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

`underset("base")((CH_(3))_(2)NH_(2))+H_(2)O hArr underset("acid")((CH_(3))_(2)NH_(3)^(+))+OH^(-)`
`K_(b)= ([(CH_(3))_(2)NH_(3)^(+)][OH^(-)])/([(CH_(3))_(2)NH_(2)])`
`= (Calpha. Calpha)/(C(1-alpha))=(Calpha^(2))/((1-alpha))`
`5.4xx10^(-4)=0.02xxalpha^(2)`
Since `alpha gt 0.05`, we will use `K_(b) = (C alpha^(2))/((1-alpha))`
or `5.4xx10^(-4)= (0.02xxalpha^(2))/((1-alpha))`
Now `alpha = 0.151`
In presence of `NaOH`, the dissociation of diethylamine will decrease due to common ion effect. Thus `0.1+Calpha = 0.1` and `1-alpha ~= 1`
Thus `K_(b)= 5.4xx10^(-4)= (Calphaxx(0.1+Calpha))/(C(1-alpha))`
or `alpha= 5.4xx10^(-3) = 0.0054`