Your beginning point right here will be to compose the well balanced chemical equation that defines the ionization that the trimethylammonium cation, #("CH"_3)_3"NH"^(+)#, the conjugate acid the trimethylamine, #("CH"_3)_3"N"#.

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

The trimethylammonium cation will react v water to reform some of the weak base and also produce hydronium cations, both in a #1:1# mole ratio.

This way that because that every mole the conjugate acid that ionizes, you get one mole that weak base and one mole the hydronium cations.

The ICE table will for this reason look favor 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 understand 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 acid dissociation consistent will be equal to

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

In your case, friend will have actually

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

Since #K_a# has actually such a tiny value when compared with the early concentration the the conjugate acid, you can use the approximation

#0.1 - x ~~ 0.1#

This will gain you

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

Solve because that #x# come find

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

Since #x# represents the equilibrium concentration the the hydronium cations, girlfriend will have

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

As friend know, the pH the the solution is identified as

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

Plug in your value to find

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