*shrug
Adding HNO3 to a solution of CaF2 and water will increase the concentration of H+ ions in the solution, according to the following reaction:
HNO3 + H2O ⇌ H3O+ + NO3-
The increased concentration of H+ ions will shift the equilibrium of the dissociation reaction of CaF2 in the opposite direction, making it less soluble. The dissociation reaction of CaF2 is as follows:
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
The solubility product constant expression for CaF2 is:
Ksp = [Ca2+][F-]^2
If the solubility of CaF2 is S, then at equilibrium, [Ca2+] = S and [F-] = 2S. Therefore,
Ksp = S * (2S)^2 = 4S^3
Now, if 0.1 M HNO3 is added to the solution, it will increase the H+ ion concentration, which will shift the dissociation equilibrium of CaF2 to the left, decreasing the solubility. The reaction can be written as:
CaF2(s) + 2H+(aq) ⇌ Ca2+(aq) + 2HF(aq)
The equilibrium expression for this reaction is:
K = [Ca2+][HF]^2/[H+]^2
At equilibrium, the concentrations of Ca2+ and HF will be less than S, and the concentration of H+ will be 0.1 M. We can use the solubility product constant and the equilibrium expression to solve for the new solubility:
K = [Ca2+][HF]^2/[H+]^2
4.0 x 10^-11 = (S - x)(2S - 2x)^2/(0.1)^2
Solving for x, we get x = 2.9 x 10^-5 M
Therefore, the new solubility of CaF2 in the presence of 0.1 M HNO3 is S - x = 1.0 x 10^-6 M.
*IG:whis.sama_ent*