Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
1-
.
Dynamic equilibrium: Is stationary state (static) apparently but dynamic system in
realty.
23-
Water vapour pressure: is the pressure of water vapour at certain temperature.
Saturated water vapour pressure: is the maximum water vapour at certain
temperature.
4-
Irreversible reaction (complete reaction): is the chemical reaction that takes
place in one direction only (forward) because the product can’t react together once more
because one of them escapes from the reaction medium as gas or ppt
Give reason:
a) Reaction of silver nitrate with sodium chloride is complete reaction?
because silver chloride escapes from the medium as white precipitate as follow
NaCl  AgNO3 
 NaNO3  AgCl 
Aq
Aq
White ppt
Aq
b) Reaction of magnesium with hydrochloric acid is irreversible
because hydrogen gas evolves (escapes from the medium)
Mg  2HCl 
 MgCl2  H 2 
s
5-
Aq
Aq
Reversible reaction: is the chemical reaction takes place in two directions (forward
and backward) because the product can react again reforming the reactant because non of the
product escapes from the medium
CH 3COOH  C2 H 5OH
liquid
liquid
CH 3COOC2 H 5  H 2O
liquid
liduid
Factors affecting the rate of reaction:
nature of reactant
i. surface area (directly proportion)
ii. kind of bonds (in ionic reaction will be more faster)
concentration
temperature
pressure
catalyst
light
The law of mass action by Waage and guldberg:
At constant temperature the rate of reaction is directly proportion to product of multiplication of
reactant concentrations, each is raised to the power of the number of molecules or ions in the
balanced equation.

 cC  dD
If a chemical reaction as aA  bB 

kc 
[C ]c [ D]d
[ A]a [ B]b
KC is the equilibrium constant
[ ] means concentration
-1-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
A, B are reactant materials
C, D are product materials
a, b, c, d is the number of molecules in the balanced equation for A, B,C and D
please note that!!!!!!
(1)
(2)
(3)
If the value of Kc is less than 1 it means that backward reaction more effective
If value of Kc is great more than 1 it means the forward reaction is more effective
until it will be complete
Concentration of solids and precipitate and water liquid constant not be written in
the law
Example 1
Calculate the equilibrium constant for the following reaction:
I2  H2
2HI given that concentration of I2, H2 and HI is 0.221, 0.221, 1.563 mole/ liter
respectively.
Solution
KC 
[ HI ]2
(1.563)2

=
[ H 2 ][ I 2 ] 0.221 0.221
50
--------------------------------------------------------------------------------------------------------------
Example 2
In the following reaction
Cu( s )  2 Ag  ( aq ) 
 Cu 2( aq )  2 Ag( s )
Kc  2 1015 explain
which reaction takes place further forward or
backward reaction?
Solution
equilibrium constant is so great, greater than one so the concentration of product will be more than
concentration of reactant and the reaction will be forward
--------------------------------------------------------------------------------------------------------------
Example 3
Explain why the following reaction takes place backward *( why AgCl is water insoluble?)

 Ag   Cl 
AgCl( s ) 

K c  1.7 1010
Solution
Since the value of the equilibrium constant is less than one so concentration of product in the
numerator is less than concentration of reactant in the denomator, this indicates the reaction takes
place backward.
--------------------------------------------------------------------------------------------------------------
Example 4
Calculate the equilibrium constant for the reaction
N2O4 ( gas )
2 NO2 (gas) Given that concentration of N2O4 is 0.213 mole/liter and concentration of
NO2 is .032 mole/liter
KC 
[ NO2 ]2 (0.0032)2

 4.18 105
[ N 2O4 ]
0.213
--------------------------------------------------------------------------------------------------------------
-2-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
3- The effect of temperature of the rate of reaction
According to the collision theory, the chemical reaction takes place faster by raising temperature,
because temperature activate the molecule increase their kinetic energy and active collision takes
place.
Activated molecules: they are molecules have kinetic energy equals to or slightly more than
the activation energy.
Activation energy: is the minimum amount of energy required to start the reaction.
--------------------------------------------------------------------------------------------------------------
4- The effect of pressure on the rate of reaction
In case of gases we represent the reactant and product by their partial pressure instead of
concentration in case of liquids
N2  3H 2
Kp =
KC 
pressure and cooling
heat
P 2 NH3
PN2  P3 H 2
2 NH3
H= -ve
But if we know concentration in mole /liter we can use Kc
[ NH 3 ]2
[ N 2 ][ H 2 ]3
--------------------------------------------------------------------------------------------------------------
Example 1
Calculate the equilibrium constant for the following gaseous reaction
Pressure of NO2 = 2 atmospheric pressure
Pressure of O2 = 1 atmosphere
Pressure of N2 = 0.2 atmosphere
Solution
KP 
P 2 NO2
PN2  P

