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`?

Answer

`alpha = 0.0054`

Explanation

`K_(b) = 5.4 xx 10^(-4)`
`c = 0.02 M`
Then, `alpha= sqrt((K_(b))/(c))`
`= sqrt((5.4 xx 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_((aq))harr,Na_((aq))^(+),+OH_((aq))^(-)),(,0.1M,0.1M):}`
And,

`{:((CH_(3))_(2)NH,+,H_(2)O,harr,(CH_(3))_(2)NH_(2)^(+),+,barOH),((0.02-x),,,,,,x),(,0.02M,,,,,,0.1M):}`

Then, `[(CH_(3))_(2)NH_(2)^(+)] = x`
`[OH^(-)] = x + 0.1 , 0.1`
`rArr K_(b) = ([(CH_(3))_(2)NH_(2)^(+)][OH^(-)])/([(CH_(3))_(2)NH])`
`5.4 xx 10^(-4) = (x xx 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.