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

Explanation

Diethy`1` amine is base and guven `overset(Θ)OH` As
`{:(,(C_(2)H_(5))_(2)NH + H_(2)O,hArr,(C_(2)H_(5))_(2)NH_(2)^(oplus) +,overset(Theta)(O)H),("Initial concentration",1,,0,0),("Equilibrium concentration",C(1-alpha),,C alpha,C alpha):}`
`:. [overset(Θ)OH] = Calpha` where `C` is concentration of base and ltbRgt `C = 0.05 M`
`{:(pH = 12,,:.pOH = 2,,),(or[overset(Θ)OH] 10^(-12)M,,:.Calpha = 10^(-2),,),(or0.05xxalpha = 10^(2),,(C = 0.05),,):}`
`:. alpha = 0.2`
Now for a base,
`K_(b) = (C alpha^(2))/((1-alpha)) (0.05 x (0.2)^(2))/((1-0.2)) = (0.05 xx 0.04)/(0.8) = 2.5 xx 10^(-3)`
Note: Do not use `K_(b) = C alpha^(2)` since `alpha = 0.2` and `I - alpha = 0.8`
If `alpha` would have been very small, then the direct formula cna be used to calculate `K_(b)` of weak base.
`pOH = (1)/(2) (pK_(b) - log C)`