rikkitikkitavi
Hazard to Others
Posts: 192
Registered: 17-6-2002
Member Is Offline
Mood: No Mood
|
|
problem with solubilities
I have this problem (work related)
I have a saturated Ca-salt solution of an organic acid, Ca concentration is about 1 M.
It is made from Ca(OH)2 + acid neutralized to pH = 7-9 , preferably 8. Since Ca(OH)2 is not 100 % pure , the solution is filtered to remove
impurities like CaCO3 , Fe(OH)3 e t c and very important , fluoride (as CaF2)
Ksp CaF2 is 5,3 *10^-9.
calculation for this gives that F- should be around sqrt(5,3 *10^-9 / 1 ) = 73 uM or about 1,3 mg / liter. ( [Ca]*[F]^2 = Ksp )
In reality concentration is much higher ? How can that be? The solution is filtered 100 % , no solids are coming through and it is an extremely fine
filter.
/rickard
|
|
|
Polverone
Now celebrating 21 years of madness
Posts: 3186
Registered: 19-5-2002
Location: The Sunny Pacific Northwest
Member Is Offline
Mood: Waiting for spring
|
|
one guess
I was able to find a number of papers dealing with supersaturated solutions of calcium fluoride. These were in the context of crystal growth
inhibition by other substances: polyvalent ionic solutes and organic acids were two common inhibitors. Many of these papers indicated that even low
concentrations of inhibitors could dramatically slow CaF2 crystal growth. So... maybe a supersaturated solution was formed, and it would take a long
time to precipitate because of other substances in solution acting as inhibitors?
PGP Key and corresponding e-mail address
|
|
|
unionised
International Hazard
Posts: 5104
Registered: 1-11-2003
Location: UK
Member Is Offline
Mood: No Mood
|
|
Calcium ions form complexes with some organic acids so, even though the total Ca concentration is about molar, the free Ca++ is rather less.
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
