Chemistry (including Biochemistry)/Question on Chemicals to Melt Ice
QUESTION: Dr Raymond:
I am an engineer with some chemistry background. Although not part of my "day job", we have been helping out some community groups handle issues, the big one right now (I live in Wisconsin) is the subject of melting ice. Had a strange question, and wondered if you could provide perspective.
As we are dealing with the crisis of depleting salt supplies, there are apparently other chemicals in the community that some companies are looking at donating. The question arose the other day as follows:
Would the addition of either calcium fluoride or Lithium Iodide work to melt the ice better, and if so, why would calcium fluoride work better than lithium iodide?
I am picking up in the middle of some conversations that were had from two individuals who are no longer with the group, and so I apologize for not being able to provide better context, but that is the question.
Can you provide any perspective?
ANSWER: Howdy Mike!
First, both with depress freezing points - however...
"Better" is a pretty ambiguous term here... Practically speaking, calcium fluoride is a great road treatment if you get it down before the first ice hits. Being from Michigan originally, I know that this is not really an option for WI.
Why one would say that CaF2 is better is based purely on freezing point depression - which is based on the number of ions generated when dissolved. However, LiI should have a higher maximum suppression due to solubility. Hear is the break down with a review of the chemistry:
Ksp is the solubility constant - the higher it is the more soluble a species is.
Basically CaF2 has a value down near 1E-11 (which is really saying a molarity of about 0.0001.
For LiI, the amount you can get into water is virtually infinite compared to this (>50% by weight).
So, CaF2 will suppress the freezing point about 50% more per molecule, but LiI is about 1E6 times more soluble.
Finally, lets consider weight: Ca (20) F2 (19*2) = ca 58 g/mole, Li (7) I (127) = ca 134 g/mole
Conceptually we can then think about it this way (using arb. units):
CaF2 will do 150 units of freezing depression work per 58 g (lets call it about 3 units per gram), but only about 0.01 percent of the 58 g can go into a liter of water.
LiI will do about 100 units of freezing depression work per 134 g (lets call it about 0.75 per gram - or 4 times lower than CaF2), but you can virtually use as much as you like.
So, pound for pound CaF2 wins (a little will saturate the system, but depresses freezing quite a lot), but if you are simply trying to melt ice, LiI may be the way to go (theoretical freezing point for water saturated is somewhere below -100 F).
Last item of note: Both F- ions and LiI have some reactivity and environmental concerns associated with them. Given the whole regions sensitivity to water issue (reasonably so), this may not be a science issue for your region, it may be a policy one where the amount of each that may be dispersed will be based on state or local regulations... In which case you may have to go with a little of both to stay below the threshold for either.
Feel free to follow up if you need to.
---------- FOLLOW-UP ----------
QUESTION: Thanks for the quick follow up. You are correct, right now our environmental folks are doing their follow up.
One question, when you indicated that CaF2 will do 150 units of freezing depression, is that because of the way the compound dissociates in water? In otherwords, it forms 3 ions, as oppose to the 2 that LiI will form? I am trying to figure out how you got the "150 units of freezing depression" and the "100 units of freezing depression"?
Correct; there is a reduced form of the system equation that governs this effect... once all the calculus is stripped away, this is what you get...
change in melting/freezing point = - i*Kf*m
hear, i is the number of ions formed, Kf is just a freezing point constant for water and m is the molality (similar to the molarity, just in units of moles/kg water instead of moles/L).
With Kf constant here, you can quickly deduce that we can convert this to relative depression per gram for comparison.
I hope that helps!