Question

The mineral fluorite is calcium fluoride, CaF_2. The goal is to calculate the molar solubility of CaF_2 in water from its solubility product constant (k_sp=3.4 times 10^-11). Write down the ICE chart: Write down K_sp in terms of the solubility (s) and solve for s. The goal now is to calculate solubility after adding 0.15 MCaCl_2. Write down the new ICE chart and solve for the new s.

Answer 1

a)
CaF2 <-------> Ca2+ + 2F-
s 2s

b)
Ksp = [Ca2+] [F-]^2
3.4*10^-11 = s*(2s)^2
4*s^3 = 3.4*10^-11
s = 2.04*10^-4 M
Answer:
2.04*10^-4 M

c)
CaCl2 being strong electrolute gives 0.15 M Ca2+

CaF2 <-------> Ca2+ + 2F-
0.15 0 (initial)
0.15+s 2s (at equilibrium)

Ksp = [Ca2+] [F-]^2
3.4*10^-11 = (0.15+s)*(2s)^2
since ksp is small, s will be small and it can be ignored as compared to 0.15
3.4*10^-11 = (0.15)*(2s)^2
3.4*10^-11 = (0.15)*4*s^2
s = 7.53*10^-6 M
Answer: 7.53*10^-6 M