# 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.

May 27, 2016

$\text{pH} = 4.40$

#### Explanation:

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, ${\text{H"_3"O}}^{+}$, 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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{K}_{a} \times {K}_{b} = {K}_{W}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where

${K}_{w} = {10}^{- 14} \to$ the ionization constant of water

Use this equation to calculate the acid dissociation constant, ${K}_{a}$, for the trimethylammonium cation

${K}_{a} = {K}_{W} / {K}_{b}$

${K}_{a} = {10}^{- 14} / \left(6.4 \cdot {10}^{- 5}\right) = 1.56 \cdot {10}^{- 10}$

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

${K}_{b} = \frac{x \cdot x}{0.1 - x} = 6.4 \cdot {10}^{- 5}$

Since ${K}_{a}$ has such a small value when compared with the initial concentration of the conjugate acid, you can use the approximation

$0.1 - x \approx 0.1$

This will get you

$1.56 \cdot {10}^{- 10} = {x}^{2} / 0.1$

Solve for $x$ to find

$x = \sqrt{\frac{1.56 \cdot {10}^{- 10}}{0.1}} = 3.95 \cdot {10}^{- 5}$

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 pH 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

$\text{pH} = - \log \left(3.95 \cdot {10}^{- 5}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4.40 \textcolor{w h i t e}{\frac{a}{a}} |}}}$