Question #76729
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 M.
Expert's answer
t the equivalence point, the methylamine and the HCl will react to completion when the volumes are the same because they have the same molarity. When this happens, you get the conjugate acid of methylamine with molarity 0.210 M and some other stuff, but that's a force so we don't worry about it.
The conjugate acid of methylamine has a Ka of (10-14) / (5*10-4).
The Ka of the conjugate acid is 2*10-11.
If we make [H3O+] = x, and assume that the final molarity of the conjugate acid is still 0.210 M, then (x2)/0.21 = 2*10-11.
x = [H3O+] = 2.05*10-6
pH = 5.69
The conjugate acid of methylamine has a Ka of (10-14) / (5*10-4).
The Ka of the conjugate acid is 2*10-11.
If we make [H3O+] = x, and assume that the final molarity of the conjugate acid is still 0.210 M, then (x2)/0.21 = 2*10-11.
x = [H3O+] = 2.05*10-6
pH = 5.69
