the concentration of

H3O+ (hydronium ions)

Accepted simplification: H+

H+ is always attached to a water

molecule.

H

O H

H

Substances that dissolved in water

to increase the concentration of

OH–(hydroxide ions)

O H


2

STRONG ACIDS AND BASES

Strong acids and bases completely ionize in aqueous solutions and are therefore strong electrolytes.

STRONG ACIDS STRONG BASES

Completely ionize in aqueous solution to form

H3O+ and an anion

Completely ionize in aqueous solution to

form OH– and a cation

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

– (aq) BOH (s) 2H O

B+ (aq) + OH

– (aq)

Examples:

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

– (aq)

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3

– (aq)

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

– (aq)

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

– (aq)

Examples:

NaOH (s) 2H O Na

+ (aq) + OH

– (aq)

KOH (s) 2H O K

+ (aq) + OH

– (aq)

Ca(OH)2 (s) 2H O Ca

2+ (aq) + 2 OH

– (aq)

Ba(OH)2 (s) 2H O Ba

2+ (aq) + 2 OH

– (aq)

NEUTRALIZATION REACTIONS BETWEEN STRONG ACIDS AND STRONG BASES

HA (aq) + H2O (l) + BOH (aq) BA (aq) + 2 H2O (l)

strong acid strong base soluble salt water

H3O+

(aq) + A

(aq) + B+

(aq) + OH

(aq) B+

(aq) + A

(aq) + 2 H2O (aq)

H3O

+ (aq) + A

(aq) + B

+ (aq) + OH

(aq) B

+ (aq) + A

(aq) + 2 H2O (aq)

Net Ionic Equation H3O

+ (aq) + OH

(aq) 2 H2O (aq)

Accepted simplification for N.I.E. H+(aq) + OH

(aq) H2O(l)

NOTE: It has been observed that all neutralization reactions between strong acids and bases:

are exothermic

have the same enthalpy of reaction per mol of H+

= –55.90 kJ

CONCLUSION:

The neutralization reaction between any strong acid and any strong base is:

H+(aq) + OH

(aq) H2O(l) H

0 = –55.90/mol H

+ ions


3

WEAK ACIDS AND BASES

Weak acids and bases partially ionize in aqueous solutions and are therefore weak electrolytes.

Weak acids and bases exist in reversible reactions (equilibrium) with the corresponding ions.

WEAK ACIDS:

HC2H3O2 (aq) + H2O (l) H3O

+ (aq) + C2O3O2

(aq)

HF (aq) + H2O (l) H3O

+ (aq) + F

(aq)

WEAK BASES:

NH3 (g) + H2O (l) NH4

+(aq) + OH

(aq)

Limitations of the Arrhenius Concept:

1. Considers acid-base reactions only in aqueous solutions

2. Singles out the OH ion as the source of base character (other species can play a similar role)


4

BRONSTED-LOWRY CONCEPT OF ACIDS AND BASES

Bronsted and Lowry defined acids and bases in terms of H

+ (proton) transfer.

ACIDS BASES

H+ (proton) donors H

+ (proton) acceptors

NEUTRALIZATION: A reaction in which a H+ (proton) is transferred

HCl (aq) + NH3 (aq) NH4Cl (aq)

strong acid weak base soluble salt

H

+ (aq) + Cl

(aq) + NH3 (aq) NH4

+ (aq) + Cl

(aq)

N.I.E. H

+

H+(aq) + NH3 (aq) NH4

+ (aq)

acid base

Bronsted-Lowry neutralization can also occur in a non-aqueous system:

H

+

HCl (benzene) + NH3(benzene) NH4Cl (s)

acid base

Examples:

Identify the Bronsted-Lowry acid and base in each of the following equations:

1. C6H5OH + C2H5O− → C6H5O

− + C2H5OH

2. H2O + Cl−

→ HCl + OH−


5

BRONSTED-LOWRY CONCEPT OF ACIDS AND BASES

In a reversible acid-base reaction, both forward and reverse reactions involve H+ transfer

H+

H+

NH3 (aq) + H2O (l) NH4+

(aq) + OH

(aq)

base acid acid base

NH4+ and NH3 H2O and OH

differ by a H+

are a conjugate acid-base pair

NH4+ is the conjugate acid of NH3

NH3 is the conjugate base of NH4+

differ by a H+


H2O is the conjugate acid of OH−

OH− is the conjugate base of H2O

H+

H+

HC2H3O2 (aq) + H2O (l) C2H3O2

(aq) H3O+

(aq)

acid base base acid

HC2H3O2 and C2H3O2

− H2O and H3O

+

differ by a H+


HC2H3O2 is the conjugate acid of

C2H3O2−

C2H3O2− is the conjugate base of

HC2H3O2

differ by a H+


H2O is the conjugate base of H3O+

H3O+ is the conjugate acid of H2O

In General: A conjugate acid-base pair consists of two species in an acid-base reaction

that differ by the loss or gain of a H+ (proton)


6

Species that can act both as an acid and a base are called amphiprotic.

Consider:

H+

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

(aq)

base acid acid base

H+

HC2H3O2(aq) + H2O (l) C2H3O2

(aq) + H3O+

(aq)

acid base base acid

H2O can

lose a H+ and act as an acid (with the weak base NH3)

OR

gain a H+ and act as a base (with the weak acid HC2H3O2)

H2O is an amphiprotic species

Bicarbonate ion(HCO3

–) is another example of amphiprotic species, as illustrated by the

following reactions:

H+

HCO3

(aq) + OH

(aq) CO32

(aq) + H2O (l)

acid base

H+

HCO3(aq) + H

+(aq) CO2 (g) + H2O (l)

base acid

The Bronsted-Lowry Concept of Acids & Bases is more general than the Arrhenius concept.

