Salt hydrolysis

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## Salt hydrolysis-

### Solution )

Sodium acetate is a salt of weak acid and strong base.

### On Salt hydrolysis-

CH3COO   +     H2O     ⇌       CH3COOH      + OH-

weak ion

Before hydrolysis      C                                                 0                         0

After hydrolysis      C ( 1-h)                                          Ch                   Ch

### Kh = Ch2/(1-h)

because,   h <<<1 ,

So,

1-h =1

### Kh = Ch2

C = 0.1 M , Kh = 5.26 x 10 -10

h = √ Kh / C

h = √ 5.6 x 10 -10  / 0.1

h = √ 56 x 10 -10

### Degree of hydrolysis ‘h’  = 7.48 x 10 -5 Ans.

[OH] = Ch =0.1 x 7.48 x 10 -5 = 7.48 x 10 -6 M

[H+ ] [OH] = Kw

[H+] = Kw / [OH] = 10 -14 / 7.48 x 10 -6

[H+] = 1.33 x 10 -9 M

pH = -log [H+]

= – log [1.33 x 10 -9]

= – [log 1.33 – 9 log 10]

= – log 1.33 + 9 log 10]

= – 0.124 +9=8.876

### i)

Kh = Kw / Kb

Given, Kb= 1.75  x 10-5 , Kw = 1 x 10-14

Kh = 1 x 10-14 / 1.75  x 10-5

### On Salt hydrolysis-

NH4 +    +      H2O     ⇌     NH4OH  +  H +

weak ion

Before hydrolysis      C                                                 0                   0

After hydrolysis      C ( 1-h)                                         Ch                 Ch

Kh = Ch x Ch /C ( 1-h) = Ch2 /(1-h)

h <<<1

So, (1-h) = 1

Kh = Ch2

C = 0.1 M

h = √ Kh / C

h = √ 5.7 x 10 -10  / 0.1

h = √ 57 x 10 -10

### h = 7.55 x 10 -5 Ans.

[H+] = Ch = 0.1 x 7.55 x 10 -5

[H+ ] = 7.55 x 10 -6

pH = -log [H+] = – log (7.55 x 10 -6)

= – (log 7.55 – 6 log 10)

= – ( 0.8779 – 6 x 1)

= -0.8779 + 6

### Alternate method for calculation of pH-

NH4Cl is a salt of strong acid and weak base . Therefore ,

pH = 7 – 1/2 pKb – 1/2 log C

Kb =1.75 x 10 -5, C = 0.1 M

pKb = -log Kb= -log (1.75 x 10 -5)

= -(log 1.75 + log 10 -5)

= -(log 1.75 – 5 log 10 )

= – ( 0.245 – 5 x 1)

= – ( – 4.755)

pKb = 4.755

pH = 7 – (4.755/2) – 1/2 log 0.1

=7 –  2.377 – (-1/2)

= 7.5 – 2.377

### iii) Calculation for [OH–]

[H+][OH] = 10-14

[OH] = 10 -14 / 7.55 x 10-6

### Solution )

Ammonium formate is a salt of weak acid and weak base.So,

pKa = 3.8 , pKb = 4.8

pH = 7 + 1/2 pKa – 1/2 pKb

= 7 + (3.8/2) – (4.8/2)

= 7 + 1.9 – 2.4 = 6.5