Final answer:
The maximum possible concentration of F- present is 5.7 x 10^-5 M.
Step-by-step explanation:
To find the maximum possible concentration of F- present, we can use the information given about the solubility product constant (Ksp) and the molar amount of Ca2+ ions present in the solution. Since the formula of CaF2 (calcium fluoride) is 1:2, we know that for every 1 mole of Ca2+ ions, there are 2 moles of F- ions. Therefore, the molar amount of F- ions is twice the molar amount of Ca2+ ions.
Given that the solution contains 6.0 x 10^-3 moles of Ca2+ per liter and the Ksp is 4.0 x 10^-11, we can set up the equation for Ksp and solve for the molar amount of F- ions:
Ksp = [Ca2+][F-]^2
4.0 x 10^-11 = (6.0 x 10^-3)(2x)^2
(2x)^2 = 4.0 x 10^-11 / (6.0 x 10^-3)
x^2 = 3.3 x 10^-9
x = sqrt(3.3 x 10^-9) = 5.7 x 10^-5 M
Therefore, the maximum possible concentration of F- present is 5.7 x 10^-5 M.