Question

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?

Answer 1

Kb = 5.4 × 10-4
c = 0.02M
Then, a = $\sqrt{\frac{K_b}{c}}$ = $\sqrt{\frac{5.4\times10^{-4}}{0.02}}$ = 0.1643
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
NaOH(aq) $\leftrightarrow$ Na + (aq) + OH-(aq)
0.1M 0.1M
And, (CH3)2NH + H2O $\leftrightarrow$ (CH3)2NH2+ + OH-
(0.02-x) x x
0.02M 0.1M
Then, [ (CH3)2NH2+] = x
[OH-] = x + 0.1 ; 0.1 ⇒ Kb = $\frac{[(CH_3)_2NH_2^+][OH]}{[(CH_3)_2NH]}$
= 5.4 × 10-4 = x × $\frac{0.1}{0.02}$
x = 0.0054
It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.