the Ksp for CaF2 is 3.9*10-11. assuming that calcium flouride dissociates completely upon dissolving and that there are no other important equilibria affecting the solubility, calculate the solubility in pure water, of calcium floride in m/L and in g/L
CaF2 ==> Ca^+2 + 2F^-
Ksp = (Ca^+)(F^-)^2
Set up an ICE chart, substitute into Ksp, and solve.
Let S = solubility CaF2 in M (moles/L)
Then S is the concn of Ca^+2
and 2S is concn F^-
3.9 x 10^-11 = (S)(2S)^2
Solve for S for moles/L
g = moles x molar mass for g/L.
To calculate the solubility of calcium fluoride (CaF2) in pure water, we need to use the concept of Ksp (the solubility product constant).
Ksp is defined as the equilibrium constant for a solid salt dissolving in water to form its constituent ions. The expression for Ksp is written as the product of the concentrations of the dissociated ions, raised to the power of their stoichiometric coefficients.
The balanced chemical equation for the dissolution of CaF2 in water is:
CaF2(s) ⇌ Ca^2+(aq) + 2F^-(aq)
From this equation, we can determine that the concentration of Ca^2+ ions is equal to the concentration of dissolved CaF2, and the concentration of F^- ions is twice the concentration of dissolved CaF2.
Let's denote the solubility of CaF2 as "S" (in moles per liter or M).
Using the information from the question, we can write the equation for the solubility product constant (Ksp):
Ksp = [Ca^2+][F^-]^2
Given that the concentration of Ca^2+ ions is equal to S, and the concentration of F^- ions is 2S:
Ksp = [S][2S]^2
3.9 x 10^-11 = 4S^3
To solve for S, we rearrange the equation:
S^3 = (3.9 x 10^-11)/4
S^3 = 9.75 x 10^-12
Taking the cube root of both sides of the equation gives us:
S ≈ 2.48 x 10^-4 M
So, the solubility of calcium fluoride in pure water is approximately 2.48 x 10^-4 moles per liter (M).
To convert this to grams per liter (g/L), we need to use the molar mass of calcium fluoride, which is 78.08 g/mol.
To convert from moles to grams, we multiply the solubility (S) by the molar mass:
Solubility in g/L = 2.48 x 10^-4 M * 78.08 g/mol
Solubility in g/L ≈ 1.94 x 10^-2 g/L
Therefore, the solubility of calcium fluoride in pure water is approximately 1.94 x 10^-2 grams per liter (g/L).