# Hydrofluoric acid, HF, has an acid dissociation constant of 6.8 * 10^(-4) at 25 °C. At equilibrium the concentration of HF is 0.025M. a. What is the equilibrium constant expression for this dissociation? b. What is the pH of this solution?

May 6, 2016

$\text{pH} = 2.39$

#### Explanation:

Hydrofluoric acid, $\text{HF}$, is a weak acid that does not ionize completely in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, and fluoride anions, ${\text{F}}^{-}$.

More specifically, the acid is partially ionized in aqueous solution, the extent of ionization depending on the value of the acid dissociation constant, ${K}_{a}$.

The equilibrium that is established when hydrofluoric acid ionizes looks like this

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

By definition, the acid dissociation constant for this equilibrium will be

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{K}_{a} = \left(\left[\text{F"^(-)] * ["H"_3"O"^(+)])/(["HF}\right]\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Keep in mind that the expression of the acid dissociation constant uses equilibrium concentrations.

Now, you know that the equilibrium concentration of hydrofluoric acid is equal to $\text{0.025 M}$.

Notice that every mole of hydrofluoric acid that dissociates produces one mole of hydronium cations and one mole of fluoride anions.

This means that if you take $x$ to be the concentration of hydrofluoric acid that ionizes, you can say that this concentration will produce a concentration of $x$ of hydronium cations and a concentration of $x$ of fluoride anions.

This means that you have

["F"^(-)] = x" " and " " ["H"_3"O"^(+)] = x

Use the expression of the acid dissociation constant to find the value of $x$

${K}_{a} = \frac{x \cdot x}{0.025} = {x}^{2} / 0.025$

This will get you

$x = \sqrt{0.025 \cdot {K}_{a}}$

$x = \sqrt{0.025 \cdot 6.8 \cdot {10}^{- 4}} = 4.12 \cdot {10}^{- 3}$

Since $x$ represents the equilibrium concentration of hydronium cations, you will have

["H"_3"O"^(+)] = 4.12 * 10^(-3)"M"

The pH of the solution is defined as

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))

In your case, you will end up with

$\text{pH} = - \log \left(4.12 \cdot {10}^{- 3}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 2.39 \textcolor{w h i t e}{\frac{a}{a}} |}}}$