### How to express amounts of solubility, mathematically.

No substance is totally insoluble in water, although many substances have very low solubility.

How to think and talk quantitatively about maximum concentrations of substances and ions.

The borderline concentrations at which precipitation will just barely occur.

91 grams of glucose will dissolve in a hundred milliliters of water at 25 degrees C.

125 grams of glucose will dissolve in a hundred milliliters of water at 30 degrees C.

If you saturate 100 ml. of water with glucose at 30 degrees, and cool this water back down to 25 degrees, the excess 34 grams of glucose will precipitate back out of solution, but the precipitation will take many hours. During that delay, the water is super-saturated with glucose. This period of supersaturation is unusually long.

But it is not at all unusual for some amount of supersaturation to occur.

When calcium sulfate dissolves in water, calcium ions separate from sulfate ions.

If there had already been some calcium ions dissolved in the water, then less calcium phosphate would be able to dissolve. We need to ask how much the solubility of calcium phosphate would be reduced for any given concentration of calcium ions that were already present. It turns out that the right way to calculate this is to multiply the concentration of calcium ions times the concentration of sulfate ions. In a saturated solution of calcium sulfate (in water, at room temperature) the concentration of calcium ions multiplied times the concentration of sulfate ions will be 4.93 times ten to minus fifth. That means that a saturated solution, with no other ions present would be considerably less than a hundredth molar. If a solubility product for a salt were ten to the minus fourth, then a saturated solution would me one hundredth molar. The solubility product of lead sulfate actually is 1 times 10 to the minus eighth. This makes the math easier. A saturated solution would therefore have a concentration of ten to the minus fourth molar concentration of lead ions, and also ten to the minus fourth concentration of sulfate ions.

If you poured sulfuric acid into a saturated solution of lead sulfate, many those extra sulfate ions would bind to sulfate ions and precipitate out as lead sulfate. The more sulfates added, the more lead sulfate crystals will precipitate out - just enough to keep the ten to the minus eight "solubility product". For example, if you added enough sulfates to raise their concentration to one molar, then the concentration of lead ions would be reduced to ten to the minus eight.

The solubility product of cuprous sulfide happens to be 10-35 ten to the minus thirty fifth. That is a very insoluble salt, as most sulfides are. Silver sulfide has a solubility product of 10 -49 That is even less soluble. But not quite as much more insoluble as it seems at first.

The equation is Ag2S. Two sugar ions per sulfide ion. Therefore the silver ion concentration squared times the sulfide ion concentration is what is equal to ten to the minus 49th power. The solubility product of calcium phosphate is said to be around 10-38 but the chemical formula is sometimes written as Ca3(PO4)2

That implies that I should multiply the cube of the calcium concentration times the square of the phosphate concentration? And that's what's going to equal 10-38 ?

Sometimes I see the chemical equation of calcium phosphate written differently, with hydroxyl groups and fluorides. I don't know whether the ten to the minus thirty eighth is based on these more complicated formulae; or if not, why not?

If we assume that -38 equals the cube of the calcium concentration times the square of the phosphate concentration, then "ball park" estimates of the concentrations of each ion that would be in equilibrium with calcium phosphate crystals would be whatever the fifth root of 38 is. (approximately ten to the 38/5), somewhat less than 10 to the minus seventh? Less than a ten millionth molar. Note the difference from the ten thousandth molar ion concentrations for lead sulfate.

I wish I knew the solubility product for strontium phosphate, technetium phosphate and phosphates of other "bone seeking" cations. Presumably they are less that 10 to the minus thirty eighth. Another good question is whether all cations whose phosphate salts have solubility products in the range of -38 , or lower are also bone seeking. Is that a sufficient as well as necessary condition for being "bone seeking"? Or are there cations with equally or more insoluble phosphates that don't get incorportated into bone?

Conversely, are there any "bone seeking" cations whose phosphate salts are appreciably more soluble? These kinds of facts ought to be in chapter one of any book on bone formation, but they are not.

Another pair of good question is whether there are any "bone seeking" anions, and what are the solubility products of their calcium salts?

"The common ion effect"

Increasing concentrations of either calcium of phosphate ions will drive precipitation of both ions into formation of more of the salt.

Solubility of all salts is increased by high ion concentrations. You probably wouldn't expect it, but calcium phosphate will be made more soluble by greatly increasing the concentration of, for example, sodium chloride. Salt water makes most salts more soluble. Effective solubility is called "activity" in this field, and was once a cutting edge subject for study and analysis. An example of the cause of this effect is that dissolved calcium ions attract a little cloud of chloride ions, and dissolved phosphates attract a little cloud of sodium ions. This increases their solubility by many percent.