2
O2
N2  2O2
2 NO2 given
that
22
 20
0.2 12
--------------------------------------------------------------------------------------------------------------
Example 2
Calculate the equilibrium constant by molar concentration for the following gaseous reaction
N2 (g)  O2 (g)
2 NO Given those concentrations 0.2, 0.1, 0.1 molar for NO, N2, and O2 respectively:
Solution
Kc 
[ NO]2
(0.2)2

4
[ N 2 ][O2 ] 0.1 0.1
--------------------------------------------------------------------------------------------------------------
5- The effect of catalyst:
The catalyst is a chemical substance changes the rat of reaction without itself being changed.
The role of catalyst in a chemical reaction decreases the activation energy and attains equilibrium in
short time.
But in equilibrium, the catalyst doesn’t change the equilibrium position? Because it increases
both forward and backward reaction by the same rate.
--------------------------------------------------------------------------------------------------------------
6- The effect of light:
-3-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
Some chemical reactions affect by light energy, as in photosynthesis reaction where green plants
convert carbon dioxide and water vapour into glucose by the effect of light.
Also, in photographic films, it is coated with gelatinous substance of silver bromide, when it is
exposed to light the silver ion accept electron from bromide ion and silver precipitate while bromine
is separated
Ag   1e 
 Ag 
--------------------------------------------------------------------------------------------------------------
Factors affecting chemical reaction already in equilibrium
These factors are controlled by Le Chatelier’s principle which states`` changing any condition of
reaction as temperature, pressure or concentration will shift the reaction in a direction that opposes
this changes’’
Applications on Le Chatelier’s principle
(1)
The effect of heat on a chemical reaction in equilibrium
a. If the reaction is exothermic (heat out with product or  ve
i) when the temperature increases, the reaction goes backward and concentration of product
decreases
ii) by cooling, the reaction goes forward and product increases
b. KC decreases by heating and increases by cooling
c. If the reaction is endothermic (heat in reactant or   ve )
i. By heating, the reaction goes forward and product increases and vice versa as in the
following reaction

 2 NO2
N2O4 

H  ve
red brown
the red brown fumes increases by heating and decreases by cooling
(2)
ii. KC increases by heating and decreases by cooling
The effect of pressure on the reaction?
a. If the volumes of reactants equal the volumes of products, the pressure has no effect on the
reaction as in the following example N2  O2 
 2 NO
b.
(3)
When volumes of both reactant and products are not equal so
i. Increasing pressure (or decrease volume of container)will shift the reaction to the
direction in the which the volume is less as in the following example
N2  3H 2
2 NH 3 the volumes of reactant is more than product so increasing
pressure, the reaction goes to the product and ammonia increases and vice versa
ii. Decreasing pressure will shift the reaction in which the volume is greater
The effect of concentration on chemical equilibrium
a. Increasing the concentration of reactant will shift the reaction forward in the direction of
products (products increases)
Decreasing concentration of reactant shifts the reaction backward
(4)
b.
Increasing volume of container means decreasing pressure and vice versa
--------------------------------------------------------------------------------------------------------------
Example 1
Show the effect of pressure, and temperature on the following reaction.

 2 NH3
N2  3H 2 

H= -92KJ
Solution
-4-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
A) Effect of pressure
Number of moles of reactant is 1+3 =4moles
Number of moles of product= 2moles
So increasing pressure will shift reaction forward and ammonia increase
B) Effect of temperature
Cooling will shift the reaction forward and ammonia increase while heating shift the reaction
backward and ammonia decomposes.
--------------------------------------------------------------------------------------------------------------
Example 2
Explain why reddish brown fumes of nitrogen dioxide disappears by cooling
Answer
2 NO
2
red  brown
N O  heat
2 4
colourless
So according to Le Chatelier principle cooling will shift the reaction forward in direction of
decomposition of nitrogen dioxide and formation of N2O4 which is colourless.
--------------------------------------------------------------------------------------------------------------
Example 3
Only one reaction is decompose by raising temperature
(1)
NO
(2)
SO3
(3)
N2 H 4
1
1
N 2  O2
2
2
1
SO2  O2
2
N 2  2H 2
H= -ve
H= +ve
H= -ve
Solution
Of course reaction number (2) as the temperature increases the reaction will be forward and SO3
decomposes faster into sulphur dioxide and oxygen
--------------------------------------------------------------------------------------------------------------
Secondly: ionic equilibrium
(1)
(2)
Ionization: is a process in which unionized molecules are changed into ions
Complete ionization: takes place in strong electrolyte in which all molecules are
changed into ions
(3)
(4)
(5)
Incomplete ionization: takes place in weak electrolyte where small fractions of
molecules change into ions
Ionic equilibrium: is the equilibrium arising between molecules of weak electrolyte
and ions resulting from it.
Ostwald law: At constant temperature ,the degree of ionization increases by dilution
Ka 
2
v
where  is the degree of ionization and (v) is the volume that contains only one
mole
Ka =  2  C where (C)is the concentration mole/ liter of the main substance acid or base
not ions

=
KA
CA
-5-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
(6)
.
Concentration of ion hydroxonium or hydroxide = 
v
=
KA
CA
V

