Theories of Acids and Bases: 1. The Arrhenius...

Ionic equilibrium 68

Chapter 7

Ionic equilibrium

Theories of Acids and Bases:

1. The Arrhenius Concept

According to this theory, an acid produces H+ (aq.) in

aqueous solution and a base produces OH- (aq.) in

aqueous solution. The reaction between an acid and a

base, in which water is produced, is called neutralization.

H+ (aq) + OH

- (aq) → H2O neutralization

HCI (aq) → H+ (aq) + Cl

- (aq) acid

NaOH (aq) → Na+ (aq) + OH

- (aq) base

H+(aq)+Cl

- (aq)+Na

+(aq) +OH

- (aq) → Na

+(aq)+ Cl

- (aq)

+ H2O

One of the most serious limitations of the theory is in

its treatment of the weak base ammonia, NH3. According

to Arrhenious a compound must contain OH" to be a base.

Ionic equilibrium Page 169

2. The Bronsted - Lowry Concept:

This concept defines acids and bases (which may be

molecules or ions) in terms of the exchange of a proton. In

an acid - base reaction, an acid donates a proton to a

base, which accepts it. In losing a proton, acid1 becomes

base1 (The conjugate base of acid1), and gaining a proton,

the original base2 becomes acid2 (the conjugate acid of

base2).

Acid1 + base2 acid2 + base1

CH3COOH + H2O H3O++ CH3COO

-

H2O + NH3 NH4+ + OH

-

H3O++ OH

- H2O + H2O

The strengths of acid and bases are based on their

tendencies to lose or gain protons. The stronger the acid

(or base), the weaker is its conjugate base (or acid). An

acid - base reaction always proceeds from the stronger

acid and base to the weaker acid and base.

e.g: HCI + H2O CI- + H3O

+

stronger acid stronger base weaker base weaker acid


3. The Lewis Concept

In the Lewis concept, the formation of a covalent bond

is the basis for defining acid - base reaction. A base (a

nucleophilic substance) supplies an unshared electron pair

for the formation of a covalent bond with an acid (an

electrophilic substance).

Compounds containing elements with incomplete

valence shells, such as BF3 or AICI3 are called Lewis

acids, while compounds or ions that have lone pairs of

electrons can behave Lewis bases.

Base donate a

pair of

electrons

Acid accept a

pair of

electrons


The dissociation of water

The dissociation of water is reversible and to a very

limited extent as illustrated by its very weak conductivity to

an electric current, and can be represented by the equation:

H2O H+ + OH

–

According to the law of mass action:

Since the fraction of water ionised is very minute or

negligible, the value of [H2O] is equal to (1) can be regarded

as equation 1, and the equation can be written therefore:

[H+] . [OH]

= Kw ………………………..(2)

Kw is known as "The ionic product of water"

Under ordinary experimental conditions and at about

25oC; the value of Kw is taken to be 1 x 10

14 and it follows

that;

[H+] . [OH

] = 10

14


Furthermore, since the dissociation of water gives rise to

equal number of hydrogen and hydroxyl ions. Equation (2)

could be written:

[H+]2 = Kw = 1 x 10

14 ………………………(3)

and in other words;

it follows that, if the [H+] = [OH

–] = 10

–7 the solution is

described as "neutral", if [H+] is more than 10

–7,

that is 10

–

6, 10

–5…, etc. the solution is said to be "acidic" and if [H

+] is

less than 10–7

, that is 10–8

, 10–9

..etc., the solution is called

"alkaline".

Example 7.1: What are [H+] and [OH-] in a 0.02 M HCI

solution.

Solution: Since HCI is a strong electrolyte

[H+] = 0.02M

[H+] [OH

-] = 1.0 x 10

-14

(2 X 10-2

) [OH-] = 1.0 x 10

-14


[OH-] = M 5.0x10

10 x 2

10 x 1.0 13

2-

-14

Notice that [OH-] is extremely small.

Example 7.2: What are [H+] and

[OH-] in a 0.005 M

solution of NaOH?

