How do you calculate the pH at the equivalence point for the titration of .190M methylamine with .190M HCl? The Kb of methylamine is 5.0x10^-4.

The net ionic equation for the titration in question is the following:

#CH_3NH_2+H^(+)->CH_3NH_3^(+)#

This exercise will be solved suing two kinds of problems: Stoichiometry problem and equilibrium problem .

Stoichiometry Problem :
At the equivalence point, the number of mole of the acid added is equal to the number o fmole of base present.

Since the concentrations of base and acid are equal, the concentration of the conjugate acid #CH_3NH_3^(+)# can be determined as follows:

Since equal volumes of the acid and base should be mixed, and since they are additive, the concentration of #CH_3NH_3^(+)# will be half the initial concentration of #CH_3NH_2#.

Thus, #[CH_3NH_3^(+)]=0.095M#

Equilibrium Problem :
The conjugate acid that will be the major species at the equivalence point, will be the only significant source of #H^(+)# in the solution and therefore, to find the pH of the solution we should find the #[H^(+)]# from the dissociation of #CH_3NH_3^(+)#:

#" " " " " " " " " "CH_3NH_3^(+)rightleftharpoons CH_3NH_2+H^(+)#
#Initial: " " " " " "0.095M" " " " "0M" " " " "0M#
#"Change": " " " " " "-xM" " " " "+xM" " "+xM#
#"Equilibrium": (0.095-x)M" " " "xM" " "xM#

#K_a=([CH_3NH_2][H^(+)])/([CH_3NH_3^(+)])#

Note that #K_w=K_axxK_b#

#=>K_a=(K_w)/(K_b)=(1.0xx10^(-14))/(5.0xx10^(-4))=2.0xx10^(-11)#

#=>K_a=([CH_3NH_2][H^(+)])/([CH_3NH_3^(+)])=(x*x)/(0.095-x)=(x^2)/(0.095-x)=2.0xx10^(-11)#

Solve for #x=1.38xx10^(-6)M=[H^(+)]#

Therefore, the pH of the solution is #pH=-log[H^(+)]#

#=>pH=-log(1.38xx10^(-6))=5.86#

Here is a video that explains in details the titration of a weak acid by a strong base:
Acid - Base Equilibria | Weak Acid - Strong Base Titration.