Bronstead-Lowry Concept introduces additional points of view:

1. A base is a species that accepts H+ ions (OH

is only one example of a base)

2. Acids and Bases can be ions as well as molecular substances

3. Acid-Base reactions are not restricted to aqueous solutions

4. Amphiprotic species can act as either acids or bases depending on what the other reactant is.


7

LEWIS CONCEPT OF ACIDS AND BASES

Gilbert Lewis defined acids and bases in terms of electron-pair transfer.

HCl (aq) + NH3 (aq) NH4Cl (aq)

strong acid weak base

H+(aq) + Cl

(aq) + NH3(aq) NH4

+(aq) + Cl

(aq)

N.I.E.: H+(aq) + NH3 (aq) NH4

+ (aq)

H H +

H+

: N H H : N H

H H electron pair electron pair

acceptor donor

LEWIS ACID LEWIS BASE

Example 1:

BF3 + NH3 BF3NH3

F H F H

F B : N H F B N H

F H F H electron

deficient molecule

electron pair electron pair

acceptor donor

LEWIS ACID LEWIS BASE


8

Example 2:

Na2O(s) + SO3 (g) Na2SO4 (s)

. . . . 2

2 : O : : O :

. . + . . . .

: O : S = O : : O S O :

. . . . . . . .

: O : : O :

. . . . electron pair electron pair

donor acceptor

LEWIS BASE LEWIS ACID

Example 3: Formation of complex ions

Al3+

(aq) + 6 H2O (l) Al(H2O)63+

electron pair electron pair hydrated ion

acceptor donor

ACID BASE

In General:

LEWIS ACIDS LEWIS BASES

electron pair acceptors

Usually positive ions or electron

deficient molecules

electron pair donors

Usually negative ions or molecules

with lone pairs

NEUTRALIZATION: A reaction in which an electron pair is transferred

SUMMARY OF ACID – BASE CONCEPTS

ACID BASE NEUTRALIZATION

Arrhenius

Concept

(least general)

Substance that

dissolved in water

increases concentration

of H3O+

Substance that

dissolved in water

increases concentration

of OH–

H+(aq) + OH

–(aq) H2O (l)

Bronsted-

Lowry

Concept

A H+ donor A H

+ acceptor A H

+ transfer

Lewis

Concept

An electron pair

acceptor

An electron pair

donor

An electron pair

transfer


9

ACID AND BASE STRENGTHS

Relative Strengths of Acids

A strong acid completely ionizes in solution, whereas a weak acid only partially ionizes.

Therefore, the strength of the acid depends on the equilibrium:

HA (aq) + H2O (l) strong acid if equibrium lies to the right

weak acid if equilibrium lies to the left

H3O+ (aq) + A

– (aq)

An acid is stronger than another acid if it lose H+ more readily.

To estimate relative acidic strength, comparisons between pairs of acids are made.

HCl (aq) and H3O+

(aq)

Recall that HCl(aq) is a strong acid and as such it ionizes completely in water:

HCl (aq) + H2O (aq) H3O+

(aq) + Cl

(aq)

The reaction goes completely to products.

“ASSUME” that a reverse reaction were possible to a very small extent:

H+

H+

HCl (aq) + H2O(aq) H3O+

(aq) + Cl

(aq)

loses H+

more loses H+

less

readily readily

HCl(aq) is a stronger acid than H3O+(aq) HCl(aq) H3O

+(aq)

Similar comparisons between other pairs of acids show the following results:

H3O+(aq) is a stronger acid than HC2H3O2(aq) H3O

+(aq) HC2H3O2(aq)

H3O+(aq) is a stronger acid than HF(aq) H3O

+(aq) HF(aq)

HF(aq) is a stronger acid than HC2H3O2 (aq) HF(aq) HC2H3O2 (aq)

HI is a stronger acid than HCl HI HCl


10


SUMMING UP:

1. HCl(aq) is a stronger acid than H3O+(aq) HCl(aq) H3O

+(aq)

2. H3O+(aq) is a stronger acid than HC2H3O2(aq) H3O

+(aq) HC2H3O2(aq)

3. H3O+(aq) is a stronger acid than HF(aq) H3O

+(aq) HF(aq)

4. HF(aq) is a stronger acid than HC2H3O2 (aq) HF(aq) HC2H3O2 (aq)

5. HI is a stronger acid than HCl HI HCl

A Table of Relative Strengths of acids can now be constructed:

Strongest Acid

HI

HCl

H3O+

HF

HC2H3O2

Weakest Acid

An acid-base reaction always goes in the direction of the weaker acid:

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

(aq)

stronger weaker

acid acid

HC2H3O2 (aq) + H2O(l) C2H3O2

(aq) + H3O

+(aq)

weaker stronger

acid acid

HF (aq) + H2O (l) F(aq) + H3O

+(aq)

weaker stronger

acid acid

Strength

of acids

increases


11


Relative Strengths of Bases

Similar to acids, a strong base completely ionizes in solution.

NaOH (aq) Na+ (aq) + OH

– (aq)

Unlike strong bases that contain OH– , weak bases produce hydroxide by reacting with water.

B (aq) + H2O (aq) BH+ (aq) + OH

– (aq)

When diprotic acids ionize, they do so in 2 different steps.

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

2–

HSO4– + H2O (l) H3O

+ (aq) + SO4

–

When diprotic bases ionize, they do so in 1 step.

Ca(OH) (aq) Ca2+

(aq) + 2 OH– (aq)

A base is stronger than another base if it gains H+ more readily

To estimate relative base strength, comparisons between pairs of bases are made.

H2O (l) and Cl

(aq)

Recall that HCl(aq) is a strong acid and as such it ionizes completely in water:

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

(aq)

The reaction goes completely to products.

“ASSUME” that a reverse reaction were possible to a very small extent:

H+

H+

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

(aq)

gains H+

more gains H+

less

readily readily

H2O(l) is a stronger base than Cl(aq) H2O(l) Cl(aq)


12


Similar comparisons between other pairs of bass show the following results:

C2H3O2(aq) is a stronger base than H2O(l) C2H3O2

(aq) H2O(l)

F (aq) is a stronger base than H2O(aq) F

(aq) H2O (aq)

C2H3O2(aq) is a stronger base than F

(aq) C2H3O2

(aq) F

(aq)

Cl is a stronger base than I

Cl

I

SUMMING UP:

1. H2O(l) is a stronger base than Cl(aq) H2O(l) Cl

(aq)

2. C2H3O2(aq) is a stronger base than H2O(l) C2H3O2

(aq) H2O(l)

3. F (aq) is a stronger base than H2O(aq) F

(aq) H2O (aq)

4. C2H3O2(aq) is a stronger base than F

(aq) C2H3O2

(aq) F

(aq)

5. Cl is a stronger base than I

Cl

I

A Table of Relative Strengths of bases can now be constructed:

Weakest Base

I

Cl

H2O

F

C2H3O2

Strongest Base

The Tables of Relative Strengths of Acids and Relative Strengths of Bases can now be combined:

HI I

HCl Cl

H3O+ H2O

HF F

HC2H3O2 C2H3O2

Strength

of acids

increases

Strength

of bases

increases

Strength

of bases

increases


13

CONCLUSIONS: 1. When an acid loses its H

+ readily, its conjugate base does not hold the H

+ very tightly.

Ex: HCl and Cl

strong weak

acid base

The strongest acids have the weakest conjugate bases

2. When a base holds the H+ very tightly, its conjugate acid does not lose the H

+ readily

Ex: C2H3O2 and HC2H3O2

strong weak

base acid

The strongest bases have the weakest conjugate acids

3. The direction of an acid-base reaction always favors the formation of the weaker acid and the

weaker base:

H+

H+

HC l(aq) + H2O (l) H3O+(aq) + Cl

(aq)

stronger stronger weaker weaker

acid base acid base

H+ H

+

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

+(aq)

weaker weaker stronger stronger

acid base base acid

H

+ H

+

HF(aq) + H2O(l) F(aq) + H3O

+(aq)

weaker weaker stronger stronger

acid base base acid


14

Examples:

For each reaction below, determine the conjugate acid-base pairs and their relative strengths:

1. HSO4– + CH3CO2

– CH3CO2H + SO4

2–

2. HCO3– + F

– HF + CO3

2–

3. NH4+ + CN

– HCN + NH3

4. The hydrogen oxalate ion (HC2O4–) is amphiprotic. Write a balanced equation showing

how it reacts as an acid towards water and another equation showing how it reacts as a

base towards water.

5. For each acid shown below, write the formula of its conjugate base:

HPO42–

CH3NH3+ NH3

6. For each base shown, write the formula of its conjugate acid:

HAsO42–

IO– O

2–


15

MOLECULAR STRUCTURE AND ACID STRENGTH

The strength of an acid depends on how easily a H+ is removed from the H Y bond.

Ease of H+ loss is determined by 2 factors:

Polarity of the H–Y bond

Strength of the H–Y bond

To be acidic, the HY bond must be polar with H atom as the positive pole.

+

+

H–Y–

As a result, compare the following bonds and their corresponding dipole moments:

Not acidic Not acidic Acidic

The Polarity of the H Y bond

The polarity of the H–Y bond depends on the EN (electronegativity difference)

between the atoms involved in the bond

The strength of the H X bond

The strength of the H Y bond depends on the size of Y (smaller atoms form stronger

bonds)

–H–Li

+ H–C

+H–F

–

The

higher

the E.N.

of Y

The

higher

the

∆E.N.

The

more

polar the

bond

The

easier to

lose H+

The

stronger

the acid

The

larger Y

is

The

weaker

the H–Y

bond

The less

tightly

the H is

held

The

easier to

lose H+

The

stronger

the acid


16


Trends among Binary Acids (HY)

1. Across a period going from left to right

The polarity of the H–Y bond determines this trend.

Electronegativity of Y increases

Polarity of the H–Y increases

Consequently:

H2O is a weaker acid than HF H2O HF

H2S is a weaker acid than HCl H2S HCl

2. Going down a group

The strength of the H–Y bond determines this trend

F H F

Atomic Strength Acidic

Cl Size H Cl of Strength

Increases Bond Increases

Br (more H Br Decreases

electron

I shells) H I

The diagram below illustrates the combined effect of bond polarity and strength on the acids of

group VI and VII.


17


Trends among Oxoacids Oxoacids are O containing acids

+

H O Y

acidic H is always attached to an O atom

The polarity of the H O bond determines the relative acidic strength

The polarity of the H O bond depends on 2 factors:

The Electronegativity of the Y atom

H O Cl H O Br H O I

EN(Cl) = 3.16 EN(Br) = 2.96 EN(I) = 2.66

(strongest) (weakest)

ACIDIC STRENGTH INCREASES

The number of O atoms bonded to Y (excluding O in OH groups)

HClO HClO2 HClO3 HClO4

(weakest) (strongest)

ACIDIC STRENGTH INCREASES

O O

H O Cl H O Cl O H O Cl O H O Cl O

O

least polar most polar

The

more EN

Y is

The more

polar the

H–O

bond

The

easier to

lose H+

The

stronger

the acid

The more

O atoms

bonded

to Y

The more

electro-

negative Y

becomes

The more polar

the H–O–Y

bond becomes

The

stronger

the acid


18


SUMMARY

BINARY ACIDS

Acidic Strength Increases

(H–Y polarity increases)

OXOACIDS (H attached to Y through O)

The more O atoms attached to Y, the stronger the acid

POLYPROTIC ACIDS

contain more than one acidic H

are stronger than their corresponding anions

H2SO4 HSO4

–

strong weak

acid acid

H3PO4 H2PO4

HPO4

2

Acidic Strength Increases

Reason: The acid strength of a polyprotic acid and its anion(s) decreases with increasing

negative charge (it becomes increasingly difficult to lose a H+)

Acidic strength

increases

(polarity of

H–O increases)

Acidic strength

increases

(H–Y bonds

weakens as size

of Y increases)


19


Examples:

1. Which member of each of the following pair is the stronger acid?

a) NH3 and PH3

b) HI and H2Te

c) H2SO3 and H2SeO3

d) HBrO4 and HBrO3

e) HBr and H2Se

2. List the following compounds in order of increasing acid strength:

a) HBr, H2Se, H2S

b) HBrO3, HClO2, HBrO


20

SELF-IONIZATION OF WATER

Pure Water is generally considered a non-electrolyte.

It is actually a very, very weak electrolyte:

H+

H2O (l) + H2O (l) H3O+

(aq) + OH

(aq)

+

3c

[H O ][OHK =

2

2

]

[H O]

Kc is very small

The equilibrium is strongly shifted to the left

[H2O]

2 = is essentially constant

[H2O] for 1L of water can be calculated as follows:

moles 1000 g 1 mole

? = 1 L x x 55 M

L 1 L 18.02 g

[H2O]2 x Kc = [H3O

+] [OH

] = Kw = 1.0 x 10

14

Kw = is called THE ION-PRODUCT OF WATER

= 1.0 x 1014

at 250C (is temperature dependent)

CONCLUSION: Kw = [H3O

+] [OH

] = 1.0 x 10

14 (25

0C)

In Pure Water: [H3O+] = [OH

] = x

Kw = x2 wx = K = 1.0x10 14

[H3O+] = [OH

] = 1.0 x 10

7 M


21

Solution of a Strong Acid:

Example: A 0.1 M solution of HCl

HCl (aq) + H2O (l) complete

ionization H3O

+ (aq) + Cl

(aq)

0 0.1 M 0.1 M

The TOTAL CONCENTRATION OF THE [H3O+] can be thought of as the sum of the:

concentration of [H3O+] provided by the acid; [H3O

+] = 1.0 x 10

1 M

concentration of [H3O+] provided by the self-ionization of water; [H3O

+] = 1.0 x 10

7 M

Total Conc. of [H3O+] = [H3O

+] + [H3O

+]

from acid from self-ionization of water

= 1.0 x 101

M + less than 1.0 x 107

M (negligible)

= 1.0 x 101

M

Concentration of the [OH] can now easily be calculated:

[H3O+] [OH

] = 1.0 x 10

14

1.0 x 10[OH] =

14

+

3

1.0 x 10=

[H O ]

14

1.0 x 10 1= 1.0 x 10 13 M

In a solution of a strong acid the concentration of the [H3O+] produced by the self-ionization of

water can be ignored.

In a solution of a Strong Base:

Example: A 0.010 M solution of NaOH

NaOH (aq) complete

ionization Na

+(aq) + OH

(aq)

0 0.010 M 0.010 M

The TOTAL CONCENTRATION OF THE [OH] can be thought of as the sum of the :

concentration of [OH] provided by the base; [OH

] = 1.0 x 10

2 M

concentration of [OH] provided by the self-ionization of water; [OH

] = 1.0 x 10

7 M

Total Conc. of [OH

–] = [OH

–] + [OH

–]

from base from self-ionization of water

= 1.0 x 102

M + less than 1.0 x 107

M (negligible)

= 1.0 x 102

M

Concentration of the [H3O+] can now easily be calculated:

[H3O+] [OH

] = 1.0 x 10

14

+

3

1.0 x 10[H O ] =

14

[OH

1.0 x 10=

]

14

1.0 x 10 2= 1.0 x 10 12

In a solution of a strong base the concentration of the [OH] produced by the self-ionization

of water can be ignored.


22

CONCLUSIONS:

1. In a neutral solution: [H3O+] = [OH

] = 1.0 x 10

7 M

2. In an acidic solution: [H3O+] [OH

]

[H3O+] 1.0 x 10

7 M

[OH] 1.0 x 10

7 M

2. In a basic solution: [H3O+] [OH

]

[H3O+] 1.0 x 10

7 M

[OH] 1.0 x 10

7 M


23

Examples:

1. What is the concentration of the [H3O+] and [OH

] in a 0.25 M solution of Ba(OH)2?

