Final answer:
To calculate the molar solubility of CaF2 in a 0.100-M HF solution, we consider the common ion effect and use the Ksp expression for CaF2. The concentration of fluoride ions from both HF and CaF2 dissolution must be accounted for, and solving the Ksp expression by incorporating these concentrations will yield the molar solubility of CaF2.
Step-by-step explanation:
The question asks about the molar solubility of CaF2 in the presence of a common ion, HF, where the dissociation constant (Ka) of HF is given. To solve for molar solubility, we use the solubility product constant (Ksp) of CaF2 and the Ka value for HF to account for the common ion effect (in this case, the fluoride ion, F-).
CaF2 (s) → Ca2+ (aq) + 2F- (aq).
Since we have a 0.100 M solution of HF, and HF dissociates according to HF (aq) → H+ (aq) + F- (aq), the concentration of F- ions from the HF solution must be taken into account when calculating the solubility of CaF2.
Let s be the molar solubility of CaF2 in the 0.100 M HF solution. The concentration of Ca2+ will be s, and the concentration of F- from the CaF2 will be 2s. However, since the solution already contains 0.100 M F- from HF, the total concentration of F- will be 0.100 + 2s.
The Ksp expression for CaF2 is: Ksp = (s)(2s + 0.100)^2.
We need to solve for s taking into account the common ion effect that reduces the solubility of CaF2 compared to its solubility in pure water.
Substituting the expressions into the Ksp equation and solving for s (either through approximation or more advanced mathematical methods such as the quadratic formula), we would find the molar solubility of CaF2 in a 0.100-M solution of HF.