Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of methylamine is 5.0× 10–4.

Answer 1

........CH3NH3^+ + H2O ==> CH3NH2 + H3O^+
I.......0.09................0........0
C........-x.................x........x
E.......0.09-x..............x.........x

Note: THE 0.09 IS 1/2 OF 1.80M. In titrations whatever amount of CH3NH2 you started with you must add an equal amount of acid to arrive at the equivalence point so you diluted the salt formed by a factor of 2 meaning that the concn is now 1/2 the starting value.

Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.09-x)
Solve for x = (H3O^+) and convert to pH.