Calculate the pH at the equivalence point for the titration of 0.120 M methylamine with 0.120 M HCl (kb of methyalamine is 5.0 x 10^-4)

Answer 1

If we call methylamine, MeNH2, then
the pH at the equivalence point is determined by the hydrolysis of the salt, MeNH3Cl formed at the equivalence point. This is Ka for MeNH3Cl
.........MeNH3^+ + H2O -->MeNH2+ + H3O^+
initial..0.06M..............0........0
change.....-x................x.......x
equil.....0.06-x.............x.......x

Ka = (Kw/Kb) = (H3O^+)(MeNH2)/(MeNH3^+)
Kw = 1E-14
Kb = Kb for MeNH2
Substitute into the Ka expression and solve for x, then convert to pH.
(Note: the concn of the salt at the equivalence point is just 1/2 the concn initially. )