NCERT Class XI Chemistry Equilibrium Solutions

Question : 52

Total: 73

What is the pH of 0.001 M aniline solution? The ionization constant of aniline is 4.27 × 10–10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Solution:

C6H5NH2+H2O ⇌ C6H5NH3++OH−
Kb =

[C6H5NH3+][OH−]

[C6H5NH2]

[OH−]2

[C6H5NH2]

[OH−] = √Kb[C6H5NH2] = √(4.27×10−10)(10−3) = 6.534 × 10−7M
pOH = - log (6.534 × 10−7) = 6.18
∴ pH = 14 - 6.18 = 7.82
C6H5NH2+H2O ⇌ C6H5NH3++OH−

Initial	C	0	0
At‌eqm.	C−Cα	Cα	Cα

Kb =

Cα.Cα

C(1−α)

Cα2

(1−α)

= Cα2 (Since 1 ⋙ α)
∴ α = √

= √

4.27×10−10

10−3

= 6.53 × 10−4
pKb = –log(4.27 × 10–10) = 9.37
pKa+pKb = 14 (for a pair of conjugate acid and base)
∴ pKa = 14 – 9.37 = 4.63
i.e. –log Ka = 4.63 or, log Ka = –4.63
or, Ka = antilog(–4.63) = 2.34 × 10–5
pH = 7.82, a = 6.53 × 10–4, Ka = 2.34 × 10–5

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