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The ionisation constant of dimethylamine...

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 ?

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Given, for dimethylamine `=5.4xx10^(-4)`, C for dimethylamine = 0.02 M
`alpha=sqrt(K_(b)/C)=sqrt((5.4xx10^(-4))/0.02)=1.62xx10^(-1)=0.164`
In the presence of 0.1 M NaOH :`(CH_(3))_(2)NH + H_(2)O to (CH_(3))_(2)^(+) + OH^(-)`
Initial conc. 0.02M 0 0
Equilibrium conc.` (0.02 –Calpha)` `Calpha` `Calpha+0.01`
`approx0.02` `approx0.1`
(0.1 M `OH^(-)` from 0.1 M NaOH
`K_(b)=([(CH_(3))_(2)NH_(2)^(+)][H^(-)])/([(CH_(3))_(2)NH]) implies 5.4xx10^(-4)` = `(Calphaxx0.1)/0.02 implies alpha=0.54%`
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