Question

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

Answer 1

`"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, `"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)|)))`