Calculate the pH at the equivalence point for the titration of 0.210 M methylamine (CH3NH2) with 0.210 M HCl. The Kb of methylamine is 5.0×10−4.

Answer 1

(CH3NH2) = (HCl) = 0.210 M; therefore, volumes will be equal which means at the equivalence point the concn of the salt will be 1/2 of 0.210 or 0.105 M.
CH3NH2 + HCl ==> CH3NH3Cl
..............CH3NH3^+ + HOH ==> CH3NH2 + H3O^+
I..............0.105................................0...............0
C..............-x....................................x................x
E..........0.105-x...............................x................x

Kb fo CH3NH3^+ = (Kw/Ka for CH3NH2) =
(x)(x)/(0.105-x). Solve for x = (H3O^+) and convert to pH.
Post your work if you get stuck.

Answer 2

5.84