Salt hydrolysis part 3

Salt hydrolysis

source : SlideServe

Salt hydrolysis-

Hydrolysis of salt of weak acid and weak base-

In such salts , both the cation of weak base and anion of weak acid will undergo hydrolysis simultaneously .

Ex –  CH₃COONH₄ (ammonium acetate), NH₄F( ammonium fluoride)

NH₄⁺ + CH₃COO^– + H₂O ———> NH₄OH + CH₃COOH

According to law of mass action,

Hydrolysis  constant ‘K_h‘ = [NH₄OH][CH₃COOH] / [NH₄⁺][CH₃COO^–] [H₂O]

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

K_h = [NH₄OH][CH₃COOH] / [NH₄⁺][CH₃COO^–]          ————eq 1

CH₃COOH   ⇌  CH₃COO^–  +  H⁺

K_a = [CH₃COO^–][H⁺] / [CH₃COOH]         ———– eq 2

NH₄OH     ⇌   NH₄⁺  + OH^–

K_b = [NH₄⁺][OH^–] / [NH₄OH ]           ———— eq 3

H₂O    ⇌  H⁺  + OH^–

K_w = [H⁺][OH^–]        ———– eq 4

By eq (2) ,(3) and (4)

K_w / K_a.K_b = [NH₄OH][CH₃COOH] [H⁺][OH^–]  / [NH₄⁺][OH^–]  [CH₃COO^–]  [H⁺]

K_w / K_a.K_b =  [NH₄OH][CH₃COOH] / [NH₄⁺][CH₃COO^–]         ——– eq 5

comparing eq (1) and (5)

K_w / K_a.K_b = K_h

Degree of hydrolysis and hydrolysis constant-

                                      NH₄ ⁺    +  CH₃COO^–   +     H₂O     ⇌     NH₄OH  +  CH₃COOH

                                    weak ion    weak ion

Before hydrolysis      C                  C                                                    0                   0

After hydrolysis      C ( 1-h)          C ( 1-h)                                        Ch                 Ch

K_h = [NH₄OH][CH₃COOH] / [NH₄⁺][CH₃COO^–]

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

K_h  = Ch x Ch / C (1-h).C (1-h)

K_h = h²/(1-h)²

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

K_h =  h²

h = √ K_h

h = degree of hydrolysis

because ,

 K_h = K_w / K_a.K_b

h = √ (K_w / K_a.K_b)

” The degree of hydrolysis of salt of weak acid and weak base does not depend upon the concentration of salt”

Expression for the pH of the salt solution –

CH₃COOH   ⇌  CH₃COO^–  +  H⁺

K_a = [CH₃COO^–][H⁺] / [CH₃COOH]

[H⁺]  = K_a [CH₃COOH]  / [CH₃COO^–]

[H⁺] = K_a. Ch/ C(1-h)

[H⁺] = K_a. h / (1-h)

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

[H⁺] = K_a. h

because ,

h = √ (K_w / K_a.K_b)

[H⁺]  = K_a√ (K_w / K_a.K_b)

[H⁺]  = √ (K_w . K_a / K_b)

Taking log and reverting the sign through out,

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

-log [H⁺] = pH , – log K_w =  pK_w , – log K_a = pK_a , – log K_b = pK_b

pH =  1/2 pK_w  + 1/2 log pK_{a  –}  1/2 log pK_b

pH =  7  + 1/2 log pK_{a  –}  1/2 log pK_b

” The pH of salt of weak acid and weak base is independent of concentration of salt.

_When  pK_{a   >} pK_b ,  than  solution will be alkaline and pH value will be more than 7. In case  pK_{b   >} pK_a ,  than e solution will be acidic and pH value will be less than 7.