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

Question:

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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 52: What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Tab....