# Question 5e0ee

Oct 24, 2015

$\text{pH} = 11.84$

#### Explanation:

This one is pretty straightforward - you need to use an ICE table and the known value of the base dissociation constant, ${K}_{b}$, to find the equilibrium concentration of hydroxide ions in solution.

So, ethylamine, ${\text{CH"_3"CH"_2"NH}}_{2}$, will accept a proton from water to form the ethylammonium ion, ${\text{CH"_3"CH"_2"NH}}_{3}^{+}$, and hydroxide ions, ${\text{OH}}^{-}$.

${\text{C"_2"H"_5"NH"_text(2(aq]) + "H"_2"O"_text((l]) -> "C"_2"H"_5"NH"_text(3(aq])^(+) " "+" " "OH}}_{\textrm{\left(a q\right]}}^{-}$

color(purple)("I")" " " " " 0.075" " " " " " " " " " " " " " " " " 0" " " " " " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " " "(+x)
color(purple)("E")" "(0.075-x)" " " " " " " " " " " " " "x" " " " " " " " " " " "x

By definition, the base dissociation constant, ${K}_{b}$, will be equal to

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

${K}_{b} = \frac{x \cdot x}{0.075 - x} = 6.4 \cdot {10}^{- 4}$

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

$0.075 - x \approx 0.075$

The equation becomes

${x}^{2} / 0.075 = 6.4 \cdot {10}^{- 4}$

$x = \sqrt{0.075 \cdot 6.4 \cdot {10}^{- 4}} = 0.006928$

The concentration of hydroxide ions will thus be

["OH"^(-)] = x = "0.006928 M"

To get the pH of the solution, calculate the $\text{pOH}$ first

"pOH" = -log(["OH"^(-)])#

$\text{pOH} = - \log \left(0.006928\right) = 2.16$

The pH of the solution will thus be

$\text{pH" = 14 - "pOH}$

$\text{pH} = 14 - 2.16 = \textcolor{g r e e n}{11.84}$