# 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?

May 16, 2017

$\left[O {H}^{-}\right] = 0.006928 \text{ M}$

$\text{pH} = 11.85$

#### Explanation:

You are given some solution of $\textcolor{b l u e}{\text{ethylamine}}$, 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 $\textcolor{b l u e}{{K}_{b}}$ provided. The ${K}_{b}$ is called the base dissociation constant. It tells you to what extent the base, $\text{ethylamine}$ will ionize in solution. Now, compare it with the color(blue)(K_a of its conjugate-acid pair, $\textcolor{b l u e}{\text{ethylammonium}}$. 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")

• ${K}_{a} = \frac{{K}_{w}}{{K}_{a}} \to {K}_{a} = \frac{\left(1 \cdot {10}^{-} 14\right)}{\left(6.4 \cdot {10}^{-} 4\right)} = 1.56 \cdot {10}^{-} 11$

$\textcolor{w h i t e}{- - - -}$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 $p H$ 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.

$- - - - - - - - - - - - - - - - - - - - -$

$\textcolor{m a \ge n t a}{\text{Step 1: Write out the reaction for the ionization of ethylamine in solution}}$

$\textcolor{w h i t e}{- - - -} {C}_{2} {H}_{5} N {H}_{2} + {H}_{2} O r i g h t \le f t h a r p \infty n s {C}_{2} {H}_{5} \stackrel{+}{N} {H}_{3} + O {H}^{-}$

$\textcolor{m a \ge n t a}{\text{Step 2: Set up an equilibrium expression}}$

Equilibrium expressions are written as products over reactants.

$\textcolor{w h i t e}{- - - -} {K}_{b} = \frac{\left[{C}_{2} {H}_{5} \stackrel{+}{N} {H}_{3}\right] \left[O {H}^{-}\right]}{\left[{C}_{2} {H}_{5} N {H}_{2}\right]}$

Now, before the reaction proceeds, we have just the reactant, $\text{ethylamine}$, and no products, meaning no $\text{ethylammonium}$ or $O {H}^{-}$ ions. As the reaction proceeds, however, the reactants will decrease and products will start forming. This is the change denoted by the $\textcolor{red}{+ | -}$ signs. The $\textcolor{red}{+}$ indicates increasing in amount and the $\textcolor{red}{-}$ indicates decreasing in amount.

${K}_{b} = \frac{\left[{C}_{2} {H}_{5} \stackrel{+}{N} {H}_{3}\right] \left[O {H}^{-}\right]}{\left[{C}_{2} {H}_{5} N {H}_{2}\right]} \to {K}_{b} = \frac{\left[0 \textcolor{red}{+ x}\right] \left[0 \textcolor{red}{+ x}\right]}{\left[0.075 \textcolor{red}{- x}\right]}$

What is the $0$ and $0.075$? Initially, we were given the concentration of $\text{ethylamine}$ as $0.075 \text{ M}$. 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.

$\textcolor{m a \ge n t a}{\text{Step 3: Find the "OH^(-) "concentration}}$

• ${K}_{b} = \left(\left[\cancel{\text{0+"x][cancel"0+"x])/([0.075cancel"-x}}\right]\right)$

• ${K}_{b} = \frac{\left[x\right] \left[x\right]}{\left[0.075\right]}$

• $\left(6.4 \cdot {10}^{-} 4\right) = \frac{\left[{x}^{2}\right]}{\left[0.075\right]}$

• $\left(6.4 \cdot {10}^{-} 4\right) \left[0.075\right] = \left[{x}^{2}\right]$

• $\sqrt{\left(6.4 \cdot {10}^{-} 4\right) \left[0.075\right]} = \sqrt{{x}^{2}}$

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

Remember when I crossed out the $\textcolor{red}{- x}$ 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%

$\textcolor{m a \ge n t a}{\text{Step 4: Find the pH}}$

Since we figured out the [C] of $O {H}^{-}$ ions, we can go ahead and find the $p O H$.

• "pOH" = -log[0.006928" M"]

• $\text{pOH} = 2.15$

Now from the following,

$\textcolor{w h i t e}{a a a a a a a a} \text{pH+pOH = 14" color(white)(aaa)"at } {25}^{\circ} C$

we can figure out the $\text{pH}$

• $p H + p O H = 14$

• $p H = 14 - 2.15 \to 11.85$

• $\textcolor{\mathmr{and} a n \ge}{p H = 11.85}$

May 16, 2017

$\left[O {H}^{-}\right] = 6.93 x {10}^{- 3} M$ => $p O H = 2.16 \implies p H = 11.84$

#### 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 = $6.93 x {10}^{- 3} M$
$p O H = - \log \left[O {H}^{_}\right] = - \log \left(6.93 x {10}^{- 3}\right)$ = $2.16$
$p H = 14 - p O H = 14 - 2.16 = 11.84$

Given: (Wk Acid Problem)
$H A r i g h t \le f t h a r p \infty n s {H}^{+} + {A}^{-}$
$\left[W k A c i d\right] = 0.075 M$; ${K}_{a} = 1.8 x {10}^{- 5}$
[H^+] = sqrt((0.075)(1.8x10^(-5))M = $1.16 x {10}^{- 3} M$
$p H = - \log \left[{H}^{+}\right] = - \log \left(1.16 x {10}^{- 3}\right) = 2.93$