The ionisation constant of dimethylamine is `5.4 xx 10^(-4)` Calculate its degree of ionisation in its 0.02 M solution. What percentage of dimethylamine is ionised if the solution is also 0.1 M in NaOH ?

Explanation

`alpha=sqrt(K_(b)//C)=sqrt((5.4xx10^(-4))//(2xx10^(-2)))=0.164` ltBRgt In presence of `0.1 M NaOH`, is x is the amount of dimethyl amine dissociated,
`{:(,(CH_(3))_(2)NH,+,H_(2)O,hArr,(CN_(3))_(2)NH^(o+),+overset(Θ)(O)H),("Initial",0.02 M,,,,,),("Final",0.02-x,,,,x,0.1+x),(,cong 0.02,,,,,cong 0.1):}`

`K_(b)=(x(0.1))/0.02 or x/0.02=K_(b)/0.1=(5.4xx10^(-4))/10^(-1)=5.4xx10^(-3)`
i.e., `alpha=5.4xx10^(-3) :. %` ionized=`0.54`.