Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | What is the pH of 0.001 M aniline soluti

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Equilibrium. This page offers a step-by-step solution to the specific question from Exercise 1, Question 52: what is the ph of 0 001 m aniline solution the io....
Question 52

What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Answer

Kb  =   4.27 x 10-10

c  =  0.001M

pH  = ?

α  =  ?

Kb  =  cα2

4.27 x 10-10  = 0.001 x α2

4270 x 10-10 = α2

α = 65.34 x 10-4

Then (anion) = cα  = 0.001 x 65.34 x 10-4

= 0.65 x 10-5

pOH = -log ( 0.65 x 10-5)

= 6.187

pH  =  7.813

Now

Ka x Kb = Kw

 ∴ 4.27 x 10-10 x Ka = Kw

Ka  =  10-14 / 4.27 x 10-10

= 2.34 x 10-5

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10-5

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