posted by Donna .
A 100.0 mL of 0.200 M methylamine, CH3NH2, is titrated with 0.100 M HCl. Calculate the pH after 70.0 mL of 0.100 M HCl has been added.
The trick to these problems is to determine what is present in the solution at the point of your calculation.
CH3NH2 + HCl ==> CH3NH2*HCl.
moles CH3NH2 = M x L = 0.200 x 0.1 = 0.02
moles HCl = M x L = 0.100 x 0.070 = 007.
So you will have an excess of CH3NH2 (all of it is not neutralized) and all of the HCl has been used. You have 0.007 moles of the salt formed. What kind of a solution is this? It is a buffered solution. You have a weak base and its salt present. Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid) = ??