Question

Consider a `"0.075 M"` solution of ethylamine (`"C"_2"H"_5"NH"_2`, `K_b = 6.4xx10^(-4)`). Calculate the hydroxide ion concentration of this solution, and calculate the pH?

Answer 1

`[OH^-] = 0.006928" M"`

`"pH" = 11.85`

Explanation:

You are given some solution of `color(blue)"ethylamine"`, an organic molecule, which gives off a basic solution in pure water. How do I know the solution will be basic and not acidic?

Well, look at the `color(blue)(K_b)` provided. The `K_b` is called the base dissociation constant. It tells you to what extent the base, `"ethylamine"` will ionize in solution. Now, compare it with the `color(blue)(K_a` of its conjugate-acid pair, `color(blue)"ethylammonium"`. We know from the following relationship

`color(blue)(K_a * K_b = K_w,color(white)(aa) K_w = 1*10^-14 ("water dissociation constant")`

• `K_a = (K_(w))/(K_(a))->K_a = ((1*10^-14))/((6.4*10^-4))=1.56*10^-11`

`color(white)(----)`that `K_b>K_a`, so the solution will be basic

`---------------------`

Now, for weak acids and weak bases we can't just plug in whatever concentration they give us to find the pH because weak acids and weak bases do not `100%` ionize in solution. That is why whenever we figure out the `pH` of a weak acid or base, we have to use something called an ICE table .

But because I don't like ICE tables, I will set up the equilibrium expression and explain it to you step-by-step.

`---------------------`

`color(magenta)"Step 1: Write out the reaction for the ionization of ethylamine in solution"`

`color(white)(----)C_2H_5NH_2 + H_2O rightleftharpoons C_2H_5stackrel(+)NH_3 + OH^-`

`color(magenta)"Step 2: Set up an equilibrium expression"`

Equilibrium expressions are written as products over reactants.

`color(white)(----)K_b = ([C_2H_5stackrel(+)NH_3][OH^-])/([C_2H_5NH_2])`

Now, before the reaction proceeds, we have just the reactant, `"ethylamine"`, and no products, meaning no `"ethylammonium"` or `OH^-` ions. As the reaction proceeds, however, the reactants will decrease and products will start forming. This is the change denoted by the `color(red)(+|-)` signs. The `color(red)(+)` indicates increasing in amount and the `color(red)(-)` indicates decreasing in amount.

`K_b = ([C_2H_5stackrel(+)NH_3][OH^-])/([C_2H_5NH_2])->K_b = ([0color(red)(+x)][0color(red)(+x)])/([0.075color(red)(-x)])`

What is the `0` and `0.075`? Initially, we were given the concentration of `"ethylamine"` as `0.075" M"`. This concentration indicates the amount you had before the reaction reached equilibrium. Notice also that the product concentrations were not initially given so the `0` indicates just that.

Note: The change should always match up with the stoichiometric values given by the balanced equation. It's a 1:1 mole ratio so we are good.

`color(magenta)("Step 3: Find the "OH^(-) "concentration")`

• `K_b = ([cancel"0+"x][cancel"0+"x])/([0.075cancel"-x"])`

• `K_b = ([x][x])/([0.075])`

• `(6.4*10^-4) = ([x^2])/([0.075])`

• `(6.4*10^-4)[0.075] = [x^2]`

• `sqrt((6.4*10^-4)[0.075]) = sqrt(x^2)`

• `x = 0.006928, "so "color(orange)([OH^-] = 0.006928`

Remember when I crossed out the `color(red)(-x)` for the change in concentration for ethylamine? I made the assumption that the change is so small that it would not make any significant difference. This is the 5% rule. If our percent ionization is less than 5%, then the assumption is valid.

`"x"/("initial "[C]")*100% ->(0.006928)/(0.075)*100%= 0.093%`

`color(white)(aaaaaaaaaaaaaaaaaaaa)0.093% < 5%`

`color(magenta)("Step 4: Find the pH")`

Since we figured out the [C] of `OH^-` ions, we can go ahead and find the `pOH`.

• `"pOH" = -log[0.006928" M"]`

• `"pOH"= 2.15`

Now from the following,

`color(white)(aaaaaaaa)"pH+pOH = 14" color(white)(aaa)"at "25^@C`

we can figure out the `"pH"`

• `pH+pOH = 14`

• `pH = 14-2.15 ->11.85`

• `color(orange)(pH = 11.85)`

Answer 2

`[OH^-] = 6.93 x 10^(-3)M` => `pOH = 2.16 => pH = 11.84`

Explanation:

As a quick-trick in working with weak bases (ammonia or ammonia derivatives in pure water)
=> `[OH^-] = sqrt((Kb)[Wk Base]` and pOH = -log[OH] & pH = 14 - pOH
and for weak acids (monoprotic and the 1st ionization step of diprotic acids) in pure water
=> `[H^+] = sqrt((Ka)[WkAcid]` => pH = -log[H]

NOTE: These are good for weak electrolytes in pure water. If system is a common ion type problem, the ICE table needs to be used.

Given:
`[WkBase] = 0.075M ... &... K_b = 6.4 x 10^(-4)`

`[OH^-] = sqrt((0.075)(6.4x10^(-4))M` = `6.93 x 10^(-3)M`
`pOH = -log[OH^_] = -log(6.93 x 10^(-3))` = `2.16`
`pH = 14 - pOH = 14 - 2.16 = 11.84`

Given: (Wk Acid Problem)
`HA rightleftharpoons H^+ + A^-`
`[WkAcid] = 0.075M`; `K_a = 1.8 x 10^(-5)`
`[H^+] = sqrt((0.075)(1.8x10^(-5))`M = `1.16 x 10^(-3)M`
`pH = -log[H^+] = -log(1.16x10^(-3)) = 2.93`