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.
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....
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