Your starting point right here will it is in to create the well balanced chemical equation that explains the ionization of the trimethylammonium cation, #("CH"_3)_3"NH"^(+)#, the conjugate acid of trimethylamine, #("CH"_3)_3"N"#.

Next, use an ICE table to determine the equilibrium concentration of the hydronium cations, #"H"_3"O"^(+)#, that result from the ionization that the conjugate acid.

The trimethylammonium cation will react with water come reform several of the weak base and produce hydronium cations, both in a #1:1# mole ratio.

This way that for every mole of conjugate acid that ionizes, you obtain one mole the weak base and also one mole the hydronium cations.

The ICE table will therefore look like this

#("CH"_ 3)_ 3"NH"_ ((aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons ("CH"_ 3)_ 3"N"_ ((aq)) + "H"_ 3"O"_ ((aq))^(+)#

#color(purple)("I")color(white)(aaaaacolor(black)(0.1)aaaaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaacolor(black)(0)##color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaaaaaaacolor(black)((+x))aaaaacolor(black)((+x))##color(purple)("E")color(white)(aaacolor(black)(0.1-x)aaaaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaacolor(black)(x)#

Now, you recognize that one aqueous equipment at room temperature has actually

#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_a xx K_b = K_W)color(white)(a/a)|)))#

where

#K_w = 10^(-14) -># the ionization constant the water

Use this equation to calculate the acid dissociation constant, #K_a#, for the trimethylammonium cation

#K_a = K_W/K_b#

#K_a = 10^(-14)/(6.4 * 10^(-5)) = 1.56 * 10^(-10)#

By definition, the mountain dissociation continuous will be equal to

#K_a = (<("CH"_3)_3"N"> * <"H"_3"O"^(+)>)/(<("CH"_3)_3"NH"^(+)>)#

In her case, girlfriend will have actually

#K_b = (x * x)/(0.1 - x) = 6.4 * 10^(-5)#

Since #K_a# has actually such a little value when compared with the early stage concentration that the conjugate acid, you have the right to use the approximation

#0.1 - x ~~ 0.1#

This will gain you

#1.56 * 10^(-10) = x^2/0.1#

Solve for #x# to find

#x = sqrt((1.56 * 10^(-10))/0.1) = 3.95 * 10^(-5)#

Since #x# to represent the equilibrium concentration that the hydronium cations, girlfriend will have

#<"H"_3"O"^(+)> = 3.95 * 10^(-5)"M"#

As you know, the pH the the equipment is characterized as

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(<"H"_3"O"^(+)>)color(white)(a/a)|)))#

Plug in your worth to find

#"pH" = - log(3.95 * 10^(-5)) = color(green)(|bar(ul(color(white)(a/a)4.40color(white)(a/a)|)))#