[Sophia7X]

Say you have K2SO4 in solution and i=2.32, how do you find percent dissociation from this?

[Borek]

I would start writing formulas for both definitions (i and α).

[Sophia7X]

percent dissociation = conc of ions dissociated/original conc
i=average # of ions dissociated

For things like HF and you have, say, i=1.08, you just subtract 1 and *100% = 8%. Not sure what to do with things that dissociate into more than 2 ions.

[Borek]

To quote wikipedia:

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of a substance as calculated from its mass.

So in this case there are 2.32 ions produced from each K₂SO₄ (instead of three). Most likely solution contains free K⁺, free SO₄^2-, and some ion pairs KSO₄^-.

Somehow it is not clear to me how to define percent dissociation for this problem. Could be I need another coffee.

[Sophia7X]

I looked up dissociation on Wikipedia and found this:

The dissociation degree is the fraction of original solute molecules that have dissociated. It is usually indicated by the Greek symbol α. There is a simple relationship between this parameter and the van 't Hoff factor i. If the solute substance dissociates into  ions, then

[Borek]

That nicely works for AB type substances. Here it is A₂B, and dissociation is a two step process. Each step has its own dissociation degree. Single number doesn't tell whole story.

Doesn't mean we can't use some definition of overall dissociation degree, its just ambiguous to me.