- #1
Max364
- 13
- 1
I have a question about calculating solubilities of sparingly soluble salts.
Eg Ksp CaF2 = 4 x 10-11
So, Saturation Index of CaF2 is:
SICaF2 = IP / Ksp
Where IP = Ionic Product = [Ca2+] x [F-]2
[Ca2+] and [F-] are molar concentrations of each ion.
Example:
We have 400 ppm Ca and 12 ppm F in a water
Ion product IP = [400/40000] x [12/19000)]2 = [0.01]. [3.99 x10-7] = 3.99 x 10-9
Saturation Index SICaF2 = 3.99 x 10-9 / (4 x 10-11 ) = 100
All is fine so far, but let's say we now have a massive amount of extra Ca, say 5,000ppm
Now, IP = [5000/40000]. [3.99 x10-7] = 4.99 x 10-8
So, Saturation Index now SICaF2 = 4.99 x 10-7 / (4 x 10-1 ) = 12.475
So massively increased due to increased Ca.
But my question is: How accurate is this new SICaF2 ?
Because we have a limited amount of F (only 12ppm) the it should not matter how much extra Ca we have? As it can only react with a fixed amount of F, ie, 1 ppm Ca reacts with 2ppm of F and once all the F is used up then excess Ca distorts the SI calculation?
Eg Ksp CaF2 = 4 x 10-11
So, Saturation Index of CaF2 is:
SICaF2 = IP / Ksp
Where IP = Ionic Product = [Ca2+] x [F-]2
[Ca2+] and [F-] are molar concentrations of each ion.
Example:
We have 400 ppm Ca and 12 ppm F in a water
Ion product IP = [400/40000] x [12/19000)]2 = [0.01]. [3.99 x10-7] = 3.99 x 10-9
Saturation Index SICaF2 = 3.99 x 10-9 / (4 x 10-11 ) = 100
All is fine so far, but let's say we now have a massive amount of extra Ca, say 5,000ppm
Now, IP = [5000/40000]. [3.99 x10-7] = 4.99 x 10-8
So, Saturation Index now SICaF2 = 4.99 x 10-7 / (4 x 10-1 ) = 12.475
So massively increased due to increased Ca.
But my question is: How accurate is this new SICaF2 ?
Because we have a limited amount of F (only 12ppm) the it should not matter how much extra Ca we have? As it can only react with a fixed amount of F, ie, 1 ppm Ca reacts with 2ppm of F and once all the F is used up then excess Ca distorts the SI calculation?