Question

I
10. At 25°C the degree of hydrolysis of 0.001 M aniline acetate 50%. Calculate the ionisation
constant of aniline, if ionisation constant of acetic acid is 1.78 x 10' and ionic product of water
is 1.01 x 10-14

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Answer 1

Given info : At 25°C the degree of hydrolysis of 0.001 M aniline acetate 50%. ionisation constant of acetic acid is 1.78 × 10¯⁵ and ionic product of water is 1.01 × 10¯¹⁴.

To find : the ionisation constant of aniline.

solution : expression of hydrolysis constant is given by, Kh = Kw/Ka × Kb ...(1)

but we also know, degree of hydrolysis is given by, h = √Kh/(1 + √Kh)

given h = 50% = 0.5

so, 0.5 = √Kh/(1 + √Kh)

⇒0.5 = 0.5√Kh

⇒Kh = 1

now using it in equation (1),

1 = Kw/(Ka × Kb)

⇒1.78 × 10¯⁵ × Kb = 1.01 × 10¯¹⁴

⇒Kb = 1.01 × 10¯¹⁴/1.78 × 10¯⁵

= 0.567 × 10^-9

= 5.67 × 10¯¹⁰

Therefore the ionisation constant of aniline is 5.67 × 10¯¹⁰