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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 ‘Kh‘ = [NH4OH][CH3COOH] / [NH4+][CH3COO] [H2O]

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

Kh = [NH4OH][CH3COOH] / [NH4+][CH3COO]          ————eq 1

CH3COOH   ⇌  CH3COO  +  H+

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

NH4OH     ⇌   NH4+  + OH

Kb = [NH4+][OH] / [NH4OH ]           ———— eq 3

H2O    ⇌  H+  + OH

Kw = [H+][OH]        ———– eq 4

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

Kw / Ka.Kb = [NH4OH][CH3COOH] [H+][OH]  / [NH4+][OH]  [CH3COO]  [H+]

Kw / Ka.Kb =  [NH4OH][CH3COOH] / [NH4+][CH3COO]         ——– eq 5

comparing eq (1) and (5)

Kw / Ka.Kb = Kh

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

 

Kh = [NH4OH][CH3COOH] / [NH4+][CH3COO]

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

Kh  = Ch x Ch / C (1-h).C (1-h)

Kh = h2/(1-h)2

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

Kh =  h2

h = Kh

h = degree of hydrolysis

because ,

 Kh = Kw / Ka.Kb

h = √ (Kw / Ka.Kb)

” 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+

Ka = [CH3COO][H+] / [CH3COOH]

[H+]  = Ka [CH3COOH]  / [CH3COO]

[H+] = Ka. Ch/ C(1-h)

[H+] = Ka. h / (1-h)

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

[H+] = Ka. h

because ,

h = √ (Kw / Ka.Kb)

[H+]  = Ka√ (Kw / Ka.Kb)

[H+]  = √ (Kw . Ka / Kb)

Taking log and reverting the sign through out,

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

-log [H+] = pH , – log Kw =  pKw , – log Ka = pKa , – log Kb = pKb

pH =  1/2 pKw  + 1/2 log pKa  –  1/2 log pKb

pH =  7  + 1/2 log pKa  –  1/2 log pKb

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

When  pKa   > pKb ,  than  solution will be alkaline and pH value will be more than 7. In case  pKb   > pKa ,  than e solution will be acidic and pH value will be less than 7.

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