Question

In the titration of 77.5 mL of 1.0 M methylamine, CH3NH2 (Kb = 4.4 10-4), with 0.38 M HCl, calculate the pH under the following conditions.

(a) after 50.0 mL of 0.38 M HCl has been added

(b) at the stoichiometric point

Answer 1

b. pH is due to the hydrolysis of the salt. You must first determine the M of the salt. That will be moles base x M base divided by the total volume (base+ acid) in liters. I will call that value x.

........CH3NH3^+ + HOH ==> CH3NH2 + H3O^+
initial...x..................0........0
change....-y.................y........y
equil....x-y..................y.......y

Ka = (Kw/Kb for CH3NH2) = (H3O^+)(CH3NH2)/(CH3NH3^+)
(1E-14/Kb) = (y)(y)/(x)
Solve for x = (H3O^+) and convert to pH.

a.
77.5 mL x 1.0M = M x L = 0.0775 mols CH3NH2.
50 mL x 0.38 = = 0.019 moles.
.........CH3NH2 + HCl ==>CH3NH3Cl+ H2O
initial...0.0775...0.......0
added...........0.019............
change.0.0775-0.019 -0.019.+0.019
equil...-0.0585...0........0.019

So you have a solution that is
(CH3NH2) = 0.0595/total volume
(CH3NH3Cl) = 0.019/total volume
This is a buffered solution. Plug these values into the buffer equation (the Henderson-Hasselbalch equation) or into the Kb expression and solve for pH.

Answer 2

Thank You