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NOTES
CLASS: 10 – CHEMISTRY
UNIT: 9 – CHEMICAL EQUILIBRIUM
A. CHOOSE THE CORRECT OPTION.
1. At dynamic equilibrium.
a. The reverse reaction stops
b. The forward reaction stops
c. Both forward and reverse reactions stop
d. Both forward and reverse reactions continue at the same rate
2. A reversible reaction has the following characteristics except
a. They proceed in both direction
b. They never complete
c. Products do not form reactants again.
d. They are represented by ⇌
3. The equilibrium constant (Kc) expression for the given reaction is,
a. !" [!$][!"$]
b. " &[$]&
[!"$]
c. ["&][$&]["$]&
d. [𝑵𝑶]𝟐
𝑵𝟐 [𝑶𝟐]
4. The equilibrium constant (Kc) is:
a. The sum of the two reactants
b. The difference of the two rate constants
c. The ratio of the two rate constants
N2(g)+O2(g) ⇌ 2NO(g)
2
d. The product of the two rate constants
5. When the value of Kcis very small it shows that
a. Reaction will go in the forward direction
b. Reaction will go in the reverse direction
c. Reaction is at equilibrium
d. Equilibrium will never establish
6. For which reaction the Kc will have no units
a. 4NH3(g) + 5O2(g) ⇌
4NO(g) +6H2O
b. N2(g)+ 3H2(g) ⇌
2NH3(g)
c. N2(g) + O2(g) ⇌
2NO(g)
d. CO(g)+ 3H2(g) ⇌
CH4(g)+ H2O(g)
7. When the value of Kc is very large it shows that
a. Reaction is at equilibrium
b. Equilibrium will never be achieved.
c. Reaction will move in the forward direction
d. Reaction will move in the reverse direction
8. Active mass means
a. The total mass of reactants
b. The total mass of products
c. The total mass of products and reactants
d. Concentration of reactants and products in moles per dm3 in a dilute
solution.
3
9. For a reversible reaction
Kc=[+],
[-].[/]&
The equation will be,
a. 4C(g) ⇌ 3A(g)+2B(g)
b. 4C(g) ⇌ A3(g)+B2(g)
c. A3(g)+ B2(g) ⇌ C4(g)
d. 3A(g)+2B(g) ⇌ 4C(g)
10. The reaction between PCI3 and CI2 will produce PCI5
PCI3(g) + CI2(g) ⇌ PCI5(g)
The units of Kc for this reaction are,
a. mol.dm-3
b. mol-1.dm-3
c. mol-1.dm3
d. mole.dm3
B. Short questions.
Q1. Define chemical equilibrium with two examples?
Ans: CHEMICAL EQUILIBRIUM:
A reversible chemical reaction is in chemical equilibrium state when the rate of
forward reaction equals the rate of reverse or backward reaction and the concentrations
of the products and reactants remain constant.
4
Examples:
a. Reaction of sulphur dioxide with oxygen.
2SO2 + O2 ⇌2SO3 b. Reaction of hydrogen with iodine:
H2 + I2 ⇌2HI
Explanation:
When liquid water is placed in a closed container at constant temperature, part of the
liquid evaporates. As water begins to evaporate, at the same time, some of the vapours
also begin to condense. Although the rate of evaporation is more than the rate of
condensation in the start, as time passes, it becomes equal to the rate of condensation
and thus a state of chemical equilibrium is established.
Water ⇌Vapours
The double half-headed arrow shows that at the equilibrium state, both the forward and
reverse processes are occurring at the same rate.
Q 2. How would you identify that dynamic equilibrium is established?
Ans) DYNAMIC EQUILIBRIUM:
Dynamic equilibrium is an equilibrium in which the two chemical processes continue
at an equal rate in opposite direction. The ratio between the concentration of products
and reactants remains constant.
Evaporationcondensation
5
Examples:
The examples of dynamic equilibrium are
a. Reaction between nitrogen and hydrogen.
N2 + 3H2 ⇌2NH3
b. Decomposition of mercuric oxide:
2HgO ⇌2Hg + O2
Ways to recognize equilibrium:
In order to recognize the equilibrium state of chemical reaction, the following
methods can be used.
1. PHYSICAL METHODS: such as Refractometry, polarimetry, spectrophotometry
etc.
2. CHEMICAL METHODS: such as titration etc.
In both methods, the equilibrium state of chemical reaction can be recognized by
knowing the concentration of reactants and products at regular intervals of time.
When the concentration of reactants and products are observed constant, then the
reaction is said to be in equilibrium state.
Q3) Compare the different macroscopic characteristics of forward and reverse reaction?
Ans: FORWARD REACTION:
It is defined as the conversion of reactants into products per unit time or the rate of
chemical reaction taking place in forward direction.
REVERSE REACTION:
It is defined as the conversion of products back into reactants per unit time or the rate of
chemical reaction taking place in reverse direction.
The comparison of macroscopic characteristics of forward and reverse reactions is
given below.
6
FORWARD REACTION REVERSE REACTION
1. It is a reaction is which reactants
react to form products.
2. It takes place from left to right
3. At initial stage, the rate of forward
reaction is very fast.
4. It slows down gradually.
1. It is a reaction in which products
react to form reactants.
2. It takes place from right to left.
3. In the beginning, the rate of revers
is negligible.
4. It speeds up gradually.
Q4) What information is required to predict the direction of a chemical reaction?
Ans) DIRECTION OF REACTION:
We can determine the direction of chemical reaction with the help of equilibrium
constant expression.
Kc=[01234567][89:56:;67]
The direction of chemical reaction at any particular time can be predicated by means of
[Products] / [reactants] ratio. By comparing the [products]/ [reactants] ratio with Kc we
have three possibilities.
I. When the ratio of [products]/ [reactants] is less than Kc it means reaction is moving
more in backward direction than forward direction. The system is not at equilibrium and
more products are required to reach the equilibrium. Therefore, the reaction will move
in the forward direction until equilibrium is reached.
II. When the ratio of [products]/ [reactants] is greater than Kc it means reaction is moving
more in forward direction than the backward direction. The system is not at equilibrium
and more reactants are required to reach equilibrium. The reaction will move in the
reverse direction until equilibrium is reached.
