Salt hydrolysis

source : Alkaline Water Plus

## 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

## 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,

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

## 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