The `pH` of `0.05M` aqueous solution of diethy`1` amine is `12.0` . Caluclate `K_(b)`.

Explanation

Diethyl amine is base and gives `OH^(-)` as, `(C_(2)H_(5))_(2)NH+H_(2)O to (C_(2)H_(5))NH_(2)^(+)+OH^(-)`
`{:(,"Initial conc.",1,0,0),(,"Equilibrium conc".(1-alpha),alpha,,):}`
`alpha" " therefore [OH^(-)]=Calpha`
Where C is conc. of bas and C=0.05M
`?pH =12 therefore pOH=2`
`or [OH^(-)]=10^(-2)M`
`therefore C alpha=10^(-2)`
`or 0.05 xx alpha= 10^(-2) (? C =0.05)`
`alpha=0.02`
Now for a base, `K_b=(Calpha^(2))/((1-alpha))=(0.05 xx (0.2)^(2))/((1-0.2))`
`=(0.05 xx 0.04)/(0.8) =2.5 xx 10^(-3)`