JEE Exam > JEE Notes > Chemistry for JEE Main & Advanced > Salt Hydrolysis
Salt Hydrolysis | Chemistry for JEE Main & Advanced PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Salt Hydrolysis
Acid + Base ? ? ? Salt + H
2
O neutralization
Salt + H
2
O ? ? ? Acid + Base ? ? ? reverse of neutralization ? Salt hydrolysis
since salt hydrolysis is an endothermic reaction hence on increasing the temperature, the extent
of hydrolysis increases.
? Types of salt
Case I Case II Case III Case IV
Page 2
Salt Hydrolysis
Acid + Base ? ? ? Salt + H
2
O neutralization
Salt + H
2
O ? ? ? Acid + Base ? ? ? reverse of neutralization ? Salt hydrolysis
since salt hydrolysis is an endothermic reaction hence on increasing the temperature, the extent
of hydrolysis increases.
? Types of salt
Case I Case II Case III Case IV
W. A + S. B S. A. + W. B. W. A + W. B. S. A. + S. B.
e.g. CH
3
COONa e.g. NH
4
Cl e.g. CH
3
COONH
4
e.g. NaCl
?Case : (I) Hydrolysis of WA + SB :
CH
3
COONa
? ? ?
CH
3
COO
–
+ Na
+
? spectator ion
?
?
Very strong Very weak
Conjugate base Conjugate acid
H – OH
CH
3
COOH NaOH
CH
3
COO
–
+ H
+
Na
+
+ OH
–
CH
3
COONa CH
3
COO
–
+ Na
+
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
__________________________________________
CH
3
COONa + H
2
O CH
3
COOH + Na
+
+ OH
–
__________________________________________________________________________
Note : (1) Only weaker part of salt undergoes hydrolysis. (2) This solution becomes basic.
?Case : (II) S.A. + W.B. (NH
4
Cl) :
NH
4
+
+ H
2
O NH
4
OH + H
+
, This solution becomes acidic
?Case : (III) W.A. + W.B. (CH
3
COONH
4
)
:
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
a x
a – x
NH
4
+
+ H
2
O NH
4
OH + H
+
a y
a – y
Can be acidic, basic or neutral
? acidic K K
b a
? ? neutral K K
b a
? ? basic K K
b a
? ?
?Case : (IV) S. A + S. B (NaCl) :
? No hydrolysis ? Solution will be neutral
? Relationship between K
a
, K
b
, and K
h
?Case (I) : W. A. + S.B. (CH
3
COONa)
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
] [OH COOH] [CH
K
–
–
–
3
?
?
? ?
K
K ?
Page 3
Salt Hydrolysis
Acid + Base ? ? ? Salt + H
2
O neutralization
Salt + H
2
O ? ? ? Acid + Base ? ? ? reverse of neutralization ? Salt hydrolysis
since salt hydrolysis is an endothermic reaction hence on increasing the temperature, the extent
of hydrolysis increases.
? Types of salt
Case I Case II Case III Case IV
W. A + S. B S. A. + W. B. W. A + W. B. S. A. + S. B.
e.g. CH
3
COONa e.g. NH
4
Cl e.g. CH
3
COONH
4
e.g. NaCl
?Case : (I) Hydrolysis of WA + SB :
CH
3
COONa
? ? ?
CH
3
COO
–
+ Na
+
? spectator ion
?
?
Very strong Very weak
Conjugate base Conjugate acid
H – OH
CH
3
COOH NaOH
CH
3
COO
–
+ H
+
Na
+
+ OH
–
CH
3
COONa CH
3
COO
–
+ Na
+
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
__________________________________________
CH
3
COONa + H
2
O CH
3
COOH + Na
+
+ OH
–
__________________________________________________________________________
Note : (1) Only weaker part of salt undergoes hydrolysis. (2) This solution becomes basic.
