Have I done this question correctly?
For fluorite, pKsp = 10.4. If 1 litre of a CaF2 solution is mixed with 1 litre of a 0.001M CaCl2 solution, what is the saturation index of fluorite for the mixture?
Ksp = 10-10.4 = [Ca2+][F-]2 = [Ca2+][2Ca2+]2 = 4[Ca2+]3
[Ca2+] = 2.15 x 10-4 mol dm-3
[F-] = 4.30 x 10-4 mol dm-3
There are 2.15 x 10-4 mol of Ca2+ in the first solution and 10-3 mol of Ca2+ in the second solution so there are 1.22 x 10-3 mol of Ca2+ in the mixture. No F- has been added so there are 4.30 x 10-4 mol F- in the mixture. Therefore, the activity of Ca2+ is 0.61 x 10-3 mol dm-3 and the activity of F- is 2.15 x 10-4 mol dm-3 (since the mixture now has a volume of 2dm3).
SI = log[AP/Ksp] = [(0.61 x 10-3)(2.15 x 10-4)2]/[10-10.4] = 0.71
So the solution is undersaturated with respect to fluorite.
And actually passed?