Question:
What is the pH of 0.001M 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
Concentration (c) = 0.001 M
Kb = ca2
4.27 x 10-10 = 0.001 x a2
a2 = 4.27 x 10-10 / 0.001 = 4270 x 10-10
a = 6.53 x 10-5
Then, anion = ca = 0.001 x 6.53 x 10-4 = 0.065 x 10-5
pOH = -log [0.065 x 10-5] = 6.18
pH = 14 - 6.18 = 7.82
Now, Ka x Kb = Kw
Ka x = 4.27 x 10-10 = 10-14
Ka = 10-14/4.27 x 10-10 = 2.34 x 10-5
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