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# Trimethylamine, (CH3)3N, is a weak base (Kb = 6.4 × 10–5) that hydrolyzes by the following equilibrium: (CH3)3N + H2O → (CH3)3NH+ + OH– What is the pH of a 0.1 M solution of (CH3)3NH+? (Enter pH to 2 decimal places; hundredth's.

##"pH" = 4.40##

Your starting point here will be to write the balanced chemical equation that describes 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 of the conjugate acid.

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

This means that for every mole of conjugate acid **that ionizes**, you get **one mole** of weak base and **one mole** of hydronium cations.

The **ICE table** will thus 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 know that an aqueous solution at room temperature has

##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** of 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 constant will be equal to

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

In your case, you will have

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

Since ##K_a## has such a small value when compared with the initial concentration of the conjugate acid, you can use the approximation

##0.1 - x ~~ 0.1##

This will get 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## represents the **equilibrium concentration** of the hydronium cations, you will have

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

As you know, the of the solution is defined 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)|)))##