Solution: NaOH is a strong

electrolyte,

NaOH → Na++ OH

-

Therefore, [OH-] = 5.0 x 10

-3 M

[H+] [OH

-] = 1.0 x 10

-14

[H+] [5.0 x 10

-3] = 1.0 x 10

-14

[H+] = M 2.0x10

105.0x

10 x 1.0 12

3

-14

The pH Concept;

The concentration of H+ in a

solution may be expressed in

terms of the pH scale. The pH of a

solution is defined as


pH = log ][

1

Hlog [H

+]

In a solution of [H+] = 10

-3 M

pH = log [H+]= -(-3)= 3

for a natural solution

[H+] = 1 x 10

-7

pH = -log 10-7

= 7

Following the definition, therefore,

pH + pOH = 14

Example 7.3: What is the pH and pOH of a solution that is

0.05 M in H+?

Solution: [H+] = 5.0 x 10

-2 M

pH = - log 5.0 x 10-2

= 1.3

pOH = 14-1.3 = 12.7

Example 7.4: What is the pH of a solution for which [OH] =

0.03M

Solution: [OH-] = 3.0 x 10

-2

pOH = - log [OH-] = - log 3.0 x 10

-2 = 1.52

pH = 14 - 1.52 = 12.48


Example 7.5: What is the [H+] of a solution with a pH of

10.6.

Solution: [H+] = antilog - 10.6 = 2.5 x 10

-11 M

Strengths of Acids and Bases;

The strength of an acid depends on how easily the

proton, H+, is lost or removed from an H - X bond in the

acid species.

a) Strong acids and strong bases:

The ionization of the strong acids e.g: HCI, HNO3 in

water goes essentially to completion. As a result we

can conclude that the [H+] in aqueous solution of

strong acid is equal to the concentration of the

dissolved acid thus, for 0.1 M HCI we have

HCl → H+ + Cl

-


0.1 M 0.1 M 0.1 M

Ca [strong acid concentration] = [H+] in its solution.

The strong bases in water are the soluble metal

hydroxides, such as NaOH and KOH. These are ionic

compounds that are completely dissociated in solution, so

we never write equilibria for their dissolution in water

NaOH → Na

+ + OH

-

0.1 M 0.1 M 0.1 M

b) Dissociation of weak electrolytes:

Weak electrolytes include weak acids and bases as well as

certain salts, such as HgCI2 and CdSO4, that are not fully

dissociated in aqueous solution. In solutions of these

substances there is equilibrium between the undissociated

species and its corresponding ions

e.g: CH3COOH + H2O H3O+ + CH3COO

-

k= ]O[H COOH]CH[

]COO[CH ]O[H

23

-

3

+

3

K [H2O] = Ka = COOH]CH[

]COO[CH ]O[H

3

-

3

+

3

Where Ka represent the acid dissociation constant

or ionization constant.


In general, for any weak acid, HA, the simplified

dissociation reaction can be written as

HA H++ A

-

and its acid dissociation constant is given by

Ka = HA][

][A ][H -+

In such a case, let us assume the partial ionization as

HA H+

+ A-

Initial

concentration

Ca

-

-

Change

-X

+x

+x

At equilibrium

Ca-X

+x

+x

Ka =

X] -[C

[X]

X] -[C

[X] [X]

a

2

a

Since Ka is small, the value of x is small. Let us assume that

x is much smaller than Ca so that Ca - x Ca, and,

Ka = a

2

C

[X] , X

2 = KaCa and X =

aaCK

Where

Ca: is the molar concentration of the weak acid, [HA]

X: equals the numerical value of the molar hydrogen -ion

concentration in solution.


[H+] =

aaCK

The latter expression is usually referred

to as Ostwald's Law of weak acids.

The same approach can also be applied

to weak bases

NH3 + H2O NH+

4+ OH-

Kb =][NH

]OH ][[NH

3

-

4

Kb is the base ionization constant.

In general, for any weak base B the ionization

equilibrium can be written as

B + H2O BH+ + OH

-

and the expression for Kb is

Kb =][B

]OH ][[BH -

also, [OH-] =

bbCK

we can write for any Ka

pKa = - log Ka and for any Kb

pKb = - log Kb

We know that the smaller the value of Ka or Kb, the

smaller the extent of ionization and the weaker the acid

or base. Relative strengths of acids and bases can also


be indicated by their pKa's. In this case the smaller the

value of pKa or pKb, the stronger is the acid or base.

For example, the pKa's for acetic, chloroacetic, and

dichloroactic acids, are

HC2H3O2 pKa = 4.74

HC2H2ClO2 pKa = 2.85

HC2HCl2O2 pKa = 1.30

The order of increasing acidity is therefore, acetic <

chloroacetic < dichloroacetic acid

The degree of ionization of weak acids

The degree of ionization of a weak acid is defined as the

fraction of molecules that react water to give ions, as

compared to the value expected for complete ionization.

This may also be expressed as a percentage.

