Question

1. Solid calcium fluoride (CaF2) establishes the following equilibrium in solution: CaF2(s) = Ca2+(aq) + 2F-(aq) Ke = 1.5 x 10-10 A solution initially contains 2.45 g of CaF2. a. In which direction will the reaction move to reach equilibrium? Explain your reasoning b. Calculate the equilibrium concentrations of Ca2+ and F. How much CaF2 (in mg) dissolves in solution? c. Another solution initially contains 2.45 g of CaF2 and 0.0025 M NaF. What are the equilibrium concentrations of Ca2+ and F-? How much CaF2 (in mg) dissolves in this solution? (Hint: Can we make any assumptions?)

Answer 1

a) The equilibrium will move to the right to form products. There are only reagents in the system.

b) The expression of Ksp:

Ksp = [Ca + 2] * [F -] ^ 2 = 4 * X ^ 3

It clears:

X = S = (Ksp / 4) ^ (1/3) = (1.5x10 ^ -10 / 4) ^ (1/3) = 3.34x10 ^ -4 M

The solubility in mg is calculated:

S = 3.34x10 ^ -4 M * (78.07 g / 1 mol) * (1000 mg / 1 g) = 26.08 mg / L

c) We have that the expression of Ksp is:

Ksp = X * (0.0025 - 2 * X) ^ 2

It is assumed that - 2 * X is negligible and clear:

X = Ksp / 0.0025 ^ 2 = 2.4x10 ^ -5 M

The mg are calculated:

mg = 2.4x10 ^ -5 M * (78.07 g / 1 mol) * (1000 mg / 1 g) = 26.08 mg / L = 1.87 mg / L

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