Ba(OH)2 (aq) Ba2+

(aq) + 2OH

(aq)

0 0.25 M 2 x 0.25 M

[OH] = 0.50 M

[H3O

+] =

2. What is the concentration of the [H3O+] and [OH

] in a 0.040 M solution of HNO3?

3. Identify each of the following solutions are acidic, basic or neutral:

a) [OH–] = 1.0 x10

–5 M

b) [H3O+] = 3.8 x10

–9 M

c) [H3O+] = 6.2 x10

–3 M

d) [OH–] = 4.5 x10

–10 M


24

pH SCALE

pH is an important chemical quantity introduced by Sorensen, to simplify handling of negative

exponents, when expressing the acidity or basicity of aqueous solutions

pH can be loosely referred to as “power of hydrogen)

By Sorensen’s definition, widely accepted today:

pH = log [H3O+] or simply: pH = log [H

+]

For a neutral solution: [H3O+] = 1.0 x 10

7 M

pH = log (1.0 x 107

) = 7.00

For an acidic solution: [H3O+] 1.0 x 10

7 M

pH 7.00

For a basic solution: [H3O+] 1.0 x 10

7 M

pH 7

pH calculations can be greatly simplified by a rearrangement of the expression:

[H3O+] [OH

] = 1.0 x 10

14

Taking logs of both sides:

log [H3O+] + log [OH

] = log (1.0 x 10

14)

pH pOH = 14.00

pH + pOH = 14.00

NOTE: pOH = (log [OH

])

A pHorseshoe

[H+] [OH

]

Up arrows indicate changing the sign

and then calculating inverse log

Down arrows indicate calculating log

and then changing the sign

pH pH + pOH = 14 pOH


25

Examples: 1. Obtain the pH corresponding to a hydroxide-ion concentration of 2.7 x10

10 M.

pOH = log (2.7 x 10

10) = 9.568

pH = 14.00 9.568 = 4.43

2. A wine was tested for acidity, and it pH was found to be 3.85 at 25

C. What is the

hydronium ion concentration?

log [H3O+] = pH = 3.85

[H3O+] = antilog (–3.85) = 10

3.85 = 1.4 x 10

4 M

3. A 1.00 L aqueous solution contains 6.78 g of Ba(OH)2. What is the pH of the solution?

First: Calculate the molarity of the Ba(OH)2 solution:

moles 6.78 g Ba(OH)2 1mol Ba(OH)2

? = x = 0.03956 M Ba(OH)2

L 1.00 L solution 171.4 g Ba(OH)2

2 moles OH

[OH] = 0.03956 M Ba(OH)2 x = 0.07912 M

1 mole Ba(OH)2

1.0 x 1014

[H3O+] = = 1.264 x 10

13 M

0.07912 M

pH = - log (1.264 x 1013

M) = 12.90

4. What is concentration of a solution of KOH with a pH of 11.89?


26

MEASUREMENT OF pH

Acid-Base Indicators

Acid-Base indicators change color within a small range.

Reason: Indicators establish an equilibrium between their acid form and their base form,

respectively:

HIn = acid form of indicator

In = base form of indicator

These two forms of the Indicator have different colors (not necessarily red and blue)

HIn (aq) + H2O (aq) H3O

+ (aq) + In

(aq)

Example: Phenolphthalein: HIn (aq) is colorless In

(aq) is pink

When a base is added to the acidic solution of phenolphthalein:

HIn (aq) + H2O (l) H3O

+(aq) + In

(aq)

OH

– reacts with H3O

+(aq) to produce H2O

Equilibrium shifts

In(aq) is produced. Solution turns pink

Phenolphthalein begins to turn pink at about pH = 8.0

Universal pH paper

This paper is impregnated with several indicators

It gives different colors to different pH ranges (to the nearest integer value)

pH-meter

Consists of two electrodes (or one combination electrode) dipped into the solution whose pH is

to be measured.

Measures the voltage that is generated between the electrodes.

This voltage depends on pH and is read on a meter calibrated directly in pH units.


27

ACID IONIZATION CONSTANT

Solutions of a Weak Acid or Base

Weak acids ionize (dissociate) partially in water

HA + H2O H3O+

(aq) + A

(aq)

????? +

3c

[H O ][AK =

2

]

[HA][H O]

[H2O] can be considered constant for dilute solutions

can be included in Kc

+

3[H O ][Ac 2

]= K x [H O]

[HA] = Ka = ACID IONIZATION CONSTANT

Ka expresses the extent of ionization of a weak acid

is a measure of the relative strength of weak acids

Weak Acids Acid-Ionization Constant (at 25

0C)

HF

HC2H3O2

HClO

HCN

Ka = 6.8 x 10–4

strongest weak acid

Ka = 1.7 x 10–5

Ka = 3.5 x 10–8

Ka = 4.9 x 10–10

weakest weak acid

Strength

Of

Weak

Acids

Increases


28

ACID IONIZATION CONSTANT

The smaller the acid ionization constant (Ka), the less the acid ionizes and the weaker the

acid.

The acid ionization constant can also be related to the pH of an acid solution. For a given

concentration, the smaller the Ka, the lower the [H3O+] and therefore the higher the pH.

Examples:

1. Rank the following 0.1 M solutions of acids in order of increasing ionization (lowest to

highest):

H2S Ka = 1.0 x 10–7

HCN Ka = 6.2 x 10–10

HNO2 Ka = 7.2 x 10–4

HOCl Ka = 2.9 x 10–8

2. Rank the following 0.1 M solutions of acids in order of decreasing pH (highest to lowest):

H2S Ka = 1.0 x 10–7

HCN Ka = 6.2 x 10–10

HNO2 Ka = 7.2 x 10–4

HOCl Ka = 2.9 x 10–8


29

CALCULATIONS WITH Ka

Examples:

1. Lactic acid, HC3H5O3, is a weak acid found in sour milk. A 0.025 M solution of lactic

acid has a pH of 2.75. What is the ionization constant, Ka for this acid? What is the

degree of ionization?

HC3H5O3 (l) + H2O (l) H3O+

(aq) + C3H5O3

(aq)

Initial 0.025 M ---- 0 0

–x ---- +x +x

Equilibrium 0.025–x ----- x x

+

3 2 3 2a

[H O ][C H OK =

2

2 3 2

] x=

[HC H O ] 0.025 x

[H3O+] = x = antilog (-pH) = antilog (- 2.75) = 1.78 x 10

3

[C3H5O3] = x = 1.78 x 10

3

[HC3H5O3] = (25 x 103

) – (1.78 x 103

) = 23.2 x 103

a

(1.78 x 10K =

3 2)

23.2 x 10 3= 1.4 x 10 4

Molar concentration of Ionized Acid

Degree of Ionization =

Total Molar concentration of Acid

1.78 x 103

Degree of Ionization = = 0.071 or 7.1%

25 103


30

Examples:

2. What are the concentrations of hydrogen ion and acetate ion in a solution of 0.10 M acetic acid?

What is the pH of the solution? What is the degree of ionization? (Ka of acetic acid = 1.7 x 105

)

HC2H3O2 (l) + H2O (l) H3O+

(aq) + C2H3O2

(aq)

Initial 0.10 ---- 0 0

–x ---- +x +x


+

3 2 3 2a

[H O ][C H OK =

2

2 3 2

] x=

[HC H O ] 0.10= 1.7 x 10

x

5

This yields a quadratic equation: x2

= 1.7 x 106

– 1.7 x 105

x

x2

+ 1.7 x 105

X – 1.7 x 106

= 0

The equation can be greatly simplified, by realizing that:

x 0.10

Therefore: 0.10 – x 0.10

can be neglected

x

2 x

2

It follows: = 1.7 x 105

x2 = 1.7 x 10

6

0.10 – X 0.10

x = [H3O+] = [C2H3O2

] = 1.30 x 10

3 M

Now we can check if the assumption: 0.10 – X 0.10 is valid or not

0.10 – 0.00130 = 0.09870 = 0.10 YES, the assumption is valid!

pH = - log (H3O

+) = - log (1.30 x 10

3) = 2.89

1.30 x 103

Degree of Ionization = = 0.013 or 1.3%

1.0 x 101

This method is commonly referred to as the “Approximation Method”


31

Examples:

3. Calculate the pH and % ionization of a 0.250 M solution of HF. (Ka = 3.5x10–4

)

Initial

Equilibrium


32

POLYPROTIC ACIDS

Polyprotic acids are acids that can release more than one H+ per molecules of acid

Examples:

1. H2SO4 (strong acid) and HSO4 (weak acid)

H2SO4 (aq) + H2O (l) complete dissociation H3O+

(aq) + HSO4

(aq)

HSO4

(aq) + H2O (l) partial dissociation

H3O+

(aq) + SO42

(aq)

2. H2CO3 (weak acid) and HCO3

(weak acid)

H2CO3 (aq) + H2O (l) H3O+

(aq) + HCO3

(aq) Ka = 4.3 x 107

HCO3

(aq) + H2O (l) H3O

+ (aq) + CO3

2 (aq) Ka = 4.8 x 10

11

3. H3PO4 (weak acid), H2PO4

(weak acid), and HPO4

2 (weak acid)

H3PO4 (aq) + H2O (l) H3O+

(aq) + H2PO4

(aq) Ka = 6.9 x 103

H2PO4

(aq) + H2O (l) H3O+

(aq) + HPO42

(aq) Ka = 6.2 x 108

HPO42

(aq) + H2O (l) H3O+

(aq) + PO43

(aq) Ka = 4.8 x 1013

NOTE: Ka1 Ka2 Ka3

Meaning: A diprotic acid - loses the first H

+ easier than the second one.

A triprotic acid: - loses the first H+ easier than the second one

- loses the second H+ easier than the third one.

Reason: The 1st H

+ separates from an ion of a single negative charge

The 2nd

H+ separates from an ion of an ion of a double negative charge

The 3rd

H+ separates from an ion of a triple negative charge


33

CALCULATING THE CONCENTRATION OF VARIOUS SPECIES

IN A SOLUTION OF A POLYPROTIC ACID

Ascorbic acid (H2Asc) is a diprotic acid with Ka1 = 7.9 x 105

and Ka2 = 1.6 x 1012

. What is the pH of a 0.10 M

solution of ascorbic acid? What is the concentration of the ascorbate ion (Asc2

), in the solution?

First Part: Calculate the pH of the solution:

Note that Ka2 (1.6 x 1012

) Ka1 (7.9 x 105

)

As such: - the 2nd

ionization can be neglected

- any H+ produced by the 2

nd ionization can be ignored.

