Salt hydrolysis – Part 2

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

K_a = [HA][OH ^–] / [A^–] [H₂O]

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

K_a [H₂O]= [HA][OH ^–] / [A^–] 

K_a[H₂O] = K_h

K_h is hydrolysis constant

K_h = [HA][OH ^–] / [A^–]         ————– eq 1

Relation among hydrolysis constant (K_h ), ionic product of water (K_w) and dissociation constant of acid(K_a)-

H₂O   ⇌  H⁺     +   OH^–

We know,

K_w = [H⁺][OH^–]         ————–eq 2

For weak acid,

H A ⇌   H⁺     +       A ^–

K_a = [H⁺][A^–]/[HA]         ——— eq 3

From eq (2) and eq ( 3)

K_w / K_a = [H⁺] [OH^–] [HA] /[H⁺][A ^–]

K_w / K_a =  [OH^–] [HA] /[A ^–]

because,

K_h = [HA][OH ^–] / [A^–] 

Thus,

K_w / K_a= K_h

K_h = K_w/K_a

putting the value of [HA], [OH ^–] and [A^–] 

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

K_h = Ch²

h = √ (Kh/C)

putting the value of K_h

h = √ (K_w/K_a.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^–] = K_w

[H⁺] = K_w / [OH^–]

putting the value of [OH^–]

[H⁺] = K_w / Ch

because ,

h = √ (K_w/K_a.C)

putting the value of ‘h’

[H⁺] = (K_w/C) √ (K_{a .} C /K_w)

[H⁺] =  √ (K_{w .}K_a/ C)

Taking log  and reverting the sign through out,

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

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

pH =  1/2 pK_w + 1/2 log C + 1/2 log pK_a

pK_w = 14

pH =  7  + 1/2 log C + 1/2 log pK_a

“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 .”