NCERT Class XI Chemistry Equilibrium Solutions

Question : 54

Total: 73

The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Solution:

Ka = 5.4 × 10–4
C = 0.02 M, Ka = Cα2
α2 =

5.4×10−4

0.02

⇒ α2 = 270 × 10−4
α = 0.164
(CH3)2NH+H2O ⇌ (CH3)2NH2++OH−

Initially:	C	0	0
At‌eqili.:	C(1−α)	Cα	Cα

But [OH−] = 0.1 M
Ka = Cα × [OH−] ⇒ 5.4 × 10−4 =

Cα×0.1

C(1−α)

But α ⋘ 1,
∴ α is neglected in the denominator.
⇒ α =

5.4×10−4

0.1

= 5.40 × 10−3 = 0.0054
% ionization = 100α = 100 × 0.0054 = 0.54

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