What is the pH of an aqueous solution of 0.364 M ethylamine (a weak base with the formula C_2H_5NH_2)?

Apr 3, 2016

$\text{pH} = 12.15$

Explanation:

In order to be able to solve this problem, you need to know the value of the base dissociation constant, ${K}_{b}$, of ethylamine, ${\text{C"_2"H"_5"NH}}_{2}$, which is listed as being equal to

${K}_{b} = 5.6 \cdot {10}^{- 4}$

http://www.bpc.edu/mathscience/chemistry/table_of_weak_bases.html

So, ethylamine is a weak base, which means that it does not ionize completely in aqueous solution to form ethylammonium cations, ${\text{C"_2"H"_5"NH}}_{3}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$.

An equilibrium will be established between the unionized ethylamine molecules and the two ions that result from its ionization.

Use this equilibrium reaction as a base for an ICE table to find the equilibrium concentration of hydroxide anions

${\text{ ""C"_ 2"H"_ 5"NH"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "C"_ 2"H"_ 5"NH"_ (3(aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

color(purple)("I")color(white)(aaaaaaacolor(black)(0.364)aaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaaaacolor(black)((-x))aaaaaaaaaaaaaaaaacolor(black)((+x))aaaaaacolor(black)((+x))
color(purple)("E")color(white)(aaaaacolor(black)(0.364-x)aaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaacolor(black)(x)

By definition, the base dissociation constant for this equilibrium will be

${K}_{b} = \left(\left[{\text{C"_2"H"_5"NH"_3^(+)] * ["OH"^(-)])/(["C"_2"H"_5"NH}}_{2}\right]\right)$

In your case, this will be equivalent to

$5.6 \cdot {10}^{- 4} = \frac{x \cdot x}{0.364 - x} = {x}^{2} / \left(0.364 - x\right)$

Rearrange to get a quadratic equation

${x}^{2} + 5.6 \cdot {10}^{- 4} \cdot x - 2.0384 \cdot {10}^{- 4} = 0$

This quadratic will produce two solutions, one positive and one negative

${x}_{1} = \textcolor{red}{\cancel{\textcolor{b l a c k}{- 0.01456}}} \text{ }$ and $\text{ } {x}_{2} = 0.014 \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$

Since $x$ represents concentration, the negative solution does not carry any physical significance. Pick the positive solution to get

$x = 0.014$

This means that the equilibrium concentration of hydroxide anions will be

["OH"^(-)] = "0.014 M"

At this point, you can calculate the pOH of the solution by using

color(purple)(|bar(ul(color(white)(a/a)color(black)("pOH" = - log(["OH"^(-)]))color(white)(a/a)|)))

You will have

$\text{pOH} = - \log \left(0.014\right) = 1.85$

In aqueous solution at room temperature, you have the following relationship between pOH and pH

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pOH " + " pH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the pH of the solution will be

$\text{pH} = 14 - 1.85 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 12.15 \textcolor{w h i t e}{\frac{a}{a}} |}}}$