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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 conc

OK. Now that we have established that tap water with natural and adjusted fluoride are identical, how do you explain the lack of problems in comminities with naturally fluoridated water?

Next, you have the solubility problem wrong. If you have a substance like CaF that isn't very soluble in water, you need to have MORE of the compound in the water to give you a high steady state ion level. That is opposed to a substance that is easily soluble. In that case you don't need to add as much.

Without going into the calculations for solubility in relation to temperature.

I'm sorry, but you MUST give at least a link to these calculations.

So your boiled water example is backwards. Although I don't think it is valid anyway.

OK. Now that we have established that tap water with natural and adjusted fluoride are identical, how do you explain the lack of problems in comminities with naturally fluoridated water?

Since a solution of calcium fluoride will be limited to a maximum solubility of 16 ppm, there will ONLY be 16 ppm of fluoride ions available in blood if the solute is calcium fluoride. With a maximum solubility of 42,200 ppm for sodium fluoride, the concentration of fluoride ions available could increase to 42,200 ppm, although a human would die well before that concentration was reached. And again, once the maximum solubility is reached, ie. the solution becomes saturated, any excess solute is simply held in aqueous suspension and is easily excreted.

So the problem with sodium fluoride is that there is virtually NO limit as to the number of fluoride ions available in the solution, where death will occur WELL before the solution reaches saturation.

Next, you have the solubility problem wrong. If you have a substance like CaF that isn't very soluble in water, you need to have MORE of the compound in the water to give you a high steady state ion level. That is opposed to a substance that is easily soluble. In that case you don't need to add as much.

It is the IONS that pose a problem, not so much the compound itself. If there are n number of ions available, as you say, a fluoride ion is a fluoride ion. Until maximum solubility is reached, ALL sodium fluoride molecules (formula units) are dissolved in the solution. So yes, there IS more calcium fluoride compound present in a solution with the same ion level as sodium fluoride, as it stops dissolving at a MUCH lower concentration.

As the fluoride ions remain bound to the calcium however, they are NOT available and pose no risk.

I'm sorry, but you MUST give at least a link to these calculations.

SOLUBILITY AS A FUNCTION OF TEMPERATURE

In a endothermic reaction such as the disassociation of sodium fluoride, energy is absorbed by the system.

The rule for solutions in relation to temperature is, if the solution process absorbs energy then the solubility will be INCREASED as the temperature is increased. If the solution process releases energy then the solubility will DECREASE with increasing temperature.

Some common salts are shown in the graph below;

I can't seem to find a chart depicting the solubility in relation to temperature for the various fluorides, but I'll see what I can find if you're interested..

So your boiled water example is backwards. Although I don't think it is valid anyway.

Nope, I didn't have it backwards, and it IS valid.

One more thing Tom. The equations provided in the link I gave you are only valid up to a point. The final equations are for non-electrolytes, but there is an interim equation that should be valid. I'll provide it below..