The ionisation constant of dimethylamine is `5.4xx10^(-4)`. Calculate its degree of ionization in its `0.02M` solution. What percentage of dimethylamine is ionized if the solution is also `0.1 M` in `NaOH`?

Explanation

Degree of dissociation `(alpha) = (K_(b)//C)^(1//2) =(5.4 xx 10^(-4)//2 xx 10^(-2))^(1//2) =0.164`
In the presence of 0.01 M NaOH if x is the amount of dimethylamine dissociated then

`K_(b) (x(0.1))/((0.02)) " or "(x)/((0.02)) =(K_(b))/((0.1)) =(5.4 xx 10^(-4))/((0.1)) =5.4 xx 10^(-4)`
Percentage of dimethyl amine ionized `=5.4 xx10^(-3) xx 100 = 0.54`