Calculate the pH at the equivalence point for the titration of 0.140 M methylamine (CH3NH2) with 0.140 M HCl. The Kb of methylamine is 5.0× 10–4.

Answer 1

The pH at the equivalence point is determined by the salt of the titration.
The salt is methylamonium chloride and its the MeNH3^+ ion. Here is the hydrolysis. The concn of the salt at the equivalence point is 0.140/2 - 0.07
.......MeNH3^+ + H2O ==> H3O^+ + MeNH2
I.......0.07..............0........0
C........-x...............x........x
E......0.07-x.............x........x

Ka for MeNH3^+ = (Kw/kb for MeNH2) = (x)(x)/(0.07-x)
Solve for x = (H3O^+) and convert to pH.

Answer 2

Building off of the last response. The pH = 5.927

Ka = Kw/Kb
Ka = (1*10^-14)/(5*10^-4) = 2*10^-11

Solving for x in the ice table:
Ka = [products]/[reactants]
2*10^-11 = (X^2)/(.07-x)
since Ka is substantially small we can disregard the fact that we are dividing by x on the other side of the equation because it will not affect our answer.
Now we have: 2*10^-11 = (X^2)/(.07).
using algebra we get x = 1.18*10^-6
pH=-log(1.18*10^-6)
pH = 5.9267

hope this helped