# The ionisation constant of dimethylamine is 5.4xx10^(-4). Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylami

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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?

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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