?Case : (II) S.A. + W.B. (NH
4
Cl) :
NH
4
+
+ H
2
O NH
4
OH + H
+
, This solution becomes acidic
?Case : (III) W.A. + W.B. (CH
3
COONH
4
)
:
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
a x
a – x
NH
4
+
+ H
2
O NH
4
OH + H
+
a y
a – y
Can be acidic, basic or neutral
? acidic K K
b a
? ? neutral K K
b a
? ? basic K K
b a
? ?
?Case : (IV) S. A + S. B (NaCl) :
? No hydrolysis ? Solution will be neutral
? Relationship between K
a
, K
b
, and K
h
?Case (I) : W. A. + S.B. (CH
3
COONa)
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
] [OH COOH] [CH
K
–
–
–
3
?
?
? ?
K
K ?
?Case (II) : S. A. + W. B. (NH
4
OH)
NH
4
+
+ H
2
O NH
4
OH + H
+
?
] [NH
] [H OH] [NH
K
4
4
h
?
?
?
?
] [OH ] [NH
] [OH ] [H OH] [NH
K
–
4
–
4
h
?
?
?
?
?
?
b
w
h
K
K
K ?
?Case (III) W. A. + W. B. (CH
3
COO NH
4
)
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
NH
4
+
+ H
2
O NH
4
OH + H
+
_______________________________________________________
CH
3
COO
–
+ NH
4
+
+ 2 H
2
O NH
4
OH + CH
3
COOH + H
+
+ OH
–
_______________________________________________________
?
] [NH ] COO [CH
COOH] [CH OH] [NH
K
4
–
3
3 4
h
?
?
?
] [H ] COO [CH
] [H ] [OH COOH] [CH
] [OH ] [NH
OH] [NH
K
–
3
–
3
–
4
4
h
?
?
?
?
?
b a
w
h
K . K
K
K ?
? pH calculation
?Case (I) : W. A. + S. B. (CH
3
COONa ? c, K
a
) , h ? degree of hydrolysis.
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
c 0 0
c (1 – h) ch ch
?
h – 1
ch
h) – (1 c
(ch)
K
2 2
h
? ?
h < < 1 ? K
h
= ch
2
?
c
K
h
h
?
?
a
w
h
–
K
K
c. K c. h c ] [OH ? ? ? ? ?
c] log – pK – [pK
2
1
pOH
a w
?
?
c] log pK [pK
2
1
pH
a w
? ? ?
?Case (II) :S. A. + W. B. (NH
4
Cl ? c, K
b
)
NH
4
+
+ H
2
O NH
4
OH + H
+
c 0 0
c (1 – h) ch ch
?
h – 1
ch
K
2
h
? h < < 1 ? K
h
= ch
2
?
c
K
h
h
?
?
K
K
c K c c.h ] [H
b
w
h
? ? ? ? ?
?
?
c] log – pK – [pK
2
1
pH
b w
?
? p
OH
=
? ? c log p p
2
1
b w
k k
? ?
?Case (III) W. A. + W. B. (CH
3
COONH
4
? K
a
, K
b
, c)
CH
3
COO
–
+ NH
4
+
+ H
2
O NH
4
OH + CH
3
COOH
c c
c (1 – h) c (1 – h) ch ch
Page 4
Salt Hydrolysis
Acid + Base ? ? ? Salt + H
2
O neutralization
Salt + H
2
O ? ? ? Acid + Base ? ? ? reverse of neutralization ? Salt hydrolysis
since salt hydrolysis is an endothermic reaction hence on increasing the temperature, the extent
of hydrolysis increases.
? Types of salt
Case I Case II Case III Case IV
W. A + S. B S. A. + W. B. W. A + W. B. S. A. + S. B.
e.g. CH
3
COONa e.g. NH
4
Cl e.g. CH
3
COONH
4
e.g. NaCl
?Case : (I) Hydrolysis of WA + SB :
CH
3
COONa
? ? ?
CH
3
COO
–
+ Na
+
? spectator ion
?
?
