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

`alpha = sqrt(K_(b)//C)=sqrt((5.4xx10^(-4))//(2xx10^(-2)))=0.164`
In presence of 0.1 M NaOH, if x is the amount of dimethyl amine dissociated,
`{:(,(CH_(3))_(2)NH,+H_(2)O,hArr,(CH_(3))_(2)NH^(+)OH,+,OH^(-),),("Initial conc.",0.02 M,,,,,,),("After disso.",0.02-x,,,x,,0.1+x,),(,~=0.02,,,,,~=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 = 5.4 xx 10^(-3) :. % ` ionized = 0.54.