# Question 6bf82

Mar 9, 2017

Here's what I got.

#### Explanation:

Propionic acid, $\text{CH"_3"CH"_2"COOH}$, is a weak acid, which means that it does not ionize completely in aqueous solution to produce hydronium cations, ${\text{H"_3"O}}^{+}$, and propionate anions, ${\text{CH"_3"CH"_2"COO}}^{-}$.

In other words, propanoic acid will exist as in aqueous solution as an equilibrium between the unionized form and the produced ions

${\text{CH"_ 3"CH"_ 2"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"CH"_ 2"COO"_ ((aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

Now, the pH of a solution is calculated by taking the negative common log of the concentration of hydronium cations

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

You can rearrange this equation to solve for the concentration of hydronium cations present in solution

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

In your case, the solution is said to have a pH of $5.5$, which corresponds to a concentration of hydronium cations equal to

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 5.5}$

["H"_3"O"^(+)] = 3.16 * 10^(-6)"M"

Now, take a look at the ionization equilibrium for propionic acid. Notice that every mole of propionic acid that ionizes produces $1$ mole of hydronium cations and $1$ mole of propionate anions.

This means that at equilibrium, the concentration of propionate anions will be equal to the concentration of hydronium cations

$\left[{\text{CH"_3"CH"_2"COO"^(-)] = ["H"_3"O}}^{+}\right] \to$ at equilibrium

Therefore, your solution will also contain

["CH"_3"CH"_2"COO"^(-)] = 3.16 * 10^(-6)"M"

Finally, you are told that 42% of the solution is in its acidic form at this pH. This means that for every $100$ moles of propionic acid added to the solution, only $42$ moles will NOT ionize.

In other words, a theoretical $\text{100 M}$ propionic acid solution will produce a

$\text{100 M" * 42/100 = "42 M} \to$ what remains in acidic form

concentration of unionized propionic acid and

$\text{100 M " - " 42 M" = "58 M} \to$ what ionizes

concentration of hydronium cations and of propionate anions at equilibrium.

This means that the equilibrium concentration of the acid will be equal to

3.16 * 10^(-6)color(red)(cancel(color(black)("M H"_3"O"^(+)))) * ("42 M CH"_3"CH"_2"COOH")/(58color(red)(cancel(color(black)("M H"_3"O"^(+)))))

$= 2.29 \cdot {10}^{- 6} \text{M CH"_3"CH"_2"COOH}$

Now, the acid dissociation constant for the above ionization equilibrium is defined as

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["CH"_3"CH"_2"COO"^(-)])/(["CH"_3"CH"_2"COOH}\right]\right)$

In your case, this will be equal to

${K}_{a} = \frac{3.16 \cdot {10}^{- 6} \cdot 3.16 \cdot {10}^{- 6}}{2.29 \cdot {10}^{- 6}} = 4.36 \cdot {10}^{- 6}$

As you know, the $\text{p} {K}_{a}$ of the acid is defined as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{p} {K}_{a} = - \log \left({K}_{a}\right)}}}$

This means that the $\text{p} {K}_{a}$ of propionic acid is equal to

$\text{p} {K}_{a} = - \log \left(4.36 \cdot {10}^{- 6}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{5.36}}}$

Mar 10, 2017

WARNING! Long answer! $\text{p"K_text(a) = 5.36}$

#### Explanation:

I tell my students that, when they see an equilibrium problem including words like "initial values" and "equilibrium values", they should immediately give in to three urges:

1. Write the chemical equation.
2. Below the equation, leave room for an ICE table.
3. Below the blank ICE table, write the equilibrium constant expression.

The chemical equation for the ionization of propionic acid is

$\text{CH"_3"CH"_2"COOH" + "H"_2"O" ⇌ "H"_3"O"^"+" + "CH"_3"CH"_2"COO"^"-}$

Let's rewrite this equation and set up an ICE table

$\textcolor{w h i t e}{m m m m m m m} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m l} c \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m} c - x \textcolor{w h i t e}{m m m m m m l} x \textcolor{w h i t e}{m m m} x$

${K}_{\textrm{a}} = \left(\left[\text{H"_3"O"^"+"]["A"^"-"])/(["HA}\right]\right)$

We don’t know the initial concentration of $\text{HA}$, so let's just call it $c$ for now.

However, we know the $\text{pH}$, so we can calculate $x = \left[\text{H"_3"O"^"+}\right]$.

["H"_3"O"^"+"] = 10^"-pH"color(white)(l) "mol/L" = 10^"-5.5" color(white)(l)"mol/L" = 3.16 × 10^"-6"color(white)(l) "mol/L"

We also know that 42 % of $\text{HA}$ is in the acidic form (i.e., as $\text{HA}$).

Thus, 58 % of $\text{HA}$ has ionized to form $\text{H"_3"O"^"+}$.

∴ The initial concentration of "HA" = 100/58 × ["H"_3"O"^"+"]

c = ["HA"]_0 = 100/58 × ["H"_3"O"^"+"] = 100/58 × 3.16 × 10^"-6"color(white)(l) "mol/L"

= 5.45 ×10^"-6" color(white)(l)"mol/L"

Our ICE table becomes

$\textcolor{w h i t e}{m m m m m m m m m} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(mm) +color(white)(mll) "A"^"-}$
$\text{I/mol·L"^"-1": color(white)(mll)5.45 × 10^"-6} \textcolor{w h i t e}{m m m m m} 0 \textcolor{w h i t e}{m m m m m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(m)"-"3.16 × 10^"-6"color(white)(mm)"+"3.16 × 10^"-6"color(white)(m)"+"3.16 × 10^"-6}$
"E/mol·L"^"-1":color(white)(ml)2.29 × 10^"-6" color(white)(mmm)3.16 × 10^"-6"color(white)(mm)3.16 × 10^"-6"

K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = (3.16 × 10^"-6" × 3.16 ×10^"-6")/(2.29 × 10^"-6") = 4.36 × 10^"-6"

"p"K_text(a) = -log(4.36 × 10^"-6") = 5.36#