Very strong Very weak
Conjugate base Conjugate acid
H – OH
CH
3
COOH NaOH
CH
3
COO
–
+ H
+
Na
+
+ OH
–
CH
3
COONa CH
3
COO
–
+ Na
+
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
__________________________________________
CH
3
COONa + H
2
O CH
3
COOH + Na
+
+ OH
–
__________________________________________________________________________
Note : (1) Only weaker part of salt undergoes hydrolysis. (2) This solution becomes basic.
?Case : (II) S.A. + W.B. (NH
4
Cl) :
NH
4
+
+ H
2
O NH
4
OH + H
+
, This solution becomes acidic
?Case : (III) W.A. + W.B. (CH
3
COONH
4
)
:
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
a x
a – x
NH
4
+
+ H
2
O NH
4
OH + H
+
a y
a – y
Can be acidic, basic or neutral
? acidic K K
b a
? ? neutral K K
b a
? ? basic K K
b a
? ?
?Case : (IV) S. A + S. B (NaCl) :
? No hydrolysis ? Solution will be neutral
? Relationship between K
a
, K
b
, and K
h
?Case (I) : W. A. + S.B. (CH
3
COONa)
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
] [OH COOH] [CH
K
–
–
–
3
?
?
? ?
K
K ?
?Case (II) : S. A. + W. B. (NH
4
OH)
NH
4
+
+ H
2
O NH
4
OH + H
+
?
] [NH
] [H OH] [NH
K
4
4
h
?
?
?
?
] [OH ] [NH
] [OH ] [H OH] [NH
K
–
4
–
4
h
?
?
?
?
?
?
b
w
h
K
K
K ?
?Case (III) W. A. + W. B. (CH
3
COO NH
4
)
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
NH
4
+
+ H
2
O NH
4
OH + H
+
_______________________________________________________
CH
3
COO
–
+ NH
4
+
+ 2 H
2
O NH
4
OH + CH
3
COOH + H
+
+ OH
–
_______________________________________________________
?
] [NH ] COO [CH
COOH] [CH OH] [NH
K
4
–
3
3 4
h
?
?
?
] [H ] COO [CH
] [H ] [OH COOH] [CH
] [OH ] [NH
OH] [NH
K
–
3
–
3
–
4
4
h
?
?
?
?
?
b a
w
h
K . K
K
K ?
? pH calculation
?Case (I) : W. A. + S. B. (CH
3
COONa ? c, K
a
) , h ? degree of hydrolysis.
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
c 0 0
c (1 – h) ch ch
?
h – 1
ch
h) – (1 c
(ch)
K
2 2
h
? ?
h < < 1 ? K
h
= ch
2
?
c
K
h
h
?
?
a
w
h
–
K
K
c. K c. h c ] [OH ? ? ? ? ?
c] log – pK – [pK
2
1
pOH
a w
?
?
c] log pK [pK
2
1
pH
a w
? ? ?
?Case (II) :S. A. + W. B. (NH
4
Cl ? c, K
b
)
NH
4
+
+ H
2
O NH
4
OH + H
+
c 0 0
c (1 – h) ch ch
?
h – 1
ch
K
2
h
? h < < 1 ? K
h
= ch
2
?
c
K
h
h
?
?
K
K
c K c c.h ] [H
b
w
h
? ? ? ? ?
?
?
c] log – pK – [pK
2
1
pH
b w
?
? p
OH
=
? ? c log p p
2
1
b w
k k
? ?
?Case (III) W. A. + W. B. (CH
3
COONH
4
? K
a
, K
b
, c)
CH
3
COO
–
+ NH
4
+
+ H
2
O NH
4
OH + CH
3
COOH
c c
c (1 – h) c (1 – h) ch ch
?
2
2
2 2
2 2
h
) h – 1 (
h
h) – (1 c
h c
K ? ?
? K
h
= h
2
1 h – 1 ?
?
h
K h ?
?
b a
w
K K
K
h ?