Degree of ionization = acid the ofmolarity Original

] A][[H -

Example 7.6: Nicotinic acid is a monoprotic acid with the

formula HC6H4NO2. A solution that is 0.012 M

nicotinic acid has a pH of 3.39 at 25°C. Calculate

the acid ionization constant, Ka, for this acid? and


what is the degree of ionization of nicotinic acid in

this solution.

Solution: 1) Ka = Nic] [H

]NiC ][[H -

or [H+] =

aaCK [H+]

2 = Ka Ca

Ka = a

2

C

][H

2) The value of [H+] can be obtained from the pH

[H+] = antilog [-pH] = antilog - 3.39 = 4.1 x 10

4 M

Ka = 5

2-4

104.10.012

]10 x [4.1 x

Degree of ionization = 034.0012.0

00041.0

C

[X]

a

The percent ionization = 0.034 x 100 = 3.4%

Problems

1. A student prepared a 0.01 M NH3 solution, and found

that NH3 has undergone 4.2% ionization. Calculate the

Kb for NH3, and the pH of the solution.

Answer Kb = 1.8 x 10-5


2. What are the concentration at 25°C of nicotinic acid,

hydrogen ion, and nicotinate ion in a solution of 0.1 M

nicotinic acid?

What is the pH the solution? What is the degree o f

ionization of the acid? Ka = 1.4 x 10-5

Answer: pH = 2.92 Degree of ionization = 0.012

Example 7. 7: What is the pH of a solution that contains,

0.1 M HCl and 0.1 M acetic acid ?

Ka (acetic acid) = 1.8 x 10-5

Solution: The solution contains 0.1 M H+ (from the strong

acid HCI) + [H+] from the weak acetic acid.

[H+] from acetic acid =

aaCK

= x0.110 x 1.8 -51.34 x 10

-3

We see that [H+] from acetic acid is very small

compared to 0.1, so in the solution [H+] 0.1 M. This gives

the pH of 1.0.


Hydrolysis:

When a salt dissolves in water, it dissociates fully to

produce cations and anions that may subsequently react

with the solvent in a process called hydrolysis. For

example, the cation of a salt may undergo the reaction

M+ + H2O MOH + H

+

NH4+ + H2O NH3 + H3O

+

While an anion may react according to

X- + H2O HX + OH

-

CH3COO- + H2O CH3COOH + OH

-

Since the H+ and OH

- ions produced influence the pH of

the salt solution, the extent to which the hydrolysis take

place determines whether the pH will be greater than, less

than, or equal to 7. In case of anions and cations of strong

acids and bases (NaCI), respectively, do not undergo

hydrolysis, and the salts derived from strong acids and

bases yield neutral solutions.

The pH of a solution of a salt can be predicted on the

basis of tine strengths of the acid and base from which the

salt is derived:


1) Salt of strong base and a strong acid. Examples

are: NaCI, KNO3/ and Ba (CIO3)2. Neither cation nor

anion hydrolyzes. The solution has a pH of 7.

2) Salt of a strong base and a weak acid. Examples

are: KNO2, Ca(C2H3O2), and NaCN. The anion

hydrolyzes to produce OH" ions.

The solution has a pH that is higher than 7.

[OH-] =

a

sw

K

CK

Cs = salt molar concentration

3) Salt of a weak base and a strong acid. Examples

are: NH4NO3, FeBr2, and AICI3. The cation hydrolyzes to

produce H+ ions. The pH of the solution is below 7

[H+] =

b

sw

K

CK

4) Salt of a weak base and a weak acid. Examples are:

NH4C2H3O2, NH4CN, and Cu(NO2)2. Both cation and anion

hydrolyze. The pH of the solution depends upon the extent

to which each ion hydrolyzes. The pH of a solution of

ammonium acetate is 7 since NH3 (Kb = 1.8 x 10-5

) and

acetic acid (Kg = 1.8 x 10-5

) are equal weak. The pH of a


solution of NH4CN, on the hand, is above 7 because HCN

(Ka = 4.0 x 10-10

) is a weaker acid than NH3 (Kb = 1.8 x 10-

5) is a base. As a consequence, the CN

- hydrolyzes to a

greater extent (producing OH-) than the NH4

+ dose

(producing H+).

