Question

Question 7.54 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?

Class XI Equilibrium Page 229

Answer 1

given, Kb of (CH3)2NH = 5.4 *10⁻⁴
concentration of solution is 0.02M, e.g., C = 0.02M
use Ostwald's theory,
    Cα² = Kb
    0.02 × α² = 5.4 *10⁻⁴
         α² = 5.4*10⁻⁴ /0.02
         α = 0.164

in the presence of 0.1 M of NaOH,
             (CH3)2NH + H2O  ⇔ (CH3)2NH2⁺  + OH⁻
at t = 0     0.02M                         0                       0
at eqlm    (0.02 - Cα)                 Cα                    Cα

now, Kb = [(CH3)NH2⁺][OH⁻]/[(CH3)2NH]
5.4 × 10⁻⁴ = Cα × O.1/0.02   [ because conc. of NaOH = 0.1M and  (0.02M-Cα) ≈ 0.02M ]
Cα = 5.4 × 10⁻⁴ *0.02/0.1 = 108 * 10⁻⁶
α = 108 * 10⁻⁶/0.02 = 5.4 × 10⁻³
    hence, percentage of α = 5.4 * 10⁻³ *100 = 0.54%