# If the pH of an aqueous solution of "0.15 M" "HF" is 2, what is the percent ionization of "HF"?

Nov 27, 2015

The percent "ionization" is the same as the percent dissociation. Since you know the pH is $2$, and you can find the pKa in your book or online ($\approx 3.15$), and since this acid is a weak acid when put into water, it is a buffer problem.

${\text{HF" rightleftharpoons "H"^(+) + "F}}^{-}$

We can use the Henderson-Hasselbalch equation.

pH = pKa + log \frac{["A"^(-)]}{["HA"]}

where ${\text{A}}^{-}$ is the conjugate base (${\text{F}}^{-}$) and $\text{HA}$ is the weak acid ($\text{HF}$).

2 = 3.15 + log \frac{["A"^(-)]}{["HA"]}

-1.15 = log \frac{["A"^(-)]}{["HA"]}

10^(-1.15) = \frac{["A"^(-)]}{["HA"]}

But we know that as $\text{HF}$ dissociates into ${\text{F}}^{-}$, the same number of $\setminus m a t h b f \text{mol}$s of $\setminus m a t h b f \left({\text{F}}^{-}\right)$ is created as the number of $\setminus m a t h b f \text{mol}$s of $\setminus m a t h b f \text{HF}$ dissociated, because there is no other source that gives ${\text{F}}^{-}$.

Additionally, both species are within the same solution, so their concentrations are determined by the same total volume. When we compare concentrations, we simply get:

\frac{["A"^(-)]}{["HA"]} = (("mol A"^(-))/cancel"L soln") / ("mol HA"/cancel"L soln") = ("mol A"^(-)) / ("mol HA")

Furthermore, we are looking for a percent. If we let $x$ be the $\text{mol}$s of ${\text{F}}^{-}$ produced, then we can normalize the right half of the equation to say that the $\text{mol}$s of $\text{HF}$ leftover is $1 - x$:

${10}^{- 1.15} = \frac{x}{1 - x}$

${10}^{- 1.15} - {10}^{- 1.15} x = x$

${10}^{- 1.15} = \left(1 + {10}^{- 1.15}\right) x$

$x = \frac{{10}^{- 1.15}}{1 + {10}^{- 1.15}}$

$= 0.066 \implies \textcolor{b l u e}{\text{6.6% dissociated}}$

Nov 27, 2015

6.7%

#### Explanation:

Here's an alternative approach you can use to find the percent ionization of the weak acid.

When placed in aqueous solution, hydrofluoric acid, $\text{HF}$, will dissociate to give hydrogen cations, ${\text{H}}^{+}$, and fluoride anions, ${\text{F}}^{-}$.

${\text{HF"_text((aq]) rightleftharpoons "H"_text((aq])^(+) + "F}}_{\textrm{\left(a q\right]}}^{-}$

Notice that you have $1 : 1$ mole ratios between all the species that take part in this reaction. This means that every mole of hydrofluoric acid that dissociates will produce one mole of hydrogen cations and one mole of fluriode anions.

Simply put, one molecule of hydrofluoric acid that ionizes will split into one hydrogen cation and one fluoride anion.

Use the pH of the solution to determine the concentration of hydrogen cations

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

["H"^(+)] = 10^(-2) = "0.010 M"

So, if the ionization of the acid produced $\text{0.010 M}$ of ${\text{H}}^{+}$, it follows that it also produced $\text{0.010 M}$ of ${\text{F}}^{-}$.

This means that out of the initial $\text{0.15 M}$ of weak acid, only $\text{0.010 M}$ ionized.

Percent ionization is defined as

$\textcolor{b l u e}{\text{% ionization" = "amount that ionized"/"initial amount} \times 100}$

In this case, you would have

"% ionization" = (0.010 color(red)(cancel(color(black)("M"))))/(0.15color(red)(cancel(color(black)("M")))) xx 100 = 6.667% ~~ color(green)(6.7%)