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Ionic equilibrium Page 168
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
Ionic equilibrium Page 170
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
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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
Ionic equilibrium Page 172
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
Ionic equilibrium Page 173
[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
Ionic equilibrium Page 174
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
Ionic equilibrium Page 175
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
-
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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.
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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.
Ionic equilibrium Page 178
[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
Ionic equilibrium Page 179
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
Ionic equilibrium Page 180
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
Ionic equilibrium Page 181
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.
Ionic equilibrium Page 182
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:
Ionic equilibrium Page 183
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
Ionic equilibrium Page 184
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
Ionic equilibrium Page 185
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)……………………………………………………………………
Ionic equilibrium Page 186
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
Ionic equilibrium Page 187
[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
Ionic equilibrium Page 188
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
Ionic equilibrium Page 189
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
Ionic equilibrium Page 190
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.
Ionic equilibrium Page 191
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)
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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):
……………………………………………………………………………
……………………………………………………………………………
Ionic equilibrium Page 193
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)
Ionic equilibrium Page 194
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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Ionic equilibrium Page 196
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