- - - - - (1)
? CH
3
COOH CH
3
COO
–
+ H
+
?
COOH] [CH
] COO [CH ] [H
K
3
–
3
a
?
?
?
] COO [CH
COOH] [CH
K ] [H
–
3
3
a
?
?
?
h) – (1 c
ch
K
a
? ?
(h < < 1) ? [H
+
] = K
a
. h ?
1) (from
K . K
K
. K ] [H
b a
w
a
?
?
?
K
K
. K
] [H
b
a
w
?
?
?
] pK – pK [pK
2
1
H p
b a w
? ?
? If , K
a
> K
b
? pK
a
< pK
b
? acidic,
K
a
= K
b
? neutral, K
a
< K
b
? basic
? Summary of hydrolysis
1. W. A. + S. B.
c] log pK [pK
2
1
pH
K
K
K
a w
a
w
h
? ? ? ?
2. W. B. + S. A.
c] log – pK – [pK
2
1
pH
K
K
K
b w
b
w
h
? ?
3. W. A. + W. B.
] pK – pK [pK
2
1
pH
K . K
K
K
b a w
b a
w
h
? ? ?
? Hydrolysis of salt of polyprotic acid / base (Na
2
CO
3
)
?
– 2
3
CO + H
2
O
–
3
HCO + OH
–
2
1
a
w – –
3 h
K
K
) (CO K ?
a
a – x x – y x + y
?
?
x x
?
–
3
HCO + H
2
O H
2
CO
3
+ OH
–
1
2
a
w –
3 h
K
K
) (HCO K ?
x – y y x + y
?
?
?
x y x
2 1
a a
K K ? ?
?
K K
h h
2 1
? ?
? Mainly hydrolysis is governed by
– 2
3
CO .
? ?
2
1
a
w
h
–
K
K
c
K c ] [OH
?
? ? ?
, ? ?
1
a
w – –
3
K
K
c
] [OH ] [HCO
?
? ?
, ? ? ?
x
y x
K
K
1
a
w
?
?
? ?
1
a
w
3 2
K
K
] CO [H ?
Page 5
Salt Hydrolysis
Acid + Base ? ? ? Salt + H
2
O neutralization
Salt + H
2
O ? ? ? Acid + Base ? ? ? reverse of neutralization ? Salt hydrolysis
since salt hydrolysis is an endothermic reaction hence on increasing the temperature, the extent
of hydrolysis increases.
? Types of salt
Case I Case II Case III Case IV
W. A + S. B S. A. + W. B. W. A + W. B. S. A. + S. B.
e.g. CH
3
COONa e.g. NH
4
Cl e.g. CH
3
COONH
4
e.g. NaCl
?Case : (I) Hydrolysis of WA + SB :
CH
3
COONa
? ? ?
CH
3
COO
–
+ Na
+
? spectator ion
?
?
Very strong Very weak
Conjugate base Conjugate acid
H – OH
CH
3
COOH NaOH
CH
3
COO
–
+ H
+
Na
+
+ OH
–
CH
3
COONa CH
3
COO
–
+ Na
+
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
__________________________________________
CH
3
COONa + H
2
O CH
3
COOH + Na
+
+ OH
–
__________________________________________________________________________
Note : (1) Only weaker part of salt undergoes hydrolysis. (2) This solution becomes basic.
?Case : (II) S.A. + W.B. (NH
4
Cl) :
NH
4
+
+ H
2
O NH
4
OH + H
+
, This solution becomes acidic
?Case : (III) W.A. + W.B. (CH
3
COONH
4
)
:
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
a x
a – x
NH
4
+
+ H
2
O NH
4
OH + H
+
a y
a – y
Can be acidic, basic or neutral
? acidic K K
b a
? ? neutral K K
b a
? ? basic K K
b a
? ?