III. When the ratio of [products]/[reactants] is equal to Kc. The system is at equilibrium.
7
Q5) Relate the active mass with rate of a chemical reaction?
Ans) ACTIVE MASS:
The term “active mass” represents the concentration of reactants and products in
mol.dm-3 for a dilute solution and is expressed in terms of square brackets [ ].
RELATIONSHIP BETWEEN ACTIVE MASS AND RATE OF CHEMICAL
REACTION:
According to the law of mass action, the rate of chemical reaction is directly
proportional to the products of the molar concentrations (active masses) of the reacting
substances. In other words, the rate of chemical reaction increases with the increase in
active masses of the reacting substances and vice versa.
Q6) At equilibrium a mixture of N2, H2, and NH3 gas at 5000C is determined to consist of
0.602 mol/dm3of N2, 0.420 mol/dm3 of H2, and 0.113 mol/dm3 of NH3. What is the
equilibrium constant for the reduction at this temperature?
N2 + 3H2 ⇌2NH3
Given Data:
[N2] = 0.602 mol/dm3
[H2] = 0.420 mol/dm3
[NH3] = 0.113 mol/dm3
Required Data:
Kc=?
8
Solution:
The equilibrium constant expression for the above reaction is:
Kc = ["<.]&
"& [<&].
= (?.AABC2D/3C.)&
(?.G?!C2D/3C.)(?.H!?C2D/3C.).
= ?.?A!IGJC2D&/3CK
?.G?!C2D/3C.L?.?IH?MMC2D./3CN
= ?.?A!IGJC2D&/3CKL3CK/C2DL3CN/C2D.
?.?HHG??JIG
= 0.286294183𝑚𝑜𝑙!/𝑑𝑚G𝑋𝑑𝑚A!/𝑚𝑜𝑙H
= 0.28630 mol-2/ dm6
Q7) State conditions necessary for chemical equilibrium?
Ans) CONDITIONS FOR EQUILIBRIUM:
The following conditions are necessary for chemical equilibrium:
1. Concentration of reactants and products.
2. Temperature of the system.
3. Pressure of the system
4. Volume of the system.
5. Catalyst, if used in the system remain unchanged.
Explanation:
Equilibrium cannot be attained in open containers because in open containers, the
gaseous reactants and products may escape into the atmosphere leaving behind no
possibility of attaining equilibrium. When the system reaches equilibrium, it will remain
in the same state forever, if the conditions do not change. When the conditions are
disturbed, the equilibrium will also be disturbed. The system always tries to regain its
9
equilibrium. The system will move either in the forward direction or in the reverse
direction to attain equilibrium.
Q8) write equilibrium constant expression for the following reactions.
1. N2+ 3H2⇌2NH3
2. 2H2+O2⇌2H2O
3. 4NH3+5O2⇌ 4NO+6H2O
Ans:
i. N2+ 3H2⇌2NH3
The equilibrium constant expression for this reaction is:
Kc= 0123456789:56:;67
Kc= "<. &
"& [<&].
ii. 2H2+O2⇌2H2O
The equilibrium constant expression for this reaction is:
Kc= 0123456789:56:;67
Kc= <&$ &
<& &[$&]
iii. 4NH3+5O2⇌ 4NO+6H2O
The equilibrium constant expression for this reaction is:
Kc= 0123456789:56:;67
Kc=["$],[<&$]K
["<.],[$&]T
10
Q9) A reaction between gaseous sulphur dioxide and oxygen gas to produce gaseous
sulphurtrioxide takes place at 6000C. At this temperature, the concentration of SO2
is found to be 1.50 mol/dm3, the concentration of O2is 1.25 mol/dm3 and the
concentration of SO3 is 3.50 mol/dm3. Using the balanced chemical equation,
calculate the equilibrium constant for this system.
GIVEN DATA:
Concentration of [SO2] =1.50 mol/dm3
Concentration of [O2] =1.25 mol/dm3
Concentration of [SO3] = 3.50 mol/dm3
REQUIRED DATA:
Kc=?
Solution:
The balanced chemical equation is:
2SO2+O2 ⇌2SO3
The equilibrium constant expression is:
Kc= [U$.]&
[U$&]&[$&]
Substituting the equilibrium concentrations of reactants and products.
Kc= [B.V?C2D/3C.]&
[A.V?C2D/3C.]&[A.!VC2D/3C.]
Kc= A!.!VC2D&/3CK
!.!VC2D&/3CKLA.!VC2D/3C.
Kc = A!.!VC2D&/3CK
!.MA!VC2D./3CN
Kc= A!.!V!.MA!V
C2D&
3CK 𝑋𝑚𝑜𝑙B/𝑑𝑚J
Kc= 4.355 𝑑𝑚B/𝑚𝑜𝑙
11
Kc= 4.355 mol-1.dm3
Q10) Describe the effect of temperature on equilibrium constant?
Ans: The numerical value of the equilibrium constant for a particular reaction is constant as
long as the temperature is kept constant.
The rate of a chemical reaction increases with increase in temperature. The extent of
this increase in rate depends upon the energy of activation of the reaction.
Since the energy of activation for the forward and backward reaction are different so a
given increase in temperature will increase the rate of forward and backward reaction
to different extends.
The values of the velocity constant for forward and backward reaction will change
differently with a given rise or fall in temperature.
Since K= kf /kb therefore, the value of equilibrium constant will change i.e. the state
of equilibrium is altered.
The value of equilibrium constant of an endothermic reaction increases and that of an
exothermic reaction decreases with rise in temperature.
For reaction having zero heat of reaction, the temperature has on effects the value of
K.
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C. LONG QUESTIONS.
1. SO3 decomposes to form SO2 and O2. For this reaction write, i. Chemical equation,
ii. Kc expression and iii. Derive the units of Kc for this reaction.
Ans)
i. Chemical Equation:
2SO3 ⇌2SO2 +O2
ii. Kc Expression:
Kc= U$& &[$&][U$.]&
iii. Units of Kc: The equation constant expression for this reaction is:
Kc= U$& &[$&][U$.]&
Putting the Values:
Kc= [C2D/3C.]&[C2D/3C.]
