[5] POLYVALENT ION HYDROLYSIS:

(1) POLY ANION HYDROLYSIS:(Poly acidic base):

For example PO₄^-3 (n=3) , C₂O₄^-2 (n=2), S^-2 (n=2) , N^-3 (n=3) etc.

Na2S, NH3PO4, Na3PO4   (Hydrolysis is study just like hydrolysis of polybasic acids / polyacidic base) we know that dissociation constant of phosphoric acid is given as:

DISSOCIATION CONSTANT OF PHOSPHORIC ACID:

Step-1

Step-2

Step-3

HYDROLYSIS OF ANION (PO₄^-3):

Step wise illustration of hydrolysis of poly basic acids and polyacidic base is given as - 

Step-1

Experimentally we know that   Ka₁>>Ka₂>>Ka_3, hence x >>y>>Z   so y and z can be neglected with respect to x 

Special Case (1):  

Special Case (2)

          then x can be calculate by quadratic equation. 

Step-2

Experimentally we know that   Ka₁>>Ka₂>>Ka₃   hence   x>>y>>Z    So y and z can be neglected with respect to x and the x present in denominator and numerator both  are cancelled .

Step-3

Experimentally we know that   Ka₁>>Ka₂>>Ka₃   hence   x>> y>>Z  So y and z present in numerator can be neglected with respect to x and the z present in denominator is also neglected with respect to y.

Finally -  

And concentration of different species  

 (2) POLY CATION HYDROLYSIS: (Poly basic acid):
For example Ca⁺² , Fe⁺³ , Al⁺³ , Mn⁺² , Cd⁺²  , Zn⁺³  etc
ILLUSTRATIVE EXAMPLE:
Calculate pH and concentration of all species in 0.1 M solution of FeCl₃ given Fe(OH)₃ have  Kb₁=10^-3 , Kb₂=10^-7 and Kb₃=10^-12.

SOLUTION:

  Fe⁺³ àundergo hydrolysis while, Cl^-  à do not under goes hydrolysis.

   We know that 

Experimentally know that Kb₁>>Kb₂>>Kb₃      so x>>Y and Y>>Z

Hence y and z neglected with respect to x

HYDROLYSIS OF FeCl₃:

Experimentally we know that   Ka₁>>Ka₂>>Ka₃ ,   hence  x>>y>>Z    So  y and  z can be neglected with respect to x

TITRATION BASED POLYVALENT ION HYDROLYSIS:

(1) TITRATION OF Na₂CO₃ Vs HCl:

(2) TITRATION OF NA₃PO₄ Vs HCl:

(1) TITRATION OF Na₂CO₃ Vs HCl:

ILLUSTRATIVE EXAMPLE:
The following volume of 0.1M HCl Solution is added to the in20 ml 0.1M Na₂CO₃ Solution the pH of resulting solution in each case.
(H₂CO₃, Ka₁= 4×10^-6, Ka₂= 5×10^-11)
(1) 0.0 ml (No HCl added)
(2) 10 ml HCl is added
(3) 20 ml HCl is added
(4) 30 ml HCl is added
(5) 40 ml HCl is added