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

In order to be able to solve this problem, you need to know the value of the base dissociation constant, #K_b#, of ethylamine, #"C"_2"H"_5"NH"_2#, which is listed as being equal to

#K_b = 5.6 * 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, #"C"_2"H"_5"NH"_3^(+)#, and hydroxide anions, #"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

#" ""C"_ 2"H"_ 5"NH"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "C"_ 2"H"_ 5"NH"_ (3(aq))^(+) + "OH"_((aq))^(-)#

#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 = ( ["C"_2"H"_5"NH"_3^(+)] * ["OH"^(-)])/(["C"_2"H"_5"NH"_2])#

In your case, this will be equivalent to

#5.6 * 10^(-4) = (x * x)/(0.364 - x) = x^2/(0.364 - x)#

Rearrange to get a quadratic equation

#x^2 + 5.6 * 10^(-4) * x - 2.0384 * 10^(-4) = 0#

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

#x_1 = color(red)(cancel(color(black)(-0.01456)))" "# and #" "x_2 = 0.014 color(white)(a)color(green)(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

#"pOH" = - log(0.014) = 1.85#

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pOH " + " pH" = 14)color(white)(a/a)|)))#

This means that the pH of the solution will be

#"pH" = 14 - 1.85 = color(green)(|bar(ul(color(white)(a/a)12.15color(white)(a/a)|)))#