H2Asc (aq) + H2O (l) H3O+

(aq) + HAsc

(aq)

Initial 0.10 M ---- 0 0

–x ---- +x +x


+

3a1

[H O ][HAscK =

2

2

] x=

[H Asc] 0.10= 7.9x10

x

5

x2

= (0.10)(7.9x10–5

) = 7.9x10–6

x = [H3O+] = 2.8x10

–3

pH = - log (2.8x10–3

) = 2.55

Second Part: Calculate the concentration of the sulfite ion (SO32

) :

H2Asc(aq) + H2O(l) HAsc(aq) + H3O

+(aq) Ka1 = 7.9 x10

5

HAsc(aq) + H2O(l) Asc

2 (aq) + H3O

+(aq) Ka2 = 1.6 x10

12

HAsc– (aq) + H2O (l) H3O

+ (aq) + Asc

2– (aq)

Initial 0.0028 ---- 0.0028 0

–y ---- +y +y

Equilibrium 0.0028 – y ----- 0.0028 + y y

+ 2

3a2

[H O ][AscK =

]

[HAsc

(0.0028 + y)(y)=

] 0.0028= 1.6x10

y

12


34

Assume that: y <<<<<< 0.0028

Therefore: 0.0028 + y 0.0028 and 0.0028 – y 0.0028

a2

(0.0028 + y)(y)K =

0.0028

(0.0028)(y)= = 1.6x10

0.0028y

12

y = [Asc2

] = 1.6 x 1012

M

General Rule: The concentration of the ion A2–

equals the second ionization constant, Ka2

Examples:

1. Carbonic acid (H2CO3) is a weak acid with Ka1= 4.3x10–7

and Ka2= 4.8x10–11

. Calculate the pH

and the carbonate ion concentration of a 0.050M solution of carbonic acid.


35

BASE–IONIZATION EQUILIBRIA

Weak bases ionize (dissociate) partially in water.

NH3 (aq) + H2O (l) NH4

+ (aq) + OH

(aq)

+

4c

[NH ][OHK =

3 2

]

[NH ][H O]

[H2O] can be considered constant for dilute solutions

can be included in Kc

+

4[NH ][OHc 2

3

]= K x [H O]

[NH ] = Kb = BASE IONIZATION CONSTANT

Kb expresses the extent of ionization of a weak base

is a measure of the relative strength of weak bases

Weak Bases Formula Base-Ionization Constant (at 25

C)

Dimethylamine

Ammonia

Aniline

Urea

(CH3)2NH

NH3

C6H5–NH2

NH2–CO–NH2

Kb = 5.1 x 10–4

strongest weak base

Kb = 1.8 x 10–5

Kb = 4.2 x 10–10

Kb = 1.5 x 10–14

weakest weak base

Strength

Of

Weak

Bases

Increases


36

CALCULATIONS WITH Kb

Examples:

1. Quinine is an alkaloid, or naturally occurring base, used to treat malaria. A 0.0015 M

solution of quinine has a pH of 9.84. What is Kb of quinine?

Step 1: Convert the pH to pOH, and then to [OH]

pOH = 14.00 – pH = 14.00 – 9.84 = 4.16

[OH] = antilog (–4.16) = 6.92 x 10

5 M

Step 2: Using the symbol Qu for quinine, assemble reaction table

Qu (aq) + H2O (l) HQu+

(aq) + OH

(aq)

Initial 0.0015 M ---- 0 0

–x ---- +x +x


+

b

[HQu ][OHK =

2] x=

[Qu] 0.0015

(6.92 x 10=

x

5 2)

(0.0015 6.92 x 10 5= 3.3 x 10

)

6

2. What is the pH of a 0.20 M solution of ammonia in water (Kb = 1.8 x 105

)?

NH3 (aq) + H2O (l) NH4+

(aq) + OH

(aq)

Initial 0.20 M ---- 0 0

–x ---- +x +x


Assume that x is small enough to neglect compared with 0.20.

+

4b

[NH ][OHK =

2

3

] x=

[NH ] 0.20

2x= = 1.8 x 10

x 0.20

5 x2 = 3.6 x 10

6

x = [OH] = 1.89 x 10

3 M (Note that x is negligible compared to 0.20)

+

3

1.0 x 10[H O ]=

14

1.89 x 10 3= 5.3 x 10 12

pH = –log (5.3 x 1012

) = 11.28


37

RELATIONSHIP BETWEEN Ka AND Kb

The quantitative relationship between an acid and its conjugate base can be seen in the

examples below:

NH4+ (aq) + H2O (l) H3O

+ (aq) + NH3 (aq)

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

– (aq)

Each of these equilibria can be expressed by an ionization constant:

+ +

3 3 4a b+

4

[H O ][NH ] [NH ][OHK = and K =

[NH ] 3

]

[NH ]

Addition of the two equations above results the following:

NH4+ (aq) + H2O (l) H3O

+ (aq) + NH3 (aq) Ka

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

– (aq) Kb

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

– (aq) Kw

Recall that when two reactions are added to form a third reaction, the equilibrium constant

for the third reaction is the products of the two added reactions. Therefore,

Kw = Ka x Kb

Note that Ka and Kb are inversely proportional for an acid-base conjugate pair. This is

expected, since as the strength of an acid increases (larger Ka), the strength of its conjugate

base decreases (smaller Kb).

Examples:

1. Hydrofluoric acid (HF) has Ka = 6.8 x 10–4

. What is Kb for the fluoride ion?

wb

a

KK = =

K

2. Which of the following anions has the largest Kb value: NO3–, PO4

3–, HCO3

– ?


38

ACID-BASE PROPERTIES OF SALT SOLUTIONS

A salt is an ionic substance resulting from a neutralization reaction:

Acid + Base Salt + Water

The aqueous solution of a salt may be: basic, acidic, or neutral

Reason: Soluble salts derived from weak acids or weak bases undergo hydrolysis (reaction

with

H2O) in aqueous solution.

