Chemistry question over pH?

I got #"pH" = 5.68#. But if you did NOT neglect the #"HCl"# leftover, then you would get a #"pH"# of #4.72#, so the assumption your professor told you to make is not great.

Well, first find out if #"HCl"# neutralized all of the methylamine or only some of it.

#"0.5800 mols"/"L" xx "0.2414 L" = "0.140012 mols HCl"#

#"0.3500 mols"/"L" xx "0.4000 L" = "0.140000 mols methylamine"#

To four sig figs, these are equal, so we assume that all the methylamine was protonated. That means the protonated methylamine will dissociate in water.

The current concentration of protonated methylamine is:

#"0.140000 mols protonated methylamine"/("0.2414 + 0.4000 L") = "0.2183 M"#

And the #"HCl"# is ASSUMED negligible:

#"0.000012 mols HCl"/("0.2414 + 0.4000 L") = 1.87 xx 10^(-5) "M"# #"<<"# #"0.2183 M"#

(It really isn't, because #["HCl"]# #">>"# #K_a#.)

The ICE table becomes:

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

#"I"" "0.2183" "" "" "" "" "0" "" "" "" "" "0#
#"C"" "-x" "" "" "" "" "+x" "" "" "" "+x#
#"E"" "0.2183-x" "" "" "x" "" "" "" "" "x#

So, given that #K_b = 5.0 xx 10^(-4)# at #25^@ "C"# and #"1 atm"#, the acid dissociation constant is

#K_a = K_w/K_b = 10^(-14)/(5.0 xx 10^(-4)) = 2.0 xx 10^(-11)#

for protonated methylamine. Therefore:

#2.0 xx 10^(-11) = (["CH"_3"NH"_2]["H"^(+)])/(["CH"_3"NH"_3^+])#

#= x^2/(0.2183 - x)#

But since #K_a# is much smaller than #10^(-5)#, #x# is also small.

#2.0 xx 10^(-11) ~~ x^2/0.2183#

#=> x = sqrt(0.2183 cdot 2.0 xx 10^(-11)) "M"#

#= 2.09 xx 10^(-6) "M"#

And since #x = ["H"^(+)]#,

#color(blue)("pH") = -log["H"^(+)] = color(blue)(5.68)#

NOTE: The #"HCl"# is not actually negligible. The new ICE table would be:

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

#"I"" "0.2183" "" "" "" "" "0" "" "" "" "" "1.87 xx 10^(-5)#
#"C"" "-x" "" "" "" "" "+x" "" "" "" "+x#
#"E"" "0.2183-x" "" "" "x" "" "" "" "" "1.87 xx 10^(-5) + x#

This new mass action expression would be:

#2.0 xx 10^(-11) ~~ (x(1.87 xx 10^(-5) + x))/(0.2183)#

Solving this into a quadratic form gives

#4.58085x^2 + 8.56619xx10^(-5)x - 2.0 xx 10^(-11) = 0#

which solves to give #x = 2.306 xx 10^(-7)# #"M"# as the only physical answer. This would give:

#["H"^(+)] = 1.89 xx 10^(-5) "M"#

and #color(blue)("pH") = -log["H"^(+)] = color(blue)(4.72)#