Salt Hydrolysis Class 11 Notes | EduRev

Chemistry for JEE

JEE : Salt Hydrolysis Class 11 Notes | EduRev

 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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