Originally Posted by e.lark
You really want to know??
For "insoluble" solids, we set up an equilibrium expression:
Fe3(PO4)2*8H2O(s) <----> 3 Fe+2(aq) + 2 PO4-3(aq) + 8 H2O(l)
Ksp = [Fe+2]^3 * [PO4-3]^2 = 1*10^-36
Let x be the amount of the solid that would dissolve (assuming excess solid in a solution), then the amount of Fe+2 = 3x, the amount of phosphate = 2x. These are the MAXIMUM amounts that would be in solution (though there can be an excess of ONE of the components).
(3x)^3 * (2x)^2 = 1*10^-36
so x = 2.4*10^-8M
This equals an iron concentration of 7.2*10^-8M = 0.004 ppm
and a phosphate concentration of 0.0045 ppm
Above these concentrations the iron phosphate precipitate would be expected to form.
HOWEVER - the above calculation assumes: Ion-free water except for the iron and phosphate, 25C temp, AND no EDTA or other complexing agent for the iron.
CAN you keep 0.1ppm iron and 1.0ppm phosphate in the water column? Sure! But we do know it is difficult to keep the iron there - this is probably one of the contributors (oxidation to Fe+3 - an even less soluble ion - is the another contributor).
SHOULD you mix the micro solution with a phosphate stock solution? NO! The much higher concentrations of iron and phosphate in stock solutions just about guarantees a precipitate.
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