- #1

Max364

- 13

- 1

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?