The ionization constant of dimethylamine is $5.4 \times 10^{-4}$. 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 $NaOH ?$

$K_{b}=5.4 \times 10^{-4}$

$c=0.02\, M$

Then, $\alpha=\sqrt{\frac{K_{b}}{c}}$

$=\sqrt{\frac{5.4 \times 10^{-4}}{0.02}}$

$=0.1643$

Now, if $0.1 \,M$ of $NaOH$ is added to the solution, then $NaOH$ (being a strong base) undergoes complete ionization.

$NaOH _{(a q)} \longleftrightarrow Na ^{+}_{(aq)}+ OH _{(aq)}^{-}$

                               $0.1\,M$           $0.1\,M$

And,

${\left( {C{H_3}} \right)_2}NH\quad  + \quad {H_2}O \leftrightarrow \quad {\left( {C{H_3}} \right)_2}NH_2^ +  + \quad OH$

$(0.02-x)$                                                  $x$                              $x$

$;0.02\,M$                                                                                $;0.1\,M$

Then, $\left[\left( CH _{3}\right)_{2} NH _{2}^{+}\right]=x$

$\left[ OH ^{-}\right]=x+0.1 ; 0.1$

$\Rightarrow K_{b}=\frac{\left[\left( CH _{3}\right)_{2} NH _{2}^{+}\right]\left[ OH ^{-}\right]}{\left[\left( CH _{3}\right)_{2} NH \right]}$

$5.4 \times 10^{-4}=\frac{x \times 0.1}{0.02}$

$x=0.0054$

It means that in the presence of $0.1\, M\, NaOH , 0.54 \%$ of dimethylamine will get dissociated.