# Question #27b26

##### 1 Answer

#### Answer:

#### Explanation:

**!! LONG ANSWER !!**

For starters, you should know that the **van't Hoff factor**, *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

#"HF"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "F"_ ((aq))^(-)#

This implies that if the acid is at **ionization**, then all the molecules of hydrofluoric acid are **unionized**, i.e. you don't get any

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

In other words, hydrofluoric acid behaves as a **non-electrolyte** at

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

#"1 mole H"^(+) + "1 mole F"^(-) = "2 moles ions"#

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

#i = (2 color(red)(cancel(color(black)("moles ions"))))/(1color(red)(cancel(color(black)("mole HF")))) = 2#

In other words, hydrofluoric acid behaves as a **strong electrolyte** at

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

#-># between#0%# and#50%# ionized

in this solution. In other words, you should have

#1 < i < 1.5#

with **closer** to

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

Your tool of choice here will be the equation

#color(blue)(ul(color(black)(DeltaT = i * K_f * b)))#

Here

Water has a cryoscopic constant equal to

#K_f = 1.86^@"C kg mol"^(-1)#

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

Since water has a normal freezing point of **freezing-point depression** is equal to

#DeltaT = T_f^@ - T_"f sol"#

#DeltaT = 0^@"C" - (-0.0433^@"C") = +0.0433^@"C"#

The **molality** of the solution is calculated by taking the number of moles of solute present in **of solvent**. In your case, this will be

#b = "0.0200 moles"/"1.00 kg" = "0.0200 mol kg"^(-1)#

Finally, rearrange the equation to solve for

#i = (DeltaT)/(K_f * b)#

and plug in your values to find

#i = (0.0433 color(red)(cancel(color(black)(""^@"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)))))#

#i = color(darkgreen)(ul(color(black)(1.16)))#

The answer is rounded to three **sig figs**.

Notice that, as predicted, we found

#1 < 1.16 < 1.5#

with **ionized** in this solution.