7.54. The ionization constant of dimethylamine is 5.4 * 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
7.54. The ionization constant of dimethylamine is 5.4 * 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
Kb= 5.4*10−4
c= 0.02M
Then, α= (Kb /c)1/2
α= (5.4*10−4 / 2 x 10-2)1/2 =0.1643
(CH3)2NH+H2O ↔ (CH3)2NH+2+OH-
[ (CH3)2NH] = 0.02 – x ≈ 0.02
[ (CH3)2NH+2] = x
[OH-] = 0.1 + x
≈ 0.1
Now, Kb= [ (CH3)2NH+2] [OH−]/ [ (CH3)2NH] = (x * 0.1) / (0.025).
x = 1.08 x 10-4
% of dimethylamine ionised = (1.08 x 10-4) x (10
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