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The ionisation constant os 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`?

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`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`.

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