?Case : (IV) S. A + S. B (NaCl) :
? No hydrolysis ? Solution will be neutral
? Relationship between K
a
, K
b
, and K
h
?Case (I) : W. A. + S.B. (CH
3
COONa)
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
] [OH COOH] [CH
K
–
–
–
3
?
?
? ?
K
K ?
?Case (II) : S. A. + W. B. (NH
4
OH)
NH
4
+
+ H
2
O NH
4
OH + H
+
?
] [NH
] [H OH] [NH
K
4
4
h
?
?
?
?
] [OH ] [NH
] [OH ] [H OH] [NH
K
–
4
–
4
h
?
?
?
?
?
?
b
w
h
K
K
K ?
?Case (III) W. A. + W. B. (CH
3
COO NH
4
)
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
NH
4
+
+ H
2
O NH
4
OH + H
+
_______________________________________________________
CH
3
COO
–
+ NH
4
+
+ 2 H
2
O NH
4
OH + CH
3
COOH + H
+
+ OH
–
_______________________________________________________
?
] [NH ] COO [CH
COOH] [CH OH] [NH
K
4
–
3
3 4
h
?
?
?
] [H ] COO [CH
] [H ] [OH COOH] [CH
] [OH ] [NH
OH] [NH
K
–
3
–
3
–
4
4
h
?
?
?
?
?
b a
w
h
K . K
K
K ?
? pH calculation
?Case (I) : W. A. + S. B. (CH
3
COONa ? c, K
a
) , h ? degree of hydrolysis.
CH
3
COO
–
+ H
2
O CH
3
COOH + OH
–
c 0 0
c (1 – h) ch ch
?
h – 1
ch
h) – (1 c
(ch)
K
2 2
h
? ?
h < < 1 ? K
h
= ch
2
?
c
K
h
h
?
?
a
w
h
–
K
K
c. K c. h c ] [OH ? ? ? ? ?
c] log – pK – [pK
2
1
pOH
a w
?
?
c] log pK [pK
2
1
pH
a w
? ? ?
?Case (II) :S. A. + W. B. (NH
4
Cl ? c, K
b
)
NH
4
+
+ H
2
O NH
4
OH + H
+
c 0 0
c (1 – h) ch ch
?
h – 1
ch
K
2
h
? h < < 1 ? K
h
= ch
2
?
c
K
h
h
?
?
K
K
c K c c.h ] [H
b
w
h
? ? ? ? ?
?
?
c] log – pK – [pK
2
1
pH
b w
?
? p
OH
=
? ? c log p p
2
1
b w
k k
? ?
?Case (III) W. A. + W. B. (CH
3
COONH
4
? K
a
, K
b
, c)
CH
3
COO
–
+ NH
4
+
+ H
2
O NH
4
OH + CH
3
COOH
c c
c (1 – h) c (1 – h) ch ch
?
2
2
2 2
2 2
h
) h – 1 (
h
h) – (1 c
h c
K ? ?
? K
h
= h
2
1 h – 1 ?
?
h
K h ?
?
b a
w
K K
K
h ?
- - - - - (1)
? CH
3
COOH CH
3
COO
–
+ H
+
?
COOH] [CH
] COO [CH ] [H
K
3
–
3
a
?
?
?
] COO [CH
COOH] [CH
K ] [H
–
3
3
a
?
?
?
h) – (1 c
ch
K
a
? ?
(h < < 1) ? [H
+
] = K
a
. h ?
1) (from
K . K
K
. K ] [H
b a
w
a
?
?
?
K
K
. K
] [H
b
a
w
?
?
?
] pK – pK [pK
2
1
H p
b a w
? ?
? If , K
a
> K
b
? pK
a
< pK
b
? acidic,
K
a
= K
b
? neutral, K
a
< K
b
? basic
? Summary of hydrolysis
1. W. A. + S. B.
c] log pK [pK
2
1
pH
K
K
K
a w
a
w
h
? ? ? ?
2. W. B. + S. A.
c] log – pK – [pK
2
1
pH
K
K
K
b w
b
w
h
? ?