KA
KA
 CA
 K A  CA
CA
CA
1
V
[ H3O]  K A  CA
[OH  ]  Kb  Cb
(7)
Hydroxonium ion: is the hydrated proton
H   H 2 O 
 H 3O 
(8)
Ionic product of water Kw : is the product of multiplication concentration of
hydrogen ion and hydroxide ion
KW  [ H  ][OH  ]  1014
(9)
(10)
PH : is the degree for measuring acidity or alkalinity of medium and equals negative
logarithm of concentration of hydrogen ions
PH= -log [ H3O ]
Note that PH + POH = 14
Solubility product Ksp is the product of multiplication of the concentration of ions
of sparingly soluble substance in its saturated solution each is raised to power number of
moles in balanced equation
Remember the following
The salt
Composition
1- NaCl
2- NH4Cl
3- CH3COONa
FeCl3
Strong acid HCl and strong base NaOH
Strong acid HCl and weak base NH4OH
Weak acid CH3COOH and strong base
Strong acid HCl and weak base Fe(OH)3
(1)
Its effect
on litmus
Neutral
Acidic
Basic
Acidic
pH value
=7
less than 7
More than 7
Less 7
The pH value of sodium carbonate is more than 7(has alkaline effect in its
solution)?
because it dissolves in water to produce weak acid (carbonic) and strong alkali (sodium
hydroxide) as follow
2 HOH
Na2 CO3
2 H   OH 
2 Na   CO32

Na   OH   H 2 CO3
 Na2 CO3  2 HOH
weak  acid
since hydrogen ion disappears from the medium, so hydroxide ion accumulate which is
responsible for alkaline effect
(2)
Ammonium chloride solution has acidic effect to litmus?
because during hydrolysis it produce strong acid and weak base as follow
HOH
NH 4 OH
H   OH
NH 4  Cl 


 H   Cl   NH 4OH
NH 4 Cl  HOH 

strong  acid
weak  base
since hydroxide ion disappears from the medium, so hydrogen ion accumulate which is
responsible for acidic effect
-6-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
Ammonium acetate has neutral effect to litmus solution?
(3)
because it produces weak acid and weak base
CH3COONH 4  HOH
CH3COOH  NH 4OH
weak  base
weak  acid
both hydrogen ions and hydroxide ions disappear from the medium so the solution will be neutral
Sodium chloride has neutral effect to litmus solution?
(4)
because on dissolving it produces strong acid and strong alkali as follow
NaCl  HOH
Na   OH   H   Cl 
strong  alkali
strong  acid
both hydrogen ions and hydroxide ions remain in the medium
The chemical reaction takes faster in case of finely divided substance
(5)
Because in case of finely divided substance the surface area increases, and the rate of
reaction is directly proportion to surface area
There are no free hydrogen ions in the solution?
(6)
Because proton has a vacant orbital makes coordinate bond with the lone pair of electrons
on oxygen in water as follow
H   H 2 O 
 H 3O 
------------------------------------------------------------------------------------------Problems
Before solving problems note that
(1)
(2)
(3)
The degree of ionization  
(4)
(5)
(6)
(7)
Concentration of H3O+ =
ka
c
where C is the concentration
The degree of dissociation (ionization)
concentration =
=
No of ionized moles
total No of moles before ionization
number of moles
volumes (liters)
Concentration of
K a  Ca
OH   Kb  Cb
where Ca is the concentration of the original weak acid
where Cb is the concentration of original weak base
PH = - log [H+]
PH+POH=14
--------------------------------------------------------------------------------------------------------------Example 1
Find PH of a solution of acetic acid of concentration 0.003M and its Ka 1.8x10-5
Solution
[H3O+] =
K aC  0.3 1.8 105  0.00232379000772445

PH=  log[H 3O ]   log0.00232379000772445  2.63
--------------------------------------------------------------------------------------------------------------Example 2
The hydrogen ion concentration is 5 104 mole/L. Calculate the concentration of hydroxide ion, PH,
POH
Solution
[ H  ][OH  ]  1014
[OH  ] 
1014
1014

 2 1010

[ H ] 5 104
-7-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
b) PH = -Log[H+] =  log5 104  3.301
c) POH +PH =14
POH = 14 –PH
= 14 – 3.301=10.699
--------------------------------------------------------------------------------------------------------------Example 3 find Ph of 0.001molar of HCl?
HCl is strong acid so it is completly ionized
[C A ]  [H 3O  ]
 PH   log 0.001  3
--------------------------------------------------------------------------------------------------------------Example 3
Calculate the concentration of hydroxide ion in 0.1M of ammonium hydroxide at 25C given that
equilibrium constant = 1.7 105
Solution
[OH  ]  Kb .Cb  0.11.7 105  1.33 103
---------------------------------------------------------------------------------------------------------------
Example 4
2CO( gas )
K p  1.67 103 at1203C
In the reaction C( S )  CO2( gas )
a) Find the partial pressure of carbon monoxide at equilibrium given that partial pressure of
CO2 is 18.275 atmosphere
b) Calculate equilibrium constant Kc given that concentrations of CO2 &CO are 0.05 and 0.83
mole/liter respectively
c) If CO2 of 0.01mole/liter with CO of 0.1mole/liter are mixed with excess of carbon, does the
reaction in equilibrium? If not what is the predominant direction
Solution
a)
KP 
P 2 (CO)
P(CO2 )
1.67 103 
P 2 CO
18.275
 P(CO)  1.67 103 18.275  174.7 Atmosphere
b)
KC 
[CO2 ] [0.83]2

