Assuming that there is a concentration of 0.05M HF and 0.0010M of CaCl2 is added, would the precipitate CaF2 form?

Answer 1

Firsts of all, to approach this problem, we need to know the Ksp for CaF₂. I looked it up and found it to be 3.9x10^-11.

Looking at the reaction, we have ...

CaCl₂(aq) + 2HF(aq) ==> CaF₂(s) + 2HCl(aq)

[HF] = 0.05 M

[CaCl₂] = 0.001 M

[CaF₂] formed = 0.001 M

CaF₂(s) ==> Ca2+(aq) + 2F-(aq)

Q = [Ca²⁺][F^-]² = (0.001)(0.002)²

Q = 4.0x10^-9

This value of Q is greater the Ksp (3.9x10^-11) so a precipitate will form