[C2D/3C.]&
Kc = C2D&/3CKLC2D/3C.
C2D&/3CK
Kc= C2D&
3CK 𝑋C2D3C. 𝑋
3CK
C2D&
Kc= C2D3C.
2. (a). Describe the equilibrium state with the help of a graph and an example.
(b). Define and explain the law of mass action.
Ans) A. CHEMICAL EQUILIBRIUM:
A reversible chemical reaction is in chemical equilibrium state when the rate of
forward reaction equals to the rate of reverse reaction and the concentrations of its products
and reactants remain constant.
13
EXPLANATION:
Let us consider a general chemical reaction in which reactant A reacts with reactant B
in the gaseous state, in a closed container to form products C and D.
A + B ⇌ C + D
The detail of this reversible reaction is given below:
CONCENTRATIONS OF SUBSTANCES AT THE START:
Initially the concentrations of reactants A and B are maximum and the concentrations
of products C and D are zero.
CONCENTRATIONS OF SUBSTANCES AFTER SOMETIME:
With the passage of time, the concentrations of reactants (A and B) decrease and the
concentrations of products (C and D) increase simultaneously.
VARIATION IN RATE OF FORWARD AND REVERSE REACTIONS:
In the beginning, the rate of forward reaction is maximum while the rate of reverse
reaction is almost zero. As the time passes and sufficient amounts of products C and D
are formed. At this stage, the rate of reverse reaction increases while that of forward
reaction decreases until the rate of forward reaction becomes equal to the rate of reverse
reaction. Thus, a state of dynamic equilibrium is established.
CONCENTRATIONS OF SUBSTANCES AT EQUILIBRIUM STATE:
The concentrations of reactants and products remain constant at equilibrium state.
14
GRAPHICAL REPREZENTTION
\
As long as we do not change the conditions of the system, such as pressure, temperature,
concentrations of the gases, the system will remain in equilibrium.
b) LAW OF MASS ACTION:
History:
This law was presented by two Norwegian Chemists, Cato Maximilian Guldberg and
Peter Waage in 1864.
Statement:
The rate or speed of chemical reaction is directly proportional to the product of the molar
concentrations (active masses) of the reacting substances.
Active Masses:
The term “Active Mass” means the molar concentration of reactants and products in
mol.dm-3 for a dilute solution and is expressed in terms of square brackets [ ].
15
Example:
Consider the following general chemical reaction.
A+B ⇌ C+D
The active mass of A= [A]
The active mass of B = [B]
The active mass of C = [C]
The active mass of D = [D]
According to the law of mass action,
Rate of forward reaction (Rf) is directly proportional to the product of the concentrations
of the reactants A and B. Applying the law.
Rf α [A] [B]
Rf = Kf [A] [B]
Where ‘Kf’ is the rate constant for the forward reaction.
For the rate of reverse reaction,
Rr α [C][D]
Rr= Kr[C ][D]
Where ‘Kr’ is the rate constant for the reverse reaction.
At equilibrium state,
Rate of forward reaction = Rate of reverse reaction
Therefore, we can write,
Rf = Rr
Kf [A] [B] = Kr [C][D]
Thus, it is cleared that the rate of a reaction is proportional to the molar concentrations
of the reactants.
16
3. a) Derive an expression for the equilibrium constant and explain its units?
b) How can you predict the direction of reaction for the Kc value.
Ans) Equilibrium constant (Kc):
The ratio of the mathematical product of the concentrations of products at equilibrium
to the mathematical product of the concentrations of reactants is called equilibrium
constant.
Representation:
It is represented by ‘Kc’, where the subscript ‘C’ indicates the equilibrium
concentrations of various species in term of mole/dm3.
Example:
Consider the following general chemical reaction.
Aa + bB ⇌cC + dD We know that
Kf [A] [B] = Kr [C][D]
Rearranging the above equation, we get:
WXWY
= + [Z]- [/]
Kc =+ [Z]- [/]
( WXWY
= Kc)
Where ‘Kc’ is called the equilibrium constant. In ‘Kc’, the subscript ‘C’ denotes the
molar concentration at equilibrium.
For more general chemical reaction,
Kc is written as,
Kc= + [[Z]\
- ][/]^
Where ‘a, b, c and d’ are the molar concentrations of A, B, C and D respectively.
17
Units of equilibrium constant:
The unit of equilibrium constant (Kc) depends on the equilibrium constant expression
for the given chemical reaction, we have three cases regarding the units of equilibrium
constant.
Case I:
Equilibrium constant has no units, if the number of moles of reactants and products are
equal in the balanced chemical equation.
Example:
H2 + I2 ⇌ 2 HI Kc for this chemical reaction is
Kc= [<_]&
<& [_&]
Kc= [`ab\`.]
&
`ab\`. [
`ab\`.]
Kc= C2D&/3CK
C2D/3C.LC2D/3C.
Kc= C2D&/3CK
C2D&/3CK
Kc= C2D&
3CK 𝑋3CK
C2D&
Kc= No unit
Case II:
Equilibrium constant has units, if the number of moles of products is greater than
the reactants in a balanced chemical equation.
Example:
N2O4 ⇌2NO2
Kc for this chemical reaction is,
18
Kc= ["$&]&
["&$,]
Kc= [`ab\`.]
&
[`ab\`.]
Kc= C2D&/3CK
C2D /3C.
Kc= C2D&
3CK 𝑋3C.
C2D
Kc= C2D3C.
Kc= mol.dm-3
19
Case III:
Equilibrium constant has units, if the number of moles of products is less than the
reactants in a balanced chemical equation.
Example:
N2+3H2⇌ 2NH3
Kc= ["<.]&
"& [<&].
Kc= [C2D/3C.]&
`ab\`.
`ab\`.
.
Kc= C2D&/3CK
C2D/3C.LC2D./3CN
Kc=C2D&/3CK
C2D,/3Cc&
Kc= C2D&
3CK 𝑋3Cc&
C2D,
Kc= 3CK
C2D&
Kc= AC2D&.3CdK
Kc= A[C2D .3Cd.]&
Therefore, it is concluded that the units of Kc depend upon the equation. It varies from
equation to equation.