 13.77
[CO] [0.05]
C)
KC 
0.12
1
0.01
So the reaction doesn’t in equilibrium in this case because Kc is (1) which doesn’t equal Kc at
equilibrium 13.77
The predominant direction is forward because Kc is not less than one
--------------------------------------------------------------------------------------------------------------
Section 1answeredproblems
(1)
Calculate equilibrium Kc in the following reaction:
CH3OH ( gas )
CO2( gas )  2H 2( gas ) at 700K (Kelvin) given that concentration of CH3OH, CO2,
H2 is 0.05, 0.03, 0.075 mole/liter respectively
-8-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
(2)
(3)
(4)
(5)
(6)
.
Calculate equilibrium Kc for the following reaction
N2O4
2 NO2 at 25C given that concentrations of N2O4 and NO2 is 0.095 and 0.0161
mole/liter respectively
Calculate concentration of NO2 when the following reaction is in equilibrium
N2O4
2 NO2 given that Kc is = 4.55 103 and concentration of N2O4is 0.16mole/liter
Find concentration of hydrogen iodide in the equilibrium
Kc = 50.53 given that concentration of hydrogen and iodine at
H 2( g )  I 2( g )
2HI ( g )
3
equilibrium is 12 10 mole /liter for each
In the reaction H 2( g )  I 2( g ) 2HI( g ) KC = 55.16
if concentration of H2, I2, and HI equal 1103 ,1.5 103 ,5 103 mole /liter at temp 425C does
this reaction is in equilibrium or not? Why?
We can represent reaction between hydrogen and nitrogen to form ammonia by the
following equations
KC
a. N2( g )  3H 2( g ) 2NH3( g )
1
b.
c.
d.
1
3
N 2( g )  H 2( g )
NH 3( g )
2
2
1
2
N 2( g )  H 2( g )
NH 3( g )
3
3
1
3
NH 3( g )
N 2( g )  H 2( g )
2
2
K C2
KC3
K C4
i. write the equilibrium constant for each reaction
ii. determine the relationship between theses four constant
(7)
1mole of hydrogen is mixed with one mole of iodine at certain temperature, if the
quantity of hydrogen and iodine remains at equilibrium is 0.2mole for each. Find the
equilibrium constant if the volume of the reaction is one liter
(8)
In equilibrium for the reaction N2( g )  3H 2( g ) 2NH3( g ) .the volume of the mixture was 1liter
and contains 0.3mole nitrogen ,0.2 hydrogen , 0.6 mole ammonia calculate equilibrium
Kc
(9)
Calculate equilibrium Kc for decomposition of PCl5 at 25C0 according to equation
PCl5( g )
PCl3( g )  Cl2( g ) .Given that capacity of the container is 6liters and contains at
equilibrium 0.021, 0.32, 0.32 moles for PCl5, PCl3 and Cl2 respect.
(10) Given that at temperature 823K
a. CuO( s)  H2( g ) Cu( s)  H2O(v) KC  67
b. CuO( s )  CO( g ) Cu( s )  CO2( g ) KC  490
i. write the equation for equilibrium constant for each reaction
ii. calculate equilibrium constant at 832K for the following reaction
c. CO2( g )  H2( g ) CO( g )  H2O(v)
SECONDLY problems for Kp
(11) a gaseous mixture in equilibrium consists of PCl5, PCl3, Cl2 their partial pressure is 0.15,
18.75, 0.2 atmosphere respectively, calculate equilibrium constant Kp at 500K for the
reaction PCl5( g ) PCl3( g )  Cl2( g )
(12) calculate Kp for the reaction
NH 4 HS( s )
NH3( g )  H 2 S( g ) Given that partial pressure form ammonia and hydrogen sulphide
is 0.54, 0.17spectively at certain temperature
1
2
-9-
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
(13)
The following reaction at equilibrium
partial pressure of chlorine gas knowing
that partial for SbCl5, SbCl3 is 0.15, 0.2 atmosphere
The following reaction in equilibrium
PCl5( g )
PCl3( g )  Cl2( g ) Kp = 25 at 298 K .Find partial pressure for PCl3 knowing that
partial pressure for PCl5, Cl2 is 0.0021 , 0.48 atmosphere respectively.
Equilibrium constant Kp for the following reaction
N2( g )  3H 2( g )
2 NH3( g ) is 41 at 400 Kelvin. Calculate Kp for the following reactions
a. 2NH3( g ) N2( g )  3H 2( g )
SbCl5( g )
(14)
(15)
.
b.
SbCl3( g )  Cl2( g )
1
3
N 2( g )  H 2( g )
2
2
2 N2( g )  6H 2( g )
K P  305 104 At500K .find
NH 3( g )
c.
4 NH3( g )
---------------------------------------------------------------------------------------------------------------
Third section
Explain the effect of the following different factors on the equilibrium in the following equations
(1)
Addition more hydrogen to the reaction N2  3H2 2NH3
(2)
Addition potassium super oxide in presence of catalyst to the reaction C( s )  CO2( g ) 2CO( g )
(3)
Addition concentrated sulphuric acid to the following
reaction CH3COOH(aq)  C2 H5OH(aq) CH3COOC2 H5(aq)  H2O(l )
(4)
1
SO2( g )  O2( g ) H=+
2
2 NH3 H  ve
Deceasing temperature in the following reaction
SO3( g )
(5)
Raising temperature in the reaction N2  3H2
(6)
Decrease pressure in the reaction 2NOCl( g ) 2NO( g )  Cl2( g )
(7)
Increase pressure in the reaction SO3( g )  H2( g ) SO2( g )  H 2O(liq )
---------------------------------------------------------------------------------------------------------------
Section 4
Explain the effect:
(1)
Pressure and temperature on formation of ammonia in the reaction
N2( g )  3H 2( g )
(2)
(5)
N2( g )  2H 2( g ) H= -ve
Effect of pressure and temperature and catalyst on the direction of reaction
SO3( g )
(4)
1
SO2( g )  O2( g ) H=+
2
Effect of iron III chloride on the red blood colour in the reaction
FeCl3( aq )  3NH 4 SCN
3NH 4Cl( aq )  Fe(SCN )3( aq ) FE
Effect of pressure , temperature and concentration of reaction on the reaction
1
1
N 2( g )  O2( g )
2
2
(6)
(7)
H=-92KJ
Effect of pressure and temperature on formation of nitrogen according to
N2 H 4( g )
(3)
2 NH3
NO( g )
H=+
In the following reaction CoCl2 .6H2O( solid ) CoCl2( solid )  6H2O(v) .What is the colour of cobalt
chloride if it is place in open container in
a. Dry air
b. moisture air
Explain practical experiment to show:
a. The effect of surface area on the rate of reaction.
b. The law of mass action (effect the concentration on the rate of reaction
c. Effect of heat on the rate of reaction
- 10 -
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
---------------------------------------------------------------------------------------------------------------
Section 5
Varity of questions
Question1
State the factors affecting the rate of reaction
--------------------------------------------------------------------------------------------------------------Question 2
On mixing one mole of CO2 with one mole of H2 then heating the mixture to 200C we reach
equilibrium when 0.6mole of each CO and H2O, and 0.4 of each CO2 and H2 remain without
reaction, please answer the following
(a)
Write the balanced chemical equation for the equilibrium
(b)
What is the effect of adding more carbon dioxide on equilibrium
(c)
Calculate equilibrium constant
---------------------------------------------------------------------------------------------------------------
Answer of problems in section 1
(1)
Kc =
[CO2 ][ H 2 ]
(0.075)2  0.03

 0.3375 102
[CH 3OH ]
0.05
2
 NO2   0.016
KC =
 N2O4  0.095
2
 NO2 
KC =
 N 2O4 
2
NO2 

3
 4.55 10 
2
(2)
(3)
 0.2728 102
0.16
  NO2   0.455 103  0.16  0.0269 mole
liter
 HI 
 H 2  I 2 
2
(4)
KC 
 HI 
2
 50.53 
 HI  
12 103 12  103
50.53 12  103  12  103  85.30  103 mole
 HI 
 H 2  I 2 
2
(5)
KC 
2
5 103 

 16.66
1.5 103  1103 
(6)
 NH 3 
i. (a) K C 
3
 N 2  H 2 
 NH 3 
(b) K 
2
1
C2
1
3
 N2 2  H 2 2
2
(c) K C3 
 NH 3 3
1
 N 2 3  H 2 
1
( d ) K C4
3
 N 2  H 2 2
 2
 NH 3 
- 11 -
liter
.
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
ii.
(7)
1
KC4 
KC1
1
K C3  K C1
3
1
K C2  K C1
2
Write the balanced equation
H 2( g )  I 2( g )
2HI ( g )
1mole  1mole
2moles
2  0.8mole
0.8mole 0.8mole
(8)
Since the volume is one liter so that number of mole is considered as the concentration
2
2
HI 
2  0.8


therefore KC 

 64
 H 2  I 2  0.2  0.2
Concentration of gases equals their moles because the volume is 1 liter
2
NH 3 

0.62
KC 

 150
3
3
 N2  H 2  0.3  0.2
0.21
 0.0035 mole
liter
6
concentration of PCl3 = 0.32  0.053 mole liter
6
0.32
Concentration of Cl2 =
 0.053 mole
liter
6
 PCl3 Cl2   0.053  0.053  0.08026
KC 
0.0035
 PCl5 
(9)
Concentration of PCl5 =
(10)
i-
CO2 
CO
ii. The equation (c) is a result of subtraction of a and b
so KC  KC  KC  67  490  423
P  PCl3   P Cl2  18.75  0.2
KP =

 25
P  PCl5 
0.15
Since NH4HS is solid therefore KP =P[NH3]  H 2 S 
KC1 
1
,
H2 
3
(11)
(12)
KC2 
1
2
KP  0.17  0.54  0.0918
(13)
(14)
(15)
.
TRY to answer
Try to answer
First
a.
b.
1
41
41  6.4
1
c. 2  5.95 104
41
- 12 -
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
Lesson 2 chapter 7
Ionic equilibrium
(1)
Ostwald law applied only on weak electrolyte because strong electrolyte is completely
ionized
Assume that we have 1 mole of weak mono protic acid as HA
H  A
HA
1  mole
 mole +  mole
1 
v
Ca 
 

v v
1
v
 (1 )C a
Ca  Ca
[H  ][A  ]
Ka 
[HA ]
2 C a 2
2 .C a
K a 

(1 )C a (1 )
(1 ) is so small
 K a 2 .C a
number of dissocated moles
total moles before dissociation
Ka   2  Ca
(2)

(3)
We can calculate ionic equilibrium Ka for weak acid using the relation  
Ka
Ca
where C
is the concentration of the acid
example 1 calculate ionic equilibrium for formic acid of concentration 0.1 molar given
that degree of ionization of this acid is 4.5% at 20C
solution
4.5
 0.045
100
Ka   2  C  (0.045)2  0.1  2 104
degree of ionization  
Concentration of hydroxonium ion H3O+ = ka  ca
Example:
calculate concentration of hydrogen ions in 0.1molar solution of acetic acid at 25C, given
ionic equilibrium constant is 1.8 105
solution
(4)
 H3O   Ka  Ca  1.8 105  0.1  1.3 103 M
(5)
Concentration of hydroxide ion [OH-] =
Example
- 13 -
Kb  Cb
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
calculate concentration of hydroxide ion (OH-) of a solution 0.2molar of methyl amine,
given that ionic equilibrium constant is 3.6 104
solution
(6)
(7)
(8)
(9)
OH    Kb  Cb  3.6 104  0.2  8.4 103 M
Ionic product of water KW   H   OH    1014
PH=  Log  H3O 
POH= - Log[OH-]
PH+POH= 14
Example
calculate the value of pOH for 0.1 molar solution of carbonic acid. Ka = 4.4 107
solution
carbonic acid is weak acid we can determine at first concentration of hydroxonium ion
 H 3O    K a  C A  4.4 107  0.1  2 104
PH   log 4  104  4.7
pH  pOH  14
 pOH  14.4.7  9.3
(10)
Solubility constant Ksp for sparingly soluble substance is the product of multiplication of
concentration of its ions
Example:
calculate solubility product of silver sulphate given that the degree of solubility is
1.4 102 mole
liter
solution
2 Ag   SO4 2
Ag 2 SO4
1 mole  2mole Ag +  1mole SO4 2
2
K SP   Ag    SO4 2 
2
 Ag    [2 1.4 102 ]2  7.84 104
concentration of
 SO4 2   1.4 102
2
K SP   Ag    SO4 2   7.84  104  1.4 102  10.976 106
---------------------------------------------------------------------------------------------------------------
Section 1
Write the scientific expression for:
(1) Substances that conduct electricity through their ions
(2) Substances their degree of ionization is 100%
(3) Ion that can not be found in water free
(4) The hydrated proton
(5) Converting unionized molecules into ions
(6) Converting molecules of ionic compounds into mobile ions
(7) The ionization that happens in strong electrolyte
(8) Ionization takes place in weak electrolyte
(9) The equilibrium that takes place in weak electrolyte between the molecules and ions from
it
(10) The relation between the degree of ionization and concentration
- 14 -
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
(11)
(12)
(13)
(14)
(15)
(16)
(17)
.
Acids that have small ionic equilibrium constant
The alkali that partially ionized in water
The product of multiplication of concentration hydrogen ion and hydroxide ion of water
The negative logarithm hydroxide ion
The measure of degree acidity or alkalinity of the medium
Dissolving a salt in water to form acid and alkali from which the salt is derived
The product of multiplication of ions concentration in their saturated solution
--------------------------------------------------------------------------------------------------------
Section 2
(1)
A
Choose from column (B) what suits column (A)
B
(1) r
(2) k
(3) k
(4) 
(5) k
(6) k
(7) k
(8) pH
(9) k
1
c
p
a
( ) The constant that represents gases by their partial pressure
( ) ionic equilibrium constant for weak acid
( ) ionic product of water
( ) rate of forward reaction
( ) solubility product
( ) degree of ionization
( ) ionic equilibrium constant for weak base
( ) Chemical equilibrium expressed using concentration
( ) The hydrogen exponential
b
w
sp
(2)
Arrange the following acids according to their strength using the values of Ka in the table
a. Phosphoric acid H3PO4
Acid Ka
b. Carbonic acid H2CO3
c. Hybobromus acid HBrO
H 3 PO4
7.6  10 3
d. Iodic acid HIO3
H 2 CO3
4.3  10 7
e. Hydrofluoric acid HF
HBrO
2.0  10 9
f. Hydrocyanic HCN
1
HIO3
1.7  10
HF
3.5  10 4
----------------------------------------------------------------------------------------------------------------HCN
4.9  10 10
Section 3
Choose the correct answer
(1)
( pure acetic acid – boric acid – sodium chloride solution) conducts electricity
(2)
the degree of ionization of (carbonic acid – sulphuric acid – nitrous acid) doesn’t increase
by dilution
(3)
the equilibrium that takes place in weak electrolyte is called (dynamic – covalent – ionic )
(4)
we can apply the law of mass action on solution of (sodium chloride – hydrofluoric acid –
hydrochloric acid)
- 15 -
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
(5)
(6)
Ostwald performed the relation between ----------a. The rate of evapouration and condensation
b. The rate of forward and backward reaction
c. Degree of ionization and and degree of dilution
d. Rate of reaction and concentration
The degree of ionization can be calculated from the relation ----------a.
(7)
.
C
1
V
b. Ka   2  C
c. Ka   2  C
The concentration of ions can be calculate form the relation
a.
C
Ka
b. KaxKb
c. Kb  C
d. Ka  C
(8)
The ionic product of water is equal to (7 – 14 – 10-14 – 10-7)
(9)
The solution will be basic when pH is (0 – 7 – 3 – 9)
(10) A solution its pH is 1 it will be (strong acid – weak acid – strong base – weak base)
(11) A solution of pOH is 8 it will be (weak acid – strong acid – weak base – strong base)
(12) Hydrolysis of sodium carbonate produces (H+, Na+/ Na+ and OH-/ CO3-2, Na+)
(13) On adding few drops of phenolphthalein to a solution of sodium acetate it become
(colourless – red – blue)
-----------------------------------------------------------------------------------------------------------------
Section 4
Complete the tables
Weak acid
Sulphorus
----------Nitrous acid
---------Carbonic
Boric acid
2- Given that Kw =
Molecular
formula
------------HF
-------------Acetic acid
--------------------10
Equi constant
Ka
----------1.6 10
4
5.1104
------------4.4 107
5.8 10
10
Concentration of the
acid C
0.5
0.1
---------------1
2.1
--------------
Conc of H+
0.86
-----------0.45 102
4.2 103
---------0.5 105
14
Kind of
solution
12
----------------------------------Acidic
10
4
---------------------------------------------10
-------------------------7
-----------------------------------------------------------13
-----------------------------------5
--------------------------------------------------------------------------------------------------------------------------------------H

OH 
pH
pOH
Section 5
Problems
- 16 -
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
First Ostwald law
(1) Calculate the degree of ionization of 0.01 molar solution of weak acid at 25C given Ka =
1.8 105
(2) Penicillin is weak acid used as antibiotic; the degree of dissociation is 2 102 in solution of
one liter that contains 0.25mole of penicillin. Calculate ionic equilibrium constant for
penicillin
(3) If the degree of dissociation of monoprotic weak acid is 0.3% in a solution of concentration
0.2 mole /liter. Find Ka of this weak acid.
Secondly: calculation of concentration of H3O+ for weak acid
(4) Given that Ka for nicotinic acid HC6H4O2 is 1.4 105 calculate [H+] in a solution of one liter
contains 0.1 mole of the acid
(5) Calculate [H3O+] for aqueous solution of volume 500ml contains 0.48 mole of HCl
Thirdly: problems of the pH &pOH
(6) Calculate the pH and pOH for lemon juice given that  H    5 103 mole liter
(7) 6 grams of acetic was dissolved in one liter of water calculate pOH of the solution given
that Ka= 1.8 105
(8) 1 gram of NaOH was dissolved in 500ml water. calculate the pH given that Na= 23 O =
16 h= 1
(9) calculate the concentration of hydrocyanic acid HCN given that Ka = 4.9 1010 and pH =
5.3
(10)
Calculate concentration of ions H+&OH- in the blood, given that its pH = 7.4
(11)
Calculate concentration of OH- in a solution ,the concentration of H+ in this solution
3 107 mole
liter
(12)
Calculate the value pH of NaOH completely ionized on dissolving 10gm in water to
form one liter of the solution
(13)
Calculate the value pH for a solution of concentration 0.2mole/liter of methyl amine
CH 3NH2 given that equilibrium constant Kb = 3.6 104
(14)
Calculate the value of Ka for a solution its concentration 0.1 molar of chlorus acid
HClO2 given that pH= 1.5
(15)
Calculate the value of Ka for a solution its concentration 0.015 molar of nitrous acid
HNO2 given that pH of this acid 2.63
(16)
The molecular formula of aspirin C9H8O4 which is weak acid, the aqueous solution of
this acid can be prepared by dissolving 3.65gm in water to form 1 liter of the solution. Given
pH of aspirin 2.6 calculate equilibrium constant Ka
(17)
Calculate both Ka and pH for benzoic acid given that its concentration 0.11molar and
its degree of ionization 2.4%
(18)
Calculate both Ka and pH for organic acid of concentration 0.02molar and degree of
ionization is 0.14%
Fourth solubility product Ksp
(19)
Calculate the degree of solubility of silver bromide AgBr given that Ksp = 7.7 1013
(20)
Calculate the degree of solubility CuS given that KSP
= 1.3 1036
(21)
Calculate Ksp of PbCl2, given that concentration of Pb+2 is 1.6 102 molar
(22)
Calcium phosphate dissolve in water according to the equation
Ca3 ( PO4 )2( s )
K SP  11033
3Ca 2  2 PO4 3
calculate concentration of phosphate ion when concentration of
calcium ions is 10-9molar
- 17 -
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
.
Answer of section 5
(1)

ka
1.8 10

C
0.01
5
 0.042
0.25
 0.25molar
1
Ka   2  C  (2 102 )2  0.25  104
(2)
concentration of the acid =
0.3
 0.003
100
K a  (0.003)2  0.2  1.8 105
(3)
de gree of ioniZation  =
(4) Concentration of the acid =
0.1
 0.1molar
1
 H3O   K A  CA  1.4 105  0.1  1.18 103 MOLAR
(5) Concentration of the acid = 0.48  0.96molar
0.5
since HCl is strong electrolyte and completely ionized so concentration of H3O+ is equal to
concentration of the acid
+
 concentration of H3O = 0.96 molar
(6) pH = -log 5 102  1.3
pOH = 14-1.3 = 12.7
(7) one mole of acetic acid CH3COOH= 12+3+12+16+16+1=60gm
6
 0.1mole
60
No of moles = concentration of the acid =
0.1
 0.1molar
1
concentration of  H 3O    1.8 105  0.1  1.34 103
pH = - log 1.34 103  2.87
pOH= 14- 2.87 =11.3
(8) one mole of NaOH = 23+16+1= 40gm
No of moles of NaOH =
1
 0.025mole
40
concentration of NaOH =
0.025
 0.05molar
0.5
since NaOH is completely ionized so concentration of OH- is the same concentration of NaOH
= 0.05 molar
pOH = -log 0.05= 1.3
pH = 14 – 1.3 = 12.7
(9) PH = - log [H3O+]
 H 3O    5  106 molar
 H 3O    K a  C
2
 H 3O  
(5  106 ) 2
C  

 5.1 102 molar
KA
4.9  1010
- 18 -
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
(10)
pH = -log [H3O+]
  H 3O    3.9  108 molar
POH  14  7.4  6.6
POH   Log OH  
OH    2.51 107 molar
(11)  H3O  OH    1014
1014
OH   
 3.33 108 mole
liter
3 107
(12)
One mole of NaOH = 40gm
10
 0.25mole
40
= No of moles  0.25  0.25molar
volume
1
No of moles of NaOH =
concentration of NaOH
 concentration of OH -   0.25
POH   log 0.25  0.6
pH  14  0.6  13.4
(13)
OH    3.6 104  0.2  0.848 102
pOH   log 0.848 102  2.07
pH  14  2.07  11.93
(14)
pH= -log [H3O+]
 H 3O    0.0316molar
 H 3O    K A  C A
2
 H 3O  
(0.0316) 2
 KA 

 0.0099 102
CA
0.1
(15)
(16)
similar to the above
mass of one mole of aspirin = 180gm
3.6
 0.02mole
180
0.02
concentration of the acid =
 0.02molar
1
pH   log  H 3O  
number of moles =
 H 3O    2.5  103 molar
2
 H 3O  
(2.5 103 ) 2
KA  

 3  104
CA
0.02
(17)
degree of ionization   2.4  .0024
100
Ka   2  C  0.0242  0.11  6.3 105
 H 3O    K A  C A  6.3 105  0.11  2.6 103
 pH   log 2.6 103  2.59
(18)
(19)
The same method of (17)
AgBr
Ag   Br 
K sp   Ag    Br    X 2  7.7 1012
X  7.7 1012  0.87 106 molar
(20)
by the same method
- 19 -
.
Chapter 7 prepared by Mr. Khaled Nasr Tel: 01005679458
(21)
PbCl2
Pb2  2Cl 
[Pb  ]  [1.6  102 ]  X
[Cl  ]  2X  3.2  102 mole / L
2
K SP  Pb 2  Cl    (1.6 102 )  (3.2 102 )2  16.38 106
----------------------------------------------------------------------------------------------------------------Thanks for all of You
With great success
Mr. Khaled Nasr
- 20 -
.