Talk:Van 't Hoff factor

This is beyond erroneous and unnecessarily complicated

What the hell The van't Hoff factor is literally just how many particles a formula breaks up into! The equations are insanely useless, and also how it's described makes 0 sense. Whoever wrote this left 2 unknowns in the equation. i = 1 + alpha??? That doesn't even make sense. Whoever wrote this needs to re-evaluate their life. — Preceding unsigned comment added by 128.110.66.69 (talk) 22:54, 30 May 2018 (UTC)[reply]

Only when the solution is so dilute that complete dissociation occurs. For most practical solutions this is not the case - the degree of dissociation diverges from ideal at pretty low concentrations (e.g. below 0.01 mol/kg). For instance for a 1 mol/kg solution of NaCl, the degree of dissociation has been measured to be 0.789, leading to an actual van't Hoff factor of 1.789. Using a value of 2 would lead to a significant error in this case. See ^[1] for more info. --137.44.3.81 (talk) 13:00, 20 June 2018 (UTC)[reply]

References

^ Heyrovska, R (1997). "Equations for Densities and Dissociation Constant of NaCI(aq) at 25°C from "Zero to Saturation" Based on Partial Dissociation" (PDF). J. Electrochem Soc. 144 (7).

Origin of the name

Arrgh! -- not near any Organic Chem. books. ... Did Jacobus Henricus van 't Hoff actually discover this factor? or did someone else discover it and named it after him? As it is now, I left it as the factor being named after him. Vonkje 14:27, 18 August 2005 (UTC)[reply]

H2SO4 example needs to be changed

H2SO4 dissociation is two step, assumption that 64% means 64 molecules out of 100 are completely ionized is horribly wrong. It'll all depend on concentration, but for the 0.05M solution something like 100% dissociation for the first step and 15% for the second step will be closer to reality. It'll be better to use acetic acid as an example.

The example calculates 64& of the acid dissociated to sulfate + oxonium ions and 36% undissociated which is absolutely impossible. --FK1954 (talk) 06:03, 24 April 2009 (UTC)[reply]

Too fancy

I'm no chemist, so excuse me, but the formula is way too fancy. I was taught that Van't hoff was observed / calculate. No a - 1/ a- whatever stuff Tourskin 04:55, 10 February 2007 (UTC)[reply]

Too simplistic

The van't Hoff factors quoted for the inorganic materials assume complete dissociation in water, which only occurs in the theoretical case of infinite dilution. Using i=2 for NaCl is OK at very low concentrations, but even at 0.0010 molal, the factor is 1.97. At higher concentrations, the factor drops even more. At 1.0 molal, i = 1.81. At this concentration, assuming i = 2 yields about 10% error for any calculations.

This effect is even more pronounced for smaller ions and for ions with larger charges. For instance, MgSO₄ has i = 1.09 at 1.0 molal , even though the infinite dilution value is 2(about 80% error), just like NaCl. Even at 0.0010 molal, i = 1.82 for MgSO₄.

Unity

Could someone provide or link to an explanation of unity, referenced within the first section, for those of us without chemistry degrees? —Preceding unsigned comment added by 70.105.186.115 (talk) 15:06, 23 January 2010 (UTC)[reply]

Definition of van't Hoff factor

In this Wikipedia article the van't Hoff factor is defined as a quotient between actual number of solute particles and theoretical molecular number of particles. Contrary to this, my old chemistry book, the one I used when studying chemistry, defines it as a quotient of actual value of the colligative property, like depression on freezing temperature, to the theoretical one. The difference is substantial. Can someone else help?Auró (talk) 11:12, 29 May 2011 (UTC)[reply]

A confusion seems to have occurred between this factor and the osmotic coefficient or dissociation degree.--5.2.200.163 (talk) 14:46, 18 June 2018 (UTC)[reply]

[1] Heyrovska, R (1997). "Equations for Densities and Dissociation Constant of NaCI(aq) at 25°C from "Zero to Saturation" Based on Partial Dissociation" (PDF). J. Electrochem Soc. 144 (7).

[1]