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Hydrolysis

Definition: When a salt of acid or base is dissolved in water, interaction occurs between ions of a salt and water is called hydrolysis.

Salts are divided into four types.


  1. Salts obtained from strong acid and strong base like KCl, Na2SO4, and KNO3.
  After hydrolysis the solution remains neutral.
  1. Salts obtained from weak acid and strong base like CH3COONa, Na2CO3, and K3PO4.
  After hydrolysis the solution remains alkaline.
  1. Salts obtained from strong acid and weak base like NH4Cl, CuSO4, and Al (NO3)3.
  After hydrolysis the solution remains acidic.
  1. Salts obtained from weak acid and weak base like aluminium acetate and ammonium formate.

Hydrolysis of salt obtained from Weak acid and Strong base:
Let we have MA a salt of weak acid and strong base. When MA is dissolved in water, the interaction occurs between A- ion and water molecules.

A
- + H2O ↔ HA + OH-
 
At equilibrium Keq =
[HA][OH-]
                                 [A-][H2O]
 
Keq [H
2O] = Kh = [HA][OH-]
                                   [A-]  (Kh = hydrolysis constant)………………… (1)

We have ionic product of water Kw = [H
+] [OH-]
And ionization constant of weak acid Ka =
[H+] [A-]
                                                                    [HA]

Kw/Ka = [H
+] [OH-]/ [H+] [A-]
                                     [HA]

            =
[HA] [OH-]
                    [A-]
            = Kh   …………………………………………………………..(from equation 1)

      Kh = Kw/Ka

Hydrolysis constant is related to ionic product of water and ionization constant of the acid.
Thus Kh =
Kw = [HA] [OH-] = [OH-] 2
                 Ka           [A-]              C

As the amount of [HA] and [OH
-] formed by hydrolysis is equal and [A-] = original concentration of salt in moles/lit

Kh =
Kw = [OH-] 2
         Ka         C


Hydrolysis of salt obtained from Strong acid and Weak base:
Let we have MA a salt of strong acid and weak base. When MA is dissolved in water, the interaction occurs between M+ ion and water molecules.

M
+ + H2O ↔ H+ + MOH
 
At equilibrium Keq =
[H+][MOH]
                                 [M+][H2O]
 
Keq [H
2O] = Kh = [H+][MOH]
                                   [M+]                                   (Kh = hydrolysis constant)………………… (2)

We have ionic product of water Kw = [H
+] [OH-]
And ionization constant of weak base Kb =
[M+] [OH-]
                                                                     [MOH]

Kw/Kb = [H
+] [OH-]/ [M+] [OH-]
                                     [MOH]

             =
[H+] [MOH]
                    [M+]
             = Kh  (from equation 2)

       Kh = Kw/Kb

Hydrolysis constant is related to ionic product of water and ionization constant of the acid.
Thus Kh =
Kw = [H+] [MOH] = [H+] 2
                 Kb           [M+]             C

As the amount of [H
+] and [MOH] formed by hydrolysis is equal and [M+] = original concentration of salt in moles/lit

Kh =
Kw = [H+] 2
         Ka       C



Hydrolysis of salt obtained from Weak acid and Weak base:
Let we have MA a salt of weak acid and weak base. When MA is dissolved in water, the interaction occurs between M +, A- ion and water molecules.

M
+ + A- + H2O  ↔  HA + MOH
 
At equilibrium Keq =
[HA][MOH]
                                [M+][A-][H2O]
 
Keq [H
2O] = Kh = [HA][MOH]
                                 [M+][A-]                         (Kh = hydrolysis constant)…………………(3)

We have ionic product of water Kw = [H
+] [OH-]
And ionization constant of weak acid Ka =
[H+] [A-]
                                                                    [HA]
And ionization constant of weak base Kb =
[M+] [OH-]
                                                                     [MOH]



  Kw      [H+] [OH-] [HA] [MOH]
Ka x Kb       [H+] [A-] [M+] [OH-]  

             =
[HA] [MOH]
                   [A-] [M+]
             = Kh  (from equation 3)

       Kh = __
Kw__
                Ka x Kb


Calculation:
Exp 1: Calculate the hydrolysis constant of acetic acid when its ionization constant is 1.75 x 10-5

Reason:
Ka = 1.75 x 10
-5
Kw = 1.0 x 10-14
Kh = ?

Kh =
Kw
         Ka
      =
1.0 x 10-14
         1.75 x 10-5
      = 5.7 x 10-10


Exp 2: Calculate the pH of an aqueous solution of 0.01M acetic acid when its ionization constant is 1.75 x 10-5

Reason:
Ka = 1.75 x 10
-5
Kw = 1.0 x 10-14
C = 0.01M
pH = ?

Kh =
Kw = [OH-]2
         Ka         C


1.0 x 10-14   = [OH-]2
1.75 x 10-5      0.01

[OH
-]2 = 5.7 x 10-12

[OH-] = 2.39

pOH = - log [OH
-]
         = - log [2.39]
         = - (-5.62)
         = 5.62
 
pH = 14- pOH
      = 14 – 5.62
      = 8.38


Exp 3: Calculate the pH of 0.02M solution of methyl amine when its ionization constant is 5.00 x 10-4

Reason:
Kb = 5.00 x 10
-4
Kw = 1.0 x 10-14
C = 0.02M
pH = ?

Kh =
Kw = [H+]2
         Kb       C


1.0 x 10-14   = [H+]2
5.0 x 10-4        0.02

 [H
+]2 = 0.4. x 10-12
 [H+] = 6.32 x 10-7
pH = - log [H+]
      = - log [6.32 x 10
-7]
      = 6.19



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