Salt hydrolysis

source : Greater minds Tutors

## Salt hydrolysis-

The phenomenon involving interaction of cations or anions ( or both) of a salt with the H+ or OH ions of water leading to the formation of acidic or alkaline or sometimes even neutral solution is called salt  hydrolysis .

The nature of cation or anion of the salt determines the nature of solution which may be acidic or basic . Salts are divided into four categories-

## 1) Hydrolysis of the salt of strong acid and weak base –

The solution of such a salt is acidic in nature . The cation of salt which has come from weak base is reactive . It reacts  with water to form weak base and H+ ions.

#### Ex .  NH4Cl , CuSO4 , AlCl3 , FeCl3 , CuCl2  etc.

NH4Cl        ———->   NH4+   + Cl

NH4+        +     H2O    ⇌       NH4OH    +      H+

weak base

Thus H+ concentration increases and the solution becomes acidic .

Consider a salt ‘ MA ‘ formed by weak base ‘MOH’ and strong acid ‘HA’.

M+    +     H2O     ⇌     MOH  +    H+

weak ion

Before hydrolysis      C                                      0                  0

After hydrolysis      C ( 1-h)                              Ch               Ch

Kb = [MOH][H+] / [M+] [H2O]

Water is present in excess hence can be regarded as constant.

Kb[H2O] = [MOH][H+] / [M+]

Kb[H2O] = Kh

### 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

H2O   ⇌  H+     +   OH

We know,

### Kw = [H+][OH–]         ————–eq 2

For weak base,

MOH  ⇌  M+     +       OH

### Kb = [M+][OH–]/[MOH]         ——— eq 3

From eq (2) and eq ( 3)

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

= [H+]  [MOH] /[M+]

because,

Kh = [MOH][H+] / [M+]

Thus,

Kw / Kb = Kh

### Kh = Kw/Kb

Kh = Ch2

h = √ (Kh/C)

putting the value of Kh

### h = √ (Kw/Kb.C)

[H+] = h x C

putting the value of ‘h’

[H+] = C √ (Kw/Kb.C)

### [H+] =  √ (Kw.C/Kb)

Taking log ,

log [H+] = 1/2 log Kw + 1/2 log C – 1/2 log Kb

because , -log [H+] = pH , -log Kw = pKw,

-pH = – 1/2 pKw + 1/2 log C + 1/2 log pKb

pH =  1/2 pKw  – 1/2 log C – 1/2 log pKb

pKw = 14