Analytical Chemistry Department web site

www.analytical2010.webs.com

http://www.analytical2010.webs.com/

Ionic equilibrium 5

Faculty of Pharmacy Final Exam Class Year: First Subject Name: Physical and inorganic Chemistry Subject Code: Department: Analytical Chemistry Department

Date: 31/1/2010 Times: 3 Hours No of Parts: 2 No of Papers: 10 No of Questions:4 Full Mark: 70

Question No. (1): [25 Degrees]

A) Complete the following sentences:

1. “Dalton” atomic theory may be summed up as follows:

a)……………………………………………………………………

b)……………………………………………………………………

c)……………………………………………………………………

d)……………………………………………………………………

2. Rutherford gold foil experiment may be summed up as follows:

a)……………………………………………………………………

b)……………………………………………………………………

c)……………………………………………………………………

d)……………………………………………………………………

3. The four quantum numbers that specify the energy and probable

location of each electron in atoms are

a)……………………………………………………………………

b)……………………………………………………………………

c)……………………………………………………………………


d)……………………………………………………………………

4. Isotopes of the element is defined as

………………………………………………………………………….

and an example of isotopes of an element are ……………………….

………………………………………….

B) For the following compounds:

1. Write the Lewis structure

2. Mention the bond type

NaCl

Al2O3

Na2O

H2

HF


[NH4]+

C) Write short notes on the following:

1. Valence shell electron pair theory

……………………………………………………………………………

……………………………………………………………………………

……………………………………………………………………………

……………………………………………………………………………

……………………………………………………………………………

……………………………………………………………………………

……………………………………………………………………………

2. What are the kinds hybridizations that the central atom exhibit

in the following compounds

BeH2

CH4

NH3


BF3

PCl5

Water

Question No. (2): In tabulated form Choice ONLY one correct

answer: [8 Degrees]

1 (…..) 5 (…..) 9 (…..) 13 (…..) 2 (…..) 6 (…..) 10 (…..) 14 (…..)

3 (…..) 7 (…..) 11 (…..) 15 (…..)

4 (…..) 8 (…..) 12 (…..) 16 (…..) 1. Which of the following is fundamental (or base) unit:

a) Mol b) m/S c) g d) All of them

2. The significant figures for the result of addition of 12.152 and 5.1 is

a) 64 b) 63.8 c) 63.81 d) 63.811

3. "Pressure" of gas

a) Is a measure of the collisions of the atoms with the container?

b) Pascal (Pa), N/m2 is an unit of its measurement

c) Can be measured using manometer.

d) all of the above

4. The measurements which are close to each other but not close to the

"correct" value are:

a)accurate and precise b) precise and inaccurate

c) accurate and not precise d) neither accurate nor precise

5. Ideal gas constant, R:

A. 0.0821 liter torr K -1mol-

1 c) 0.0821 ml atm K

-1mol-

1


B. 0.0821 liter atm K -1mmol-

1 d) 0.0821 liter atm K

-1mol-

1

6. What is the density of gas which its diffusion is 1.414 times of the rate

of diffusion of CO2 at STP?

a) 2.5 c) 1

b) 1.77 d) none of the above

7. Which of the following compounds is more miscible in water

a)CH3OH c)CH3CH(OH)2

b) CH3CH2OH d) CHCl3

8. Its solubility in water decreases by elevation of temperature :

a) CO2 b) glucose c) NaCl d) KMnO4

9. Carbonated beverages is an example of:

a) liquid / liquid solution mixture c) solid /liquid solution mixture

b) gas /liquid solution mixture d) gas/gas solution mixture

10. The unit of heat capacity is:

a) JoC

-1 c) J

og

-1

b) Jog

-1 C

-1 d) Kj

11. The apparatus used for determination of heat of combustion is :

a) coffee –cup calorimeter c) Bomb calorimeter

b) both a and b d) none of the above

12. which of the following COMPLETE thermochemical equation :

a) C2H4 + 3 O2 → 2CO2 (g) + 2H2O

b) H2 (g) + ½ O2 (g) → H2O (l) ΔH = -68.32 Kcal

c) ½H2 + ½Cl2 → HCl Δ H = -44.0 Kcal

d) CO (g) + O2 (g) → CO2 (g) Δ H = -284.5 Kj

13. If [H+] is less than 10

-7, the solution is:

a) Alkaline b)acidic c) neutral d)unionizable

14. What are [H+] and [OH

-] in 0.005 M solution of NaOH

a) [H+] = 2.0 x 10-12 and [OH-] =5.0 x 10-3


b) [H+] = 5.0 x 10-2 and [OH-] =1.0 x 10-14

c) [H+] = 2.0 x 10-10 and [OH-] =1.0 x 10-14

d) [H+] = 2.0 x 10-10 and [OH-] =1.0 x 10-7

15. For the reaction N2O4(g) 2 NO2(g), If the concentrations of the substances present in an equilibrium mixture at 25°C are [N2O4] = 4.27 x 10-2 mol/L and [NO2] = 1.41 x 10-2 mol/L , what is the value of Kc for this temperature. a) 1.41 x 10-2 mol/L b) 4.66 x 10-3 mol/L

c) 4.16 x 10-3 mol/L d) 4.16 x 10-5 mol/L

16. For the reactions involving gases; the partial pressures of the reactants

and products are proportional to their:

a) Temperature

b) Ionic state

c) Molar concentrations

d) All of the above

Question No. (3): Complete the following: [17 Degrees]

1. The reaction between an acid and base in which water is

produced is called ……………………..

