Question #27b26
1 Answer
Explanation:
!! LONG ANSWER !!
For starters, you should know that the van't Hoff factor,
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
In this case, every mole of hydrofluoric acid added to the solution will produce
In other words, hydrofluoric acid behaves as a non-electrolyte at
At
#"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
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
#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
#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