(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>>Ka3,
hence x >>y>>Z so y and z can be neglected with respect to x
Special Case (1):
Special Case (2)
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
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
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