2. An acid base reaction always proceeds from the

……………. acid and base to the weaker ………………..

acid and base.

3. In the Lewis concept, the formation of …………………is

the basis for defining acid – base reaction

4. All …………… processes tend to attain a state of

equilibrium

5. The addition of a catalyst causes a system to achieve

……………… ………but does not alter the position of

equilibrium.


6. Le Chatelier's Principle, states that:

……………………………………………………………………………

……………………………………………………………………………

7. Theory can be defend as "……………

……………………………………………………………………………

8. 1st Law of Thermodynamics state that:

………………………………………………………………………

9. Kirchoff's equations at constant at constant pressure =

…………………… And at constant volume=

10.Thermochemical equations It must essentially:

a)………………………………………………………………………

b)………………………………………………………………………

c)………………………………………………………………………

11.Compare between types of solution deviate from ideal

behavior

Negative deviation Positive deviation

a)

b)


c)

d)

12. Describe what a gas is according to kinetic theory:

a)………………………………………………………………………

b)………………………………………………………………………

c)………………………………………………………………………

d)………………………………………………………………………

e) ………………………………………………………………………

11. Define and mention laws of:

a)Graham’s Law (Molecular Effusion and Diffusion):

……………………………………………………………………………

……………………………………………………………………………

b) Daltons Law (Gas Mixtures and Partial Pressures):

……………………………………………………………………………

……………………………………………………………………………


c) Henry’s Law –

……………………………………………………………………………

……………………………………………………………………………

d) Raoult’s Law

……………………………………………………………………………

……………………………………………………………………………

……………………………………………………………………………

e) Hess's law of constant heat summation

……………………………………………………………………………

……………………………………………………………………………

Question No. (4): Calculate: [20 Degrees]

1. Convert the quantity from 14 m/s to miles per hour (mi/hr).

2. How many grams of N2F4 can theoretically be prepared from 4.0g of NH3 and 14.0g of F2? 2NH3 + 5F2 N2F4 + 6HF and if 4.8g of N2F4 is obtained from the experiment, what is the percent yields? (At.Wt. of F= 19)

3. Calcium hydride, CaH2, reacts with water to form hydrogen gas: CaH2(s) + 2 H2O (l) Ca(OH)2(aq) + 2 H2(g) How many grams of CaH2 are needed to generate 10.0L of H2 gas if the pressure of H2 is 740 torr at 23oC? (At.wt. of Ca2+=40, H+ =1.00 and O =16)


4. 6.4g of naphthalene C10H8 when burned under constant

volume gave 123 KJ at 20°C, calculate E and H. C10H8(s) + 12 O2(g) → 10CO2(g) + 4H2O(l)

5. Given that energies for H-H, O = O and O - H bonds are 104, 118 and 111 kcal mol-1 respectively, calculate the heat of the reaction

H2 (g)+ ½O2 (g) → H2O(g)

6. Assuming ideality, calculate the vapor pressure of 1.0 m solution of a non - volatile, on dissociating solute in water at 50°C. The vapor pressure of water 50°C is 0.122 atm.

7. Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile nonelectrolyte. Calculate the boiling point of a 25.0 mass percent solution of ethylene glycol in water (Kb for water= 0.52 oC/m).

8. A 1.00 g sample of a biological material was dissolved in enough water to give 1.00 x 102 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25oC. Calculate the molarity and approximate molecular weight of the material.

9. What is the pH and pOH of a solution that is 0.05 M in H+?

10. At 500 K. 1.0 mol of ONCI(g) is introduced into a one - liter container. At equilibrium the ONCI(g) is 9.0% dissociated: 2 ONCI(g) 2 NO(g) + CI2(g) Calculate the value of Kc for equilibrium at 500 K.

مع تمنيانتا بالنجاح أ.م وفاء السيد حسن * أ.م مرفت محمد حسني * أ.د هشام عزت عبداللطيف


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Theories of Acids and Bases: 1. The Arrhenius...

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