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11th
Chemistry
Equilibrium
Hydrolysis of Salts and the pH of their Solutions
Hydrolysis of Salts and the pH of their Solutions
Equilibrium
Chemistry
Class 11
Overview
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Videos
Hydrolysis of Salts of Strong Acids and Weak Bases
21 mins
Hydrolysis of Salts of Weak Acids and Strong Bases
29 mins
Hydrolysis of Salts of Strong Acids and Strong Bases
7 mins
Hydrolysis of Salts of Weak Acids and Weak Bases
16 mins
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Study Materials
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Buffer Solutions: Definition, Types, Preparation, Examples and Videos
Chemical Reactions of Ether: Various Types of Reactions, Videos, Q&A
Quick summary
with Stories
Hydrolysis-Salt Hydrolysis - I
2 mins read
Hydrolysis-Salt Hydrolysis - II
3 mins read
Revise
with Concepts
Hydrolysis of Salts
Example
Definitions
Formulaes
Related questions
K
a
for hydrofluoric acid is
6
.
9
×
1
0
−
4
. What is the equilibrium constant
K
for the following reaction?
F
−
+
H
2
O
⇌
H
F
+
(
O
H
)
−
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Which of the following are correct for
0
.
3
M
solution of calcium lactate in water if
K
a
of lactic acid is
8
.
6
×
1
0
−
4
?
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In a mixture of weak acid and its salt, the ratio of concentration of acid to salt is increased ten fold. The pH of the solution is ________.
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p
k
a
of acetic acid and
p
K
b
of ammonium hydroxide are 4.76 and 4.75 respectively.
Calculate the pH of ammonium acetate solution.
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This section contains questions each with two columns-I and II. Match the column I with column II.
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The degree of hydrolysis of a salt of weak acid and strong base is
≈
0
.
5
. The equation to be used to calculate the accurate value of the degree of hydrolysis
(
h
)
is
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The correct increasing order of extent of hydrolysis is :
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What will be the amount of
(
N
H
4
)
2
S
O
4
(in g) which must be added to 500 mL of 0.2 M
N
H
4
O
H
to yield a solution of pH 9.35? [Given,
p
K
a
of
N
H
4
+
= 9.26,
p
K
b
(
N
H
3
)
= 14 -
p
K
a
(
N
H
4
+
)
]
View solution
A
0
.
0
1
0
M
solution of
P
u
O
2
(
N
O
3
)
2
was found to have a
p
H
of
4
.
0
. What is the hydrolysis constant,
K
h
, for
P
u
O
2
2
+
and what is
K
b
for
P
u
O
2
O
H
+
?
View solution
p
H
of
0
.
0
0
1
M
solution of
Z
n
C
l
2
is :
Z
n
2
+
+
H
2
O
⇌
Z
n
(
O
H
)
+
+
H
+
Use : [
K
a
=
1
.
0
×
1
0
−
9
]
View solution
View more
More topics
Equilibrium: Basic Definitions
Equilibrium in Physical Processes
Equilibrium in Chemical Processes
Law of Chemical Equilibrium and Equilibrium Constant
Homogeneous and Heterogeneous Equilibria
Applications of Equilibrium Constants
Relationship between Equilibrium Constant, Reaction Quotient and Gibbs Energy
Factors Affecting Equilibria
Ionic Equilibrium in Solution
Acids, Bases and Salts
Ionization of Acids and Bases
More About Ionization of Acids and Bases
Hydrolysis of Salts and the pH of their Solutions
Buffer Solutions
Solubility Equilibria
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