Salt hydrolysis
source : Alkaline Water Plus
Salt hydrolysis-
Hydrolysis of salt of strong base and weak acid-
The solution of such a salt is basic in nature . The anion of salt reacts with water to form weak acid and OH- ions.
A– + H2O ⇌ H A + OH –
weak ion
Before hydrolysis C 0 0
After hydrolysis C ( 1-h) Ch Ch
Ka = [HA][OH –] / [A–] [H2O]
Water is present in excess hence can be regarded as constant.
Ka [H2O]= [HA][OH –] / [A–]
Ka[H2O] = Kh
Kh is hydrolysis constant
Kh = [HA][OH –] / [A–] ————– eq 1
Relation among hydrolysis constant (Kh ), ionic product of water (Kw) and dissociation constant of acid(Ka)-
H2O ⇌ H+ + OH–
We know,
Kw = [H+][OH–] ————–eq 2
For weak acid,
H A ⇌ H+ + A –
Ka = [H+][A–]/[HA] ——— eq 3
From eq (2) and eq ( 3)
Kw / Ka = [H+] [OH–] [HA] /[H+][A –]
Kw / Ka = [OH–] [HA] /[A –]
because,
Kh = [HA][OH –] / [A–]
Thus,
Kw / Ka= Kh
Kh = Kw/Ka
putting the value of [HA], [OH –] and [A–]
Kh = Ch x Ch /C (1-h) = Ch2/(1-h)
because (1-h) is nearly equal to one therefore ,
Kh = Ch2
C = concentration of salt
h = degree of hydrolysis
Kh = Ch2
h = √ (Kh/C)
putting the value of Kh
h = √ (Kw/Ka.C)
Expression for the pH of the salt solution –
HA ⇌ H+ + A–
A– + H2O ⇌ H A + OH –
weak ion
Before hydrolysis C 0 0
After hydrolysis C ( 1-h) Ch Ch
[OH–] = h x C
[H+] [OH–] = Kw
[H+] = Kw / [OH–]
putting the value of [OH–]
[H+] = Kw / Ch
because ,
h = √ (Kw/Ka.C)
putting the value of ‘h’
[H+] = (Kw/C) √ (Ka . C /Kw)
[H+] = √ (Kw .Ka/ C)
Taking log and reverting the sign through out,
-log [H+] = -1/2 log Kw – 1/2 log Ka + 1/2 log C
because , -log [H+] = pH , -log Kw = pKw
pH = 1/2 pKw + 1/2 log C + 1/2 log pKa
pKw = 14
pH = 7 + 1/2 log C + 1/2 log pKa
“According to above expression, pH of the solution will always be greater than 7, it means an aqueous solution of a salt of weak acid and strong base is always alkaline .”