3. W. A. + W. B.
] pK – pK [pK
2
1
pH
K . K
K
K
b a w
b a
w
h
? ? ?
? Hydrolysis of salt of polyprotic acid / base (Na
2
CO
3
)
?
– 2
3
CO + H
2
O
–
3
HCO + OH
–
2
1
a
w – –
3 h
K
K
) (CO K ?
a
a – x x – y x + y
?
?
x x
?
–
3
HCO + H
2
O H
2
CO
3
+ OH
–
1
2
a
w –
3 h
K
K
) (HCO K ?
x – y y x + y
?
?
?
x y x
2 1
a a
K K ? ?
?
K K
h h
2 1
? ?
? Mainly hydrolysis is governed by
– 2
3
CO .
? ?
2
1
a
w
h
–
K
K
c
K c ] [OH
?
? ? ?
, ? ?
1
a
w – –
3
K
K
c
] [OH ] [HCO
?
? ?
, ? ? ?
x
y x
K
K
1
a
w
?
?
? ?
1
a
w
3 2
K
K
] CO [H ?
? Solubility of sparingly soluble salt
(aq.) AgCl (s) AgCl
Solubility
? ? ? ? ? (aq) Cl (aq) Ag
–
?
?
? In ionic equilibrium all the components of equilibrium should be in same phase.
? In case of solubility equilibrium of sparingly soluble salt equilibrium is a heterogeneous equilibrium.
?Example (1) : Solubility of AgCl
AgCl (s) Ag
+
(aq) + Cl
–
(aq)
[AgCl]
] [Cl ] [Ag
K
–
eq
?
?
? K
sp
= K
eq
. [AgCl] = [Ag
+
] [Cl
–
]
?
It is a endothermic reaction on increasing temperature T
?
? solubility
?
?
?
?
?
?
?
? ?
?
2 1 sp
sp
T
1
–
T
1
R 2.303
H
K
K
log
1
2
?Example (2) : Solubility of Ag
2
CO
3
Ag
2
CO
3
(s) 2 Ag
+
+
– 2
3
CO
? ] [CO ] [Ag K
– 2
3
2
sp
?
? , ? In general , if salt is M
x
N
y
type, M
x
N
y
x M
y+
+ y N
x–
?
y – x x y
sp
] [N ] [M K
?
?
let , Q = [M
y+
]
x
[N
x–
]
y
Q = ionic product
If , Q = K
sp
? Equilibrium (Saturated solution) Q > K
sp
? PPt (Super saturated)
Q < K
sp
? Unsaturated solution.
? Various case in solubility
? Case (1) : Ag
2
SO
4
(S = 10
–3
mol/L)
Ag
2
SO
4
2 Ag
+
+ SO
– 2
4
S 0 0
0 2 S S K
sp
= [Ag
+
]
2
[SO
4
- -
] = (2 S)
2
(S) = 4 S
3
= 4 × (10
–3
)
3
? Case (2) : S (moL / L)
? ? ?
S (gm / L)
10
–3
4 2
SO Ag
S
(gm/L) = 10
–3
× (108 × 2 + 96)
? Case (3) : K
sp
given , Solubility = ?
Ag
2
SO
4
2 Ag
+
+
– 2
4
SO
S 0 0
0 2 S S
Read More
|
352 videos|596 docs|309 tests
|
FAQs on Salt Hydrolysis - Chemistry for JEE Main & Advanced
| 1. What is salt hydrolysis? |  |
Ans. Salt hydrolysis refers to the chemical reaction between water and a salt, resulting in the formation of an acidic or basic solution. It occurs when the cation or anion of the salt reacts with water molecules, leading to the release of either hydrogen ions (H+) or hydroxide ions (OH-), respectively.