I. Salt of a strong base (NaOH) and a weak acid (HC2H3O2)

completely

NaC2H3O2 (aq) Na+

(aq) + C2H3O2

(aq)

dissociated

Na+

(aq) + H2O (l) No Reaction

H

+ H

+

C2H3O2

(aq) + H2O (l) H C2H3O2 (aq) + OH

(aq)

Bronsted Bronsted Bronsted Bronsted

base acid acid base

NOTE: The resulting solution is basic

This reaction is referred to as the Hydrolysis of the acetate (C2H3O2) ion


39

II. Salt of a weak base (NH4OH) and a strong acid (HCl)

completely

NH4Cl (aq) NH4+

(aq) + Cl

(aq)

dissociated

Cl


H

+ H

+

NH4+

(aq) + H2O (l) NH3 (aq) + H3O+

(aq)

Bronsted Bronsted Bronsted Bronsted

acid base base acid

NOTE: The resulting solution is acidic

The reaction is referred to as the Hydrolysis of the ammonium(NH4+)ion

III. Salt of a strong base (NaOH) and a strong acid (HCl)

completely

NaCl (aq) Na+

(aq) + Cl

(aq)

dissociated

Na+


Cl


The resulting solution is neutral


40

IV. Salt of a weak base (NH4OH) and a weak acid (HC2H3O2)

completely

NH4C2H3O2(aq) NH4+(aq) + C2H3O2

(aq)

dissociated

Both ions undergo hydrolysis:

C2H3O2

(aq) + H2O (l) HC2H3O2 (aq) + OH

(aq)

Is the solution basic?

NH4 (aq) + H2O (l) NH3 (aq) + H3O+

(aq)

Is the solution acidic?

Acidity or Basicity of solution depends on the relative acid-base strength of the two ions.

Ka of NH4+

must be compared with Kb of C2H3O2

If: Ka Kb Solution is acidic

If: Ka Kb Solution is basic

CONCLUSIONS:

Solution of a salt of a strong base and weak acid is basic

Solution of a salt of a weak base and a strong acid is acidic

Solution of a salt of a strong acid and a strong base is neutral.

Solution of a salt of a weak base and a weak acid is:

neutral, if: Ka(cation) Kb(anion)

acidic, if: Ka(cation) Kb(anion)

basic, if: Ka(cation) Kb(anion)


41

Examples:

1. For each of the following salts, indicate whether the aqueous solution will be acidic, basic or neutral.

Salt Parent Base Parent Acid Solution of salt is:

Fe(NO3)3 Fe(OH)3

insoluble

HNO3

strong acid acidic

Ca(CN)2 Ca(OH)2

strong base

HCN

weak acid Basic

Na2CO3

Na2S

KClO4

Al(NO3)3

2. Write hydrolysis equation(s) for each of the following salts:

a) Na2CO3

Na2CO3 2 Na+ + CO3

2–

Na+ + H2O no reaction (cation of a strong base, NaOH)

CO32–

+ H2O HCO3– + OH

– (conjugate base of a weak acid HCO3

–)

b) Na2S

c) KClO4


42

THE pH OF A SALT SOLUTION

What is the pH of a 0.10 M NaCN (sodium cyanide) solution?

NOTE: NaCN is the salt of a strong base (NaOH) with a weak acid (HCN). Therefore:

we expect the solution to be basic

we know beforehand that the pH 7

Consider the hydrolysis of NaCN:

NaCN (aq) Na+

(aq) + CN

(aq)

Na+


CN

(aq) + H2O (l) HCN (aq) + OH

(aq) Kb(CN) = ? (not readily available)

Solution: Consider the ionization equilibrium of HCN:

HCN (aq) + H2O (l) H3O+

(aq) + CN(aq) Ka = 4.9 x 10

10

Ka (HCN) x Kb (CN) = Kw bK (CN w

a

K 1.0 x 10) = =

K (HCN)

14

4.9 x 10 10= 2.0 x 10 5

Now we can calculate the pH of the solution from equilibrium data:

CN–

(aq) + H2O (l) HCN(aq) + OH

(aq)

Initial 0.10 ---- 0 0

–x ---- +x +x


b

[HCN][OHK =

]

[CN

2x=

] 0.10= 2.0 x 10

x

5

Consider the usual assumption: 0.10 – x 0.10

x2

2.0 x 105

= x = [OH] = 1.4 x 10

3

0.10

pOH = – log (1.4 x 10

3) = 2.85 pH = 14.00 – 2.85 = 11.15


43

Examples:

1. Household bleach is a 5% solution of sodium hypochlorite (NaClO). This corresponds to a

molar concentration of about 0.70 M NaClO. Ka for HClO (hypochlorous acid) is 3.5 x 108

.

What is the pH of household bleach?

NaClO(aq) Na+(aq) + ClO

(aq)

NaClO is the salt of a _____________base with a ____________ acid.

Therefore the solution of this salt is expected to be _____________

ClO

(aq) + H2O (l) HClO (aq) + OH

(aq) Kb = ?

Kb (ClO) =

ClO–

(aq) + H2O (l) HClO(aq) + OH

(aq)

Initial 0.70 ---- 0 0

–x ---- +x +x


b

[HClO][OHK =

]

[ClO

2x=

] 0.70=

x

Consider the usual assumption: 0.70 – x 0.70

x2

Kb = = x = [OH] =

0.70

pOH = – log [OH

–] =

pH = 14.00 –

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