You said:
It depends on whether or not the chemical is SOLUBLE. There is also a characteristic known as SOLUBILITY. A soluble compound will DISSOLVE in a SOLVENT such as water, but only until the solvent reaches SATURATION for that soluble compound.
A you can see, calcium fluoride is MUCH less soluable than sodium fluoride. And that is EXACTLY where the problem lies.
Absolutely. BUT, when we measure the concentration of fluoride in someone's water, we are measuring the amount of FLUORIDE IONS in that water. Which means that the ion has already "dissolved" in the water. Whether that means that there had to be a whole lot of CaF in the water, or less of the NaF, doesn't matter. Solubility will give you the amount of compound you need to get a specific ion level, but it doesn't affect the ion itself.
Once again, you and I go out and measure some water supplies. We go to one house that has a well with a natural fluoride level of 1ppm. And then we go to another house that has a supplemented fluoride supply and measure it at the same 1ppm. The fact that the one has fluoride ions from CaF and the other from NaF is irrelevant, they are the same thing. And the only time solubility comes into play is in determining the amount of fluoride containing compounds to put into the water to get that level.
You keep saying:
The NUMBER of free ions is DRASTICALLY less when the solute is calcium fluoride as opposed to sodium fluoride.
But when you are measuring the fluoride levels in a water supply, you are SPECIFICALLY counting free fluoride ions.
there would be no such thing as salt water, it would simply be a sea of free ions. We know that isn't the case.
Ummm, what happens to salt when it DISSOLVES in water then?
(Incidently, I did mean FLUORIDE ION in our past discussions, NOT fluorine).
Ok, you appear to have a valid point. That is the common argument in this matter. When the testing is done from a water supply, yes, the number of flouride ions are measured, and yield a certain value. BUT....
The argument falls apart when you consider the solubility of the compounds, sodium fluoride and calcium fluoride. It is when that water is BOILED that the difference between the two come into play. Water that has been treated with sodium fluoride has both Na+ and F- ions dissolved, whereas water with naturally occuring calcium fluoride has Ca+ and F- ions. As the water evaporates, the concentration of fluoride increases. Without going into the calculations for solubility in relation to temperature, where solubility of fluorides increase in relation to an increase in temperature, let's take a look at what happens in each case...
In the sodium fluoride solution, as water boils, the concentration can reach 42,200 ppm, whereas in the solution of calcium fluoride, the MAXIMUM concentration is 16 ppm. SO, the availability of fluoride ions is virtually unlimited in the case of sodium fluoride solutions, where it is limited to concentrations of 16 ppm in calcium fluoride solutions. THAT is the difference.
Now of course, one wouldn't usually boil water for so long as to reach those extreme concentrations, but it IS theoretically possible.
SO, that doesn't explain anything yet, does it. Well, maybe it does...
It is in the body that we see a major problem. As over the course of a day, we consume fluoride from ALL of our food sources. Every bottle of soda, every item of food, every thing we ingest has fluoride. IF the ONLY source of fluoride was from calcium fluoride, the HIGHEST concentration we'd ever see in our bloodstream is 16 ppm. With sodium fluoride on the other hand, the concentration can theoretically go up to 42,200. Now THAT is not a good thing....
When the solution reaches SATURATION, the remaining NaCl simply forms a HETROGENEOUS MIXTURE. Just as in when food coloring is added to water, the food coloring might combine with the water and form a MIXTURE, but it does not form an ionic solution. When a solution of salt is saturated, any extra salt is simply held in aqueous suspension.