| 2. How does salt hydrolysis occur? | |
Ans. Salt hydrolysis occurs when the dissociated ions of a salt react with water molecules. If the cation of the salt is a strong acid, it does not react with water, but if it is a weak acid, it will donate a proton to water, forming an acidic solution. On the other hand, if the anion of the salt is a strong base, it does not react with water, but if it is a weak base, it will accept a proton from water, resulting in a basic solution.
| 3. What are the factors that influence salt hydrolysis? | |
Ans. Several factors can influence salt hydrolysis. One major factor is the strength of the acid or base formed by the dissociated ions of the salt. If the acid or base is strong, the hydrolysis reaction will be more significant. The concentration of the salt solution, temperature, and the presence of catalysts or inhibitors can also affect the extent of salt hydrolysis.
| 4. What are the consequences of salt hydrolysis? | |
Ans. The consequences of salt hydrolysis depend on whether an acidic or basic solution is formed. If the hydrolysis results in an acidic solution, it will increase the concentration of hydrogen ions (H+), leading to a decrease in pH. Conversely, if a basic solution is formed, the concentration of hydroxide ions (OH-) will increase, resulting in an increase in pH. These changes in pH can have various effects on chemical reactions and biological systems.
| 5. How is salt hydrolysis useful in everyday life? | |
Ans. Salt hydrolysis has several practical applications in everyday life. One common example is the use of buffering agents, which are salts that can resist changes in pH by undergoing hydrolysis. Buffers play a crucial role in maintaining the pH balance in many products, such as food and beverages, cosmetics, and pharmaceuticals. Additionally, understanding salt hydrolysis is essential in fields like chemistry, biology, and environmental science, as it helps explain various chemical reactions and the behavior of substances in aqueous solutions.
About this Document
|
4.70/5 Rating |
|
Dec 26, 2024 Last updated |
Document Description: Salt Hydrolysis for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation.
The notes and questions for Salt Hydrolysis have been prepared according to the JEE exam syllabus. Information about Salt Hydrolysis covers topics
like and Salt Hydrolysis Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Salt Hydrolysis.
Introduction of Salt Hydrolysis in English is available as part of our Chemistry for JEE Main & Advanced
for JEE & Salt Hydrolysis in Hindi for Chemistry for JEE Main & Advanced course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: Salt Hydrolysis | Chemistry for JEE Main & Advanced
Description
Full syllabus notes, lecture & questions for Salt Hydrolysis | Chemistry for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Chemistry for JEE Main & Advanced | Best notes, free PDF download
Information about Salt Hydrolysis
In this doc you can find the meaning of Salt Hydrolysis defined & explained in the simplest way possible. Besides explaining types of
Salt Hydrolysis theory, EduRev gives you an ample number of questions to practice Salt Hydrolysis tests, examples and also practice JEE
tests
|
352 videos|596 docs|309 tests
|
Download as PDF
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Summary
,Extra Questions
,MCQs
,Objective type Questions
,Viva Questions
,practice quizzes
,Semester Notes
,shortcuts and tricks
,study material
,Sample Paper
,video lectures
,Important questions
,Salt Hydrolysis | Chemistry for JEE Main & Advanced
,Free
,Previous Year Questions with Solutions
,Exam
,Salt Hydrolysis | Chemistry for JEE Main & Advanced
,ppt
,mock tests for examination
,past year papers
,Salt Hydrolysis | Chemistry for JEE Main & Advanced
;
Additional Information about Salt Hydrolysis for JEE Preparation
Salt Hydrolysis Free PDF Download
The Salt Hydrolysis is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Salt Hydrolysis now and kickstart your journey towards success in the JEE exam.
Importance of Salt Hydrolysis
The importance of Salt Hydrolysis cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Salt Hydrolysis Notes
Salt Hydrolysis Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Salt Hydrolysis.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Salt Hydrolysis Notes on EduRev are your ultimate resource for success.
Salt Hydrolysis JEE Questions
The "Salt Hydrolysis JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Salt Hydrolysis on the App
Students of JEE can study Salt Hydrolysis alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Salt Hydrolysis,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Salt Hydrolysis is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup