# What is the pH of (CH3)3N solution, in a titration of 25mL of 0.12M (CH3)3N with 0.1M HCl, and Kb = 6.3 x 10^-5?

##### 1 Answer
Nov 28, 2015

$\text{pH} = 5.03$

#### Explanation:

First thing first, I assume that you're interested in finding the pH of the solution at equivalence point, that is, when all the weak base has been neutralized by the strong acid.

The reaction between trimethylamine, ("CH"_3)_3"N", a weak base, and hydrochloric acid, $\text{HCl}$, a strong acid, will produce trimethylammonium ions, ("CH"_3)_3"NH"^(+), and chloride ions, ${\text{Cl}}^{-}$.

For simplicity, I'll use $\text{Me}$ to symbolize a methyl group, ${\text{-CH}}_{3}$. So, the reaction will look like this

${\text{Me"_3"N"_text((aq]) + "HCl"_text((aq]) -> "Me"_3"NH"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

Notice that the weak base and the strong acid react in a $1 : 1$ mole ratio. Moreover, the trimethylammonium ion, which is the conjugate acid of trimethylamine, is produced in a $1 : 1$ mole ratio with both reactants.

This means that for every mole of weak base and strong acid consumed by the reaction, one mole of conjugate acid will be produced.

Use the molarity and volume of the trimethylamine solution to determine how many moles of weak base you start with

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

$n = \text{0.12 M" * 25 * 10^(-3)"L" = "0.0030 moles Me"_3"N}$

According to the balanced chemical equation, at equivalence point you need equal numbers of moles of strong acid and weak base. This means that you need to add

$\textcolor{b l u e}{c = \frac{n}{V} \implies V = \frac{n}{c}}$

V = (0.0030 color(red)(cancel(color(black)("moles"))))/(0.1color(red)(cancel(color(black)("moles")))/"L") = "0.030 L" = "30 mL"

of hydrochloric acid solution. The total volume of the resulting solution will be

${V}_{\text{total" = "25 mL" + "30 mL" = "55 mL}}$

Now, the weak base and the strong acid will neutralize each other. This means that the reaction will produce

$n = {\text{0.0030 moles Me"_3"NH}}^{+}$

The cocnentration of the trimethylammonium ions will be

["Me"_3"NH"^(+)] = "0.0030 moles"/(55 * 10^(-3)"L") = "0.545 M"

Now, the conjugate acid will react with water to reform the weak base and produce hydronium ions, ${\text{H"_3"O}}^{+}$. Use an ICE table to determine the equilibrium concentration of the hydronium ions

${\text{ " "Me"_3"NH"_text((aq])^(+) + "H"_2"O"_text((l]) " "rightleftharpoons" " "Me"_3"N"_text((aq]) + "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

color(purple)("I")" " " " " " "0.545" " " " " " " " " " " " " " " " " " "0" " " " " " " " "0
color(purple)("C")" " " " " "(-x)" " " " " " " " " " " " " " " "(+x)" " " " "(+x)
color(purple)("E")" " " "0.545-x" " " " " " " " " " " " " " " "x" " " " " " " " "x

Now, to get the acid dissociation constant for the trimethylammonium ion, use the equation

$\textcolor{b l u e}{{K}_{a} \cdot {K}_{b} = {K}_{W}} \text{ }$, where

${K}_{w}$ - the self-ionization constant for water, equal to ${10}^{- 14}$ at room temperature.

In your case, you will have

${K}_{a} = {K}_{W} / {K}_{b} = {10}^{- 14} / \left(6.3 \cdot {10}^{- 5}\right) = 1.6 \cdot {10}^{- 10}$

By definition, the acid dissociation constant will be

${K}_{a} = \left(\left[{\text{Me"_3"N"] * ["H"_3"O"^(+)])/(["Me"_3"NH}}^{+}\right]\right)$

${K}_{a} = \frac{x \cdot x}{0.545 - x}$

Because ${K}_{a}$ is so small, you can say that

$0.545 - x \approx 0.545$

This means that you have

${K}_{a} = {x}^{2} / 0.545 = 1.6 \cdot {10}^{- 10}$

The value of $x$ will be

$x = \sqrt{0.545 \cdot 1.6 \cdot {10}^{- 10}} = 9.3 \cdot {10}^{- 6}$

This means that the concentration of hydronium ions will be

["H"_3"O"^(+)] = x = 9.3 * 10^(-6)"M"

The pH of the solution will thus be

"pH" = -log( ["H"_3"O"^(+)])

$\text{pH} = - \log \left(9.3 \cdot {10}^{- 6}\right) = \textcolor{g r e e n}{5.03}$