20
b) Direction of reaction:
We can determine the direction of reaction with the help of equilibrium constant
expression.
Kc= [01234567]89:56:;67
The direction of chemical reaction at any particular time can be predicted by means of
[products] / [reactants] ratio. The value of [products]/ [reactants] ratio leads to one of
the following three possibilities.
a. When the ratio of [products]/ [reactants] is less than Kc. The system is not at
equilibrium and more products are required to reach the equilibrium. Therefore, the
reaction will move in the forward direction until equilibrium is reached.
b. When the ratio of [products]/[reactants] is greater than Kc. The system is not at
equilibrium and more reactants are required to reach equilibrium. The reaction will
move in the reverse direction until equilibrium is reached.
c. When the ratio of [products] / [reactants]is equal to Kc. The system is at equilibrium.
4. a) Kc has different units in different reactions. Prove it with suitable example.
b) How can you predict the extent of reaction from the Kc value
Ans) a) UNITS OF EQUILIBRIUM CONSTANT:
The unit of equilibrium constant (Kc) depends on the equilibrium constant expression
for the given chemical reaction, we have three cases regarding the units of equilibrium
constant.
Case I:
Equilibrium constant has no units, if the number of moles of reactants and products are
equal in the balanced chemical equation.
21
Example:
H2+ I2 ⇌ 2 HI Kc for this chemical reaction is
Kc= [<_]&
<& [_&]
Kc= C2D&/3CK
C2D/3C.LC2D/3C.
Kc= C2D&/3CK
C2D&/3CK
Kc= C2D&
3CK 𝑋3CK
C2D&
Kc= No unit
Case II:
Equilibrium constant has units, if the number of moles of products is greater than
the reactants in a balanced chemical equation.
Example:
N2O4 ⇌2NO2
Kc for this chemical reaction is:
Kc= ["$&]&
["&$,]
Kc= [`ab\`.]
&
[`ab\`.]
Kc= C2D&/3CK
C2D /3C.
Kc= C2D&
3CK 𝑋3C.
C2D
Kc= C2D3C.
Kc= mol.dm-3
Case III:
Equilibrium constant has units, if the number of moles of products is less than the
reactants in a balanced chemical equation.
22
Example:
N2+3H2⇌ 2NH3
Kc= ["<.]&
"& [<&].
Kc= [`ab\`.]
&
`ab\`.
`ab\`.
.
Kc= C2D&/3CK
C2D/3C.LC2D./3CN
Kc= C2D&/3CK
C�D,/3Cc&
Kc= C2D&
3CK 𝑋3Cc&
C2D,
Kc= 3CK
C2D&
Kc= AC2D&.3CdK
Kc= A[C2D .3Cd.]&
Therefore, it is concluded that the units of Kc depend upon the equation. It varies from
equation to equation.
b) THE EXTENT OF A CHEMICAL REACTION:
The value of Kc tells us about the extent of chemical reaction from which the quantities
of reactants and products can also be predicted. Consider the general reaction.
aA + bB ⇌ cC+dD
Kc=[+]�[Z]\
[-]][/]^
The extent of reaction depends upon the magnitude of ‘Kc’ so when,
i. Kc is very small: When the concentration of [A] and [B] is large and that of [C] and [D] is
small, the equilibrium mixture will contain the reactants mainly and only a small
23
amount of products will be present. It reflects that the reaction does not
proceed appreciably in the forward direction.
ii. Kc is very large:
When the concentration of [A] and [B] is small and that of [C] and [D] is
large, the equilibrium mixture will consist almost entirely of products and a small
amount of reactants will be present. This indicates that the reaction is completed
in forward direction.
iii. Kc is Neither very Large Nor very small: The concentration of the products and reactants will be very close and hence, Kc is close to 1.0. Thus, equilibrium mixture will contain appreciable amount of products and reactants.
5. a) Kc expression for a reaction is given below,
Kc=[+e&]&[<&$]&
[<+e],[$&]
For this reaction write,
i) Reactants and products ii) Derive the units of Kc.
b) Explain the importance of equilibrium constant, support your answer with examples
and reasons.
Ans ) a)i. CHEMICAL EQUITATION:
The balanced chemical equation for this Kc expression is,
4HCI + O2 ⇌ 2CI2 + 2H2O Reactants:
HCI and O2
Products: CI2 and H2O
ii. Units of Kc:
As Kc= [+e&]&[<&$]&
[<+e],[$&]
24
Putting the values:
Kc=[C2D/3C.]&[C2D/3C.]&
[C2D/3C.],[C2D/3C.]
Kc= C2D&/3CKLC2D&/3CK
C2D,/3Cc&LC2D/3C.
Kc= C2D,/3Cc&
C2DTL3CcT
Kc= C2D,
3Cc& 𝑋3CcT
C2DT
Kc= 3C.
C2D
Kc= AC2D.3Cd.
b) Importance of equilibrium constant:
The value of equilibrium constant is specific and remains constant at a particular
temperature. The value of Kc helps us to predict.
1) Direction of reaction:
We can determine the direction of reaction with the help of equilibrium constant
expression.
Kc= [01234567]89:56:;67
The direction of chemical reaction at any particular time can be predicted by means
of [Products] / [reactants] ratio. The value of [products] / [reactants] ratio leads to one
of the following three possibilities.
a. When the ratio of [products]/ [reactants] is less than Kc. The system is not at
equilibrium and more products are required to reach the equilibrium. Therefore, the
reaction will move in the forward direction until equilibrium is reached.
25
b. When the ratio of [products] / [reactants] is greater than Kc. The system is
not at equilibrium and more reactants are required to reach equilibrium. The reaction
will move in the reverse direction until equilibrium is reached.
c. When the ratio of [products] / [reactants]is equal to Kc. The system is at equilibrium.
