Salt hydrolysis – Part 1

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

K_b = [MOH][H⁺] / [M⁺] [H₂O]

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

K_b[H₂O] = [MOH][H⁺] / [M⁺]

K_b[H₂O] = K_h

K_h = [MOH][H⁺] / [M⁺]        ————– eq 1

K_h  = Ch x Ch /C (1-h) = Ch²/(1-h)

because (1-h) is nearly equal to one therefore ,

K_h = Ch²

C = concentration of salt

h = degree of hydrolysis

H₂O   ⇌  H⁺     +   OH^–

We know,

K_w = [H⁺][OH^–]         ————–eq 2

For weak base,

MOH  ⇌  M⁺     +       OH^–

K_b = [M⁺][OH^–]/[MOH]         ——— eq 3

From eq (2) and eq ( 3)

K_w / K_b = [H⁺] [OH^–] [MOH] /[M⁺][OH^–]

= [H⁺]  [MOH] /[M⁺]

because,

K_h = [MOH][H⁺] / [M⁺]

Thus,

K_w / K_b = K_h

K_h = K_w/K_b

K_h = Ch²

h = √ (Kh/C)

putting the value of K_h

h = √ (K_w/K_b.C)

[H⁺] = h x C

putting the value of ‘h’

[H⁺] = C √ (K_w/K_b.C)

[H⁺] =  √ (K_w.C/K_b)

Taking log ,

log [H⁺] = 1/2 log K_w + 1/2 log C – 1/2 log K_b

because , -log [H⁺] = pH , -log K_w = pK_w,

-pH = – 1/2 pK_w + 1/2 log C + 1/2 log pK_b

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

pK_w = 14

pH =  7  – 1/2 log C – 1/2 log pK_b

“According to above expression, pH of the solution will always be smaller than 7, it means an aqueous solution of a salt of strong acid and weak base is always acidic.”