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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
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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.
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