2) THE EXTENT OF A CHEMICAL REACTION:
The value of Kc tells us about the extent of chemical reaction from which the quantities
of reactants and products can also be predicted. Consider the general reaction.
aA + bB ⇌ cC + dD
Kc=[+][[Z]\
[-]][/]^
The extent of reaction depends upon the magnitude of ‘Kc’ so when,
i. Kc is very small: When the concentration of [A] and [B] is large and that of [C] and [D] is
small, the equilibrium mixture will contain the reactants mainly and only a small
amount of products will be present. It reflects the reaction does not proceed
appreciably in the forward direction.
ii. Kc is very large:
When the concentration of [A] and [B] is small and that of [C] and [D] is
large, the equilibrium mixture will consist almost entirely of products and a small
amount of reactants will be present. This indicates that the reaction is completed
in forward direction.
iii. Kc is Neither very Large Nor very small: The concentration of the products and reactants will be very close and hence, Kc is close to 1.0. Thus, equilibrium mixture will contain appreciable amount of products and reactants.
26
3) THE EFFECT OF EXTERNAL CONDITIONS ON THE POSITION OF EQUILIBRIUM:
When a system reaches equilibrium it will remain in the same state indefinitely, if the
conditions do not change. However, the equilibrium state of a system is disturbed if external
conditions are changed, e.g. change of pressure, temperature and concentrations of reactants
and products alter the position of the equilibrium. Whenever, the equilibrium is disturbed by
changes in the external conditions, the system always tends to restore equilibrium.
27
UNIT: 10
ACIDS, BASES AND SALTS
A. Choose the correct option. 1. According to the Arrhenius concept, which of the Following is not an acid,
a) HCI b) H2SO4 c) CO2 d) HN3 2. AICI3 is an acid according to,
a) Arrhenius b) Lowery and Bronsted c) Lewis d) All of these 3. Which of the Following is a Lewis base
a) HCI b) AICI3 c) BF3 d) F- 4. Neutral Solution has a pH Value of:
a) 3 b) 5 c) 7 d)14 5. The pOH of 0.001M Solution of nitric acid is:
a) 0.001 b) 10.0 c) 11 d)14 6. When a strong base and weak acid reacts, the only products are,
a) Neutral salt and water b) basic salt and water c) Acidic salt and water d) Acidic, basic salt and water
7. NH3 is a base according to
a) Arrhenius b) Lowery and Bronsted c) Lewis d) Both b and c 8. Which one of the following is a basic salt?
a) KCI b) NaCI c) Pb(OH)CI d) KHSO4 9. The bases which are soluble in water are called:
a) Acids b) Alkalis c) Salts d) Amphoteric substances
10. The example of weak acid is: a) HNO3 b) CH3COOH c) HCI d) H2SO4
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B. SHORT QUESTIONS. 1. When a clear liquid is placed in a beaker. How can you identify it as an
acid, base or neutral (water)?
Ans. The following methods are used to identify that whether the given liquid is acid, base or neutral (water)
1. Litmus Paper
2. pH Scale
1. LITMUS PAPER:
In this method, litmus paper is dipped in a beaker which contains liquid. If it turns
blue litmus paper red then the given liquid will be acid. If it turns red litmus paper blue then
the given liquid will be base. If the liquid does not change the colour of red and blue litmus
papers then it will be neutral (water).
2. pH SCALE:
We can also use pH scale to identify that whether the given liquid is acid, base or
neutral (water). The pH scale measures the acidic, basic or neutral nature of solution. It ranges
from 0 (zero) to 14. If the pH value of liquid is smaller than 7, then it is acid. If it is equal to
7, then it is neutral (water). If it is greater than 7, then it is base.
2. Justify H+ ion as a lewis acid?
Ans) According to Lewis Concept, acid is a species that accepts or tends to accept a
pair of electrons. As H+ ion is a positively charged ion that does not have any electron
in its 1st shell. So, it has the tendency to accept a pair of electrons. Therefore, it acts
as a lewis acid.
For Example:
An acid base reaction involves the donation of an electron pair from the base to
an acid to form a co-ordinate covalent bond.
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H2O
H
H N HH
h
H+ +:NH3
3. Distinguish Strong acids from weak acids? Give two examples of each?
Ans) STRONG ACID:
A Strong acid is one that ionizes completely in an aqueous solution and gives a higher concentration of H+ ions. A strong acid is a strong electrolyte.
For example:
Hydrochloric acid (HCI) and sulphuric acid (H2SO4) are the examples of strong
acids because when both are added into water, all the molecules of HCl and H2SO4
are dissociated completely into positively and negatively charged ions.
HCI H+ + CI-
H2SO4 2H+ + SO4
-2
WEAK ACID:
The acids which do not completely dissociate in an aqueous solution and give lower
concentration of H+ ions are called weak acids.
For Example:
Acetic acid (CH3 COOH) and carbonic acid (H2CO3) are weak acids because
when both are added into water, very few molecules of both acids are dissociated.
CH3COOH ⇌CH3COO- + H+
H2CO3 ⇌2 H+ + CO3-2
Acid Base
Ammoniumion
H2O
H2O
H2O
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4. Compare the physical properties of acids and bases.
Ans) Comparison between the physical properties of acids and bases is given
below.
PHYSICAL PROPERTIES OF ACIDS:
1. Taste:
Acids have sour taste.
2. Litmus paper:
They turn blue litmus paper red.
3. pH Value:
Acidic solutions have pH values less than 7.
4. Ions: They produce H+ ions when dissolved in water.
5. Conductivity:
Aqueous solutions of acids conduct electric current.
6. Corrosive Action:
Strong acids are very corrosive (destroy fabric and cause burns on the skin).
7. Indicators: They change the colour of acid-base indicators.
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PHYSICAL PROPERTIES OF BASES:
1. Taste: Bases have bitter taste.
2. Litmus paper: They turn red litmus paper blue.
3. pH value: Basic solution has pH values greater than 7.
4. Ions: They produce OH- ions when dissolved in water.
5. Conductivity: Aqueous solutions of bases conduct electricity.
6. Corrosive Action: Strong bases are corrosive to skin.
7. Indicators: The change the colour of acid-base indicators.
5. A carbonated drink has [H+] =3.2 X 10-3M, classify the drink as neutral, acidic or basic with reason.
Ans) Solution:
Hydrogen ion concentration [H+] = 3.2X10-3M
We know that:
pH= - log [H+]
Putting the value of [H+]
pH= - log [3.2X10-3]
pH= - [log (3.2X10-3)]
pH= -[log 3.2 +log10-3]
pH=-[log 3.2 + (-3) log10]
pH=-[0.5051-3(1)] ( log10=1)
pH= - [0.5051-3]
pH= - [-2.4949]
pH= 2.4949 or 2.5
As the pH value is less than 7. Therefore, the carbonated drink is acidic in nature.
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6. Write the chemical name of an acid present in the following: a) Apple b) Grape juice c) lemon juice d) sour milk
Ans) a) Apple contains malic acid (C4H6O5)
b) Grape juice contains tartaric acid (C4H6O6)
c) Lemon Juice contains Citric acid (C6H8O7)
d) Sour milk contains lactic acid (C3H6O3)
7. What determines the strength of a base? Give one example of each solution that is strongly and weakly basic.
Ans) Hydroxide ion (OH-) donation power of a substance in an aqueous solution
determines the strength of a base.
STRONG BASES: The bases which completely dissociate in an aqueous solution and give a higher
concentration of hydroxide ions (OH-) are called strong bases.
Example: Sodium hydroxide (NaOH)
NaOH Na+ + OH-
WEAK BASES: The bases which do not completely dissociate in an aqueous solution and give a
lower concentration of hydroxide ions (OH-) are called weak bases.
Example: Ammonium hydroxide (NH4OH)
NH4OH ⇌ NH+4 + OH-
8. Calculate the pH and pOH of 0.5 M solution of HCI. (pH=0.301, pH=13.69)
Solution:
HCI H+ + CI-
0.5 M 0.5M Hydrogen ion concentration [H+] = 0.5 M
= 5/10 M
= 5X10-1M
H2O
H2o
H2O
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We know that:
pH= -log [H+]
Putting the value of [H+]
pH= - log[5X10-1]
pH=- [log 5x 10-1]
pH= - [log 5+ log 10-1]
pH= - [log5 + (-1) log10]
pH=- [0.6989-1(1)] ( log10=1)
pH= - [0.6989-1]
pH=- [-0.3011]
pH= 0.3011
We also know that:
pH + pOH=14
Putting the value of pH.
0.3011+ pOH=14
Subtracting 0.3011 from both sides:
0.3011-0.3011+pOH=14-0.3011
pOH= 13.6989
9. Calculate the pOH of 0.005M H2SO4. (pOH=11.699) Solution:
H2SO4 2H+ + SO−2
0.005M 0.005M
Hydrogen ion concentration [H+] =0.005M
= 5/1000 M
= 5/103M
= 5X10-3M
H2o
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We know that:
pH=-log[H+]
Putting the value of [H+].
pH= -log[5X10-3]
pH=- [log5 Xlog10-3]
pH= - [log 5+ log 10-3]
pH=-[log5 + (-3)log10)
pH= -[log5-3log10]
pH= - [0.6989 – 3(1)] ( log10=1)
pH=-[0.6989-3]
pH=-[-2.3011]
pH= 2.3011
We also know that
pH + pOH=14
Putting the value of pH.
2.3011+pOH=14
Subtracting 2.3011 form both sides.
2.3011-2.3011+pOH=14-2.3011
pOH = 11.6989
10. Calculate the pH of 0.2 M solution of NaOH. (pH=13.30)
Solution:
NaOH Na+ + OH- 0.2 M 0.2M Hydroxide ion concentration [OH-] =0.2 M
= 2/10 M
= 2X10-1M
H2O
35
We know that:
pOH= -log[OH-]
Putting the value of [OH-].
pOH=-log[2X10-1]
pOH= -[log2 X 10-1]
pOH=-[log2+log10-1]
pOH= -[log2+(-1) log10]
pOH= - [log2 -1log10]
pOH= -[0.3010-1(1)] ( log10=1)
pOH= - [0.3010-1]
pOH=-[-0.699]
pOH = 0.699
We also know that
pH + pOH =14
Putting the value of pOH.
pH + 0.699=14
Subtracting 0.699 from both sides.
pH + 0.699 -0.699=14-0.699
pH =13.301
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C. Long Questions.
1. (a) What is salt? (b) Write down the different types of salts with examples.
Ans) a) SALT:
A compound formed due to the neutralization reaction of an acid and a base is
called a salt.
COMPOSITION OF SALT:
A salt consists of positive ions and negative ions. Positive ions come from a base while
negative ions come from an acid. For example, sodium chloride (NaCI) is composed of
sodium positive ions (Na+) from a base (NaOH) and Chloride negative ions (CI-) from
an acid (HCI).
NaOH + HCI NaCI + H2O
Base Acid salt water
Examples of salts:
Some examples of salts are given below:
i. Silver bromide (AgBr) ii. Potassium sulphate (K2SO4) iii. Ferric phosphate (FePO4)
B) Types of salts:
The three main types of salts are,
i. Neutral salts ii. Acidic salts
iii. Basic salts
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I. NEUTRAL SALTS: Neutral salts are formed when the hydrogen ions of an acid are completely
replaced by metal ions or a group of atoms, behaving like metal ions. A neutral
salt is formed, when a strong acid reacts with a strong base.
Example: HCI + NaOH NaCI + H2O Strong Strong salt water Acid base
Some other examples of neutral salts, are potassium sulphate (K2SO4),
sodium sulphate (Na2SO4), Sodium Carbonate (Na2CO3), ammonium sulphate [(NH4)2 SO4], and sodium phosphate (Na3PO4).
II. Acidic Salts:
Acidic Salts are formed when hydrogen ions of a Polyprotic acid are partially
replaced by metal ions or a group of atoms behaving like metal ions. These salts
can further react with bases to form neutral salts. These salts are formed by
polybasic or polyprotic acids only.
Example: H2SO4 + KOH KHSO4 + H2O Acidic salt KHSO4 + KOH K2SO4 + H2O
Neutral salt Some other examples of acidic salts are carbonic acid (H2CO3) and phosphoric
acid (H3PO4).
III. Basic Salts:
Basic salts are formed when hydroxide ions (OH-) of a poly-acid base are
partially neutralized by an acid. These salts can further react with acids to form
neutral slats. These salts are produced by poly-acid bases only.
Example:
Pb (OH) 2 + HCI Pb (OH) CI + H2O
Basic salt
Pb(OH) CI + HCI PbCI2 + H2O
Neutral salt
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2. a) Define the auto-ionization of water. How can you find the pH of water? b) Why some acids are called monoprotic, diprotic and polyprotic acids. Explain your answer with suitable examples.
Ans) a) AUTO-IONIZATION OF WATER:
The reaction in which two water molecules produce ions is called the auto-
ionization or self-ionization of water.
Explanation:
During auto-ionization of water, two molecules of water react to produce Hydronium
ion (H3O+) and hydroxide ion (OH-) by the transfer of a proton.
H2O + H2O H3O+ + OH-
In order to understand the concept of self-ionizaiton of water, we take one molecule of
water and its dissociation at 250C, as
H2O (I) H+ (aq) + OH-
(aq)
The equilibrium expression of this reaction may be written as,
Kc= <m [$<d][<&$]
As concentration of water (H2O) is almost constant, so the above equation may be
written as,
Kc. [H2O] = [H+] [OH-]
Where,
Kc. [H2O] = Kw
The mathematical product of [H+] and [OH-] remains constant in water and dilute
aqueous solutions at a constant temperature. This constant mathematical product is
called the ionization constant of water, Kw, and is expressed by the following equation.
So, we can write the above equation as,
Kw= [H+] [OH]
39
Conductivity experiments show that the product of concentrations of H+ and OH- ions
in pure water is always 1.0 X 10-14 mol/dm3 of water at 250C and is called water
dissociation constant ‘Kw’ .
Kw=[H+][OH-]= 1.0 x 10-14mol/dm3of water at 250C
As we know that one molecule of water produces one H+ and one OH- ion on
dissociation.
Therefore, we can say that,
[H+][OH-] = 1.0 x 10-14
[H+] = [OH-]
Or [H+] [H+] = 1.0 X 10-14
[H+] 2 =1.0 X 10-14
[𝐻h]! = 1.0𝑋10qAH
Therefore, [H+] = 1.0 X 10-7
and [OH-]= 1.0 x 10-7
In water at 250C, [H+] = 1.0 x 10-7 M and [OH-] =1.0 x 10-7M.
Kw = [H+] [OH-]
Kw= (1.0 X 10-7 M)(1.0 X 10-7 M)
Kw= 1.0 X10-14M2
pH of water:
The pH value of water can be found by the following equation:
pH= -log[H+]
For pure water [H+] = 1.0 x 10-7M
Putting the value of [H+],
pH= -log[1.0 x 10-7]
pH= -log [10-7]
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pH= -(-7) log10
pH =7 log 10
pH =7x1 ( log10= 1)
pH= 7
b) MONOPROTIC ACID:
The acids which produce one proton per molecule are called monoprotic acids.
Example:
HCI, HNO3, HCN and CH3COOH are the examples of monoprotic acids. Their
dissociation in water is as follows:
H2O HCI H+ + CI-
H2O HNO3 H+ + NO3
- H2O
HCN H+ + CN- H2O
CH3COOH H+ + CH3COO-
DIPROTIC ACID: The acids which give two protons per molecule are called diprotic acids.
Examples: H2SO4 and H2CO3 are the examples of diprotic acids. Their dissociation in water is as
follows,
H2O
H2SO4 2H+ + SO4-2
H2O H2CO3 2H+ + CO3
-2
POLYPROTIC ACIDS:
The acids which give more than one proton per molecule are called polyprotic acids.
Examples:
H2CO3 and H3PO4 are the examples of polyprotic acids. Their dissocation in water is
41
as follows;
H2O H2CO3 2H+ + CO3
-2 H2O
H3PO4 3H+ + PO4-3
3. a) Discuss the concept of Lewis acids and bases with examples. b) Give the Bronsted-Lowry definition of acids and bases. Write equation that explains the definition.
Ans) a) LEWIS CONCEPT OF ACIDS AND BASES:
Introduction:
G. N. Lewis presented his own concept about acids and bases in 1923.
Lewis Acid:
Lewis acid is a species (ion or molecule) that accepts or tends to accept a pair of
electrons.
Lewis Base:
Lewis base is a species (ion or molecule) that donates or tends to donate a pair of electrons.
Species which Act as Lewis Acids:
Compounds having less than eight electrons in their valence shell or positively charged
ions that can accept a pair of electrons can act as Lewis acids. For example, BF3 and H+
etc.
Species which act as Lewis Bases:
Compounds having lone pair of electrons in the valence shell or negatively
charged ions that can donate an electron pair can behave as Lewis bases. For example,
NH3, CI- and H2O etc.
Examples:
42
Consider the following chemical reactions.
F H
BF3 + :NH3 F B N H
Acid Base F H
Complex
H +
H+ + :NH3 H N N
Acid Base H
Ammonium Ion
+
H+ + H2𝑂 H O H
Acid Base H
Hydronium ion
An acid- base reaction involves the donation of electrons pair from a base to an acid and forming a coordinate covalent bond.
B) BRONSTED- LOWRY CONCEPT OF ACIDS AND BASES:
Introduction:
This concept was presented by the Danish Chemist J. N. Bronsted and an English Chemist T.M. Lowry in 1923.
Bronsted-lowry Acid:
According to Bronsted-Lowry concept, an acid is a substance (ion or molecule)
that is a proton (H+) donor.
Bronsted- Lowry Base:
According to Bronsted- lowry concept, a base is a substance (ion or molecule) that is a
proton (H+) acceptor.
Examples:
..
43
i. HCI + NH3 NH4+ + CI-
Acid Base
In the above reaction, ammonia gas (NH3) accepts a proton (H+) from hydrochloric acid
(HCI). Therefore, HCI is an acid and NH3 is a base.
ii. H2O + NH3 NH4+ + OH-
Acid Base
In this reaction, water (H2O) donates a proton (H+) to ammonia gas (NH3). Therefore, H2O acts as an acid and NH3acts as a base.
iv. HCI + H2O ⇌ H3O+ + CI-
Acid Base
In this chemical reaction, hydrochloric acid (HCI) donates a proton (H+) to
water (H2O). Therefore, HCI acts as an acid and H2O acts as a base.
4. Below are two equations showing how two alkalis react with water: NaOH + H2O ⇌ Na+ + OH-
NH3 + H2O ⇌NH4+ + OH-
a. Name both the alkalis. b. Which is classified as weak alkali and why? c. What is the likely pH of each alkali?
Ans) a: ALKALIS:
The two alkalis used in the above two reactions are sodium hydroxide (NaOH) and ammonia gas (NH3).
B: WEAK ALKALI:
Ammonia gas (NH3) is a weak alkali.
Reason:
Weak alkali is an alkali which partially ionizes in water. Since, ammonia gas
(NH3) partially ionizes in water (H2O), therefore, it is a weak alkali.
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C: pH OF ALKALI:
According to pH scale, the pH value of an alkali (NaOH) is 14 while the pH value of an alkali (NH3) is 11.9.
5. a) Write the balanced neutralization reaction of: i. Strong acid and strong base.
ii. Strong acid and weak base. iii. Weak acid and strong base. iv. Weak acid and weak base.
b) Define pH and pOH.
Ans) a:
i) Strong Acid and Strong Base:
HCI + NaOH NaCI + H2O
Strong Strong Salt water Acid Base
ii) Strong Acid and weak Base:
HNO3 + NH4OH NH4NO3 + H2O
Strong weak Salt water Acid Base iii) Weak acid and strong base:
H2CO3 + NaOH NaHCO3 + H2O
weak strong Salt water Acid Base
iv) Weak acid and weak base:
CH3COOH + NH4OH CH3COONH4 + H2O
weak weak Salt water Acid Base
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B) The pH:
The negative logarithm of the molar hydrogen ion concentration [H+] is called the pH
of a solution. Here ‘p’ stands for “protenz” (the potential to be) and ‘H+’ stands for “Hydrogen
ion”.
Mathematical form:
pH= -log[H+]
pH of water:
The pH of water is calculated as,
pH= -log[H+]
We know that the concentration of hydrogen ion [H+] for pure water is
1.0 x 10-7M.
Putting the value of [H+]
pH= -log[1.0 x 10-7]
pH= - log[10-7]
pH= - (-7) log 10
pH= 7 log 10
pH= 7x1 ( log 10=1)
pH=7
The pOH:
The negative logarithm of the molar hydroxide ion concentration [OH-] is called the pOH of a
solution. Here ‘p’ stands for “Protenz” (the potential to be) and ‘OH-’ stands for “hydroxyl
ion”.
46
Mathematical form:
pOH= -log [OH-]
pOH of water:
The pOH of water can be calculated as:
pOH= - log [OH-]
We know that the concentration of hydroxyl ion [OH-] for pure water is 1.0 x 10-7 M.
Putting the value of [OH-]
pOH= -log[1.0 x 10-7]
pOH= - log[10-7]
pOH= - (-7) log 10
pOH= 7 log 10
pOH= 7x1 ( log 10=1)
pOH =7
We know that the sum of the pH and the pOH of a neutral solution at 250C is equal to 14.
pH+ pOH=14
6. a) According to your understanding which one of the three acid definitions is the broadest? Explain. b) Write the uses of any three salts.
Ans ) A) ARRHENIUS CONCEPT OF ACIDS:
According to Arrhenius concept, an acid is a chemical substance that dissociates in
aqueous solution to give hydrogen ions (H+).
BRONSTED AND LOWRY CONCEPT OF ACIDS:
According to Bronsted- Lowry concept, an acid is a substance (ion or molecule) that is
a proton (H+) donor.
47
LEWIS CONCEPT OF ACIDS:
According to Lewis concept, an acid is a species (ion or molecule) that accepts or tends
to accept, a pair of electrons.
BROADEST DEFINITION OF ACID:
Among the three acid definitions, Lewis definition is the broadest one.
EXPLANATION:
The Arrhenius concept limits the acids to aqueous medium which cannot explain the
acidic behaviour of acids in non-aqueous medium.
Similarly, the Bronsted-Lowry concept restricts acids to protons which cannot
explain the behaviour of acids having no protons like AICI3 and BF3 etc.
The Lewis concept about acids is the broadest of all the three concepts because it
completely explains the Arrhenius and Bronsted-Lowry definitions of acids. But they
did not explain the acid definition of Lewis concept.
Example:
BF3 is a Lewis acid; it cannot be explained to be an acid on the bases of Arrhenius
concept because it does not produce hydrogen ion in aqueous solution. Similarly, it
cannot be explained, to be an acid on the basis of Bronsted-lowry concept because it
does not donate proton. However, it can be explained, to be an acid on the basis of Lewis
concept, to be an acid as it accepts electron pair.
b) USES OF SALTS:
Salts are used in different ways ranging from household to industry. Some
important uses of different salts are given below;
1) USES OF SODIUM CHLORIDE (NAC1):
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Sodium Chloride is used for seasoning and preserving food. In industry, it
is used as basic raw material for the extraction of sodium metal, preparation of caustic
soda (NaOH), washing soda (Na2CO3) etc. One of the major application of sodium
chloride is de-icing of roadways in sub-freezing weather.
2) USES OF SODIUM CARBONATE (Na2CO3): Sodium carbonate is also known as Soda Ash or washing soda. Sodium Carbonate
is used in Laundries as a cleaning agent and as water softener. It is used as a raw
material in glass manufacturing. It is also used in the paper, petroleum refining and
leather industry.
3) Uses of Sodium bicarbonate (NaHCO3): Sodium bicarbonate or Sodium hydrogen carbonate is also known as baking soda
as it is used in the preparation of cakes and other confectionaries. It is also used as a
medicine, as an Antacid, and also in toothpastes etc.
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UNIT: 11
ORGANIC CHEMISTRY
(Notes will be uploaded soon.)
