# Question 27b26

Jun 11, 2017

$i = 1.16$

#### Explanation:

For starters, you should know that the van't Hoff factor, $i$, is simply a measure of the ratio that exists between the number of moles of solute added to a solution and the number of moles of particles of solute produced in solution.

Now, hydrofluoric acid is a weak acid, which implies that it does not ionize completely in aqueous solution to produce hydrogen cations and fluoride anions

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

This implies that if the acid is at 0% ionization, then all the molecules of hydrofluoric acid are unionized, i.e. you don't get any ${\text{H}}^{+}$ and ${\text{F}}^{-}$.

In this case, every mole of hydrofluoric acid added to the solution will produce $1$ mole of unionized molecules, so the van't Hoff factor will be equal to $1$.

In other words, hydrofluoric acid behaves as a non-electrolyte at 0% ionization.

At 100% ionization, all the molecules of hydrofluoric acid are ionized. This implies that every mole of hydrofluoric acid added to the solution will produce $2$ moles of ions

$\text{1 mole H"^(+) + "1 mole F"^(-) = "2 moles ions}$

This time, the van't Hoff factor is equal to

$i = \left(2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles ions"))))/(1color(red)(cancel(color(black)("mole HF}}}}\right) = 2$

In other words, hydrofluoric acid behaves as a strong electrolyte at 100% ionization.

Now, since hydrofluoric acid is a weak acid, you should expect the ionization equilibrium to lie to the left, meaning that most of the molecules will remain unionized in aqueous solution.

So even without doing any calculation, you should say that the acid will be

$\to$ between 0% and 50% ionized

in this solution. In other words, you should have

$1 < i < 1.5$

with $i$ closer to $1$ than to $1.5$ because of the fact that hydrofluoric acid is a weak acid.

To find the value of the van't Hoff factor, use this equation

Your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta T = i \cdot {K}_{f} \cdot b}}}$

Here

• $\Delta T$ is the freezing-point depression
• $i$ is the van't Hoff factor
• ${K}_{f}$ is the cryoscopic constant of the solvent, which in your case is water
• $b$ is the molality of the solution

Water has a cryoscopic constant equal to

${K}_{f} = {1.86}^{\circ} {\text{C kg mol}}^{- 1}$

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

Since water has a normal freezing point of ${0}^{\circ} \text{C}$, you can say that the freezing-point depression is equal to

$\Delta T = {T}_{f}^{\circ} - {T}_{\text{f sol}}$

$\Delta T = {0}^{\circ} \text{C" - (-0.0433^@"C") = +0.0433^@"C}$

The molality of the solution is calculated by taking the number of moles of solute present in $\text{1 kg}$ of solvent. In your case, this will be

$b = {\text{0.0200 moles"/"1.00 kg" = "0.0200 mol kg}}^{- 1}$

Finally, rearrange the equation to solve for $i$

$i = \frac{\Delta T}{{K}_{f} \cdot b}$

and plug in your values to find

$i = \left(0.0433 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{^@"C"))))/(1.86color(red)(cancel(color(black)(""^@"C"))) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.0200 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg}}^{- 1}}}}\right)$

$i = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.16}}}$

The answer is rounded to three sig figs.

Notice that, as predicted, we found

$1 < 1.16 < 1.5$

with $i$ closer to $1$ than to $1.5$. This means that the acid is 16%# ionized in this solution.