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)|)))#