InterviewSolution
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InterviewSolution
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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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Answer» `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` |
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