# Question fcf52

Mar 13, 2016

$7.2 \cdot {10}^{- 4}$

#### Explanation:

The first thing to note here is that the concentration of the acid is said to be $\text{0.1 F}$, or $\text{0.1 formal}$.

Formality is simply a measure of concentration that does not take into account the form in which the solute exists in solution.

In this case, you can say that a $\text{0.1-F}$ solution will be equivalent to a $\text{0.1-M}$ solution.

Now, hydrofluoric acid, $\text{HF}$, is a weak acid that does not ionize completely in aqueous solution. The problem tells you that in your solution, only 8.13% of the acid is ionized.

If you think about this at a molecular level, a 8.13% ionization basically tells you that out of $10000$, or ${10}^{4}$, molecules of hydrofluoric acid present in solution, only $813$ will ionize, i.e. give off their proton.

The equilibrium reaction for the ionization of hydrofluoric acid looks like this

color(red)("H")"F"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "F"_text((aq])^(-) + "H"_3"O"_text((aq])^(color(red)(+)

Notice that every molecule of $\text{HF}$ that ionizes forms one molecule of fluoride anions, ${\text{F}}^{-}$, the conjugate base of the acid, and one molecule of hydronium ions, ${\text{H"_3"O}}^{+}$.

By definition, the acid dissociation constant, ${K}_{a}$, will be equal to

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

Now, if you start with $\text{0.1 M}$ of hydrofluoric acid, and you know that 8.13% of the molecules ionize, you can calculate the equilibrium concentration of the fluoride and hydronium ions by using a proportion

$\left[\text{F"^(-)] = ["H"_3"O"^(+)] = 8.13/100 * ["HF}\right]$

In this case, you will have

["F"^(-)] = ["H"_3"O"^(+)] = 8.13/100 * "0.1 M" = 8.13 * 10^(-3)"M"

But remember, if $8.13 \cdot {10}^{- 3} \text{M}$ is ionized, that must mean that $8.13 \cdot {10}^{- 3}$ of the acid is no longer in molecular form. Therefore, the equilibrium concentration of the acid will be

overbrace("0.1 M")^(color(purple)("what you start with")) - overbrace(8.13 * 10^(-3)"M")^(color(blue)("what ionizes")) ~~ "0.09187 M"#

Now you're ready to plug in your values into the equation for ${K}_{a}$

${K}_{a} = \frac{8.13 \cdot {10}^{- 3} \cdot 8.13 \cdot {10}^{- 3}}{0.09187} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 7.2 \cdot {10}^{- 4} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The actual value for the acid dissociation constant of hydrofluoric acid is $7.2 \cdot {10}^{- 4}$, so this is an excellent result!

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf