(CH3)2NH + H2O = (CH3)2NH2{+} + OH{-}
Kb = [(CH3)2NH2{+}] [OH{-}]/[(CH3)2NH]
At 11.98 the contribution of OH{-} concentration from dissociated water is negligible.
[OH{-}] = 10^(11.98-14.00)
[(CH3)2NH2{+}] = [OH{-}]
[(CH3)2NH] = 0.164M - [OH{-}]
Kb = [OH{-}]^2 / (0.164M - [OH{-}])
= (0.009549{9})^2 / (0.164M - 0.009549{9})
pKa = 14 - ( -log{10}(Kb) )
The pH of a 0.164 M aqueous solution of dimethylamine is 11.98. Write the ionization equation,
calculate the values of Kb and pKa
and comment on strength of the
1 answer
1 answer