What is the #"pH"# when #"50 mL"# of #"0.15 M"# #"HCl"# neutralizes an equal volume of methylamine? #K_b = 7.52 xx 10^(-4)# for methylamine.

I got a #"pH"# of about #5.9# to two sig figs. You really only gave one though.

At the equivalence point, the acid completely neutralizes the original form of the base (#"CH"_3"NH"_2#). As a result, we can know that since the volumes that reacted are equal, the starting concentration of #"CH"_3"NH"_2# was also #"0.15 M"#.

Therefore, since #"HCl"# is a strong acid, #100%# of the #"CH"_3"NH"_2# was converted to #"CH"_3"NH"_3"Cl"#, i.e.

#"CH"_3"NH"_2(aq) + "HCl"(aq) -> "CH"_3"NH"_3"Cl"(aq)#,
#"CH"_3"NH"_3"Cl"(aq) -> "CH"_3"NH"_3^(+)(aq) + "Cl"^(-)(aq)#,

and it now acts as a weak acid in solution.

#"CH"_3"NH"_3^(+)(aq) rightleftharpoons "CH"_3"NH"_2(aq) + "H"^(+)(aq)#

#"I"" ""0.15 M"" "" "" "" "" ""0 M"" "" "" "" ""0 M"#
#"C"" "-x" "" "" "" "" "+x" "" "" "" "+x#
#"E"" "(0.15 - x) "M"" "" "x" M"" "" "" "" "x" M"#

As a result, we can then solve for the #"pH"# through the #["H"^(+)]#.

#K_a = K_w/K_b = 10^(-14)/(7.52 xx 10^(-4))#

#= 1.33 xx 10^(-11) = x^2/(0.15 - x)#

This #K_a# is smaller than #10^(-5)# or so. Thus, we can make the small #x# approximation...

#1.33 xx 10^(-11) = x^2/(0.15 - cancel(x)^"small")#

#=> x = ["H"^(+)] = ul(1.4_12 xx 10^(-6) "M")#

And that makes the #"pH"#:

#color(blue)("pH") = -log["H"^(+)]#

#= -log(1.4_12 xx 10^(-6))#

#= 5.8_500 ~~ color(blue)(5.9)#
(where subscripts indicate digits past the last significant digit.)