Solubility Rules vs. Vant Hoff Factors

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

ak21

Full Member
10+ Year Member
Joined
Mar 16, 2010
Messages
188
Reaction score
3
Hey everyone,

I just had a quick question. So one of the practices questions I did asked me the vant hoff factor of AlCL3. And I said it was 1 because according to the solubility rules for the MCAT it shouldn't be soluble since aluminum isn't a group I metal or a NH4+, NO3-, ClO4-, or an acetate ion. But the correct answer is 4 according to the book...does this mean solubility rules don't apply when you consider an ionic compounds vant hoff factor? Thanks for the help!

Members don't see this ad.
 
I'm not sure where you got your solubility rules (or why - I never memorized any of them except NH4+ and NO3-), but AlCl3 like most chlorides is very soluble. That's why chemistry labs use AgCl so much, it is one of the rare exceptions to the chloride rule.

The theoretical van't Hoff factor of AlCl3 is 4, because in theory it ought to break up into 4 ions when dissolved in water. In reality it is 3.something, depending on its concentration. The lower the concentration, the closer it approaches 4.
 
Also, you do ignore solubility rules for the theoretical van't hoff number. So the van't hoff number for AgCl would be 2. While AgCl is insoluable, a small amount of AgCl would dissociate, and the van't hoff number tells you that it the concentration that is dissolved should be doubled what would be expected when calculated from its mass.

As mentioned above, the real van't hoff number is usually a little less than the theoretical/expected van't hoff number. The reality is that ions sometimes pair up even when they are solvated. That doesn't mean it paired up to become a solid. What it means is that the ions join together to form a "super solute" which brings the concentration and van't hoff number down.
 
Top