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)Orarr,(C_(2)H_(5))_(2)NH_(2)^(+),+OH^(-)),("Intitial conc"., 1,0,0),("Equilibrium. conc".,(1-alpha),alpha,alpha):}`
`:. [OH^(-)]=Calpha`
`:. [OH^(-)]=Calpha`
Where C is conc. of base and C `= 0.05M`
`:. pOH=2 , ( :. pH=12)`
or `[OH^(-)]=10^(-2)M`
`:. Calpha=10^(-2)`
or `0.05xxalpha=10^(-2)`
or `0.05xxalpha=10^(-2) , (.: C=0.05)`
`:. alpha = 0.2`
Now for abase,
`K_(b)=(Calpha^(2))/((1-alpha))=(0.05xx(0.2)^(2))/((1-0.2))`
`=(0.05xx0.04)/(0.8)=2.5xx10^(-3)`